Existence of Extremals for a Fourier Restriction Inequality on the One-Sheeted Hyperboloid

Abstract

We prove the existence of functions that extremize the endpoint $L^2$ to $L^4$ adjoint Fourier restriction inequality on the one-sheeted hyperboloid in Euclidean space $\mathbb {R}^4$ and that, taking symmetries into consideration, any extremizing sequence has a subsequence that converges to an extremizer.

1 Introduction

In seminal paper [43] R. Strichartz addressed the adjoint restriction problem of the Fourier transform to $d-1$ dimensional quadric submanifolds of Euclidean space $\mathbb {R}^d$, establishing the necessary and sufficient conditions on p such that an $L^2\rightarrow L^p$ estimate holds. Recently, there has been interest in studying the existence of extremizers and the sharp $L^2\rightarrow L^p$ estimates for adjoint Fourier restriction operators and progress has been made in the case of quadric curves and surfaces: the paraboloid and parabola [22, 28], the cone [5, 22, 40], the sphere and circle [6, 8, 13, 23, 26, 36, 42], the two-sheeted hyperboloid and hyperbola [9, 10, 39] and the saddle [7, 18, 19] (see also [1, 15, 16, 25] for the case of power curves and surfaces). The study of such sharp $L^2$ to $L^p$ estimates is intimately related to the study of extremizers and sharp constants for Strichartz estimates for classical partial differential equations, such as the Schrödinger, hyperbolic Schrödinger, wave and Klein–Gordon equations. In this note we address the case of the one-sheeted (or hyperbolic) hyperboloid in $\mathbb {R}^4$, which is related to the so called Klein–Gordon equation with imaginary mass.

1.1 Setting

Let ${\mathcal {H}}^3$ denote the upper half of the three dimensional one-sheeted hyperboloid in $\mathbb {R}^4$,

$$\begin{aligned} {\mathcal {H}}^3=\bigg \{(x,\sqrt{\vert x\vert ^2-1}):x\in \mathbb {R}^3,\, \vert x\vert \geqslant 1\bigg \}, \end{aligned}$$

equipped with the measure $\mu$ with density

$$\begin{aligned} \,\textrm{d}\mu (x,t)=\mathbb {1}_{\{\vert x\vert > 1\}}(x)\,\varvec{\delta }\! \begin{pmatrix} t-\sqrt{\vert x\vert ^2-1} \end{pmatrix}\!\frac{\,\textrm{d}t\,\textrm{d}x}{\sqrt{\vert x\vert ^2-1}}, \end{aligned}$$

(1.1)

so that for all $g\in {\mathcal {S}}(\mathbb {R}^4)$ it holds that

$$\begin{aligned} \int \limits _{{\mathcal {H}}^3} g(x,t)\,\textrm{d}\mu (x,t)=\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert > 1\}}g(y,\sqrt{\vert y\vert ^2-1})\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}. \end{aligned}$$

(1.2)

A function $f:{\mathcal {H}}^3\rightarrow \mathbb {R}$ can be identified with a function from $\mathbb {R}^3$ to $\mathbb {R}$, using the orthogonal projection from¹${\mathcal {H}}^3$ to $\mathbb {R}^3\times \{0\}$, and in what follows we do so. We denote the $L^p({\mathcal {H}}^3,\mu )$ norm of a function f on ${\mathcal {H}}^3$ by $\Vert f \Vert _{L^p({\mathcal {H}}^3)}$, $\Vert f \Vert _{L^p(\mu )}$ or simply $\Vert f \Vert _{L^p},\,\Vert f \Vert _{p}$ if it is clear from context.

The Fourier extension operator on the hyperboloid ${\mathcal {H}}^3$, also known as the adjoint Fourier restriction operator, is given by

$$\begin{aligned} Tf(x,t)=\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert > 1\}} e^{ix\cdot y} e^{it\sqrt{\vert y\vert ^2-1}} f(y,\sqrt{\vert y\vert ^2-1})\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}, \end{aligned}$$

(1.3)

where $(x,t)\in \mathbb {R}^3\times \mathbb {R}$ and $f\in {\mathcal {S}}(\mathbb {R}^4)$. Note that $Tf(x,t)=\widehat{f\mu }(-x,-t)$, with the Fourier transform in $\mathbb {R}^{4}$ defined by ${\hat{g}}(x,t)=\int \limits _{\mathbb {R}^3\times \mathbb {R}}e^{-i(x\cdot y+ts)}g(y,s)\,\textrm{d}y\,\textrm{d}s$.

Strichartz proved in [43] that for all $\frac{10}{3}\leqslant p\leqslant 4$ there exists $C_p<\infty$ such that for all $f\in L^2({\mathcal {H}}^3)$ the following estimate for Tf holds

$$\begin{aligned} \Vert Tf \Vert _{L^p(\mathbb {R}^4)}\leqslant {{\textbf {H}} }_{p}\Vert f \Vert _{L^2({\mathcal {H}}^3)}, \end{aligned}$$

(1.4)

where ${{\textbf {H}} }_{p}<\infty$ denotes the best constant in (1.4),

$$\begin{aligned} {{\textbf {H}} }_{p}=\sup _{0\ne f\in L^2(\mathcal H^3)}\frac{\Vert Tf \Vert _{L^p(\mathbb {R}^4)}}{\Vert f \Vert _{L^2({\mathcal {H}}^3)}}. \end{aligned}$$

(1.5)

The (full) one-sheeted hyperboloid is defined by

$$\begin{aligned} \overline{{\mathcal {H}}}^3:=\big \{(x,t)\in \mathbb {R}^3\times \mathbb {R}:t^2=\vert x\vert ^2-1,\, \vert x\vert \geqslant 1\big \},\end{aligned}$$

and we endow it with the Lorentz invariant measure $\bar{\mu }=\mu _++\mu _{-}$ where $\mu _+=\mu$ as in (1.1)–(1.2) is supported on ${\mathcal {H}}^3$, and $\mu _{-}$ is given by

$$\begin{aligned} \,\textrm{d}\mu _{-}(x,t)=\mathbb {1}_{\{\vert x\vert >1 \}}\,\varvec{\delta }\! \begin{pmatrix} t+\sqrt{\vert x\vert ^2-1} \end{pmatrix}\!\frac{\,\textrm{d}t\,\textrm{d}x}{\sqrt{\vert x\vert ^2-1}}, \end{aligned}$$

so that $\mu _-$ equals the reflection of $\mu$ via the reflection map $(x,t)\mapsto (-x,-t)$ and is supported on $-{\mathcal {H}}^3$. The adjoint Fourier restriction operator on $\overline{{\mathcal {H}}}^3$ is

$$\begin{aligned} \begin{aligned} {\overline{T}}f(x,t)&=\widehat{f{{\bar{\mu }}}}(-x,-t)=\int \limits _{{\overline{{\mathcal {H}}}}^3} e^{i(x\cdot y+ts)} f(y,s)\,\textrm{d}{\bar{\mu }}(y,s)\\&=\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert> 1\}} e^{ix\cdot y} e^{it\sqrt{\vert y\vert ^2-1}} f_{+}(y)\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}\\&\quad +\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert > 1\}} e^{ix\cdot y} e^{-it\sqrt{\vert y\vert ^2-1}} f_{-}(y)\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}, \end{aligned} \end{aligned}$$

(1.6)

where $f=f_++f_-$, the function $f_+$ is supported on the upper half of the one-sheeted hyperboloid, ${\mathcal {H}}^3$, and the function $f_-$, on the lower half, $-{\mathcal {H}}^3$, and we have identified their domains with $\mathbb {R}^3$ via the orthogonal projection as stated before. We see that ${\overline{T}}f(x,t)=Tf_+(x,t)+Tf_-(x,-t)$.

The triangle inequality and (1.4) imply that for $\frac{10}{3}\leqslant p\leqslant 4$ the following estimate holds

$$\begin{aligned} \Vert {\overline{T}}f \Vert _{L^p(\mathbb {R}^4)}\leqslant \overline{{\textbf {H}}}_{p}\Vert f \Vert _{L^2(\overline{{\mathcal {H}}}^3)}, \end{aligned}$$

(1.7)

where $\overline{{{\textbf {H}} }}_{p}<\infty$ is the sharp constant

$$\begin{aligned} \overline{{{\textbf {H}} }}_{p}=\sup _{0\ne f\in L^2(\overline{\mathcal {H}}^3)}\frac{\Vert {\overline{T}}f \Vert _{L^p(\mathbb {R}^4)}}{\Vert f \Vert _{L^2(\overline{{\mathcal {H}}}^3)}}. \end{aligned}$$

(1.8)

The Lorentz group on $\mathbb {R}^4$, denoted ${\mathcal {L}}$, preserves $\overline{{\mathcal {H}}}^3$, ${\bar{\mu }}$, and acts on functions on $\overline{{\mathcal {H}}}^3$ by composition: $L^*f(x,t):=f(L(x,t))$, $L\in {\mathcal {L}}$ (see Sect. 2 for more details). In particular, we have $\Vert f \Vert _{L^q(\overline{{\mathcal {H}}}^3)}=\Vert L^*f \Vert _{L^q(\overline{{\mathcal {H}}}^3)}$ and $\Vert Tf \Vert _{L^p(\mathbb {R}^4)}=\Vert T(L^*f) \Vert _{L^p(\mathbb {R}^4)}$, for all $p,q\in [1,\infty ]$.

Definition 1.1

An extremizer (or maximizer) for (1.4) is a function $0\ne f\in L^2({\mathcal {H}}^3)$ that satisfies $\Vert Tf \Vert _{L^p(\mathbb {R}^4)}={{\textbf {H}} }_p\Vert f \Vert _{L^2({\mathcal {H}}^3)}$. An $L^2$-normalized extremizing sequence for (1.4) $\{f_n\}_n\subset L^2({\mathcal {H}}^3)$ is such that $\Vert f_n \Vert _{L^2({\mathcal {H}}^3)}=1$ and $\Vert Tf_n \Vert _{L^p(\mathbb {R}^4)}\rightarrow {{\textbf {H}} }_p$, as $n\rightarrow \infty$. A corresponding definition holds for extremizers and extremizing sequences for (1.7).

1.2 Main Results

This paper is devoted to the study of the sharp instances of (1.4) and (1.7) in the endpoint case $p=4$, that is, the inequalities

$$\begin{aligned} \Vert {T}f \Vert _{L^4(\mathbb {R}^4)}&\leqslant {{\textbf {H}}}_{4}\Vert f \Vert _{L^2({{\mathcal {H}}}^3)}, \end{aligned}$$

(1.9)

$$\begin{aligned} \Vert {\overline{T}}g \Vert _{L^4(\mathbb {R}^4)}&\leqslant \overline{{\textbf {H}}}_{4}\Vert g \Vert _{L^2(\overline{{\mathcal {H}}}^3)}, \end{aligned}$$

(1.10)

and our main results concern the existence of extremizers as well as the precompactness of extremizing sequences. The statements of the main results of this paper are as follows.

Theorem 1.2

There exists an extremizer in $L^2({\mathcal {H}}^3)$ for inequality (1.9). Moreover, for every $L^2$-normalized complex valued extremizing sequence $\{f_n\}_{n}$ for (1.9), there exist a subsequence $\{f_{n_k}\}_{k}$ and a sequence $\{(x_k,t_k)\}_{k}\subset \mathbb {R}^3\times \mathbb {R}$ such that $\{e^{ix_k\cdot y}e^{it_k\sqrt{\vert y\vert ^2-1}}f_{n_k}\}_k$ is convergent in $L^2({\mathcal {H}}^3)$.

Theorem 1.3

There exists an extremizer in $L^2({\overline{{\mathcal {H}}}}^3)$ for inequality (1.10). Moreover, for every $L^2$-normalized complex valued extremizing sequence $\{f_n\}_{n}$ for (1.10), there exist a subsequence $\{f_{n_k}\}_{k}$ and sequences $\{\xi _k\}_{k}\subset \mathbb {R}^4$ and $\{L_k\}_k\subset {\mathcal {L}}$ such that $\{e^{i\xi _k\cdot \xi }L_k^*f_{n_k}\}_k$ is convergent in $L^2(\overline{{\mathcal {H}}}^3)$.

In the statement of the theorems we are writing $e^{ix_k\cdot y}e^{it_k\sqrt{\vert y\vert ^2-1}}f_{n_k}$ for the function $y\mapsto e^{ix_k\cdot y}e^{it_k\sqrt{\vert y\vert ^2-1}}f_{n_k}(y)$ and $e^{i\xi _k\cdot \xi }L_k^*f_{n_k}$ for the function $\xi \mapsto e^{i\xi _k\cdot \xi }f_{n_k}(L_k\xi )$.

Remark 1.4

Note the qualitative difference regarding existence of extremizers between the one-sheeted hyperboloid and the two-sheeted hyperboloid (or their upper sheets) equipped with its Lorentz invariant measure, which are defined respectively by

$$\begin{aligned} \{(x,t)\in \mathbb {R}^3\times \mathbb {R}:t^2=\vert x\vert ^2+1\},\, \bigr (\,\varvec{\delta }\! \begin{pmatrix} t-\sqrt{\vert x\vert ^2+1} \end{pmatrix}\!+\,\varvec{\delta }\! \begin{pmatrix} t+\sqrt{\vert x\vert ^2+1} \end{pmatrix}\!\bigr ) \frac{\,\textrm{d}t\,\textrm{d}x}{\sqrt{\vert x\vert ^2+1}}, \end{aligned}$$

both of which can be considered as “perturbations” of the cone. It was shown in [39] that for the $L^2$ to $L^4(\mathbb {R}^4)$ adjoint Fourier restriction inequality on the two-sheeted hyperboloid and on its upper sheet, extremizers do not exist and the best constant was calculated explicitly. On the other hand, for the $L^2$ to $L^4(\mathbb {R}^4)$ adjoint Fourier restriction inequality on the cone, extremizers exist, their exact form was obtained and the best constant was calculated (see [5]).

We note that the results in [21] do not apply to the case of the hyperboloid due to the lack of scale invariance, but information can be obtained from the arguments therein, although we will not go that route. See the discussion in [39, Sect. 2] for some details in the related context of the two-sheeted hyperboloid.

We take this opportunity to indicate a correction to [39, Thm. 1.2, Prop. 7.5], where the value of the best constant for the $L^2\rightarrow L^6$ adjoint Fourier restriction inequality on the two-sheeted hyperboloid in $\mathbb {R}^2$, there denoted $\bar{{{\textbf {H}} }}_{2,6}$, is incorrect. Details can be found in version 3 of [39] available at www.arxiv.org.

The convolution form of inequalities (1.9) and (1.10), obtained via Plancherel’s theorem, tells us that in both cases, ${\mathcal {H}}^3$ and $\overline{{\mathcal {H}}}^3$, there exist nonnegative real valued extremizers, and the symmetrization method used in [23], or the one in [35], can be adapted to show that if a function f is a nonnegative real valued extremizer for ${\overline{T}}$ on $\overline{{\mathcal {H}}}^3$ then f is necessarily an even function: $f(x,t)=f(-x,-t)$, for ${\bar{\mu }}$-a.e. $(x,t)\in {\overline{{\mathcal {H}}}}^3$. We discuss the details in Sect. 2.

It would be of interest to treat the endpoint $p=\frac{10}{3}$ as well, and more generally to study the existence of extremizers at the endpoint and non-endpoint cases for all²$d\geqslant 2$, as was recently done for non-endpoint cases of the two-sheeted hyperboloid in [9, 10]. Our analysis here extends the known results on sharp Fourier extension inequalities for quadric manifolds as studied in Strichartz paper [43].

1.3 Organization of the Paper and Outline of the Proofs of the Main Theorems

From now on, references to the sharp inequalities (1.4) and (1.7) are understood with $p=4$, unless it is explicitly said otherwise.

An important tool in our proofs is [20, Prop. 1.1] which we include next for the convenience of the reader.

Proposition 1.5

Let $\mathbb {H}$ be a Hilbert space and $S:\mathbb {H}\rightarrow L^p(\mathbb {R}^d)$ be a continuous linear operator, for some $p\in (2,\infty )$. Let $\{f_n\}_{n}\subset \mathbb {H}$ be such that:

• (i) $\Vert f_n \Vert _{\mathbb {H}}=1$,

• (ii) $\lim \limits _{n\rightarrow \infty }\Vert Sf_n \Vert _{L^p(\mathbb {R}^d)}=\Vert S \Vert _{\mathbb {H}\rightarrow L^p(\mathbb {R}^d)}$,

• (iii) $f_n\rightharpoonup f$ and $f\ne 0$,

• (iv) $Sf_n\rightarrow Sf$ a.e. in $\mathbb {R}^d$.

Then $f_n\rightarrow f$ in $\mathbb {H}$. In particular, $\Vert f \Vert _{\mathbb {H}}=1$ and $\Vert Sf \Vert _{L^p(\mathbb {R}^d)}=\Vert S \Vert _{\mathbb {H}\rightarrow L^p(\mathbb {R}^d)}.$

To prove Theorem 1.2 we apply Proposition 1.5 with $p=d=4$, $\mathbb {H}$ equals to $L^2({\mathcal {H}}^3)$ and S equals T. We need to verify (iii) and (iv), as (i) and (ii) are immediate for a normalized extremizing sequence. We handle (iv) as in [38, Prop. 8.3] and [21]. To prove (iii) we will see that the only way it can fail, the failure being that a weak limiting function equals zero, is that the extremizing sequence concentrates at infinity, which is defined as follows for ${\mathcal {H}}^3$, with an analogous definition for ${\overline{{\mathcal {H}}}}^3$.

Definition 1.6

We say that the sequence $\{f_n\}_{n}\subset L^2({\mathcal {H}}^3)$ concentrates at infinity if $\inf \limits _n\Vert f_n \Vert _{L^2({\mathcal {H}}^3)}>0$ and for every $\varepsilon ,R>0$ there exists $N\in \mathbb {N}$ such that for all $n\geqslant N$

$$\begin{aligned} \Vert f_n\mathbb {1}_{\{\vert y\vert \leqslant R\}} \Vert _{L^2({\mathcal {H}}^3)}<\varepsilon , \end{aligned}$$

where, as mentioned before, we are identifying a function on ${\mathcal {H}}^3$ with a function on $\{y\in \mathbb {R}^3:\vert y\vert \geqslant 1\}$.

Finally, to discard the possibility of concentration at infinity we will use a comparison argument with the cone.

In the case of the full one-sheeted hyperboloid ${\overline{{\mathcal {H}}}}^3$ there is the addition of Lorentz invariance, and our proof of Theorem 1.3 will require additional steps when compared to the case of the upper half, ${\mathcal {H}}^3$. Because of this, in addition to the use of Proposition 1.5 and a comparison to the double cone, we will use a concentration-compactness argument to be able to discard concentration at infinity.

More in detail, the organization of the paper is as follows. In Sect. 3 we explicitly calculate the double convolution $\mu *\mu$ which we use in Sect. 4 to prove the strict lower bounds

$$\begin{aligned} {{\textbf {H}} }_4>(2\pi )^{5/4},\qquad \overline{{{\textbf {H}} }}_4>\Bigl (\frac{3}{2}\Bigr )^{1/4}(2\pi )^{5/4}, \end{aligned}$$

(1.11)

which tell us that the best constant for the adjoint Fourier restriction operator on the (resp. full) one-sheeted hyperboloid is strictly greater than that for the (resp. double) cone.

In Sect. 5 we prove Theorem 1.2 by collecting the necessary ingredients to use Proposition 1.5. Here the first inequality in (1.11) is used to show that the $L^2$ mass of an extremizing sequence can not tend to infinity (i.e. there is no concentration at infinity).

From Sect. 6 onward we focus on the full one-sheeted hyperboloid ${\overline{{\mathcal {H}}}}^3$. As mentioned before, the existence of Lorentz invariance adds complexity to the proof of Theorem 1.3, compared to the much simpler proof of Theorem 1.2. We will use a concentration-compactness type argument that we discuss in Sect.  9. In short, given an $L^2$ normalized extremizing sequence $\{f_n\}_n$ for ${\overline{T}}$, three possibilities hold (possibly after passing to a further subsequence): compactness, vanishing or dichotomy. In Sect. 10 we prove bilinear estimates at (radial) dyadic scales and show that they imply that dichotomy can not occur. In Sect. 11 we obtain a (radial) dyadic refinement of (1.7) and use it to show that vanishing can not occur.

To treat the compactness case, it will be necessary to study so called “cap bounds” or refinements of the $L^2\rightarrow L^4$ estimate for the adjoint Fourier restriction operators T and ${\overline{T}}$ and this we achieve in Sect. 8 by “lifting” to the hyperboloid the results for the sphere in $\mathbb {R}^3$, as proved in [13], and recalled in Sect. 7 (more precisely we study so called $\delta$-quasi-extremals and their relationship with caps). By doing this lifting of the cap refinements available for the sphere, we do not have to develop bilinear estimates in the angular variable, but only in the radial variable.

In Sect. 12 we study some limiting relationships between the hyperboloid and the cone. The results of this section together with the second strict inequality in (1.11) are used to study the compactness alternative in the case of an extremizing sequence concentrating at infinity, discarding some possible behaviors.

Finally, in Sect. 13 we put together all the preliminary results of previous sections to show that if an extremizing sequence satisfies compactness then it is precompact in $L^2({\overline{{\mathcal {H}}}}^3)$, modulo multiplication by characters and composition with Lorentz transformations, completing the proof of Theorem 1.3.

Although our approach to the proof of Theorem 1.3 depends on the Lebesgue exponent “4” being an even integer, which for other works in this field has meant to restrict to nonnegative (and possibly symmetric) extremizing sequences, we point out that we are able to handle the case of general complex valued extremizing sequences. Besides the fact that some arguments are simpler if one works with an even integer as we can multiply out some expressions, they could (in principle) be reworked for general real Lebesgue exponents. In the view of the author, the crucial step where evenness is used is in the inequality $\Vert T(f) \Vert _{L^4(\mathbb {R}^4)}\leqslant \Vert T(\vert f\vert ) \Vert _{L^4(\mathbb {R}^4)}$, which may not hold for non even exponents. This is used in the proof of Theorem 1.3, Case 1.

Having explained our methods, we now mention a different possible path to two aspects of our proof. As stated earlier, in this work we obtain a relationship between quasi-extremals and caps by lifting the known results for the sphere but we mention that there is the alternative route through bilinear estimates to obtain cap refinements of inequalities (1.9) and (1.10). The works [9, 10] treat the related two-sheeted hyperboloid in the non-endpoint cases and of particular interest is the development of bilinear estimates in the angular and radial variables which offer a template to obtain similar results for the one-sheeted hyperboloid (see also [2,3,4]).

A second aspect of our proof is the use of a concentration-compactness type argument. There is a different possible approach, the missing mass method³ (MMM). This is a general framework to address the problem of existence in optimization problems; in this particular setting of maximizers for adjoint Fourier restriction inequalities it was first introduced by Frank et al. [26] for the case of the sphere, and later also successfully applied to power curves and (hyper-)surfaces [15, 25]. It has the advantage of allowing complex valued functions in the setting of general Lebesgue exponents, which could be useful when addressing the remaining cases (specially the endpoint cases) of (1.4) and (1.7), that is, when $10/3\leqslant p<4$ and the ambient space is $\mathbb {R}^4$, as well as the remaining Strichartz estimates for the one-sheeted hyperboloid in $\mathbb {R}^{d+1}$, where $2(d+2)/d\leqslant p\leqslant 2(d+1)/(d-1)$, $d\geqslant 2$, and $6\leqslant p<\infty$ if $d=1$.

1.4 Notation and Some Definitions

The set of natural numbers is $\mathbb {N}=\{0,1,2,\dotsc \}$ and $\mathbb {N}^*=\{1,2,3,\dotsc \}$.

For $s>0$, we let ${\mathcal {H}}^3_s:=\{(x,t):x\in \mathbb {R}^3,\, t=\sqrt{\vert x\vert ^2-s^2}\}$, equipped with the measure

$$\begin{aligned} \,\textrm{d}\mu _s(x,t)=\mathbb {1}_{\{\vert x\vert >s \}}\,\varvec{\delta }\! \begin{pmatrix} t-\sqrt{\vert x\vert ^2-s^2} \end{pmatrix}\!\frac{\,\textrm{d}x\,\textrm{d}t}{\sqrt{\vert x\vert ^2-s^2}}, \end{aligned}$$

(1.12)

and adjoint Fourier restriction operator $T_s$,

$$\begin{aligned} T_sf(x,t)=\widehat{f\mu _s}(-x,-t)=\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert >s\}}e^{ix\cdot y}e^{it\sqrt{\vert y\vert ^2-s^2}} f(y)\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-s^2}}.\nonumber \\ \end{aligned}$$

(1.13)

There are corresponding definitions of ${\overline{{\mathcal {H}}}}^3_s,\,{\bar{\mu }}_s$ and ${\overline{T}}_s$ in analogy with the case $s=1$.

The cone in $\mathbb {R}^4$ is denoted $\Gamma ^3:=\{(y,\vert y\vert ):y\in \mathbb {R}^3\}$ which comes with its Lorentz and scale invariant measure $\sigma _c$,

$$\begin{aligned} \int \limits _{\Gamma ^3}f\,\textrm{d}\sigma _c=\int \limits _{\mathbb {R}^3}f(y,\vert y\vert )\frac{\,\textrm{d}y}{\vert y\vert }. \end{aligned}$$

The adjoint Fourier restriction operator on the cone, $T_c$, is given by the expression

$$\begin{aligned} T_cf(x,t)=\int \limits _{\mathbb {R}^3} e^{ix\cdot y} e^{it\vert y\vert } f(y)\frac{\,\textrm{d}y}{\vert y\vert }, \end{aligned}$$

(1.14)

which acts, a priori, on functions $f\in {\mathcal {S}}(\mathbb {R}^3)$. The adjoint Fourier restriction operator on the double cone, ${\overline{\Gamma }}^3:=\Gamma ^3\cup -\Gamma ^3$, denoted by ${\overline{T}}_c$, is given by the expression

$$\begin{aligned} {\overline{T}}_cf(x,t)=\int \limits _{\mathbb {R}^3} e^{ix\cdot y} e^{it\vert y\vert } f(y,\vert y\vert )\frac{\,\textrm{d}y}{\vert y\vert } +\int \limits _{\mathbb {R}^3} e^{ix\cdot y} e^{-it\vert y\vert } f(y,-\vert y\vert )\frac{\,\textrm{d}y}{\vert y\vert },\qquad \end{aligned}$$

(1.15)

$f\in {\mathcal {S}}(\mathbb {R}^4)$. We let ${{\textbf {C}} }_{4},\overline{{{\textbf {C}} }}_{4}<\infty$ denote the best constants in the inequalities

$$\begin{aligned} \Vert T_c f \Vert _{L^4(\mathbb {R}^4)}\leqslant {{\textbf {C}} }_{4}\Vert f \Vert _{L^2(\Gamma ^3)},\quad \Vert {\overline{T}}_c f \Vert _{L^4(\mathbb {R}^4)}\leqslant \overline{{{\textbf {C}} }}_{4}\Vert f \Vert _{L^2({\overline{\Gamma }}^3)}, \end{aligned}$$

respectively. We sometimes use the alternative notation $\Vert T \Vert ={{\textbf {H}} }_4$, $\Vert {\overline{T}} \Vert =\overline{{{\textbf {H}} }}_4$, $\Vert T_c \Vert ={{\textbf {C}} }_{4}$ and $\Vert {\overline{T}}_c \Vert =\overline{{{\textbf {C}} }}_{4}$.

The sphere of radius $r>0$ in $\mathbb {R}^3$ is ${\mathbb {S}}_r^2:=\{y\in \mathbb {R}^3:\vert y\vert =r\}$. The sphere of radius 1 is denoted simply ${\mathbb {S}}^2$. On ${\mathbb {S}}_r^2$ we consider the measure $\sigma _r$,

$$\begin{aligned} \int \limits _{{\mathbb {S}}_r^2}f\,\textrm{d}\sigma _r=\int \limits _{{\mathbb {S}}^2}f(r\omega )r\,\textrm{d}\sigma (\omega ), \end{aligned}$$

(1.16)

where $\sigma$ is the surface measure on ${\mathbb {S}}^2$. With this choice, $\sigma _r({\mathbb {S}}_r^2)=r\sigma ({\mathbb {S}}^2)$, for all $r>0$. For $r>0$ and a function $f:\mathbb {R}^3\rightarrow {\mathbb {C}}$ we set $f_r:{\mathbb {S}}^2\rightarrow \mathbb {R}$ by $f_r(\cdot )=f(r\,\cdot )$.

We let ${{\textbf {S}} }$ denote the best constant in the convolution form of the Tomas–Stein inequality for the sphere ${\mathbb {S}}^2$,

$$\begin{aligned} \Vert f\sigma *f\sigma \Vert _{L^2(\mathbb {R}^3)}\leqslant {{{\textbf {S}}} }^2\Vert f \Vert _{L^2({\mathbb {S}}^2)}^2.\end{aligned}$$

We also use the following convention. For $f:\overline{{\mathcal {H}}}^3\rightarrow \mathbb {R}$ we write $f=f_++f_-$, where $f_+$ is supported on ${\mathcal {H}}^3$ and $f_-$ on the reflection of ${\mathcal {H}}^3$ with respect to the origin, $-{\mathcal {H}}^3$, and we further identify their domains with $\mathbb {R}^3$ via the orthogonal projection. For $A\subseteq \mathbb {R}^3$ we define

$$\begin{aligned} \int \limits _A f\,\textrm{d}\mu :=\int \limits _{\{(x,t)\in {\mathcal {H}}^3:x\in A\}}f\,\textrm{d}\mu . \end{aligned}$$

$f\in L^1({\mathcal {H}}^3)$, while for ${\overline{{\mathcal {H}}}}^3$,

$$\begin{aligned} \int \limits _{A} f \,\textrm{d}{{\bar{\mu }}}:=\int \limits _{\{(x,t)\in {\overline{{\mathcal {H}}}}^3:x\in A\}} f\,\textrm{d}{{\bar{\mu }}}: \end{aligned}$$

$f\in L^1({\overline{{\mathcal {H}}}}^3)$, so that in both cases the integral over $A\subset \mathbb {R}^3$ equals to the integral over the “lift” of A to ${\mathcal {H}}^3$ or ${\overline{{\mathcal {H}}}}^3$, as it corresponds.

An element $R\in SO(4)$ that preserves the t-axis, $R(0,0,0,1)=(0,0,0,1)$, is canonically identified with an element of SO(3), and as such we will just write $R\in SO(3)$.

We let $\psi _s(r)=\sqrt{r^2-s^2}\mathbb {1}_{\{r\geqslant s\}}$, $\phi _s(t)=\psi _s^{-1}(t)=\sqrt{t^2+s^2}\mathbb {1}_{\{t\geqslant 0\}}$. The (closed) ball of radius $r>0$ centered at $y\in \mathbb {R}^3$ is B(y, r). For a set A, $\mathbb {1}_A$ denotes the characteristic function of A and $A^\complement$, the complement of A with respect to a set containing A that will be understood from context, usually $\mathbb {R}^3$, ${\mathcal {H}}^3$ or ${\overline{{\mathcal {H}}}}^3$. We sometimes slightly abuse notation and use $\vert A\vert$ to denote the measure of a set A, where the measure used must be understood from context, for instance, if A is an interval it refers to the Lebesgue measure, if $A\subseteq {\mathbb {S}}^2$, it refers to the surface measure, etc. The support of a function f is denoted ${\text {supp}}(f)$.

We will use the usual asymptotic notation $X\lesssim Y,\,Y\gtrsim X$ if there exists a constant C (independent of X, Y) such that $\vert X\vert \leqslant CY$; we use $X\asymp Y$ if $X\lesssim Y$ and $Y\lesssim X$; when such constants depend on parameters of the problem that we want to make explicit, such as $\alpha ,\dotsc$ etc., we write $\lesssim _{\alpha ,\dotsc },\gtrsim _{\alpha ,\dotsc }$ and $\asymp _{\alpha ,\dotsc }$. At times we will use the common asymptotic notation $o(\cdot )$ and $O(\cdot )$. Thus, $g_n=o(f_n)$ if $g_n/f_n\rightarrow 0$ as $n\rightarrow \infty$, while $g_n=O(f_n)$ if $\vert g_n\vert \leqslant C\vert f_n\vert$ for all n. If there is more than one parameter, say $n\in \mathbb {N}$ and $s>0$, then $g_n(s)=o_n(f_n(s))$ means the limit of $g_n/f_n\rightarrow 0$ is taken with respect to n and is uniform in s, that is $\sup _s\vert g_n(s)\vert /\vert f_n(s)\vert \rightarrow 0$ as $n\rightarrow \infty$.

2 Lorentz Invariance, Symmetrization and Caps

2.1 Lorentz Invariance

Recall that the Lorentz group on $\mathbb {R}^4$, denoted ${\mathcal {L}}$, is defined as the group of invertible linear transformations in $\mathbb {R}^4$ that preserve the bilinear form

$$\begin{aligned} B(x,y)=x_{4}y_{4}-x_{3}y_{3}-x_2y_2-x_1y_1, \end{aligned}$$

for $x=(x_1,x_2,x_3,x_4)\in \mathbb {R}^4$ and $y=(y_1,y_2,y_3,y_4)\in \mathbb {R}^4$. If $L\in {\mathcal {L}}$ then $\vert \det L\vert =1$. Given that we can write ${\overline{{\mathcal {H}}}}^3=\{(x,t)\in \mathbb {R}^{3+1}:B((x,t),(x,t))=-1\}$ it is direct that ${\mathcal {L}}$ preserves the hyperboloid: $L({\overline{{\mathcal {H}}}}^3)={\overline{{\mathcal {H}}}}^3$, for every $L\in {\mathcal {L}}$. Moreover, ${\mathcal {L}}$ preserves the measure ${\bar{\mu }}$, in the sense that for every $f\in L^1({\overline{{\mathcal {H}}}}^3)$ and $L\in {\mathcal {L}}$

$$\begin{aligned} \int \limits _{{\overline{{\mathcal {H}}}}^3} f(x,t)\,\textrm{d}{\bar{\mu }}(x,t)=\int \limits _{{\overline{{\mathcal {H}}}}^3} f(L(x,t))\,\textrm{d}{\bar{\mu }}(x,t). \end{aligned}$$

(2.1)

To see this, note that a simple calculation shows that we can write

$$\begin{aligned} \,\textrm{d}\bar{\mu }(x,t)=\,\varvec{\delta }\! \begin{pmatrix} t^2-\vert x\vert ^2+1 \end{pmatrix}\!\,\textrm{d}x \,\textrm{d}t\end{aligned}$$

so that

$$\begin{aligned} \int \limits _{\mathbb {R}^4} f(x,t) \,\textrm{d}{{\bar{\mu }}}(x,t)= \int \limits _{\mathbb {R}^4} f(x,t)\,\varvec{\delta }\! \begin{pmatrix} t^2-\vert x\vert ^2+1 \end{pmatrix}\!\,\textrm{d}t\,\textrm{d}x.\end{aligned}$$

Then, if L is a Lorentz transformation and $f\in L^1(\overline{{\mathcal {H}}}^3)$, (2.1) can be seen to hold by the change of variable formula.

For $t\in (-1,1)$ the Lorentz boost $L^t\in {\mathcal {L}}$ is the linear map

$$\begin{aligned} L^t(\xi _1,\xi _2,\xi _3,\tau )=\biggl (\frac{\xi _1+t\tau }{\sqrt{1-t^2}},\xi _2,\xi _3,\frac{t\xi _1+\tau }{\sqrt{1-t^2}}\biggr ), \end{aligned}$$

(2.2)

while $L_t$ denotes the rescaling $L_t:=(1-t^2)^{1/2}L^t$, so that $(L_t)^{-1}=(1-t^2)^{-1/2}L^{-t}$.

2.2 Convolution Form

With the Fourier transform in $\mathbb {R}^d$ normalized as ${\widehat{F}}(x)=\int \limits _{\mathbb {R}^d}e^{-ix\cdot y}F(y)\,\textrm{d}y$ we have the identities

$$\begin{aligned} \widehat{F*G}={\widehat{F}}\,{\widehat{G}},\quad \Vert {\widehat{F}} \Vert _{L^2(\mathbb {R}^d)}=(2\pi )^{d/2}\Vert F \Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

so that using $Tf(x,t)=\widehat{f\mu }(-x,-t)$ and ${\overline{T}}g(x,t)=\widehat{g{\bar{\mu }}}(-x,-t)$ we find the equalities

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}=2\pi \Vert f\mu *f\mu \Vert _{L^2(\mathbb {R}^4)}^{1/2},\quad \Vert {\overline{T}}g \Vert _{L^4(\mathbb {R}^4)}=2\pi \Vert g{{\bar{\mu }}}*g{{\bar{\mu }}} \Vert _{L^2(\mathbb {R}^4)}^{1/2}.\nonumber \\ \end{aligned}$$

(2.3)

Using this convolution form of the $L^4$ norm and the triangle inequality we see that $\Vert Tf \Vert _{L^4(\mathbb {R}^4)}\leqslant \Vert T\vert f\vert \Vert _{L^4(\mathbb {R}^4)}$ and $\Vert {\overline{T}}g \Vert _{L^4(\mathbb {R}^4)}\leqslant \Vert {\overline{T}}\vert g\vert \Vert _{L^4(\mathbb {R}^4)}$, so that if f is an extremizer for (1.4) (resp. g for (1.7)), then so is $\vert f\vert$ (resp. $\vert g\vert$), showing that if extremizers exist then there are nonnegative real valued extremizers.

2.3 Symmetrization

Let $f\in L^2({\overline{{\mathcal {H}}}}^3)$ be a complex valued function. Denote the reflection of f by $\widetilde{f}(x,t)=f(-x,-t)$ and the nonnegative $L^2$-symmetrization of f by

$$\begin{aligned}f_\sharp (x,t)=\biggl (\frac{\vert f(x,t)\vert ^2+\vert f(-x,-t)\vert ^2}{2}\biggr )^{1/2}.\end{aligned}$$

Regarding the relationship between f and $f_\sharp$ we have the following lemma.

Lemma 2.1

Let $f\in L^2({\overline{{\mathcal {H}}}}^3)$ be a complex valued function. Then

$$\begin{aligned} \Vert f{\bar{\mu }}*f{\bar{\mu }} \Vert _{L^2(\mathbb {R}^4)}\leqslant \Vert f_\sharp {\bar{\mu }}*f_\sharp {\bar{\mu }} \Vert _{L^2(\mathbb {R}^4)}. \end{aligned}$$

(2.4)

Proof

As in [23, Proof of Prop. 3.2] we write

$$\begin{aligned} f{\bar{\mu }}*{\widetilde{f}}{\bar{\mu }}(\xi ,\tau )&=\int f(y,s)f(-x,-t)\,\varvec{\delta }\! \begin{pmatrix} (\xi ,\tau )-(y,s)-(x,t) \end{pmatrix}\!\,\text {d}{{\bar{\mu }}}(y,s)\,\text {d}{{\bar{\mu }}}(x,t)\\&=\frac{1}{2}\int (f(y,s)f(-x,-t)+f(-y,-s)f(x,t))\\&\quad \times \,\varvec{\delta }\! \begin{pmatrix} (\xi ,\tau )-(y,s)-(x,t) \end{pmatrix}\!\,\text {d}{{\bar{\mu }}}(y,s)\,\text {d}{{\bar{\mu }}}(x,t), \end{aligned}$$

and apply the Cauchy–Schwarz inequality

$$\begin{aligned} \vert f(y,s)f(-x,-t)+f(-y,-s)f(x,t)\vert \leqslant 2f_\sharp (y,s)f_\sharp (x,t), \end{aligned}$$

to obtain that for all $(\xi ,\tau )\in \mathbb {R}^4$

$$\begin{aligned} \vert f{\bar{\mu }}*{\widetilde{f}}{\bar{\mu }}(\xi ,\tau )\vert \leqslant f_\sharp {\bar{\mu }}*f_\sharp {\bar{\mu }}(\xi ,\tau ). \end{aligned}$$

Then

$$\begin{aligned} \Vert f{\bar{\mu }}*f{\bar{\mu }} \Vert _{L^2(\mathbb {R}^4)}=\Vert f{\bar{\mu }}*\widetilde{f}{\bar{\mu }} \Vert _{L^2(\mathbb {R}^4)}\leqslant \Vert f_\sharp {\bar{\mu }}*f_\sharp {\bar{\mu }} \Vert _{L^2(\mathbb {R}^4)}. \end{aligned}$$

$\square$

Since we also have

$$\begin{aligned} \Vert f \Vert _{L^2({\bar{\mu }})}=\Vert f_\sharp \Vert _{L^2({\bar{\mu }})}, \end{aligned}$$

it follows that if extremizers exist for ${\overline{T}}$, then there exist real valued extremizers for ${\overline{T}}$ which are nonnegative even functions on $\overline{{\mathcal {H}}}^3$. Moreover, any nonnegative real valued extremizer is necessarily even. This can be explained by studying the cases of equality in (2.4) by following the proof of the inequality (see [8] for a detailed discussion in the case of the sphere) or, alternatively, by using the same method as in the proof of [35, Lemma 6.1] where a different approach to symmetrization is used and the cases of equality were studied. Therefore, we have the following result.

Proposition 2.2

If $f\in L^2({\overline{{\mathcal {H}}}}^3)$ is a nonnegative real valued extremizer for (1.7), then $f(x,t)=f(-x,-t)$ for ${{\bar{\mu }}}$-a.e. $(x,t)\in {\overline{{\mathcal {H}}}}^3$.

There are some interesting problems that we do not address in this article:

• (i) the nonnegativity of all real valued extremizers,

• (ii) the relationship between complex and real valued extremizers,

• (iii) the smoothness of extremizers.

We provide the following comments in the context of the $L^2({\mathbb {S}}^{d-1})\rightarrow L^p(\mathbb {R}^d)$ adjoint Fourier restriction inequality on the sphere. Christ and Shao [14] showed that for the case of the the sphere ${\mathbb {S}}^2$ in $\mathbb {R}^3$ and $p=4$ each complex valued extremizer is of the form $x\mapsto ce^{ix\cdot \xi }F(x)$, for some $\xi \in \mathbb {R}^3$, some $c\in {\mathbb {C}}$ and some nonnegative extremizer F, and that extremizers are of class $C^{\infty }$; these results were later expanded to all dimensions $d\geqslant 2$ and even integers p in [36, Lemma 2.2, Thm. 1.2] and [37]. Note that the answer obtained for (ii) resolves (i). By using the outline in [14, 36, 37], the Euler–Lagrange equation, which can be obtained as in [12], and the results in [11] we expect similar relationships for the case of ${\mathcal {H}}^3$ and ${\overline{{\mathcal {H}}}}^3$, but have not investigated the extent to which the arguments would need to be changed.

A related question is that of the rate of decay at infinity of an extremizer for which the argument in [27] gives a possible route; see also [35].

We remark that Theorems 1.2 and 1.3 are stated for general (possibly complex valued) extremizing sequences, that is, we do not assume nonnegativity and/or symmetry.

2.4 Caps

A (closed) spherical cap is a set of the form for some $x_0\in {\mathbb {S}}^2$ and $t>0$. If we want to be explicit about the dependence on $x_0$ and t we write .

A cap ${\mathcal {C}}$ of ${\mathcal {H}}_s^3$ is a set of the form

(2.5)

where $s\leqslant a<b\leqslant \infty$ and is a spherical cap. When $a=s\,2^k$ and $b=s\,2^{k+1}$ for some $k\in \mathbb {Z}$ we say that ${\mathcal {C}}$ is a dyadic cap. We identify a cap ${\mathcal {C}}$ as before with its orthogonal projection to $\mathbb {R}^3\times \{0\}$, and moreover we use spherical coordinates and write the cap in (2.5) as , where the hyperboloid it belongs to will be understood from context. A cap ${\mathcal {C}}$ of $\overline{{\mathcal {H}}}_s^3$ is such that either ${\mathcal {C}}\subseteq {\mathcal {H}}^3_s$ or its reflection with respect to the origin $(-{\mathcal {C}})\subseteq {\mathcal {H}}^3_s$ is a cap on ${\mathcal {H}}_s^3$.

The $\mu _s$-measure of a cap is easily calculated

(2.6)

For a cap in ${\mathcal {H}}^3_s$ and $t>0$ we define the rescaled cap as the cap in ${\mathcal {H}}^3_{ts}$ given by

and note that

$$\begin{aligned} \mu _{ts}(t{\mathcal {C}})=t^2\mu _s({\mathcal {C}}). \end{aligned}$$

(2.7)

We also note that for such a cap ${\mathcal {C}}\subset {\mathcal {H}}^3_s$ there exist $R\in SO(3)$ and $\varepsilon \in [0,\pi ]$ such that

$$\begin{aligned}{} & {} R^{-1}({\mathcal {C}})=\{(r\omega ,\sqrt{r^2-s^2}):a\leqslant r\leqslant b,\nonumber \\{} & {} \omega =(\cos \varphi ,\cos \theta \sin \varphi ,\sin \theta \sin \varphi ),\, \theta \in [0,2\pi ],\,\varphi \in [0,\varepsilon ] \}.\nonumber \\ \end{aligned}$$

(2.8)

Keeping this notation in mind for the rest of the section we study the use of Lorentz transformations and scaling in the regimes where ${\bar{\mu }}({\mathcal {C}})$ is large and small. The following two lemmas will be useful in Sect. 13 when dealing with the full one-sheated hyperboloid ${\overline{{\mathcal {H}}}}^3$. To motivate them, let us see how their need arises as we try to prove the precompactness of an extremizing sequence. Let $\{f_n\}_n\subset L^2({\overline{{\mathcal {H}}}}^3)$ be an extremizing sequence for ${\overline{T}}$. Because of “cap refinements” of (1.10) (Lemma 8.1), for each $f_n$ we can find a dyadic cap , , such that

$$\begin{aligned} \int \limits _{{\mathcal {C}}_n}\vert f_n\vert \,\textrm{d}{\bar{\mu }}\gtrsim {\bar{\mu }}({\mathcal {C}}_n)^{1/2}. \end{aligned}$$

If we could find Lorentz transformations $L_n\in {\mathcal {L}}$ such that for each $n\in \mathbb {N}$, $L_n^{-1}({\mathcal {C}}_n)$ is contained in a fixed ball of $\mathbb {R}^4$, independent of n, then $\{L_n^*f_n\}_n$ does not concentrate at infinity and then its precompactness modulo multiplication by characters $\xi \mapsto e^{i\xi _n\cdot \xi }$ would easily follow (this is the content of Proposition 5.2 below). For this reason, it is useful to study when such Lorentz transformations can be found. As noted in [9, Lemma 4] for the two-sheeted hyperboloid, there are Lorentz transformations that can map certain caps of bounded measure into a ball whose radius depends only on the value of the measure of the cap. We record this property for the one-sheeted hyperboloid in the next lemma.

Lemma 2.3

Let $s>0$, $k\in \mathbb {N}$ and ${\mathcal {C}}_k\subset {\overline{{\mathcal {H}}}}^3_s$ be a dyadic cap of the form , for some spherical cap . Let R and $\varepsilon$ be associated to ${\mathcal {C}}_k$ as in (2.8), then:

• (i) The ${{\bar{\mu }}}_s$-measure of ${\mathcal {C}}_k$ satisfies

  $$\begin{aligned} \begin{aligned} {{\bar{\mu }}}_s({\mathcal {C}}_k)&=3\pi s^2 (1+o_k(1))2^{2k}(1-\cos \varepsilon )\\&=\frac{3\pi s^2}{1+\cos \varepsilon }(1+o_k(1))2^{2k}\sin ^2\varepsilon . \end{aligned} \end{aligned}$$

  (2.9)

• (ii) Suppose $\varepsilon \in [0,\frac{\pi }{2}]$. Then, there exists $t\in [0,1)$ such that the orthogonal projection of $L^{-t}R^{-1}({\mathcal {C}}_k)\subset {\overline{{\mathcal {H}}}}^3_s$ to $\mathbb {R}^3$ is contained in a ball of $\mathbb {R}^3$ of radius comparable to $s+s^{-1}{{\bar{\mu }}}_s({\mathcal {C}}_k)+{{\bar{\mu }}}_s({\mathcal {C}}_k)^{1/2}$.

Proof

Without loss of generality, we may assume that ${\mathcal {C}}_k$ is contained in the upper half ${\mathcal {H}}^3_s$. For part (i), (2.6) implies that the ${{\bar{\mu }}}_s$-measure of ${\mathcal {C}}_k$ is given by the expression

$$\begin{aligned} {{\bar{\mu }}}_s({\mathcal {C}}_k)=\pi s^2(1-\cos \varepsilon )\Bigl (\ln \Bigl (\frac{2^{k+1}+\sqrt{2^{2(k+1)}-1}}{2^k+\sqrt{2^{2k}-1}}\Bigr ) +2^{k+1}\sqrt{2^{2(k+1)}-1}-2^k\sqrt{2^{2k}-1}\Bigr ). \end{aligned}$$

The expression involving the logarithm converges to $\ln (2)$ as $k\rightarrow \infty$, while

$$\begin{aligned} 2^{k+1}\sqrt{2^{2(k+1)}-1}-2^k\sqrt{2^{2k}-1} =3\cdot 2^{2k}(1+o_k(1)). \end{aligned}$$

For part (ii), let $R\in SO(3)$ and $\varepsilon \in [0,\frac{\pi }{2}]$ be such that (2.8) holds. The image of $R^{-1}({\mathcal {C}}_k)$ under the Lorentz boost $L^{-t}$ is

$$\begin{aligned} L^{-t}R^{-1}({\mathcal {C}}_k)= & {} \Bigl \{\Bigl (\frac{r\cos \varphi \, -\, t\sqrt{r^2\,-\,s^2}}{\sqrt{1\,-\,t^2}},r\cos \theta \sin \varphi ,r\sin \theta \sin \varphi ,\frac{\sqrt{r^2\,-\,s^2}\,-\,tr\cos \varphi }{\sqrt{1\,-\,t^2}}\Bigr ){:}\nonumber \\{} & {} r \in [s2^k,s2^{k+1}],\theta \in [0,2\pi ],\varphi \in [0,\varepsilon ]\Bigr \}. \end{aligned}$$

(2.10)

Let $t=\sqrt{1-2^{-2(k+1)}}$, so that the first coordinate of a point in the set on the right hand side of (2.10) is bounded as follows

$$\begin{aligned} \biggl \vert \frac{r\cos \varphi -t\sqrt{r^2-s^2}}{\sqrt{1-t^2}} \biggr \vert&=2^{k+1}r\vert \cos \varphi -\sqrt{1-2^{-2(k+1)}}\sqrt{1-(s/r)^2}\vert \\&\leqslant 2^{2(k+1)}s(1-\cos \varphi )+2^{2(k+1)}s(1-(1-2^{-2(k+1)}))\\&= 2^{2(k+1)}s(1-\cos \varepsilon )+s\\&\lesssim \frac{{{\bar{\mu }}}_s({\mathcal {C}}_k)}{s}+s, \end{aligned}$$

where in the last line we used (2.9). The second and third coordinates are bounded as follows

$$\begin{aligned} \vert r\cos \theta \sin \varphi \vert ,\,\vert r\sin \theta \sin \varphi \vert \leqslant 2^{k+1}s\sin \varepsilon \lesssim \sqrt{{{\bar{\mu }}}({\mathcal {C}}_k)}. \end{aligned}$$

Then $L^{-t}R^{-1}({\mathcal {C}}_k)$ is contained in the set

$$\begin{aligned} \Bigl \{(x,t)\in {\overline{{\mathcal {H}}}}^3_s:\vert x\vert \leqslant C\Bigl (\sqrt{{{\bar{\mu }}}_s({\mathcal {C}}_k)}+\frac{{{\bar{\mu }}}_s({\mathcal {C}}_k)}{s}+s\Bigr ) \Bigr \}, \end{aligned}$$

for some constant C independent of k and s. $\square$

Continuing with the comment before Lemma 2.3, suppose now that the measure of the caps ${\mathcal {C}}_n$ is such that $\lim _{n\rightarrow \infty }{\bar{\mu }}({\mathcal {C}}_n)=\infty$, and set $s_n=2^{-N_n}\rightarrow 0$ as $n\rightarrow \infty$, so that if $R_n,\,\varepsilon _n$ are related to ${\mathcal {C}}_n$ as in (2.8), then, (2.9) implies ${\bar{\mu }}({\mathcal {C}}_n)\asymp s_n^{-2}\sin ^2(\varepsilon _n)\rightarrow \infty$ as $n\rightarrow \infty$. We rescale and define so that ${\bar{\mu }}_{s_n}({\widetilde{{\mathcal {C}}}}_n)=s_n^{2}{\bar{\mu }}({\mathcal {C}}_n)$. We may also rescale the sequence $\{f_n\}_n$ by setting $g_n:=s_n^{-1}f_n(s_n^{-1}\cdot )\in L^2({\overline{{\mathcal {H}}}}^3_{s_n})$, which then satisfies

$$\begin{aligned} \int \limits _{{\widetilde{{\mathcal {C}}}}_n}\vert g_n\vert \,\textrm{d}{\bar{\mu }}_{s_n}\gtrsim {\bar{\mu }}_{s_n}({\widetilde{{\mathcal {C}}}}_n)^{1/2}. \end{aligned}$$

(2.11)

If the sequence $\{{\bar{\mu }}_{s_n}({\widetilde{{\mathcal {C}}}}_n)\}_n$ (possesses a subsequence that) is bounded below away from zero, then we will be able to use a comparison argument with the cone, as in a sense the ${\overline{{\mathcal {H}}}}^3_{s_n}$’s are approaching the cone ${\overline{\Gamma }}^3$, as $n\rightarrow \infty$. In this way, it will be established that this possibility does not arise and here the strict inequality between the best constants of this two manifolds comes into play. We are then lead to consider the complementary case, that is, when $\{{\bar{\mu }}_{s_n}({\widetilde{{\mathcal {C}}}}_n)\}_n$ converges to zero. In this scenario we would like to use Lorentz transformations together with dilations in the following way. We want to find a sequence $\{L_n\}_n\subset {\mathcal {L}}$ such that ${\tilde{f}} _n:=L_n^*f_n$ can be appropriately rescaled so that ${\tilde{g}}_n:=a_n^{-1}{\tilde{f}}_n(a_n^{-1}\cdot )\in L^2({\overline{{\mathcal {H}}}}^3_{a_n})$, for some sequence $a_n\rightarrow 0$ as $n\rightarrow \infty$, satisfies (2.11) with the corresponding sequence $\{{\bar{\mu }}_{a_n}({\widetilde{{\mathcal {C}}}}_n)\}_n$ bounded below away from zero. In this way we will also be able to rule out this scenario. The following lemma will tell us how to find the $L_n$’s and the $a_n$’s.

Lemma 2.4

Let $s\leqslant \frac{1}{2}$, be a spherical cap and be a cap in the hyperboloid ${\mathcal {H}}^3_s$. Let R and $\varepsilon$ be as in (2.8) and suppose that $\varepsilon \in [0,\frac{\pi }{2}]$ and $s^{-2}\sin ^2\varepsilon \geqslant 8$. Then there exist $0\leqslant t<1$ such that $L_t^{-1}R^{-1}({\mathcal {C}})\subset {\mathcal {H}}^3_{\frac{s}{\sqrt{1-t^2}}}$ satisfies

$$\begin{aligned} \mu _{\frac{s}{\sqrt{1-t^2}}}(L_t^{-1}R^{-1}({\mathcal {C}}))\geqslant \tfrac{\pi }{2}\text { and } L_t^{-1}R^{-1}({\mathcal {C}})\subseteq [\tfrac{7}{16},\tfrac{33}{16}]\times {\mathbb {S}}^2. \end{aligned}$$

(2.12)

Moreover, if $\varepsilon \in [0,\frac{\pi }{3}]$, we can take $t=\cos \varepsilon$, while if $\varepsilon \in (\frac{\pi }{3},\frac{\pi }{2}]$ we can take $t=0$.

We point out that the value “8” in the inequality $s^{-2}\sin ^2\varepsilon \geqslant 8$ is meant to mean “large” and can be change to any other positive constant with the understanding that the values in (2.12) will change accordingly. Note that in the comment before the statement of the previous lemma we had $s_n^{-2}\sin ^2(\varepsilon _n)\rightarrow \infty$ as $n\rightarrow \infty$ so that in the application that condition will surely be fulfilled. We will then take $t_n=\cos \varepsilon _n$ and $a_n=s_n/\sqrt{1-t_n^2}=(s_n^{-1}\sin \varepsilon _n)^{-1}\rightarrow 0$, as $n\rightarrow \infty$.

Proof of Lemma 2.4

With $R\in SO(3)$ and $\varepsilon \in [0,\frac{\pi }{2}]$ satisfying (2.8), note that $L_t^{-1}R^{-1}({\mathcal {C}})=(1-t^2)^{-1/2}L^{-t}R^{-1}({\mathcal {C}})\subseteq {\mathcal {H}}^3_{s(1-t^2)^{-1/2}}$, for every $t\in (-1,1)$. According to (2.6), the $\mu _s$-measure of ${\mathcal {C}}$ satisfies

$$\begin{aligned} \mu _s({\mathcal {C}})&=2\pi (1-\cos \varepsilon )\Bigl (\frac{{s}^{2}}{2}\ln \left( \sqrt{{r}^{2}-{s}^{2}}+r\right) +\frac{r}{2}\sqrt{{r}^{2}-{s}^{2}}\Bigr )\Big \vert _{1}^2\\&\geqslant \pi (1-\cos \varepsilon )(\sqrt{4-s^2}-\sqrt{1-s^2})\geqslant \pi (1-\cos \varepsilon ), \end{aligned}$$

so that in what follows we can assume $\cos \varepsilon \geqslant 1/2$, otherwise we are done by taking $t=0$. From (2.7), for $t\in (0,1)$,

$$\begin{aligned} \mu _{\frac{s}{\sqrt{1-t^2}}}(L_t^{-1}R^{-1}({\mathcal {C}}))=(1-t^2)^{-1}\mu _s({\mathcal {C}}), \end{aligned}$$

so that choosing $t=\cos \varepsilon$ gives $\mu _{s(1-t^2)^{-1/2}}(L_t^{-1}R^{-1}({\mathcal {C}}))\geqslant \frac{\pi }{1+\cos \varepsilon }\geqslant \frac{\pi }{2}$. On the other hand, we have

$$\begin{aligned} L_t^{-1}R^{-1}({\mathcal {C}})= & {} {} \Bigl \{(1-t^2)^{-1/2}\Bigl (\frac{r\cos \varphi \,-\,t\sqrt{r^2\,-\,s^2}}{\sqrt{1\,-\,t^2}},\\{} & {} r\cos \theta \sin \varphi ,r\sin \theta \sin \varphi , \frac{\sqrt{r^2\,-\,s^2}\,-\,tr\cos \varphi }{\sqrt{1\,-\,t^2}}\Bigr ):r\in [1,2],\\{} & {} \theta \in [0,2\pi ],\varphi \in [0,\varepsilon ]\Bigr \}, \end{aligned}$$

and since $\cos \varphi \geqslant \cos \varepsilon$ and $1\leqslant r\leqslant 2$ we obtain that the fourth coordinate of any point in $L_t^{-1}R^{-1}({\mathcal {C}})$ is bounded as follows

$$\begin{aligned} \frac{\sqrt{r^2-s^2}-tr\cos \varphi }{1-t^2}=\frac{r(\sqrt{1-(s/r)^2}-t\cos \varphi )}{1-t^2}\leqslant 2\frac{1-\cos ^{2}\varepsilon }{1-\cos ^2\varepsilon }= 2, \end{aligned}$$

and

$$\begin{aligned} \frac{r\left( \sqrt{1-(s/r)^2}-t\cos \varphi \right) }{1-t^2}&= \frac{r}{\sin ^2\varepsilon }\left( \sqrt{1-(s/r)^2}-\cos \varepsilon \cos \varphi \right) \\&\geqslant \frac{r}{\sin ^2\varepsilon }\left( \sqrt{1-(s/r)^2}-\cos \varepsilon \right) \\&=\frac{r}{\sqrt{1-(s/r)^2}+\cos \varepsilon }\Bigl (1-\frac{1}{r^2s^{-2}\sin ^2\varepsilon }\Bigr )\\&\geqslant \frac{r}{2}\Bigl (1-\frac{1}{8r^2}\Bigr )\geqslant \frac{7}{16}. \end{aligned}$$

Therefore

$$\begin{aligned} L_t^{-1}R^{-1}({\mathcal {C}})\subseteq [\phi _{\frac{s}{\sqrt{1-t^2}}}(\tfrac{7}{16}),\phi _{\frac{s}{\sqrt{1-t^2}}}(2)]\times {\mathbb {S}}^2. \end{aligned}$$

Now, from the definition of t and the assumption that $s^{-2}\sin ^2\varepsilon \geqslant 8$ we obtain

$$\begin{aligned} \frac{s}{\sqrt{1-t^2}}=\frac{s}{\sin \varepsilon }\leqslant \frac{\sqrt{2}}{4}, \end{aligned}$$

so that the following inequalities hold

$$\begin{aligned}r\leqslant \phi _{\frac{s}{\sqrt{1-t^2}}}(r)=\sqrt{r^2+s^2(1-t^2)^{-1}}\leqslant \sqrt{r^2+1/8},\end{aligned}$$

from where $\phi _{\frac{s}{\sqrt{1-t^2}}}\left( \tfrac{7}{16}\right) \geqslant \frac{7}{16}$ and $\phi _{\frac{s}{\sqrt{1-t^2}}}(2)\leqslant \frac{33}{16}$ and then we find $L_t^{-1}R^{-1}({\mathcal {C}})\subseteq \left[ \tfrac{7}{16},\tfrac{33}{16}\right] \times {\mathbb {S}}^2$. $\square$

3 Calculation of a Double Convolution

In previous studies of quadric surfaces and curves and their perturbations it has become clear the importance of the double or triple, and more generally the n-th fold, convolution of the underlying measure. Its properties may determine existence or nonexistence of extremizers and in some cases it can be used to find their explicit form and/or the value of the best constant in the corresponding adjoint Fourier restriction inequality. In the case of the one-sheeted hyperboloid and its upper half, the double convolution will be used to prove that extremizing sequences do not concentrate at infinity.

Let $\mu _s*\mu _s$ denote the double convolution of $\mu _s$ with itself, defined by duality

$$\begin{aligned} \langle \mu _s*\mu _s,f\rangle =\int \limits _{(\mathbb {R}^4)^2} f(x+x',t+t') \,\text {d}\mu _s(x,t)\,\text {d}\mu _s(x',t'), \end{aligned}$$

for all $f\in {\mathcal {S}}(\mathbb {R}^4)$. It is not difficult to see that $\mu _s*\mu _s$ is absolutely continuous with respect to the Lebesgue measure in $\mathbb {R}^4$, indeed this follows from (1.4) since $e^{-\tau }(\mu _s*\mu _s)\in L^2(\mathbb {R}^4)$, it being the (inverse) Fourier transform of the $L^2(\mathbb {R}^4)$ function $(\widehat{e^{-\tau }\mu _s})^2$ (see also [34, Prop. 2.1]). In what follows we identify $\mu _s*\mu _s$ with its Radon–Nicodym derivative with respect to the Lebesgue measure in $\mathbb {R}^4$.

Proposition 3.1

Let $\mu _s$ be the measure on ${\mathcal {H}}^3_s$ defined in (1.12). Then

• (i) The support of the convolution measure $\mu _s*\mu _s$ is

  $$\begin{aligned} {\textrm{supp}}(\mu _s*\mu _s)=\{(\xi ,\tau )\in \mathbb {R}^4:\tau \geqslant 0,\,\vert \xi \vert \leqslant \sqrt{\tau ^2+s^2}+s\}. \end{aligned}$$

• (ii) For every $(\xi ,\tau )\in \mathbb {R}^4$ with $\tau \geqslant 0$ we have the formula

  

  (3.1)

When $\xi =0$ and $\tau > 0$ we understand that in (3.1) $\mu _s*\mu _s(0,\tau )\,{=}\, 2\pi \bigg (1\,{+}\,\frac{4s^2}{\tau ^2}\bigg )^{1/2}$.

We postpone the proof of Proposition 3.1 and study the behavior of $\mu _s *\mu _s(\xi ,\tau )$ for large $\tau$.

Lemma 3.2

For all $\tau >0$,

$$\begin{aligned} 2\pi \Bigl (1+\frac{4s^2}{\tau ^2}\Bigr )^{1/2}\leqslant \sup _{\xi \in \mathbb {R}^3}\mu _s*\mu _s(\xi ,\tau )\leqslant 2\pi \Bigl (1+\frac{2s}{\tau }\Bigr ). \end{aligned}$$

In particular

$$\begin{aligned} \lim _{\tau \rightarrow \infty } \sup _{\xi \in \mathbb {R}^3}\mu _s*\mu _s(\xi ,\tau )=2\pi .\end{aligned}$$

Proof

We start by noting that

$$\begin{aligned} \mu _s*\mu _s(s\xi ,s\tau )=\mu *\mu (\xi ,\tau ),\end{aligned}$$

hence it is enough to consider the case $s=1$. We analyze the different cases in formula (3.1).

Case 1: $\vert \xi \vert <\sqrt{\tau ^2+1}-1$. Then

$$\begin{aligned} \Bigl (1+\frac{4}{\tau ^2}\Bigr )^{1/2}\leqslant \Bigl (1+\frac{4}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2}\leqslant \Bigl (\frac{\sqrt{\tau ^2+1}+1}{\sqrt{\tau ^2+1}-1}\Bigr )^{1/2}=\frac{\tau }{\sqrt{\tau ^2+1}-1}.\end{aligned}$$

Case 2: $\sqrt{\tau ^2+1}-1\leqslant \vert \xi \vert \leqslant \sqrt{\tau ^2+4}$. Then

$$\begin{aligned} \frac{\tau }{\sqrt{\tau ^2+4}}\leqslant \frac{\tau }{\vert \xi \vert }\leqslant \frac{\tau }{\sqrt{\tau ^2+1}-1}. \end{aligned}$$

Case 3: $\sqrt{\tau ^2+4}< \vert \xi \vert \leqslant \sqrt{\tau ^2+1}+1$. Then $\vert \xi \vert ^2-\tau ^2>4$ and

$$\begin{aligned} \vert \xi \vert \mapsto \frac{\tau }{\vert \xi \vert }-\Bigl (1+\frac{4}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2}, \end{aligned}$$

is a decreasing function of $\vert \xi \vert$. Then

$$\begin{aligned} \frac{\tau }{\vert \xi \vert }-\Bigl (1+\frac{4}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2}\leqslant \frac{\tau }{\sqrt{\tau ^2+4}},\end{aligned}$$

and

$$\begin{aligned} \frac{\tau }{\vert \xi \vert }-\Bigl (1+\frac{4}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2}\geqslant \frac{\tau }{\sqrt{\tau ^2+1}+1}- \Bigl (1-\frac{2}{\sqrt{\tau ^2+1}+1}\Bigr )^{1/2}=0. \end{aligned}$$

As a conclusion, for all $\tau >0$ and $\xi {\in }\,\mathbb {R}^3$

$$\begin{aligned} \mu *\mu (\xi ,\tau )\leqslant \frac{2\pi \tau }{\sqrt{\tau ^2+1}-1}=2\pi \Bigl (\Bigl (1+\frac{1}{\tau ^2}\Bigr )^{1/2}+\frac{1}{\tau }\Bigr ) \leqslant 2\pi \Bigl (1+\frac{2}{\tau }\Bigr ), \end{aligned}$$

and for $\tau >0$

$$\begin{aligned} \sup _{\xi \in \mathbb {R}^3}\mu *\mu (\xi ,\tau )\geqslant 2\pi \Bigl (1+\frac{4}{\tau ^2}\Bigr )^{1/2}. \end{aligned}$$

$\square$

We now turn to the proof of Proposition 3.1.

Proof of Proposition 3.1

Part (i) is a simple calculation and is left to the reader. For part (ii) we start by discussing a change of coordinates that was used in the proof of [22, Lemma 5.1] in the arxiv’s second version of [22]; see also Appendix 3 on the arxiv’s version of [39] where an outline of the computation of the double convolution of the Lorentz invariant measure on the two-sheeted hyperboloid was given using the same technique.

For each fixed $\xi \ne 0$ we consider a spherical coordinate system with axis $\xi$, that is, each $\eta \in \mathbb {R}^3$ is described as $\eta =(\rho \cos \theta \sin \varphi ,\rho \sin \theta \sin \varphi ,\rho \cos \varphi )$, where $\rho =\vert \eta \vert \geqslant 0$, $\varphi \in [0,\pi ]$ is the angle between $\xi$ and $\eta$ and $\theta \in [0,2\pi ]$ is a polar coordinate angle on the plane orthogonal to $\xi$. Then $\,\textrm{d}\eta =\rho ^2\sin \varphi \,\textrm{d}\rho \,\textrm{d}\theta \,\textrm{d}\varphi$.

Define the new variable $\varsigma =\vert \xi -\eta \vert$, which corresponds to the size of the side opposite to the origin, 0, in the triangle whose vertices are located at $0,\,\xi$ and $\eta$. Then

$$\begin{aligned} \varsigma ^2=\vert \xi \vert ^2+\rho ^2-2\vert \xi \vert \rho \cos \varphi . \end{aligned}$$

Changing variables from $\varphi$ to $\varsigma$, gives $\varsigma \,\textrm{d}\varsigma =\vert \xi \vert \rho \sin \varphi \,\textrm{d}\varphi$, so that in the variables $(\rho ,\varsigma ,\theta )$ we have $\,\textrm{d}\eta =\frac{\rho \varsigma }{\vert \xi \vert }\,\textrm{d}\rho \,\textrm{d}\varsigma \,\textrm{d}\theta$. The range of $\varsigma$ can be seen by using that $\varsigma$, $\vert \xi \vert$ and $\rho$ are the sizes of the sides of a triangle, so $\vert \rho -\varsigma \vert \leqslant \vert \xi \vert \leqslant \rho +\varsigma$, which translates into $\vert \vert \xi \vert -\rho \vert \leqslant \varsigma \leqslant \vert \xi \vert +\rho$.

Using delta calculus (see for instance the survey article [24]) and the previous change of variables we have

where we changed variables $u=\sqrt{\rho ^2-s^2},\, v=\sqrt{\varsigma ^2-s^2}$ and $R_s=R_s(\xi )$ is the image of the region $\{(\rho ,\varsigma ):\vert \rho -\varsigma \vert \leqslant \vert \xi \vert ,\rho + \varsigma \geqslant \vert \xi \vert ,\rho \geqslant s,\varsigma \geqslant s\}$ under the transformation $(\rho ,\varsigma )\mapsto (u,v)$. Using the change of variables $a=u-v,\, b=u+v$, so that $2\,\textrm{d}u\,\textrm{d}v=\,\textrm{d}a\,\textrm{d}b$, we obtain

(3.2)

where ${\widetilde{R}}_s={\widetilde{R}}_s(\xi )$ is the image of $R_s(\xi )$ under the map $(u,v)\mapsto (a,b)$, $\tilde{\ell }_\tau$ is the horizontal line $\{(a,b)\in \mathbb {R}^2:b=\tau \}$, $\ell _\tau$ is the line $\{(u,v)\in \mathbb {R}^2:u+v=\tau \}$ and $\vert R_s\cap \ell _\tau \vert$ denotes the measure of $R_s\cap \ell _\tau$ as a subset of $\ell _\tau$ with the induced Lebesgue measure. In order to calculate $\vert R_s\cap \ell _\tau \vert$ we divide the analysis into two cases.

Case 1: $\vert \xi \vert \leqslant 2s$. The boundary of the region

$$\begin{aligned} \bigg \{(\rho ,\varsigma ):\vert \rho -\varsigma \vert \leqslant \vert \xi \vert ,\rho + \varsigma \geqslant \vert \xi \vert ,\rho \geqslant s,\varsigma \geqslant s\bigg \}, \end{aligned}$$

consists of two (bounded) line segments and two half lines. Its image in the (u, v)-plane, $R_s$, is bounded by two line segments and two curves and is symmetric with respect to the diagonal $u=v$. The line segments have equations

$$\begin{aligned} \bigg \{(u,v):u=0,\,0\leqslant v\leqslant \sqrt{(\vert \xi \vert +s)^2-s^2}\bigg \}, \bigg \{(u,v):0\leqslant u\leqslant \sqrt{(\vert \xi \vert +s)^2-s^2},\,v=0\bigg \}, \end{aligned}$$

and the curves have equations

$$\begin{aligned} \begin{aligned}&\bigg \{(u,v):u\geqslant 0,\, v=\bigg ((\sqrt{u^2+s^2}+\vert \xi \vert )^2-s^2\bigg )^{1/2}\bigg \},\\&\bigg \{(u,v):u\geqslant \bigg ((\vert \xi \vert +s)^2-s^2\bigg )^{1/2},\, v=\bigg ((\sqrt{u^2+s^2}-\vert \xi \vert )^2-s^2\bigg )^{1/2}\bigg \}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.3)

Then $\vert R_s\cap \ell _\tau \vert$ is given by

$$\begin{aligned} \vert R_s\cap \ell _\tau \vert ={\left\{ \begin{array}{ll} \sqrt{2}\tau &{}\text {, if }0\leqslant \tau \leqslant \sqrt{(\vert \xi \vert +s)^2-s^2}\\ \sqrt{2}\vert u-v\vert &{}\text {, if } \tau >\sqrt{(\vert \xi \vert +s)^2-s^2}, \end{array}\right. } \end{aligned}$$

where in the last expression u and v are related to $(\xi ,\tau )$ by the equations $u+v=\tau$ and $v=\bigl ((\sqrt{u^2+s^2}+\vert \xi \vert )^2-s^2\bigr )^{1/2}$. Therefore

where $u_1(\xi ,\tau )$ and $(\xi ,\tau )$ are related by the expression

$$\begin{aligned} \tau =u_1(\xi ,\tau )+\bigl ((\sqrt{u_1(\xi ,\tau )^2+s^2}+\vert \xi \vert )^2-s^2\bigr )^{1/2}, \end{aligned}$$

(3.4)

and $0\leqslant u_1(\xi ,\tau )\leqslant \frac{\tau }{2}$.

Case 2: $\vert \xi \vert >2s$. Now the boundary of the region $\{(\rho ,\varsigma ):\vert \rho -\varsigma \vert \leqslant \vert \xi \vert ,\rho + \varsigma \geqslant \vert \xi \vert ,\rho \geqslant s,\varsigma \geqslant s\}$ consists of three (bounded) line segments and two half lines and the region $R_s$ is now bounded by two line segments and three curves. The line segments have equations

$$\begin{aligned}{} & {} \{(u,v):u=0,\, \sqrt{(\vert \xi \vert -s)^2-s^2}\leqslant v\leqslant \sqrt{(\vert \xi \vert +s)^2-s^2} \},\\{} & {} \{(u,v):\sqrt{(\vert \xi \vert -s)^2-s^2}\leqslant u\leqslant \sqrt{(\vert \xi \vert +s)^2-s^2},\, v=0 \}. \end{aligned}$$

The next two curves have equations as in (3.3). The last boundary curve is the image of the segment $\{(\rho ,\varsigma ):\rho +\varsigma =\vert \xi \vert ,\, s\leqslant \rho \leqslant \vert \xi \vert -s \}$. Its equation is

$$\begin{aligned} \bigg \{(u,v):0\leqslant u\leqslant \bigl ((\vert \xi \vert -s)^2-s^2\bigr )^{1/2},\,v=\bigl ((\vert \xi \vert -\sqrt{u^2+s^2})^2-s^2\bigr )^{1/2} \bigg \}, \end{aligned}$$

and note that it is the graph of a strictly decreasing and concave function of u. It follows that

$$\begin{aligned} \vert R_s\cap \ell _\tau \vert ={\left\{ \begin{array}{ll} \sqrt{2}(\tau -\vert u_2-v_2\vert ) &{}\text {, if }\sqrt{(\vert \xi \vert -s)^2-s^2}\leqslant \tau \leqslant \sqrt{\vert \xi \vert ^2-(2s)^2},\\ \sqrt{2}\tau &{}\text {, if }\sqrt{\vert \xi \vert ^2-(2s)^2}\leqslant \tau \leqslant \sqrt{(\vert \xi \vert +s)^2-s^2},\\ \sqrt{2}\vert u_1-v_1\vert &{}\text {, if } \tau \geqslant \sqrt{(\vert \xi \vert +s)^2-s^2}, \end{array}\right. } \end{aligned}$$

where $(u_1,v_1),\,(u_2,v_2)$ are the solutions to the equations $u_1+v_1=\tau ,\,u_2+v_2=\tau$, $v_1=\biggl (\bigg (\sqrt{u_1^2+s^2}+\vert \xi \vert \bigg )^2-s^2\biggr )^{1/2}$ and $v_2=\biggl (\bigg (\vert \xi \vert -\sqrt{u_2^2+s^2}\bigg )^2-s^2\biggr )^{1/2}$.

Then

where $u_1(\xi ,\tau )$ is as in (3.4) and $u_2(\xi ,\tau )$ and $(\xi ,\tau )$ are related by the expression

$$\begin{aligned} \tau =u_2(\xi ,\tau )+\biggl (\bigg (\sqrt{u_2(\xi ,\tau )^2+s^2}-\vert \xi \vert \bigg )^2-s^2\biggr )^{1/2}, \end{aligned}$$

and $0\leqslant u_2(\xi ,\tau )\leqslant \frac{\tau }{2}$. Algebraic manipulation shows that for $(\xi ,\tau )$ in their respective domains of definition

$$\begin{aligned} \tau -2u_i(\xi ,\tau )=\vert \xi \vert \Bigl (1+\frac{4s^2}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2},\quad i=1,2. \end{aligned}$$

(3.5)

Collecting all in one expression we have

Replacing $u_1(\xi ,\tau )$ and $u_2(\xi ,\tau )$ using (3.5) we obtain using (3.2)

$$\begin{aligned} \begin{aligned} \mu _s*\mu _s(\tau ,\xi )&=\frac{2\pi }{\vert \xi \vert }\biggl (\tau \mathbb {1}_{\{\tau \leqslant \sqrt{(\vert \xi \vert +s)^2-s^2}\}} \mathbb {1}_{\{\vert \xi \vert \leqslant 2s\}}\\&\quad +\Bigl (\tau -\vert \xi \vert \Bigl (1+\frac{4s^2}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2}\Bigr )\mathbb {1}_{\{\sqrt{(\vert \xi \vert -s)^2-s^2} \leqslant \tau <\sqrt{\vert \xi \vert ^2-(2s)^2}\}}\mathbb {1}_{\{\vert \xi \vert>2s\}}\\&\quad +\tau \mathbb {1}_{\sqrt{\vert \xi \vert ^2-(2s)^2}\leqslant \tau \leqslant \sqrt{(\vert \xi \vert +s)^2-s^2}}\mathbb {1}_{\{\vert \xi \vert>2s\}}\\&\quad +\vert \xi \vert \Bigl (1+\frac{4s^2}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{1/2}\mathbb {1}_{\{\tau >\sqrt{(\vert \xi \vert +s)^2-s^2}\}}\biggr ). \end{aligned} \end{aligned}$$

(3.6)

Rearranging (3.6) we find that $\mu _s *\mu _s$ can be written in the equivalent form (3.1). $\square$

More generally, the same method used in the proof of Proposition 3.1 allows us to write an explicit formula for $\mu _s*\mu _t$, for any $s,\,t\geqslant 0$. For instance, as it will be useful in Sect. 12, we have

$$\begin{aligned} \mu _s*\mu _0(\xi ,\tau )=\frac{\pi }{\vert \xi \vert }\int \limits _{\widetilde{Q}_s(\xi )}\,\varvec{\delta }\! \begin{pmatrix} \tau -b \end{pmatrix}\!\,\text {d}a\,\text {d}b, \end{aligned}$$

(3.7)

where $\widetilde{Q}_s(\xi )$ is the image of the set $\{(\rho ,\varsigma ):\vert \rho -\varsigma \vert \leqslant \vert \xi \vert ,\,\rho +\varsigma \geqslant \vert \xi \vert ,\,\rho \geqslant 0,\,\varsigma \geqslant s \}$ under the transformations $(\rho ,\varsigma )\mapsto (u,v)=(\rho ,\sqrt{\varsigma ^2-s^2})\mapsto (a,b)=(u-v,u+v)$. Here $\mu _0$ equals $\sigma _c$, the Lorentz invariant measure on the cone. A calculation similar to the one for $\mu _s*\mu _s$ gives the following explicit formula

(3.8)

Using (3.8) we see that for each $\tau \geqslant 0$

$$\begin{aligned} \sup _{\xi \in \mathbb {R}^3}\mu _s*\sigma _c(\xi ,\tau )=2\pi \cdot {\left\{ \begin{array}{ll} \frac{\tau }{\sqrt{\tau ^2+s^2}}&{}, 0\leqslant \tau \leqslant \frac{s}{2}(2(\sqrt{5}-1))^{1/2},\\ 1+\frac{(s-\sqrt{s^2-\tau ^2})^2}{\tau ^2}{}&{},\frac{s}{2}(2(\sqrt{5}-1))^{1/2}\leqslant \tau \leqslant s,\\ 1+\frac{s}{2\tau -s}&{}{},\tau \geqslant s, \end{array}\right. }\nonumber \\ \end{aligned}$$

(3.9)

and $\Vert \mu _s*\sigma _c \Vert _{L^\infty (\mathbb {R}^4)}=4\pi$.

The methods introduced in this section allow us to write explicit formulas for double convolutions of the form $f\mu _s*f\mu _s$, for f regular enough (continuous, for instance) similar to those for the sphere [13, pp. 282]. Indeed, unwinding the changes of variables leading to (3.2) in the proof of Proposition 3.1 (as well as the ones leading to (3.7)), for $\xi \in \mathbb {R}^3\setminus \{0\}$ we let

$$\begin{aligned} \alpha _s(a,b,\theta )&=\frac{\vert \xi \vert ^2+ab}{\vert \xi \vert \sqrt{(a+b)^2+4s^2}},\quad \beta _s(a,b,\theta )=\frac{\vert \xi \vert ^2+ab-s^2}{\vert \xi \vert (a+b)}, \\ \omega _{s}(a,b,\theta )&=\bigl (\sqrt{1-\alpha _s(a,b,\theta )^2}\cos \theta ,\sqrt{1-\alpha _s(a,b,\theta )^2}\sin \theta , \alpha _s(a,b,\theta )\bigr ),\\ \vartheta _s(a,b,\theta )&=\bigl (\sqrt{1-\beta _s(a,b,\theta )^2}\cos \theta ,\sqrt{1-\beta _s(a,b,\theta )^2}\sin \theta ,\beta _s(a,b,\theta )\bigr ), \end{aligned}$$

and

$$\begin{aligned} F_s(a,b)&=\int \limits _0^{2\pi }f\Bigl (\xi -\sqrt{(\tfrac{a+b}{2})^2+s^2}\,\omega _s(a,b,\theta )\Bigr )\, f\Bigl (\sqrt{(\tfrac{a+b}{2})^2+s^2}\,\omega _s(a,b,\theta )\Bigr )\,\textrm{d}\theta ,\\ G_s(a,b)&=\int \limits _0^{2\pi }f\bigl (\xi -\tfrac{a+b}{2}\,\vartheta _s(a,b,\theta )\bigr )\, f\bigl (\tfrac{a+b}{2}\,\vartheta _s(a,b,\theta )\bigr )\,\textrm{d}\theta ,\\ H_0(a,b)&=\int \limits _0^{2\pi }f\bigl (\xi -\tfrac{a+b}{2}\,\omega _0(a,b,\theta )\bigr )\, f\bigl (\tfrac{a+b}{2}\,\omega _0(a,b,\theta )\bigr )\,\textrm{d}\theta . \end{aligned}$$

Recalling the sets ${\widetilde{R}}_s(\xi )$ and $\widetilde{Q}_s(\xi )$ from (3.2) and (3.7) we have

$$\begin{aligned} f\mu _s*f\mu _s(\xi ,\tau )&=\frac{1}{2\vert \xi \vert }\int \limits _{\widetilde{R}_s(\xi )}F_s(a,b)\,\varvec{\delta }\! \begin{pmatrix} \tau -b \end{pmatrix}\!\,\text {d}a\,\text {d}b=\frac{1}{2\vert \xi \vert }\int \limits _{\mathbb {R}}F_s(a,\tau )\mathbb {1}_{\widetilde{R}_s(\xi )}(a,\tau )\,\text {d}a, \end{aligned}$$

(3.10)

$$\begin{aligned} f\mu _s*f\sigma _c(\xi ,\tau )&=\frac{1}{2\vert \xi \vert }\int \limits _{ \widetilde{Q}_s(\xi )}G_s(a,b)\,\varvec{\delta }\! \begin{pmatrix} \tau -b \end{pmatrix}\!\,\text {d}a\,\text {d}b=\frac{1}{2\vert \xi \vert }\int \limits _{\mathbb {R}}G_s(a,\tau )\mathbb {1}_{\widetilde{Q}_s(\xi )}(a,\tau )\,\text {d}a, \end{aligned}$$

(3.11)

and

$$\begin{aligned} f\sigma _c*f\sigma _c(\xi ,\tau )=&{} \frac{1}{2\vert \xi \vert }\int \limits _{\{(a,b):\vert a\vert \leqslant \vert \xi \vert \leqslant b\}} H_0(a,b)\,\varvec{\delta }\! \begin{pmatrix} \tau -b \end{pmatrix}\!\,\text {d}a\,\text {d}b\nonumber \\=&{} \frac{1}{2\vert \xi \vert }\int \limits _{\mathbb {R}} H_0(a,\tau )\mathbb {1}_{\{(a,b):\vert a\vert \leqslant \vert \xi \vert \leqslant b \}}(a,\tau )\,\text {d}a. \end{aligned}$$

(3.12)

It is worth noting that $\mathbb {1}_{{\widetilde{Q}}_s(\xi )}\rightarrow \mathbb {1}_{\{(a,b):\vert a\vert \leqslant \vert \xi \vert \leqslant b \}}$ and $\mathbb {1}_{{\widetilde{R}}_s(\xi )}\rightarrow \mathbb {1}_{\{(a,b):\vert a\vert \leqslant \vert \xi \vert \leqslant b \}}$ pointwise in $\mathbb {R}^2$ as $s\rightarrow 0^+$. Moreover, when f is continuous, $F_s\rightarrow H_0$ and $G_s\rightarrow H_0$ pointwise in the region $\{(a,b):a+b\geqslant 0\}\subset \mathbb {R}^2$, as $s\rightarrow 0^+$.

4 Comparison with the Cone

Recall that $\sigma _c$ denotes the scale and Lorentz invariant measure on the cone $\Gamma ^3$ and $T_c$ denotes its associated adjoint Fourier restriction operator. From [5] we know the value of the sharp constant

$$\begin{aligned} \sup _{0\ne f\in L^2(\sigma _c)}\frac{\Vert f\sigma _c*f\sigma _c \Vert _{L^4(\mathbb {R}^4)}^2}{\Vert f \Vert _{L^2(\sigma _c)}^4}= 2\pi . \end{aligned}$$

(4.1)

We had defined the numerical constants

$$\begin{aligned} {{{\textbf {C}}} }_{4}&=\sup _{0\ne f\in L^2(\sigma _c)}\frac{\Vert T_cf \Vert _{L^4(\mathbb {R}^4)}}{\Vert f \Vert _{L^2(\sigma _c)}}=2\pi \sup _{0\ne f\in L^2(\sigma )}\frac{\Vert f\sigma _c*f\sigma _c \Vert _{L^4(\mathbb {R}^4)}^{1/2}}{\Vert f \Vert _{L^2(\sigma _c)}},\\ {{{\textbf {H}}} }_{4}&=\sup _{0\ne f\in L^2(\mu )}\frac{\Vert Tf \Vert _{L^4(\mathbb {R}^4)}}{\Vert f \Vert _{L^2(\mu )}}=2\pi \sup _{0\ne f\in L^2(\mu )}\frac{\Vert f\mu *f\mu \Vert _{L^4(\mathbb {R}^4)}^{1/2}}{\Vert f \Vert _{L^2(\mu )}}. \end{aligned}$$

The next proposition gives a comparison between ${{\textbf {C}} }_{4}$ and ${{\textbf {H}} }_4$ and its role is the analog of the comparison of the best constant for the sphere and the paraboloid in $\mathbb {R}^3$ as used in [13] where a strict inequality was needed to rule out concentration at a pair of antipodal points. In our present case, a strict inequality will rule out concentration at infinity.

Proposition 4.1

$\qquad \displaystyle {{\textbf {H}} }_{4}>{{\textbf {C}} }_{4}.$

Proof

For $s>0$ we consider the family of trial functions $f_a(y)=e^{-\frac{a}{2}\sqrt{\vert y\vert ^2-s^2}}$, $a>0$, and claim that

$$\begin{aligned} \sup _{a>0}\frac{\Vert T_sf_a \Vert _{L^4(\mathbb {R}^4)}}{\Vert f_a \Vert _{L^2({\mathcal {H}}^3_s)}}> \sup _{0\ne f\in L^2(\sigma _c)}\frac{\Vert T_cf \Vert _{L^4(\mathbb {R}^4)}}{\Vert f \Vert _{L^2(\sigma _c)}}. \end{aligned}$$

Using spherical coordinates, the $L^2({\mathcal {H}}^3_s)$-norm of $f_a$ is given by the expression

$$\begin{aligned} \Vert f_a \Vert _{L^2({\mathcal {H}}^3_s)}^2&=\int \limits _{\mathbb {R}^3} e^{-a\sqrt{\vert x\vert ^2-s^2}}\frac{\,\textrm{d}x}{\sqrt{\vert x\vert ^2-s^2}}=4\pi \int \limits _s^\infty e^{-a\sqrt{r^2-s^2}}\frac{r^2}{\sqrt{r^2-s^2}}\,\textrm{d}r\\&=4\pi \int \limits _0^\infty e^{-a\tau }\sqrt{\tau ^2+s^2}\,\textrm{d}\tau . \end{aligned}$$

It is easier to estimate $\Vert T_sf_a \Vert _{L^4(\mathbb {R}^4)}$ if we use the convolution form (2.3),

$$\begin{aligned} \Vert T_sf_a \Vert _{L^4(\mathbb {R}^4)}=2\pi \Vert f_a\mu _s *f_a\mu _s \Vert _{L^2(\mathbb {R}^4)}^{1/2}. \end{aligned}$$

As in [39, Appendix 2], using that $f_a$ is the restriction to ${\mathcal {H}}^3_s$ of the exponential of the linear function in $\mathbb {R}^4$, $(\xi ,\tau )\mapsto e^{-\frac{a}{2}\tau }$, we obtain

$$\begin{aligned}f_a\mu _s *f_a\mu _s(\xi ,\tau )=e^{-\frac{a}{2}\tau }\bigl (\mu _s *\mu _s(\xi ,\tau )\bigr ). \end{aligned}$$

It will be enough for our purpose to use

as obtained from (3.1). In this way

$$\begin{aligned}{} & {} f_a\mu _s *f_a\mu _s(\xi ,\tau )\\{} & {} \quad \geqslant \frac{2\pi }{\vert \xi \vert }e^{-\frac{a}{2}\tau } \biggl ( \vert \xi \vert \Bigl (1+\frac{4s^2}{\tau ^2-\vert \xi \vert ^2}\Bigr )^{\frac{1}{2}}\mathbb {1}_{\{\vert \xi \vert <\sqrt{\tau ^2+s^2}-s\}}+\tau \mathbb {1}_{\{\sqrt{\tau ^2+s^2}-s\leqslant \vert \xi \vert \leqslant \sqrt{\tau ^2+(2s)^2}\}}\biggr ), \end{aligned}$$

so that using spherical coordinates we obtain

$$\begin{aligned} \Vert f_a\mu _s*f_a\mu _s \Vert _{L^2(\mathbb {R}^4)}^2&\geqslant (2\pi )^2\int \limits _{\mathbb {R}^3\times \mathbb {R}} e^{-a\tau }\biggl ( \vert \xi \vert ^2\Bigl (1+\frac{4s^2}{\tau ^2-\vert \xi \vert ^2}\Bigr )\mathbb {1}_{\{\vert \xi \vert <\sqrt{\tau ^2+s^2}-s\}}\\&\quad +\tau ^2\mathbb {1}_{\{\sqrt{\tau ^2+s^2}-s\leqslant \vert \xi \vert \leqslant \sqrt{\tau ^2+(2s)^2}\}} \biggr )\,\textrm{d}\tau \frac{\,\textrm{d}\xi }{\vert \xi \vert ^2}\\&=16\pi ^3 \int \limits _0^{\infty }e^{-a\tau }\biggl (\tau ^2\sqrt{\tau ^2+4s^2}-\frac{2}{3}(\tau ^2+4s^2)\sqrt{\tau ^2+s^2} +\frac{8s^3}{3}\\&\quad + 2s^2\tau \log \Bigl (\frac{\tau +\sqrt{\tau ^2+s^2}}{s}\Bigr )\biggr )\,\textrm{d}\tau . \end{aligned}$$

Since by scaling it is enough to consider $s=1$ (see Sect. 14) we let

$$\begin{aligned} I(a)&=16\pi ^3 \int \limits _0^{\infty }e^{-a\tau }\Bigl (\tau ^2\sqrt{\tau ^2+4}-\tfrac{2}{3}(\tau ^2+4)\sqrt{\tau ^2+1} +\tfrac{8}{3}\\&\quad + 2\tau \log \bigl (\tau +\sqrt{\tau ^2+1}\bigr )\Bigr )\,\textrm{d}\tau ,\\ II(a)&=16\pi ^2\biggl (\int \limits _0^\infty e^{-a\tau }\sqrt{\tau ^2+s^2}\,\textrm{d}\tau \biggr )^2, \end{aligned}$$

then

$$\begin{aligned} \frac{\Vert f_a\mu *f_a\mu \Vert _{L^2(\mathbb {R}^4)}^2}{\Vert f_a \Vert _{L^2(\mu )}^4} \geqslant \frac{I(a)}{II(a)}.\end{aligned}$$

From Lemma A.1 in the Appendix, we conclude that for all $a>0$ small enough

$$\begin{aligned} \frac{\Vert f_a\mu *f_a\mu \Vert _{L^2(\mathbb {R}^4)}^2}{\Vert f_a \Vert _{L^2(\mu )}^4}>2\pi . \end{aligned}$$

(4.2)

This finishes the proof in view of (4.1). $\square$

Remark 4.2

The easy lower bound we can obtain for $\Vert f_a\mu *f_a\mu \Vert _{L^2(\mathbb {R}^4)}^2\Vert f_a \Vert _{L^2(\mu )}^{-4}$ using the analog of [34, Lemma 6.1] is not good enough in this case to obtain (4.2).

Let us now move to the full one-sheeted hyperboloid ${\overline{{\mathcal {H}}}}^3$. Recall that ${\overline{T}}_c$ denotes the adjoint Fourier restriction operator on the double cone ${\overline{\Gamma }}^3$. An argument in [22] can be used to show that

$$\begin{aligned} {\overline{{{\textbf {C}} }}_4}=\Bigl (\frac{3}{2}\Bigr )^{\frac{1}{4}}{{\textbf {C}} }_4, \end{aligned}$$

(4.3)

see for instance [39, Prop. 7.3]. We now compare the best constants for ${\overline{T}}$ and ${\overline{T}}_c$.

Proposition 4.3

$\qquad \overline{{{\textbf {H}} }}_4>\overline{{{\textbf {C}} }}_4$.

Proof

Let $f_a(y)=e^{-\frac{a}{2}\sqrt{\vert y\vert ^2-1}}$ be as in the proof of Proposition 4.1 and set $g_a=f_{a,+}+f_{a,-}$, where $f_{a,+}=cf_a$ and $f_{a,-}=cf_a$ (here there are domain identifications through projections to $\mathbb {R}^3$), in other words, $g_a(\xi ,\tau )=ce^{-\frac{a}{2}\vert \tau \vert }\mathbb {1}_{\overline{{\mathcal {H}}}^3}(\xi ,\tau )$, where c is such that $g_a$ is $L^2$ normalized. Expanding and using the positivity of $f_{a,+}$ and $f_{a,-}$ (which for brevity we simply call $f_+$ and $f_-$, respectively) we see that

$$\begin{aligned} \Vert {\overline{T}}g_a \Vert _{L^4}^4&= \Vert Tf_+ \Vert _{L^4(\mathbb {R}^3)}^4+\Vert Tf_- \Vert _{L^4(\mathbb {R}^3)}^4+4\Vert (Tf_+)(Tf_-(\cdot ,-\cdot )) \Vert _{L^2}^2\\ {}&\quad +4(2\pi )^4\langle f_+\mu *f_+\mu ,f_+\mu *f_-\mu _-\rangle \\ {}&\quad +4(2\pi )^4\langle f_+\mu *f_-\mu _-,f_-\mu _-*f_-\mu _-\rangle \\ {}&\geqslant \Vert Tf_+ \Vert _{L^4(\mathbb {R}^3)}^4+\Vert T_-f_- \Vert _{L^4(\mathbb {R}^3)}^4+4\Vert (Tf_+)(Tf_-(\cdot ,-\cdot )) \Vert _{L^2}^2. \end{aligned}$$

On the other hand $Tf_-(\cdot ,-\cdot )=\overline{Tf_+}$, the complex conjugate, since $f_-(y)=f_+(-y)$. Then $\Vert (Tf_+)(Tf_-(\cdot ,-\cdot )) \Vert _{L^2}^2=\Vert Tf_+ \Vert _{L^4(\mathbb {R}^3)}^4=\Vert Tf_- \Vert _{L^4(\mathbb {R}^3)}^4$ and we obtain

$$\begin{aligned} \Vert {\overline{T}}g_a \Vert _{L^4}^4\geqslant 6\Vert Tf_{a,+} \Vert _{L^4(\mathbb {R}^3)}^4. \end{aligned}$$

If $a>0$ is small enough, then from (4.2) in the proof of Proposition 4.1 and using $\Vert f_{a,+} \Vert _{L^2(\mu )}=\sqrt{2}/2$, we obtain

$$\begin{aligned} \overline{{{\textbf {H}} }}_4^4\geqslant \Vert {\overline{T}}g_a \Vert _{L^4}^4\geqslant 6\Vert Tf_{a,+} \Vert _{L^4(\mathbb {R}^3)}^4>6{{\textbf {C}} }_4^4\Vert f_{a,+} \Vert _{L^2(\mu )}^4=\frac{3}{2}{{\textbf {C}} }_4^4. \end{aligned}$$

The conclusion follows using (4.3). $\square$

5 The Upper Half of the One-Sheeted Hyperboloid

In this section we present the proof of Theorem 1.2. The proof of precompactness of extremizing sequences, modulo multiplication by characters, is much simpler for the upper half of the one-sheeted hyperboloid as the full Lorentz invariance of $\overline{{\mathcal {H}}}^3$ is absent for ${\mathcal {H}}^3$.

In what follows we collect the necessary results to invoke Proposition 1.5 and the first such step is to show that, with enumeration as in Proposition 1.5, (i) and (iii) imply (iv), possibly after passing to a subsequence.

Proposition 5.1

Let $\{f_n\}_{n}$ be a sequence in $L^2({\mathcal {H}}^3)$ satisfying $\sup _n\Vert f_n \Vert _{L^2({\mathcal {H}}^3)}<\infty$. Suppose that there exists $f\in L^2({\mathcal {H}}^3)$ such that $f_n\rightharpoonup f$ as $n\rightarrow \infty$. Then, there exists a subsequence $\{f_{n_k}\}_{k}$ such that $Tf{_{n_k}}\rightarrow Tf$ a.e. in $\mathbb {R}^4$.

The previous result implies an analogous one for the full two-sheeted hyperboloid ${\overline{{\mathcal {H}}}}^3$. Recall the Fourier multiplier notation

$$\begin{aligned} e^{it\sqrt{-\Delta -s^2}}u(x)=\frac{1}{(2\pi )^3}\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert \geqslant s\}}e^{ix\cdot y}e^{it\sqrt{\vert y\vert ^2-s^2}}\hat{u}(y)\,\textrm{d}y, \end{aligned}$$

(5.1)

and the homogeneous $\dot{H}^{1/2}(\mathbb {R}^3)$ Sobolev norm and inner product

$$\begin{aligned} \Vert u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}^2:=\int \limits _{\mathbb {R}^3}\vert {\hat{u}}(y)\vert ^2\vert y\vert \,\textrm{d}y, \quad \langle u,v \rangle _{\dot{H}^{1/2}(\mathbb {R}^3)}:=\int \limits _{\mathbb {R}^3}{\hat{u}}(y)\overline{{\hat{v}}(y)}\vert y\vert \,\textrm{d}y. \end{aligned}$$

(5.2)

Proof of Proposition 5.1

The proof follows similar lines to the proofs of [21, Thm. 1.1] and [38, Prop. 8.3]. We start by splitting $f_n=f_n\mathbb {1}_{B(0,2)}+f_n\mathbb {1}_{B(0,2)^\complement }=:f_{n,1}+f_{n,2}$, respectively, and $f=f\mathbb {1}_{B(0,2)}+f\mathbb {1}_{B(0,2)^\complement }=:f_{1}+f_{2}$. The conclusion of the proposition will follow if we show that there exists a subsequence $\{f_{n_k}\}_{k}$ such that $Tf_{n_k,1}\rightarrow Tf_1$ and $Tf_{n_k,2}\rightarrow Tf_2$ a.e. in $\mathbb {R}^4$.

Since $f_{n,1}\rightharpoonup f_1$ in $L^2({\mathcal {H}}^3)$ and the support of $f_{n,1}$ is contained on the compact set B(0, 2), it follows that $Tf_{n,1}(x,t)\rightarrow Tf_1(x,t)$ for all $(x,t)\in \mathbb {R}^4$ provided that the function $(y,s)\mapsto e^{ix\cdot y}e^{its}\mathbb {1}_{B(0,2)}(y)$ belongs to $L^2({\mathcal {H}}^3)$, which is the case.

To study the pointwise convergence of $Tf_{n,2}$ define $g_n$ and g by their Fourier transforms as follows

$$\begin{aligned}{\hat{g}}_n(y)=\frac{f_{n,2}(y)}{\sqrt{\vert y\vert ^2-1}},\quad \quad {\hat{g}}(y)=\frac{f_{2}(y)}{\sqrt{\vert y\vert ^2-1}}.\end{aligned}$$

Because

$$\begin{aligned}\Vert f_{n,2} \Vert _{L^2({\mathcal {H}}^3)}^2=\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert \geqslant 2\}}\vert f_n(y)\vert ^2\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}\leqslant \sup _k \Vert f_k \Vert _{L^2({\mathcal {H}}^3)}^2\lesssim 1,\end{aligned}$$

we see that the norms of the $g_n$’s in the homogeneous Sobolev space ${\dot{H}}^{1/2}(\mathbb {R}^3)$ are uniformly bounded

$$\begin{aligned}\Vert g_n \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}^2=\int \limits _{\mathbb {R}^3}\vert {\hat{g}}_n(y)\vert ^2\vert y\vert \,\textrm{d}y\leqslant \frac{2}{\sqrt{3}}\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert \geqslant 2\}}\vert f_{n}(y)\vert ^2\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}\lesssim 1.\end{aligned}$$

The weak convergence of $\{f_{n,2}\}_n$ to $f_2$ in $L^2({\mathcal {H}}^3)$ easily implies $g_n\rightharpoonup g$ as $n\rightarrow \infty$ in ${\dot{H}}^{1/2}(\mathbb {R}^3)$. On the other hand

$$\begin{aligned}(2\pi )^{3}\Vert g_n \Vert _{L^2(\mathbb {R}^3)}^2=\Vert \widehat{g_n} \Vert _{L^2(\mathbb {R}^3)}^2\leqslant \frac{1}{\sqrt{3}}\int \limits _{\{y\in \mathbb {R}^3:\vert y\vert \geqslant 2\}}\vert f_{n,2}(y)\vert ^2\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}\lesssim 1,\end{aligned}$$

so $\{g_n\}_n$ is uniformly bounded in $L^2(\mathbb {R}^3)$.

The operator T applied to $f_{n,2}$ equals $(2\pi )^3e^{it\sqrt{-\Delta -1}}g_n$, where the operator $e^{it\sqrt{-\Delta -1}}$ is understood in the Fourier multiplier sense as in (5.1). Let $t\in \mathbb {R}$ be fixed. By the continuity of $e^{it\sqrt{-\Delta -1}}$ in $\dot{H}^{1/2}(\mathbb {R}^3)$ we obtain

$$\begin{aligned}e^{it\sqrt{-\Delta -1}}g_n\rightharpoonup e^{it\sqrt{-\Delta -1}}g,\end{aligned}$$

weakly in $\dot{H}^{1/2}(\mathbb {R}^3)$, as $n\rightarrow \infty$. Then, by the Rellich–Kondrashov Theorem ([17, Thm. 7.1]), for any $R>0$

$$\begin{aligned}e^{it\sqrt{-\Delta -1}}g_n\rightarrow e^{it\sqrt{-\Delta -1}}g,\end{aligned}$$

strongly in $L^2(B(0,R))$, as $n\rightarrow \infty$. Denote by

$$\begin{aligned}F_n(t):=\int \limits _{\vert x\vert <R}\biggl \vert e^{it\sqrt{-\Delta -1}}(g_n-g) \biggr \vert ^2\,\textrm{d}x=\Vert e^{it\sqrt{-\Delta -1}}(g_n-g) \Vert _{ L^2(B(0,R))}^2.\end{aligned}$$

By Hölder’s inequality and Sobolev embedding, [17, Thm. 6.5], we obtain

$$\begin{aligned} F_n(t)&=\Vert e^{it\sqrt{-\Delta -1}}(g_n-g) \Vert _{L^2(B(0,R))}^2\leqslant CR\Vert e^{it\sqrt{-\Delta -1}}(g_n-g) \Vert _{L^3(B(0,R))}^2\\&\leqslant CR\Vert e^{it\sqrt{-\Delta -1}}(g_n-g) \Vert _{{\dot{H}}^{1/2}(\mathbb {R}^3)}^2\lesssim R, \end{aligned}$$

then, by the Fubini and Dominated Convergence Theorems we have that

$$\begin{aligned}\int \limits _{-R}^R F_n(t)dt=\int \limits _{-R}^{R}\int \limits _{\vert x\vert <R}\biggl \vert e^{it\sqrt{-\Delta -1}}(g_n-g) \biggr \vert ^2\,\textrm{d}x\,\textrm{d}t\rightarrow 0,\end{aligned}$$

as $n\rightarrow \infty$. This implies that, up to a subsequence,

$$\begin{aligned}e^{it\sqrt{-\Delta -1}}g_n(x)-e^{it\sqrt{-\Delta -1}}g(x)\rightarrow 0\quad \text {a.e. }(x,t)\in B(0,R)\times (-R,R).\end{aligned}$$

Repeating the argument on a discrete sequence of radii $R_n$ such that $R_n\rightarrow \infty$, as $n\rightarrow \infty$ we conclude, by a diagonal argument, that there exists a subsequence $\{g_{n_k}\}_k$ of $\{g_n\}_n$ such that

$$\begin{aligned}e^{it\sqrt{-\Delta -1}}g_{n_k}(x)-e^{it\sqrt{-\Delta -1}}g(x)\rightarrow 0\quad \text {a.e. for }(x,t)\in \mathbb {R}^4,\end{aligned}$$

or equivalently, in terms of the sequence $\{f_{n,2}\}_{n}$ and the operator T,

$$\begin{aligned}Tf_{n_k,2}(x,t)\rightarrow Tf_2(x,t)\quad \text {a.e. }(x,t)\in \mathbb {R}^4.\end{aligned}$$

$\square$

We now show that the only obstruction to precompactness of extremizing sequences is the possibility of concentration at infinity, as in Definition 1.6.

Proposition 5.2

Let $\{f_n\}_{n}\subset L^2({\mathcal {H}}^3)$ be an $L^2$ normalized extremizing sequence for T. Suppose that $\{f_n\}_{n}$ does not concentrate at infinity. Then there exist a subsequence $\{f_{n_k}\}_k$ and a sequence $\{(x_k,t_k)\}_k\subset \mathbb {R}^4$ such that $\big \{e^{ix_k\cdot y}e^{it_k\sqrt{\vert y\vert ^2-1}}f_{n_k}\big \}_{k}$ is convergent in $L^2({\mathcal {H}}^3)$.

Proof

If $\{f_n\}_{n}$ does not concentrate at infinity, then there exist $\varepsilon ,R>0$ with the property that for all $N\in \mathbb {N}$ there exists $n\geqslant N$ such that

$$\begin{aligned}\Vert f_n\mathbb {1}_{B(0,R)} \Vert _{L^2({\mathcal {H}}^3)}\geqslant \varepsilon .\end{aligned}$$

We can generate a subsequence, $\{f_{n_k}\}_{k}$, such that $\Vert f_{n_k}\mathbb {1}_{B(0,R)} \Vert _{L^2({\mathcal {H}}^3)}\geqslant \varepsilon$ for all $k\in \mathbb {N}$. Rename the subsequence as $\{f_n\}_n$, if necessary. Writing $f_n=f_n\mathbb {1}_{B(0,R)}+f_n\mathbb {1}_{B(0,R)^\complement }=:f_{n,1}+f_{n,2}$, respectively, we have

$$\begin{aligned} \Vert Tf_{n,1} \Vert _{L^4(\mathbb {R}^4)}&=\Vert T(f_n-f_{n,2}) \Vert _{L^4(\mathbb {R}^4)}\geqslant \Vert Tf_n \Vert _{L^4(\mathbb {R}^4)}-\Vert Tf_{n,2} \Vert _{L^4(\mathbb {R}^4)}\nonumber \\&\geqslant \Vert Tf_n \Vert _{L^4(\mathbb {R}^3)}-{{\textbf {H}} }_{4}\Vert f_{n,2} \Vert _{L^2({\mathcal {H}}^3)}\nonumber \\&= \Vert Tf_n \Vert _{L^4(\mathbb {R}^3)}-{{\textbf {H}} }_{4}\bigg (1-\Vert f_{n,1} \Vert _{L^2({\mathcal {H}}^3)}^2\bigg )^{1/2}\nonumber \\&\geqslant \Vert Tf_n \Vert _{L^4(\mathbb {R}^3)}-{{\textbf {H}} }_{4}\sqrt{1-\varepsilon ^2}. \end{aligned}$$

(5.3)

As the right hand side in (5.3) converges to $c:={{\textbf {H}} }_{4}-{{\textbf {H}} }_{4}\sqrt{1-\varepsilon ^2}>0$ as $n\rightarrow \infty$ we see that

$$\begin{aligned} \Vert Tf_{n,1} \Vert _{L^4(\mathbb {R}^4)}\geqslant \frac{c}{2}>0, \end{aligned}$$

(5.4)

for all large n.

We may use the argument in the proof of [20, Thm. 1.1] to construct the sequence $\{(x_n,t_n)\}_n$. In brief, the argument goes as follows. Taking any ${\bar{p}}\in \big [\frac{10}{3},4\big )$, interpolating the $L^4$ norm of $Tf_{n,1}$ between $L^{{\bar{p}}}$ and $L^\infty$ and using (5.4) together with the boundedness of T in $L^{\bar{p}}$ imply that $\Vert Tf_{n,1} \Vert _{L^\infty (\mathbb {R}^4)}\gtrsim 1$, so that there exists a sequence $\{(x_n,t_n)\}_n\subset \mathbb {R}^4$ such that $\vert Tf_{n,1}(x_n,t_n)\vert \geqslant C>0$, that is, $\vert (T(e^{ix_n\cdot y}e^{it_n\sqrt{\vert y\vert ^2-1}}f_{n,1}))(0,0)\vert \geqslant C>0$. The compact support of $f_{n,1}$ implies that $Tf_{n,1}$ belongs to $C^\infty (\mathbb {R}^4)$ and $\Vert Tf_{n,1} \Vert _{L^\infty (\mathbb {R}^4)}\lesssim \Vert f_{n,1} \Vert _{L^1}\lesssim 1$, $\Vert \nabla _{x,t} Tf_{n,1} \Vert _{L^\infty (\mathbb {R}^4)}\lesssim \Vert f_{n,1} \Vert _{L^1}\lesssim 1$. By the Arzelá–Ascoli Theorem, it follows that $\{T(e^{ix_n\cdot y}e^{it_n\sqrt{\vert y\vert ^2-1}}f_{n,1})\}_n$ is precompact in the space of continuous functions on the unit ball of $\mathbb {R}^4$. On the other hand, passing to a subsequence, we may assume $F_n:=e^{ix_n\cdot y}e^{it_n\sqrt{\vert y\vert ^2-1}}f_{n,1}\rightharpoonup f_1$ weakly in $L^2({\mathcal {H}}^3)$, for some $f_1\in L^2({\mathcal {H}}^3)$, and then $T(F_{n})(x,t)\rightarrow Tf_1(x,t)$ for all $(x,t)\in \mathbb {R}^4$. Moreover, $T(F_n)\rightarrow Tf_1$ uniformly in the unit ball, so that $\vert Tf_1(0,0)\vert \geqslant C>0$, which implies that $f_1\ne 0$.

Compactness of the unit ball in $L^2({\mathcal {H}}^3)$ in the weak topology implies that, after passing to a further subsequence, $e^{ix_n\cdot y}e^{it_n\sqrt{\vert y\vert ^2-1}}f_n\rightharpoonup f$, for some $f\in L^2(\mathcal {H}^3)$. Since $f_1=f\mathbb {1}_{B(0,R)}$ a.e. in $\mathbb {R}^3$ we conclude that $f\ne 0$. Therefore condition (iii) of Proposition 1.5 is satisfied. Proposition 5.1 implies that condition (iv) is also satisfied. As (i) and (ii) are immediate, we conclude that $e^{ix_n\cdot y}e^{it_n\sqrt{\vert y\vert ^2-1}}f_n\rightarrow f$ in $L^2({\mathcal {H}}^3)$, and we are done. $\square$

To conclude the precompactness of extremizing sequences we need to discard the possibility of concentration at infinity. For this we use a comparison argument with the cone where the upper bound for $\mu _s*\mu _s$ as found in Lemma 3.2 will be useful.

Lemma 5.3

Let $a>1$ and $f\in L^2({\mathcal {H}}^3)$ be supported in the region where $\vert y\vert \geqslant a$. Then

$$\begin{aligned} \Vert f\mu *f\mu \Vert _{L^2(\mathbb {R}^4)}^2\leqslant 2\pi \Bigl (1+\frac{1}{\sqrt{a^2-1}}\Bigr )\Vert f \Vert _{L^2(\mu )}^4. \end{aligned}$$

Proof

If f is supported where $\vert y\vert \geqslant a$, then the support of $f\mu *f\mu$ is contained in the region $\{(\xi ,\tau )\in \mathbb {R}^4:\tau \geqslant 2\sqrt{a^2-1}\}$. The Cauchy–Schwarz inequality provides the a.e. pointwise bound

$$\begin{aligned} \vert f\mu *f\mu \vert ^2(\xi ,\tau )\leqslant \bigl (\vert f\vert ^2\mu *\vert f\vert ^2\mu \bigr )(\xi ,\tau )\bigl (\mu *\mu \bigr )(\xi ,\tau ), \end{aligned}$$

which together with the upper bound in Lemma 3.2 imply

$$\begin{aligned} \vert f\mu *f\mu \vert ^2(\xi ,\tau )\leqslant 2\pi \Bigl (1+\frac{1}{\sqrt{a^2-1}}\Bigr )\bigl (\vert f\vert ^2\mu *\vert f\vert ^2\mu \bigr )(\xi ,\tau ), \end{aligned}$$

for a.e. $(\xi ,\tau )\in \mathbb {R}^4$. Integrating over $(\xi ,\tau )\in \mathbb {R}^4$ yields the result. $\square$

It is now direct to prove our first main theorem.

Proof of Theorem 1.2

We start by noting that if an $L^2$-normalized sequence $\{f_n\}_n$ concentrates at infinity, then for any $a>1$, $\Vert f_n\mathbb {1}_{B(0,a)} \Vert _{L^2(\mu )}\rightarrow 0$ as $n\rightarrow \infty$, therefore, for such a sequence we obtain, using Lemma 5.3, that

$$\begin{aligned}\limsup _{n\rightarrow \infty }\frac{\Vert f_n\mu *f_n\mu \Vert _{L^2(\mathbb {R}^4)}^2}{\Vert f_n \Vert _{L^2(\mu )}^4}\leqslant 2\pi . \end{aligned}$$

Using Proposition 4.1 we conclude that an extremizing sequence for T does not concentrate at infinity. We apply Proposition 5.2 to conclude. $\square$

6 The Full One-Sheeted Hyperboloid

Our task in the sections to come is to prove Theorem 1.3, the existence of extremals for the adjoint Fourier restriction inequality on the one-sheeted hyperboloid ${\overline{{\mathcal {H}}}}^3$. In the $L^4$ case, there is an argument available for the cone $\Gamma ^3$ that allows to relate the best constant and extremizers for $\Gamma ^3$ with that for the double cone ${\overline{\Gamma }}^3$. It relies on the observation that the algebraic sums $\Gamma ^3+\Gamma ^3$ and $\Gamma ^3+(-\Gamma ^3)$ intersect on a null set of $\mathbb {R}^3$, namely, $(\Gamma ^3+\Gamma ^3)\cap (\Gamma ^3+(-\Gamma ^3))=\Gamma ^3$, so that for any $f_+,g_+,h_+\in L^2(\Gamma ^3)$ and $f_-\in L^2(-\Gamma ^3)$ one has

$$\begin{aligned} \bigl \langle f_+\sigma _c*g_+\sigma _c, h_+\sigma _c*f_-{\widetilde{\sigma }}_c \bigr \rangle _{L^2(\mathbb {R}^4)}=0, \end{aligned}$$

where ${\widetilde{\sigma }}_c$ denotes the reflection of $\sigma _c$, supported on $-\Gamma ^3$. An analogous property in the $L^4$ case applies to the two-sheeted hyperboloid in $\mathbb {R}^4$ and allows one to obtain its best constant from that of the upper sheet only (see [39, Prop. 7.3, Cor. 7.4]). This approach is not applicable to ${\overline{{\mathcal {H}}}}^3$ because here ${\mathcal {H}}^3+{\mathcal {H}}^3$ and ${\mathcal {H}}^3+(-{\mathcal {H}}^3)$ intersect on a set of infinite Lebesgue measure.

The argument we use to prove precompactness of extremizing sequences (modulo multiplication by characters and Lorentz transformations) is close to that of Brocchi, Oliveira e Silva and the author [1] and of [38] by the author using a concentration-compactness argument, a refined cap estimate, comparison to the cone and the use of Lorentz invariance. It borrows from the Christ–Shao argument [13] the cap refinement of the Tomas–Stein inequality for ${\mathbb {S}}^2$ to obtain a similar refinement for the hyperboloid, as well as the understanding that it will be necessary to compare to a “limiting” manifold, in our case, the cone.

In the next section we review the aforementioned cap refinement for the Tomas–Stein inequality for ${\mathbb {S}}^2$ that will be used in the subsequent section to obtain a corresponding cap refinement for the adjoint Fourier restriction inequality on the hyperboloid via a lifting method. In later sections we consider the concentration-compactness argument.

7 The Tomas–Stein Inequality for ${\mathbb {S}}^2$ and Refinements

The sharp convolution form of the Tomas–Stein inequality for ${\mathbb {S}}^2$ states that for all $f\in L^2({\mathbb {S}}^2)$ we have

$$\begin{aligned} \Vert f\sigma *f\sigma \Vert _{L^2(\mathbb {R}^3)}\leqslant {{{\textbf {S}}} }^2\Vert f \Vert _{L^2({\mathbb {S}}^2)}^2, \end{aligned}$$

(7.1)

where ${{\textbf {S}} }=(2\pi )^{1/4}$ is the sharp constant, as obtained in [22].

In this section we review some refinements of (7.1) that will be useful in the next section. The exposition here follows that of [13, Sect. 6]. We start by setting things up to define the $X_p$ spaces, $p\in [1,\infty )$, and the first step is to generate increasingly finer “grids” for the sphere ${\mathbb {S}}^2$. With this in mind, for each integer $k\geqslant 0$ choose a maximal subset $\{y_k^j\}_j\subset {\mathbb {S}}^2$ satisfying $\vert y_k^j-y_k^l\vert \geqslant 2^{-k}$, for all $j\ne l$. Then, for each $k\geqslant 0$, the spherical caps have finite overlap and cover ${\mathbb {S}}^2$, that is, , and there exists a constant C, independent of k, such that any point in ${\mathbb {S}}^2$ belongs to no more than C caps in , for every $k\geqslant 0$. For $p\in [1,\infty )$, the $X_p$ norm of f is defined by the expression

(7.2)

Moyua et al. showed in [31] that there is a continuous inclusion $L^2({\mathbb {S}}^2)\subset X_p$ for all $p\in (1,2)$ and that for any $p\geqslant \frac{12}{7}$ there exists $C<\infty$ such that for all $f\in L^ 2({\mathbb {S}}^2)$

$$\begin{aligned} \Vert \widehat{f\sigma } \Vert _{L^4(\mathbb {R}^3)}\leqslant C\Vert f \Vert _{X_p}. \end{aligned}$$

(7.3)

Let us define

which by Hölder’s inequality satisfies $\Lambda _{k,j}(f)\leqslant 1$. It was shown in [13, Lemma 6.1] that for any $p\in [1,2)$, there exists $C<\infty$ and $\gamma >0$ such that for any $f\in L^2({\mathbb {S}}^2)$,

$$\begin{aligned} \Vert f \Vert _{X_p}\leqslant C\Vert f \Vert _{L^2({\mathbb {S}}^2)}\sup _{k,j}(\Lambda _{k,j}(f))^\gamma . \end{aligned}$$

(7.4)

Combining the two results, (7.3) and (7.4), by choosing any $p\in [\frac{12}{7},2)$, we obtain the following “cap refinement” of (7.1): there exists $C<\infty$ and $\gamma >0$ such that for all $f\in L^2({\mathbb {S}}^2)$

$$\begin{aligned} \Vert \widehat{f\sigma } \Vert _{L^4(\mathbb {R}^3)}\leqslant C\Vert f \Vert _{L^2({\mathbb {S}}^2)}\sup _{k,j}(\Lambda _{k,j}(f))^\gamma . \end{aligned}$$

(7.5)

A $\delta$-quasi-extremal for the sphere is a function $f\ne 0$ that satisfies $\Vert f\sigma *f\sigma \Vert _{L^4(\mathbb {R}^3)}\geqslant \delta ^2{{\textbf {S}} }^2\Vert f \Vert _{L^2({\mathbb {S}}^2)}^2$. With the aid of the previous inequality, Christ and Shao proved the following result regarding $\delta$-quasi-extremals.

Lemma 7.1

([13, Lemma 2.9]) For any $\delta >0$ there exists $C_{\delta }>0$ and $\eta _\delta >0$ with the following property. If $f\in L^2({\mathbb {S}}^2)$ satisfies $\Vert f\sigma *f\sigma \Vert _2\geqslant \delta ^2{{\textbf {S}} }^2\Vert f \Vert _2^2$ then there exist a decomposition $f=g+h$ and a spherical cap satisfying

(7.6)

(7.7)

(7.8)

(7.9)

Moreover (7.8) and (7.9) hold with constants that satisfy $C_\delta \asymp \delta ^{-1/\gamma }$ and $\eta _\delta \asymp \delta ^{1/\gamma }$, where $\gamma >0$ is a universal constant⁴.

It will be our task in the next section to obtain an analogous result for the hyperboloid and for this it will be convenient to briefly discuss the construction of the function g and the cap in the conclusion of the previous lemma. Letting $f\in L^2({\mathbb {S}}^2)$ be a $\delta$-quasi-extremal, inequality (7.5) implies that there is a constant $c_0\in (0,\infty )$, independent of f, such that

$$\begin{aligned} \sup _{k,j}\Lambda _{k,j}(f)\geqslant 2c_0\delta ^{1/\gamma }. \end{aligned}$$

It follows that there exist k and j such that $\Lambda _{k,j}(f)\geqslant c_0\delta ^{1/\lambda }$. Let . Then,

(7.10)

Let , , $g=f\mathbb {1}_A$ and $h=f-f\mathbb {1}_A$. It is now a simple exercise to prove that $g,\,h$ and satisfy the conditions stated in the lemma.

Remark 7.2

Let us consider the following scenario: a measurable set $E\subseteq \mathbb {R}$ and a measurable function $F:E\times {\mathbb {S}}^2\rightarrow {\mathbb {C}}$ that satisfies $F\in L^2(E\times {\mathbb {S}}^2)$, $\Vert F_r\sigma *F_r\sigma \Vert _{L^2(\mathbb {R}^3)}\geqslant \delta ^2{{{\textbf {S}}} }^2\Vert F_r \Vert _{L^2({\mathbb {S}}^2)}^{2} >0$ for all $r\in E$, where $F_r(x)=F(r,x)$, $(r,x)\in E\times {\mathbb {S}}^2$. Applying Lemma 7.1 to $F_r$ for each $r\in E$ generates caps and functions $G_r$ and $H_r$, and in this way functions $G,H:E\times {\mathbb {S}}^2\rightarrow {\mathbb {C}}$, which a priori may not be measurable in the product space $E\times {\mathbb {S}}^2$. This can be overcome if we are careful with the choice of the caps as we now proceed to explain. For a collection of spherical caps satisfying (7.10) with and $f=F_r$, for all $r\in E$, denote

Then, as explained following (7.10), we can take $G=F\mathbb {1}_{{\mathcal {G}}_1}$ and $H=F-F\mathbb {1}_{{\mathcal {G}}_1}$. We need to argue that we can have ${\mathcal {G}}_0$ and ${\mathcal {G}}_1$ measurable, by choosing the collection appropriately. When $r\in E$, then $\sup _{k,j}\Lambda _{k,j}(F_r)\geqslant 2c(\delta )$, for some universal constant $c(\delta )$. The cap is to be chosen so that $\Lambda _{k,j}(F_r)\geqslant c(\delta )$, that is,

The set of caps in ${\mathbb {S}}^2$ is parametrized by indices k and j where $k\in \mathbb {N}$ and $j\in \{1,2,\dotsc ,J_k\}$, for some $J_k<\infty$. Let ${\mathcal {Z}}=\{(k,j):k\in \mathbb {N},\, j\in \{1,\dotsc ,J_k\} \}$ and define the function $\Theta :E\times {\mathcal {Z}}\rightarrow \mathbb {R}$ by

By Fubini’s theorem, for each fixed $(k,j)\in E\times {\mathcal {Z}}$, $\Theta (\cdot ,k,j)$ is a measurable function. By assumption, for each $r\in E$, $\sup _{k,j}\Theta (r,k,j)\geqslant 2c(\delta )$. We want to find a measurable function $\tau :E\rightarrow {\mathcal {Z}}$ such that

$$\begin{aligned}\Theta (r,\tau (r))\geqslant \sup _{k,j}\Theta (r,k,j)-c(\delta )\geqslant c(\delta ),\end{aligned}$$

a so called $c(\delta )$-maximizer. That this is possible is a consequence of measurable selection theorems, see for instance [41, Thm. 4.1].

For such a measurable selection function $\tau$ write $\tau (r)=(k(r),j(r))\in {\mathcal {Z}}$, then the function $E\rightarrow {\mathbb {S}}^2$, $r\mapsto y_{k(r)}^{j(r)}$, is measurable and we can write ${\mathcal {G}}_0=\{(x,r):r\in E, \vert x-y_{k(r)}^{j(r)}\vert \leqslant 2^{-k(r)+1} \}$. We can therefore assume that the sets ${\mathcal {G}}_0$ and ${\mathcal {G}}_1$ are measurable sets in $E\times {\mathbb {S}}^2$, so that G and H are measurable functions. In this way, we have the following lemma.

Lemma 7.3

Let $E\subseteq \mathbb {R}$ be a measurable set and $F:E\times {\mathbb {S}}^2\rightarrow {\mathbb {C}}$ be a measurable function satisfying $F\in L^2(E\times {\mathbb {S}}^2)$, $\Vert F_r\sigma *F_r\sigma \Vert _{L^2(\mathbb {R}^3)}\geqslant \delta ^2{{{\textbf {S}}} }^2\Vert F_r \Vert _{L^2({\mathbb {S}}^2)}^{2}>0$ for all $r\in E$, where $F_r(x)=F(r,x)$, $(r,x)\in E\times {\mathbb {S}}^2$. Then, there are spherical caps and measurable functions G, H satisfying: $F=G+H$, G and H have disjoint supports, $0\leqslant \vert G\vert ,\vert H\vert \leqslant \vert F\vert$, and for all $r\in E$:

We now prove a slight improvement of Lemma 7.1 that adds one more restriction to the function g. It tells us that we can replace a $\delta$-quasi-extremal for the sphere for a better controlled one at the expense of powers of $\delta$.

Lemma 7.4

For any $\delta >0$ there exists $C_{\delta }>0$, $\eta _\delta >0$ and $\lambda _\delta >0$ with the following property. If $f\in L^2({\mathbb {S}}^2)$ satisfies $\Vert f\sigma *f\sigma \Vert _2\geqslant \delta ^2{{{\textbf {S}}} }^2\Vert f \Vert _2^2$ then there exist a decomposition $f=g+h$ and a spherical cap satisfying (7.6), (7.7), (7.8), 7.9 and

$$\begin{aligned} \Vert g\sigma *g\sigma \Vert _2\geqslant \lambda _\delta {{{\textbf {S}}} }^2\Vert f \Vert _2^2. \end{aligned}$$

(7.11)

Moreover (7.8), (7.9) and (7.11) hold with constants that satisfy $C_\delta \asymp \delta ^{-1/\gamma },\,\eta _\delta \asymp \delta ^{1+1/\gamma }$ and $\lambda _\delta \asymp \delta ^{6+4/\gamma }$, where $\gamma >0$ is a universal constant.

Remark 7.5

It is not difficult to see (e.g. [38, Lemma 6.2]) that for a function g satisfying (7.8) and (7.9) there is a lower bound for the $L^1$ norm of the form

(7.12)

Note that the sharp estimate (7.1) for ${\mathbb {S}}^2$ implies that for g satisfying (7.11) we have

$$\begin{aligned}{{{\textbf {S}}} }\Vert g \Vert _{2}\geqslant \Vert g\sigma *g\sigma \Vert _2^{1/2}\geqslant \lambda _\delta ^{1/2}{{{\textbf {S}}} }\Vert f \Vert _2,\end{aligned}$$

so that

$$\begin{aligned} \Vert g \Vert _{L^2({\mathbb {S}}^2)}\geqslant \lambda _\delta ^{1/2}\Vert f \Vert _{L^2({\mathbb {S}}^2)}. \end{aligned}$$

(7.13)

Proof of Lemma 7.4

Take $C_\delta$ and $\eta _\delta$ as in the conclusion of Lemma 7.1. We claim that the lemma at hand holds with respective constants $C_\delta$, $\delta \eta _\delta /\sqrt{2}$ and $\lambda _\delta =(\delta ^3\eta _\delta ^2/8)^2$. To see this we employ a decomposition algorithm, reminiscent of that in [13, Sect. 8, step 6A], defined in the following inductive way.

Let $G_0=f$ and $f_0=0$ and suppose that for $N\geqslant 0$ we have defined $G_{N}$ and $f_k$, for $0\leqslant k\leqslant N$, satisfying:

$$\begin{aligned}&f=G_N+f_0+\cdots +f_{N}, \end{aligned}$$

(7.14)

$$\begin{aligned}&{\text {supp}}(G_{N}),{\text {supp}}(f_0),\dots ,{\text {supp}}(f_{N})\text { are pairwise disjoint,} \end{aligned}$$

(7.15)

$$\begin{aligned}&\Vert G_N\sigma *G_N\sigma \Vert _2\geqslant \frac{1}{2}\delta ^2{{{\textbf {S}}} }^2\Vert f \Vert _2^2. \end{aligned}$$

(7.16)

The previous conditions are satisfied if $N=0$. We now define the inductive step of the algorithm. If (7.14), (7.15) and (7.16) hold for N we define $G_{N+1}$ and $f_{N+1}$ in the following way.

Given that $\Vert G_N\sigma *G_N\sigma \Vert _2\geqslant \frac{1}{2}\delta ^2{{{\textbf {S}}} }^2\Vert f \Vert _2^2\geqslant \frac{1}{2}\delta ^2{{{\textbf {S}}} }^2\Vert G_N \Vert _2^2$ we can apply Lemma 7.1 to $G_N$ to obtain a decomposition $G_N=g_N+h_N$ and a cap . Define $G_{N+1}=h_N$ and $f_{N+1}=g_N$. The functions $G_{N+1}$ and $f_{N+1}$ therefore have disjoint supports and satisfy

(7.17)

(7.18)

where the second inequality in (7.18) follows as in (7.13).

The algorithm terminates at $N\geqslant 1$ when either $\Vert f_{N}\sigma *f_{N}\sigma \Vert _2\geqslant \lambda _\delta {{{\textbf {S}}} }^2\Vert f \Vert _2^2$ or $\Vert G_N\sigma *G_N\sigma \Vert _2 < \frac{1}{2}\delta ^2{{{\textbf {S}}} }^2\Vert f \Vert _2^2$. In the former case we say the algorithm stops in a win and set $g=f_{N}$, $h=G_N+f_0+\cdots +f_{N-1}$, and the Lemma is proved.

Let $N_\delta :=\lceil 2\eta _\delta ^{-2}\delta ^{-2}\rceil$. We claim that the algorithm stops in a win for some $N\leqslant N_\delta$. We first show that the algorithm can not run for more than $N_\delta$ steps, otherwise, using (7.18) we have

$$\begin{aligned}\Vert f \Vert _2\geqslant \biggl (\sum _{k=1}^{N_\delta +1}\Vert f_k \Vert _2^2\biggr )^{1/2}\geqslant \frac{1}{\sqrt{2}}(N_\delta +1)^{1/2}\eta _\delta \delta \Vert f \Vert _2>\Vert f \Vert _2,\end{aligned}$$

which is impossible.

Second, we show that if the algorithm has not stopped in a win during the first N steps for some $N\leqslant 2N_\delta$, then we can perform the step $N+1$. More precisely, if $\Vert f_k\sigma *f_k\sigma \Vert _2 <\lambda _\delta {{{\textbf {S}}} }^2\Vert f \Vert _2^2$ for all $1\leqslant k\leqslant N$, for some $N\leqslant 2N_\delta$, then $\Vert G_N\sigma *G_N\sigma \Vert _2\geqslant \frac{1}{2}\delta ^2{{{\textbf {S}}} }^2\Vert f \Vert _2^2$. Indeed, using Plancherel’s theorem and then the triangle inequality we obtain

$$\begin{aligned} \Vert G_{N}\sigma *G_{N}\sigma \Vert _2^{1/2}&\geqslant \Vert f\sigma *f\sigma \Vert _2^{1/2}-\sum _{k=1}^{N}\Vert f_k\sigma *f_k\sigma \Vert _2^{1/2} \geqslant \delta {{{\textbf {S}}} }\Vert f \Vert _2-N\lambda _\delta ^{1/2}{{{\textbf {S}}} }\Vert f \Vert _2\\ {}&\geqslant (\delta -2N_\delta \lambda _\delta ^{1/2}){{{\textbf {S}}} }\Vert f \Vert _2\\ {}&\geqslant \frac{1}{2}\delta {{{\textbf {S}}} }\Vert f \Vert _2. \end{aligned}$$

It follows that the algorithm stops in a win for some $N\leqslant N_\delta$. This finishes the proof. $\square$

The next topic we review is that of “weak interaction between distant caps”. For spherical caps there is a notion of distance. Let $(y,a),\,(y',a')\in {\mathbb {S}}^2\times (0,\infty )$ denote the centers and radii of the spherical caps ,

The distance between and is defined by the expression

(7.19)

where, as in [33], we can take ${\text {d}}$ to be the hyperbolic distance between (y, a) and $(y',a')$ in the upper half space model, that is⁵

The following lemma quantifies the notion of weak interaction between distant caps.

Lemma 7.6

([13, Lemma 7.6]) For any $\varepsilon >0$ there exists $\rho <\infty$ such that

An inspection of the proof of the previous statement in [13, Lemma 7.6] shows that an analog result holds if we have caps and , with $r,\,t\in [1,2]$, that is, denoting the rescale of to ${\mathbb {S}}^2$,

we have the following lemma.

Lemma 7.7

Let $r,t\in [1,2]$, and . Then for any $\varepsilon >0$ there exists $\rho <\infty$ such that , whenever .

8 Lifting to the Hyperboloid the Inequality for the Sphere

The aim of this section is to use the Tomas–Stein inequality for the sphere ${\mathbb {S}}^2$ to obtain qualitative properties of $\delta$-quasi-extremals for the hyperboloid. The connection here between the hyperboloid and the sphere is that the latter corresponds to horizontal traces of the former. This connection between the adjoint Fourier restriction operator on a hypersurface and on its traces appears, for instance, in the work of Nicola [32]. An alternative approach to the methods in this section can be developed using refined bilinear estimates, but we choose to give a different and new argument. The main result of this section is the following lemma.

Lemma 8.1

Let $0\leqslant s\leqslant \frac{1}{2}$. For any $\delta >0$ there exists $C_{\delta }>0$, $\eta _\delta >0$ and $\nu _\delta >0$ with the following property. If $f(x,t)\in L^2({\mathcal {H}}^3_s)$ supported where $1\leqslant \vert x\vert \leqslant 2$ satisfies $\Vert f\mu _s * f\mu _s \Vert _{L^2(\mathbb {R}^4)}\geqslant \delta ^2{{{\textbf {H}} }}_4^2\Vert f \Vert _{L^2}^2$ then there exist a decomposition $f=g+h$, a spherical cap and a cap satisfying

$$\begin{aligned}&0\leqslant \vert g\vert ,\vert h\vert \leqslant \vert f\vert , \end{aligned}$$

(8.1)

$$\begin{aligned}&g,h \text { have disjoint supports}, \end{aligned}$$

(8.2)

$$\begin{aligned}&{\text {supp}}(g)\subseteq {\mathcal {C}}, \end{aligned}$$

(8.3)

$$\begin{aligned}&\vert g(x,t)\vert \leqslant C_\delta \Vert f \Vert _{L^2}\mu _s({{\mathcal {C}}})^{-1/2}\mathbb {1}_{\mathcal {C}}(x,t)\;\text { for all }(x,t), \end{aligned}$$

(8.4)

$$\begin{aligned}&\Vert g \Vert _{L^2}\geqslant \eta _\delta \Vert f \Vert _{L^2}, \end{aligned}$$

(8.5)

$$\begin{aligned}&\Vert g \Vert _{L^1}\geqslant \nu _\delta \mu _s({{\mathcal {C}}})^{1/2}\Vert f \Vert _{L^2}. \end{aligned}$$

(8.6)

The constants $C_\delta ,\,\eta _\delta$ and $\nu _\delta$ are uniform in $s\leqslant \frac{1}{2}$.

Remark 8.2

The previous lemma is equivalent to the analog result for ${\overline{{\mathcal {H}}}}^3_s$. Indeed, that the result for ${\overline{{\mathcal {H}}}}^3_s$ implies a similar one for ${\mathcal {H}}^3_s$ is immediate. On the other direction, if $f\in L^2({\overline{{\mathcal {H}}}}^3_s)$ is a $\delta$-quasi-extremal for (1.10), that is

$$\begin{aligned}\Vert {\overline{T}}_sf \Vert _{L^4(\mathbb {R}^4)}^4=(2\pi )^{4}\Vert f{\bar{\mu }}_s*f{\bar{\mu }}_s \Vert _{L^2(\mathbb {R}^4)}^2\geqslant (2\pi )^{4}\delta ^{4}\overline{{{{\textbf {H}}} }}_4^4\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}^4, \end{aligned}$$

then, writing $f=f_++f_-$ so that ${\overline{T}}_sf=T_sf_++T_sf_-(\cdot ,-\cdot )$ and $\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}^2=\Vert f_+ \Vert _{L^2({\mathcal {H}}^3_s)}^2+\Vert f_- \Vert _{L^2({\mathcal {H}}^3_s)}^2$ we obtain that

$$\begin{aligned}\Vert f_\epsilon {\mu }_s*f_\epsilon {\mu }_s \Vert _{L^2(\mathbb {R}^4)}^2\geqslant 2^{-4}\delta ^4 \overline{{{{\textbf {H}}} }}_4^4\Vert f_\epsilon \Vert _{L^2({\mathcal {H}}^3_s)}^4, \end{aligned}$$

for $\epsilon =+$ or for $\epsilon =-$, so that if both $\Vert f_+ \Vert _{L^2({\mathcal {H}}^3_s)}^2\geqslant \delta ^2\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}^2$ and $\Vert f_- \Vert _{L^2({\mathcal {H}}^3_s)}^2\geqslant \delta ^2\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}^2$, then we obtain the conclusions in Lemma 8.1 for f from the ones for $f_+$ or $f_-$, as it corresponds. On the other hand, if say $\Vert f_- \Vert _{L^2({\mathcal {H}}^3_s)}^2< \delta ^2\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}^2$, then $\Vert f_+ \Vert _{L^2({\mathcal {H}}^3_s)}^2\geqslant (1-\delta ^2)\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}^2$ and

$$\begin{aligned} \Vert Tf_+ \Vert _{L^4}\geqslant \Vert {\overline{T}}f \Vert _{L^4}-\Vert Tf_- \Vert _{L^4}\geqslant 2\pi \delta (\overline{{{\textbf {H}} }}_4-{{\textbf {H}} }_4)\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)} \geqslant c\delta {{\textbf {H}} }_4\Vert f_+ \Vert _{L^2({\mathcal {H}}^3_s)},\end{aligned}$$

so that Lemma 8.1 applied to $f_+$ yields the result for f.

The support condition $1\leqslant \vert x\vert \leqslant 2$ can be changed to $a\leqslant \vert x\vert \leqslant b$ for any $a\geqslant s$ and $b<\infty$, understanding that the implicit constants may depend on a, b. We can alternatively state the previous lemma for $f\in L^2({\mathcal {H}}^3)$ supported where $2^{N}\leqslant \vert x\vert \leqslant 2^{N+1}$, $N\geqslant 1$, the implicit constants independent of N, as can be easily checked by the use of scaling.

Recall that we write $\psi _s(r)=\sqrt{r^2-s^2}\mathbb {1}_{\{r\geqslant s\}}$ and $\phi _s(t)=\psi _s^{-1}(t)=\sqrt{t^2+s^2}\mathbb {1}_{\{t\geqslant 0\}}$ and for $f\in {\mathcal {S}}(\mathbb {R}^3)$ and $r>0$ we denote by $f\sigma _r$ the measure supported on ${\mathbb {S}}_r^2:=\{y\in \mathbb {R}^3:\vert y\vert =r\}$ given by

$$\begin{aligned}\langle f\sigma _r,\varphi \rangle =\int \limits _{{\mathbb {S}}^2}f(ry)\varphi (ry)r\,\textrm{d}\sigma (y).\end{aligned}$$

We denote $f_r$ the function $x\mapsto f(rx)$, which we consider as a function from ${\mathbb {S}}^2$ to ${\mathbb {C}}$.

In the next lemma we show that we can write the double convolution of functions on the hyperboloid ${\mathcal {H}}^3_s$ as an integral of convolutions of sliced spheres.

Lemma 8.3

Let $s\geqslant 0$. For $f,g\in L^2({\mathcal {H}}^3_s)$ we have the representation formula

$$\begin{aligned} \bigl (f\mu _s *g\mu _s\bigr )(x,t)=\int \limits _0^t \bigl (f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')}\bigr )(x)\,\textrm{d}t', \end{aligned}$$

(8.7)

for a.e. $(x,t)\in \mathbb {R}^3\times \mathbb {R}_+$.

Proof

Let $\varphi \in C_c^\infty (\mathbb {R}^4)$. Using spherical coordinates we have

$$\begin{aligned} \langle f\mu _s *g\mu _s,\varphi \rangle&=\int \limits _{\vert x\vert ,\vert y\vert \geqslant s} \varphi (x+y,\psi _s(x)+\psi _s(y))f(x)g(y)\frac{\,\text {d}x\,\text {d}y}{\sqrt{\vert x\vert ^2-s^2}\sqrt{\vert y\vert ^2-s^2}}\\ {}&=\int \limits _s^\infty \int \limits _s^\infty \int \limits _{{\mathbb {S}}^2}\int \limits _{{\mathbb {S}}^2}\varphi (r\omega +r'\omega ',\psi _s(r)+\psi _s(r')) f(r\omega )g(r'\omega ')\frac{r^2r'^2\,\text {d}\omega \,\text {d}\omega ' \,\text {d}r \,\text {d}r'}{\sqrt{r^2-s^2}\sqrt{r'^2-s^2}}. \end{aligned}$$

We change variables $(r,r')\mapsto (u,u')=(\psi _s(r),\psi _s(r'))=(\sqrt{r^2-s^2},\sqrt{r'^2-s^2})$ and obtain

$$\begin{aligned} \langle f\mu _s *g\mu _s,\varphi \rangle= & {} \int \limits _0^\infty \int \limits _0^\infty \int \limits _{{\mathbb {S}}^2}\int \limits _{{\mathbb {S}}^2}\varphi (\phi _s(u)\omega +\phi _s(u')\omega ', u+u')\\ {}{} & {} \times f(\phi _s(u)\omega )g(\phi _s(u')\omega ')\phi _s(u)\phi _s(u')\,\text {d}\omega \,\text {d}\omega ' \,\text {d}u \,\text {d}u'. \end{aligned}$$

We change variables $(u,u')\mapsto (t,t')=(u+u',u)$ and obtain

$$\begin{aligned} \langle f\mu _s *g\mu _s,\varphi \rangle&=\int \limits _0^\infty \int \limits _0^t\int \limits _{{\mathbb {S}}^2}\int \limits _{{\mathbb {S}}^2}\varphi (\phi _s(t')\omega +\phi _s(t-t')\omega ', t)\\ {}&\quad \times f(\phi _s(t')\omega )g(\phi _s(t-t')\omega ')\phi _s(t')\phi _s(t-t')\,\text {d}\omega \,\text {d}\omega ' \,\text {d}t' \,\text {d}t\\ {}&=\int \limits _0^\infty \int \limits _0^t \Bigl (\int \limits _{\mathbb {R}^3} \varphi (x,t)\bigl (f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')}\bigr )(x)dx\Bigr ) \,\text {d}t' \,\text {d}t\\ {}&=\Bigl \langle \int \limits _0^t\bigl (f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')}\bigr )(x)\,\text {d}t', \varphi \Bigr \rangle , \end{aligned}$$

where we used Fubini’s Theorem and that for any $r,r'>0$,

$$\begin{aligned} \langle f\sigma _r * g\sigma _{r'},\varphi (\cdot ,t)\rangle&=\int \limits _{\mathbb {R}^3} \varphi (x,t)\bigl (f\sigma _{r} * g\sigma _{r'}\bigr )(x)\,\textrm{d}x\\&=\int \limits _{{\mathbb {S}}^2_r\times {\mathbb {S}}^2_s}\varphi (x+x',t)f(x)g(x')\,\textrm{d}\sigma _r(x)\,\textrm{d}\sigma _{r'}(x')\\&=\int \limits _{{\mathbb {S}}^2\times {\mathbb {S}}^2}\varphi (r\omega +r'\omega ',t)f(r\omega )g(r'\omega ')rr' \,\textrm{d}\sigma (\omega )\,\textrm{d}\sigma (\omega ').\end{aligned}$$

$\square$

Next, we record a formula for the $L^p({\mathcal {H}}^3_s)$ norm in terms of the $L^p$ norm of the slices.

Lemma 8.4

Let $f\in L^p({\mathcal {H}}^3_s)$. Then

$$\begin{aligned} \Vert f \Vert _{L^p({\mathcal {H}}^3_s)}^p=\int \limits _0^\infty \Vert f_{\phi _s(t)} \Vert _{L^p({\mathbb {S}}^2)}^p\phi _s(t)\,\textrm{d}t. \end{aligned}$$

(8.8)

Proof

Using spherical coordinates we have

$$\begin{aligned} \Vert f \Vert _{L^p({\mathcal {H}}^3_s)}^p&=\int \limits _s^\infty \int \limits _{{\mathbb {S}}^2}\vert f(r\omega )\vert ^p\frac{r^2}{\sqrt{r^2-s^2}}\,\textrm{d}\omega \,\textrm{d}r=\int \limits _0^\infty \int \limits _{{\mathbb {S}}^2}\vert f(\phi _s(t)\omega )\vert ^p\phi _s(t)\,\textrm{d}\omega \,\textrm{d}t\\&=\int \limits _0^\infty \Vert f_{\phi _s(t)} \Vert _{L^p({\mathbb {S}}^2)}^p\phi _s(t)\,\textrm{d}t. \end{aligned}$$

$\square$

We now analyze the dependence of $\Vert f\sigma _{r} * g\sigma _{r'} \Vert _{L^2(\mathbb {R}^3)}$ in $(r,r')$. We start with the scaling property of $\widehat{f\sigma _r}$ as a function of r. We have

$$\begin{aligned} (\widehat{f\sigma _r})(x)=\int \limits _{{\mathbb {S}}_r^2}e^{-ix\cdot y}f(y)\,\textrm{d}\sigma _r(y)=\int \limits _{{\mathbb {S}}^2}e^{-irx\cdot y}f(ry)r\,\textrm{d}\sigma (y)=r(\widehat{f_r\sigma })(rx). \end{aligned}$$

Thus

$$\begin{aligned}\Vert \widehat{f\sigma _r} \Vert _{L^4(\mathbb {R}^3)}=r^{1/4}\Vert \widehat{f_r\sigma } \Vert _{L^4(\mathbb {R}^3)}\leqslant (2\pi )^{3/4}r^{1/4}{{\textbf {S}} }\Vert f_r \Vert _{L^2({\mathbb {S}}^2)}.\end{aligned}$$

Then, the Cauchy–Schwarz inequality implies that for any $r,r'>0$

$$\begin{aligned} \Vert \widehat{f\sigma _r}\,\widehat{g\sigma _{r'}} \Vert _{L^2(\mathbb {R}^3)}\leqslant \Vert \widehat{f\sigma _r} \Vert _{L^4}\Vert \widehat{g\sigma _{r'}} \Vert _{L^4}\leqslant (2\pi )^{3/2}{\varvec{S}}^2(rr')^{1/4}\Vert f_r \Vert _{L^2({\mathbb {S}}^2)}\Vert g_{r'} \Vert _{L^2({\mathbb {S}}^2)}, \end{aligned}$$

so that

$$\begin{aligned} \Vert f\sigma _r* g\sigma _{r'} \Vert _{L^2(\mathbb {R}^3)}\leqslant {{\textbf {S}} }^2(rr')^{1/4}\Vert f_r \Vert _{L^2({\mathbb {S}}^2)}\Vert g_{r'} \Vert _{L^2({\mathbb {S}}^2)}, \end{aligned}$$

(8.9)

and in particular, when $r=r'$ we obtain

$$\begin{aligned} \Vert f\sigma _r*g\sigma _{r} \Vert _{L^2(\mathbb {R}^3)}=r^{1/2}\Vert f_r\sigma *g_r\sigma \Vert _{L^2(\mathbb {R}^3)}\leqslant {{{\textbf {S}}} }^2r^{1/2}\Vert f_r \Vert _{L^2({\mathbb {S}}^2)}\Vert g_r \Vert _{L^2({\mathbb {S}}^2)}. \nonumber \\ \end{aligned}$$

(8.10)

Definition 8.5

A quasi-cap of ${\mathcal {H}}^3_s$ is a measurable set ${\mathcal {G}}\subseteq {\mathcal {H}}^3_s$ for which there exist $E\subseteq \mathbb {R}$ and spherical caps , for $t\in E$, such that

(8.11)

We note that a cap is also a quasi-cap; the difference in a generic quasi-cap is that the spherical caps may not be the same as in the case of a cap, and the set E may not be an interval.

In our main result of the section, Lemma 8.1, we want to obtain an analog of Lemma 7.1 for a compact subset of the hyperboloid. The idea is to use the cap Lemma 7.1 for the sphere on horizontal slices of the hyperboloid via (8.7) in a measurable way (recall Remark 7.2), and show that there are enough aligned sliced caps of similar size to obtain a cap for the hyperboloid. We do it for the upper sheet as the full one-sheeted hyperboloid follows from this as already noted in Remark 8.2. The proof of Lemma 8.1 is accomplished in the following way. First, we show that on a large subset of t’s in $[\psi _s(1),\psi _s(2)]$ we can apply Lemma 7.4 to the function $x\in {\mathbb {S}}^2\mapsto f(\phi _s(t)x)$ in a measurable way. This will allow us to prove a version of Lemma 8.1 where instead of a cap we have a quasi-cap. Next, we show that a subset of the quasi-cap of large relative measure is comparable to a cap and satisfies the requirements of Lemma 8.1, which then are shown to be satisfied by the cap itself. To prove this last point, we will make use of the quantitative version of the statement that “distant spherical caps interact weakly” as stated in Lemmas 7.6 and 7.7.

Proof of Lemma 8.1

In what follows, $c(\delta )$ denotes a constant that depends only on $\delta$ and is allowed to change from line to line.⁶ Recall from Remark 7.5 that (8.6) can be obtained from (8.4) and (8.5) with $\nu _\delta =\eta _\delta ^2/C_\delta$.

We first argue that we can assume that the support of $f(\cdot ,t)$ does not contain antipodal points for each $t\in [\psi _s(1),\psi _s(2)]$. We can cover ${\mathbb {S}}^2$ as the union of finitely many spherical caps each of radius $\frac{1}{4}$, whose centers form a maximally $\frac{1}{4}$-separated set on ${\mathbb {S}}^2$, and induce a decomposition of $\mathcal {H}_s^3$ as the union of the caps . By the triangle inequality we can therefore assume that f is supported on the cap , for some k, at the expense of changing $\delta$ by $\delta /\kappa$. The reason for doing this is to ensure that there are no nearly antipodal spherical caps later on.

Let us start by noting that for (x, t) in the support of f and $s\in [0,\frac{1}{2}]$ we have $\vert x\vert \in [1,2]$ and $t=\psi _s(x)\in [\psi _s(1),\psi _s(2)]=[\sqrt{1-s^2},\sqrt{4-s^2}]\subseteq [\frac{\sqrt{3}}{2},2]$, and that from Lemma 8.4

$$\begin{aligned} \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t)} \Vert _{L^2({\mathbb {S}}^2)}^2\,\textrm{d}t\leqslant \Vert f \Vert _{L^2({\mathcal {H}}^3_s)}^2\leqslant 2 \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t)} \Vert _{L^2({\mathbb {S}}^2)}^2\,\textrm{d}t. \end{aligned}$$

On the other hand $(f\mu _s *f\mu _s)(x,t)$ is supported where $2\psi _s(1)\leqslant t\leqslant 2\psi _s(2)$. From Lemma 8.3 for a.e. $(x,t)\in \mathbb {R}^4$ we have

$$\begin{aligned} f\mu _s *f\mu _s(x,t)=\int \limits _{\psi _s(1)}^{\psi _s(2)} (f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t-t')})(x)\,\text {d}t', \end{aligned}$$

(8.12)

(recall that $\phi _s(\tau )=0$ for $\tau <0$). Let

$$\begin{aligned} E_{\gamma }=\Biggl \{t\in [{\psi _s(1)},{\psi _s(2)}]:\begin{aligned} \Vert f\sigma _{\phi _s(t)} * f\sigma _{\phi _s(t)} \Vert _{L^2(\mathbb {R}^3)}&\geqslant \gamma ^2\delta ^2{{\textbf {H}} }_{4}^2{{\textbf {S}} }^2\Vert f_{\phi _s(t)} \Vert _2^2,\\ \Vert f_{\phi _s(t)} \Vert _2&\geqslant \gamma \delta {{\textbf {H}} }_{4}\Vert f \Vert _2 \end{aligned} \Biggr \}, \end{aligned}$$

and

$$\begin{aligned} E_{\gamma ,\lambda }=\Biggl \{t\in [{\psi _s(1)},{\psi _s(2)}]:\, {\begin{aligned} \Vert f\sigma _{\phi _s(t)} *f\sigma _{\phi _s(t)} \Vert _{L^2(\mathbb {R}^3)}&\geqslant \gamma ^2\delta ^2{{{\textbf {H}}} }_{4}^2{{{\textbf {S}}} }^2\Vert f_{\phi _s(t)} \Vert _2^2,\\ \lambda \delta {{{\textbf {H}}} }_{4}\Vert f \Vert _2\geqslant \Vert f_{\phi _s(t)} \Vert _2&\geqslant \gamma \delta {{{\textbf {H}}} }_{4}\Vert f \Vert _2 \end{aligned}} \Biggr \}. \end{aligned}$$

Here, $\Vert f_{\phi _s(t)} \Vert _2=\Vert f({\phi _s(t)}\,\cdot ,t) \Vert _{L^2({\mathbb {S}}^2)}$, while $\Vert f \Vert _2=\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}$. We claim that $\vert E_{\gamma }\vert \geqslant c(\delta )$ and $\vert E_{\gamma ,\lambda }\vert \geqslant c(\delta )$ if $\gamma$ and $\lambda$ are chosen small and large enough depending on $\delta$, respectively. Let us first analyze $\vert E_\gamma \vert$. From (8.12), using Fubini’s theorem and Minkowski’s integral inequality we have

$$\begin{aligned} \delta ^2{{{\textbf {H}}} }_4^2\Vert f \Vert _2^2&\leqslant \Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)} (f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t-t')})(x)\,\text {d}t' \Bigr \Vert _{L^2_{t,x}}\\ {}&\leqslant \Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t-t')} \Vert _{L^2_x}\mathbb {1}_{E_{\gamma }^\complement }(t')\,\text {d}t' \Bigr \Vert _{L^2_t}\\ {}&\quad +\Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)} (f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t-t')})(x)\mathbb {1}_{E_{\gamma }}(t')\,\text {d}t' \Bigr \Vert _{L^2_{x,t}}. \end{aligned}$$

Plancherel’s theorem and the Cauchy–Schwarz inequality give

$$\begin{aligned} \Vert f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t-t')} \Vert _{L^2_x}\leqslant \Vert f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t')} \Vert _{L^2_x}^{1/2}\Vert f\sigma _{\phi _s(t-t')} *f\sigma _{\phi _s(t-t')} \Vert _{L^2_x}^{1/2}, \end{aligned}$$

so that using the sharp estimate for $\Vert f\sigma _{\phi _s(t-t')} *f\sigma _{\phi _s(t-t')} \Vert _{L^2_x}$ as in (8.10), recalling that $\phi _s(t'),\,\phi _s(t-t')\in [1,2]$, we obtain

$$\begin{aligned}&\Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f\sigma _{\phi _s(t')} *f\sigma _{\phi _s(t-t')} \Vert _{L^2_x}\mathbb {1}_{E_\gamma ^\complement }(t')\,\text {d}t' \Bigr \Vert _{L^2_t}\\ {}&\quad \leqslant 2\gamma \delta {{{\textbf {H}}} }_{4}{{{\textbf {S}}} }^2 \Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t')} \Vert _2\Vert f_{\phi _s(t-t')} \Vert _2\,\text {d}t' \Bigr \Vert _{L^2_t}\\ {}&\qquad +2\gamma \delta {{{\textbf {H}}} }_{4}{{{\textbf {S}}} }^2\Vert f \Vert _2\Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t-t')} \Vert _2\,\text {d}t' \Bigr \Vert _{ L^2_t}\\ {}&\quad \leqslant 8\gamma \delta {{{\textbf {H}}} }_{4}{{{\textbf {S}}} }^2\Vert f \Vert _2^2. \end{aligned}$$

Therefore, choosing $\gamma =\delta {{\textbf {H}} }_{4}/(16{{\textbf {S}} }^2)$ we obtain

$$\begin{aligned}\Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)} (f\sigma _{\phi _s(t')} * f\sigma _{\phi _s(t-t')})(x)\mathbb {1}_{E_\gamma }(t')\,\textrm{d}t' \Bigr \Vert _{L^2_{x,t}}\geqslant \frac{1}{2}\delta ^2{{\textbf {H}} }_{4}^2\Vert f \Vert _2^2.\end{aligned}$$

For this choice of $\gamma$ we then obtain

$$\begin{aligned} \begin{aligned} \frac{1}{2}\delta ^2{{\textbf {H}} }_{4}^2\Vert f \Vert _2^2&\leqslant \Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)} (f\sigma _{\phi _s(t')} * f\sigma _{\phi _s(t-t')})(x)\mathbb {1}_{E_\gamma }(t')\,\textrm{d}t' \Bigr \Vert _{L^2_{x,t}}\\&\leqslant \Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f\sigma _{\phi _s(t')} * f\sigma _{\phi _s(t-t')} \Vert _ {L^2_x}\mathbb {1}_{E_\gamma }(t')\,\textrm{d}t' \Bigr \Vert _{L^2_t}\\&\leqslant 2{{\textbf {S}} }^2\Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t')} \Vert _{L^2_x}\Vert f_{\phi _s(t-t')} \Vert _{L^2_x} \mathbb {1}_{E_\gamma }(t')\,\textrm{d}t' \Bigr \Vert _{L^2_t}\\&\leqslant 2{{\textbf {S}} }^2\vert E_\gamma \vert ^{1/2}\int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t)} \Vert _{L^2_x}^2\,\textrm{d}t \leqslant 2{{\textbf {S}} }^2\Vert f \Vert _2^2\vert E_\gamma \vert ^{1/2}, \end{aligned} \end{aligned}$$

and therefore $\vert E_\gamma \vert \geqslant {{\textbf {H}} }_{4}^4\delta ^4/(16{{\textbf {S}} }^4)$.

To analyze $\vert E_{\gamma ,\lambda }\vert$ we use

$$\begin{aligned} E_{\gamma ,\lambda }=E_\gamma \cap \{t\in [{\psi _s(1)},{\psi _s(2)}]:\Vert f_{\phi _s(t)} \Vert _2\leqslant \lambda \delta {{\textbf {H}} }_{4}\Vert f \Vert _2\}. \end{aligned}$$

Chebyshev’s and Hölder’s inequalities imply

$$\begin{aligned} \vert \{t\in [{\psi _s(1)},{\psi _s(2)}]:\Vert f_{\phi _s(t)} \Vert _2>\lambda \delta {{\textbf {H}} }_{4}\Vert f \Vert _2\}\vert&\leqslant \frac{1}{\lambda \delta {{\textbf {H}} }_{4}\Vert f \Vert _2}\int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert f_{\phi _s(t)} \Vert _2\,\textrm{d}t\\&\leqslant \frac{2}{\lambda \delta {{\textbf {H}} }_{4}}. \end{aligned}$$

Therefore, choosing $\lambda =64{{\textbf {S}} }^4/({{\textbf {H}} }_4^5\delta ^5)$ we obtain

$$\begin{aligned} \vert E_{\gamma ,\lambda }\vert \geqslant \vert E_\gamma \vert - \vert \{t\in [{\psi _s(1)},{\psi _s(2)}]:\Vert f_{\phi _s(t)} \Vert _2> \lambda \delta {{\textbf {H}} }_{4}\Vert f \Vert _2\}\vert \geqslant \frac{{{\textbf {H}} }_4^4}{32{{\textbf {S}} }^4}\delta ^4. \end{aligned}$$

From now on, let us fix such values of $\gamma$ and $\lambda$ and let $E:=E_{\gamma ,\lambda }$. From the definition of E and (8.10), we have that for $t\in E$

$$\begin{aligned} \Vert f_{\phi _s(t)}\sigma *f_{\phi _s(t)}\sigma \Vert _{L^2(\mathbb {R}^3)}\geqslant c\phi _s(t)^{-1/2}\delta ^{4}\Vert f_{\phi _s(t)} \Vert _{L^2({\mathbb {S}}^2)}^2, \end{aligned}$$

so that Lemma 7.1 imply that for $t\in E$ there are caps and a decomposition $f_{{\phi _s(t)}}=G_{{\phi _s(t)}}+H_{{\phi _s(t)}}$. In this way we obtain a decomposition $f=g+h$, where $g(\phi _s(t)x,t)=G_{\phi _s(t)}(x)\mathbb {1}_{E}(t)$, $x\in {\mathbb {S}}^2$, $t\in [{\psi _s(1)},{\psi _s(2)}]$. As argued in Remark 7.2 and recorded in Lemma  7.3, by using a measurable selection theorem we can perform this decomposition in such a way that g and h are measurable functions and is a measurable subset of ${\mathcal {H}}^3_s$, so that ${\mathcal {G}}_0$ is a quasi-cap. According to Lemma 7.1, g and h satisfy the following conditions: $f=g+h$, $0\leqslant \vert g\vert ,\vert h\vert \leqslant \vert f\vert$, g and h have disjoint supports, $g(x,t)=0$ if $t\notin E$,

(8.13)

Note that Lemma 8.4 and (8.13) imply

$$\begin{aligned} \Vert g \Vert _{2}\geqslant \eta _\delta \Vert f \Vert _2. \end{aligned}$$

Given that for $t\in E$ we have $\delta ^2{{\textbf {H}} }_4\Vert f \Vert _2\lesssim \Vert f_{\phi _s(t)} \Vert _2\lesssim \delta ^{-4}{{\textbf {H}} }_4\Vert f \Vert _2$ we conclude, possibly by changing the constants that depend on $\delta$, that the function g satisfies

(8.14)

and

(8.15)

Summing up, we can restate what has been done so far in the following way: If $f\in L^2({\mathcal {H}}_s^3)$ satisfies $\Vert f\mu _s*f\mu _s \Vert _2\geqslant \delta ^2{{{\textbf {H}}} }_4^2\Vert f \Vert _2^2$ and is supported where $1\leqslant \vert x\vert \leqslant 2$ then there exist a decomposition $f=g+h$, a set $E\subseteq [{\psi _s(1)},{\psi _s(2)}]$ satisfying $\vert E\vert \gtrsim \delta ^4$ and a quasi-cap ${\mathcal {G}}_0$ (associated to E as in (8.11)) such that g and h have disjoint supports,

and (8.15) holds. This is the analog of Lemma 8.1 with a quasi-cap instead of a cap.

Using the quasi-cap analog of Lemma 8.1, as described in the previous paragraph, we can argue exactly as in Lemma 7.4 for the sphere to ensure, possibly after changing the constants that depend on $\delta$, that there exist a quasi-cap, which we continue to denote ${\mathcal {G}}_0$, associated to a set $E\subseteq [\psi _s(1),\psi _s(2)]$ with $\vert E\vert \gtrsim \delta ^4$, and functions g and h with the properties of the previous paragraph and additionally

$$\begin{aligned} \Vert g\mu _s *g\mu _s \Vert _{L^2(\mathbb {R}^4)}\geqslant c_\delta \Vert f \Vert _2^2. \end{aligned}$$

(8.16)

The next and final step is to show that the caps , $t\in E$, which define ${\mathcal {G}}_0$ are aligned for a large fraction of the t’s, and by this we mean that they have close radii and centers, up to powers of $\delta$.

Recall that for caps there is a distance function , defined in (7.19), that is relevant in Lemmas 7.6 and  7.7. For $\rho >0$ define

Then, starting from (8.16) we have the estimate

where in the second to last line we used (8.14) and the last line holds if $\rho$ is large enough as a function of⁷$\delta$, by the use of Lemma 7.7. For such choice of $\rho$ we can therefore ensure that

$$\begin{aligned} \Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert g\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')} \Vert _{L_x^2}\mathbb {1}_{A_{\rho }}(t',t-t')\,\text {d}t' \Bigr \Vert _{L^2_t}\geqslant \frac{c_\delta }{2}\Vert f \Vert _2^2. \end{aligned}$$

(8.17)

Note that (8.14) implies $\Vert g_{\phi _s(t)} \Vert _2\leqslant C_\delta \Vert f \Vert _2$ for all $t\in E$. This and (8.17) imply that

$$\begin{aligned} \frac{c_\delta }{2}\Vert f \Vert _2^2&\leqslant 2{{\textbf {S}} }^2\Bigl \Vert \int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert g_{\phi _s(t')} \Vert _2\Vert g_{\phi _s(t-t')} \Vert _2\mathbb {1}_{A_\rho }(t',t-t')\,\textrm{d}t' \Bigr \Vert _{L^2_t} \\&\leqslant 2{{\textbf {S}} }^2C_\delta ^2\Vert f \Vert _2^2\int \limits _{\psi _s(1)}^{\psi _s(2)}\Vert \mathbb {1}_{A_\rho }(t',t-t') \Vert _{L^2_t} \,\textrm{d}t'\\&\leqslant 4{{\textbf {S}} }^2C_\delta ^2\Vert f \Vert _2^2\,\vert A_\rho \vert ^{1/2}, \end{aligned}$$

where $\rho =\rho (\delta )$ is the already fixed function of $\delta$ and $\vert A_\rho \vert$ denotes the Lebesgue measure of $A_\rho \subseteq \mathbb {R}^2$. As $\vert A_\rho \vert \leqslant 2$ we conclude that $\vert A_\rho \vert \asymp c(\delta )$. By Fubini’s theorem, the fibers are a.e. measurable, the function is measurable and $\vert A_\rho \vert \leqslant 2{\text {ess\,sup}}_{t\in E}\vert A_\rho (t)\vert$. We then obtain the following estimate

from where we conclude the existence of a spherical cap such that

Denote and . For $t\in B_\rho$, the radii and the distance between the centers of the caps and are of the same order modulo powers of $\delta$. More precisely, if we let (y, a) denote the center and radius of a cap ${\mathscr {C}}_t$, $t\in B_\rho$, then the definition of the distance function $\varrho$ ensures that

$$\begin{aligned} c(\delta )a_0\leqslant a\leqslant c'(\delta )a_0,\,\text { and } \vert y_0-y\vert \leqslant c''(\delta )a_0. \end{aligned}$$

(8.18)

This is the only place where we used the assumption that f is supported on a cap , were the radius of is $\frac{1}{4}$, because this implies that the centers of the caps associated to $g_{\phi _s(t)}$, $t\in E$, can be chosen to be at distance at most $\frac{1}{2}$ from each other and therefore any two caps for $t,\,t'\in E$ are not nearly antipodal.

From (8.18) we conclude that for $t\in B_\rho$ we have and there exists $c(\delta )\geqslant 1$ such that the $c(\delta )$-enlargement of , denoted and defined by

contains for all $t\in B_\rho$, and hence the cap contains the quasi-cap ${\mathcal {G}}_1:=\{(x,t)\in {\mathcal {G}}_0:t\in B_\rho \}$. Note also that , for all $t\in B_\rho$.

Now, for each $t\in E$, $g_{\phi _s(t)}$ is supported on and , as stated in (8.15). If in addition $t\in B_\rho$, then

and so integrating in $t\in B_\rho$ and using that $\phi _s(t)\geqslant 1$ if $t\geqslant \psi _s(1)$ gives

Given that we obtain

Then $g\mathbb {1}_{{\mathcal {G}}_1},\, f-g\mathbb {1}_{{\mathcal {G}}_1}$ and ${\mathcal {C}}$ satisfy all of our requirements, given that ${\text {supp}}(g\mathbb {1}_{{\mathcal {G}}_1})\subseteq \overline{{\mathcal {G}}_1}\subseteq {\mathcal {C}}$, ${\mathcal {G}}_1\subseteq {\mathcal {G}}_0$, for all $t\in B_\rho$, and thus

$$\begin{aligned} \vert g\mathbb {1}_\mathcal {G_1}(x,t)\vert&\leqslant c(\delta )\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}\mu _s({\mathcal {C}})^{1/2}\mathbb {1}_{{\mathcal {C}}}(x,t),\text { for all }(x,t),\\ \Vert g\mathbb {1}_{{\mathcal {G}}_1} \Vert _{L^2({\mathcal {H}}^3_s)}&\geqslant c(\delta )\Vert f \Vert _{L^2({\mathcal {H}}^3_s)},\\ \Vert g\mathbb {1}_{{\mathcal {G}}_1} \Vert _{L^1({\mathcal {H}}^3_s)}&\geqslant c(\delta )\mu _s({\mathcal {C}})^{1/2}\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}. \end{aligned}$$

$\square$

9 A Concentration-Compactness Lemma

The result of this section is stated for ${\overline{{\mathcal {H}}}}^3_s$ but a similar statement and proof also hold for ${\mathcal {H}}^3_s$.

Lemma 9.1

Let $\{\rho _n\}_{n}$ be a sequence in $L^2({\overline{{\mathcal {H}}}}^3_s)$ satisfying

$$\begin{aligned}\int \limits _{{\overline{{\mathcal {H}}}}^3_s}\vert \rho _n\vert ^2\,\textrm{d}{{\bar{\mu }}}_s=\lambda ,\end{aligned}$$

where $\lambda >0$ is fixed. Then there exists a subsequence $\{\rho _{n_k}\}_{k}$ such that $\{\vert \rho _{n_k}\vert ^2\}_k$ satisfies one of the following three possibilities:

• (i) (compactness) there exists $\ell _k\in \mathbb {N}$ such that

  $$\begin{aligned}\forall \varepsilon >0,\, \exists R<\infty ,\int \limits _{\{s2^{\ell _k-R}\leqslant \vert y\vert \leqslant s2^{\ell _k+R}\}}\vert \rho _{n_k}\vert ^2 \,\textrm{d}{{\bar{\mu }}}_s\geqslant \lambda -\varepsilon ;\end{aligned}$$

• (ii) (vanishing) $\displaystyle \lim \limits _{k\rightarrow \infty }\sup _{\ell \in \mathbb {N}}\int \limits _{\{s2^{\ell -R}\leqslant \vert y\vert \leqslant s2^{\ell +R}\}}\vert \rho _{n_k}\vert ^2\,\textrm{d}{{\bar{\mu }}}_s=0$, for all $R<\infty$;

• (iii) (dichotomy) There exists $\alpha \in (0,\lambda )$ such that for all $\varepsilon >0$, there exist $R\in \mathbb {N}$, $k_0\geqslant 1$ and nonnegative functions $\rho _{k,1},\rho _{k,2}\in L^2({\overline{{\mathcal {H}}}}^3_s)$ satisfying for $k\geqslant k_0$:

  $$\begin{aligned}{} & {} \Vert \rho _{n_k}-(\rho _{k,1}+\rho _{k,2}) \Vert _{L^2({\overline{{\mathcal {H}}}}^3_s)}\leqslant \varepsilon , \end{aligned}$$

  (9.1)
  $$\begin{aligned}{} & {} \biggl \vert \int \limits _{{\overline{{\mathcal {H}}}}^3_s}\vert \rho _{k,1}\vert ^2\,\textrm{d}{{\bar{\mu }}}_s-\alpha \biggr \vert \leqslant \varepsilon ,\,\biggl \vert \int \limits _{{\overline{{\mathcal {H}}}}^3_s}\vert \rho _{k,2}\vert ^2\,\textrm{d}{{\bar{\mu }}}_s-(\lambda -\alpha ) \biggr \vert \leqslant \varepsilon , \end{aligned}$$

  (9.2)
  $$\begin{aligned}{} & {} {\text {supp}}(\rho _{k,1})\subseteq \{y\in \mathbb {R}^3:s2^{\ell _k-R}\leqslant \vert y\vert \leqslant s2^{\ell _k+R}\}, \end{aligned}$$

  (9.3)
  $$\begin{aligned}{} & {} {\text {supp}}(\rho _{k,2})\subseteq \{y\in \mathbb {R}^3:\vert y\vert \leqslant s2^{\ell _k-R_k}\}\cup \{y\in \mathbb {R}^3:\vert y\vert \geqslant s2^{\ell _k+R_k}\},\nonumber \\ \end{aligned}$$

  (9.4)

  for certain sequences $\{\ell _k\}_k$ and $\{R_k \}_k$, where $R_k\rightarrow \infty$ as $k\rightarrow \infty$.

Proof

The proof is identical to the proof of Lemma I.1 in [30], by defining the sequence of functions

$$\begin{aligned}Q_n:[0,\infty )\rightarrow \mathbb {R}_+,\quad Q_n(t)=\sup \limits _{\ell \in \mathbb {N}}\int \limits _{\{s2^{\ell -t}\leqslant \vert y\vert \leqslant s2^{\ell +t}\}}\vert \rho _n(y)\vert ^2 \,\textrm{d}{{\bar{\mu }}}_s(y).\end{aligned}$$

We omit the details. $\square$

In the forthcoming sections, we will be working with an $L^2$ normalized extremizing sequence $\{f_n\}_n$ and will apply the preceding lemma with $\lambda =1$. We will slightly abuse notation and say that $\{f_n\}_n$ satisfies either concentration, vanishing or dichotomy, when the sequence $\{\vert f_n\vert ^2\}_n$ satisfies the respective alternative.

10 Bilinear Estimates and Discarding Dichotomy

In this section we show that an extremizing sequence for ${\overline{T}}$ can not satisfy the dichotomy condition (iii) of Lemma 9.1. This will be a consequence of bilinear estimates at dyadic scales.

Proposition 10.1

There exists a constant $C<\infty$ with the following property. Let $s>0$, $k,k'\in \mathbb {N}$ and $f,g\in L^2({\mathcal {H}}_s^3)$ supported where $2^{k}s\leqslant \vert y\vert \leqslant 2^{k+1}s$ and $2^{k'}s\leqslant \vert y\vert \leqslant 2^{k'+1}s$ respectively. Then

$$\begin{aligned}\Vert T_sf\cdot T_sg \Vert _{L^2(\mathbb {R}^4)}\leqslant C2^{-\frac{1}{4}\vert k-k'\vert }\Vert f \Vert _{L^2(\mathcal H_s^3)}\Vert g \Vert _{L^2({\mathcal {H}}_s^3)}.\end{aligned}$$

Proof

Without loss of generality we can assume $k'\geqslant k$. Using Lemma 8.3 we write

$$\begin{aligned}f\mu _s *g\mu _s(x,t)=\int \limits _0^t (f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')})(x)\,\text {d}t',\end{aligned}$$

so that by Minkowski’s integral inequality

$$\begin{aligned} \Vert f\mu _s *g\mu _s \Vert _{L^2_{x,t}}\leqslant \biggl \Vert \int \limits _0^t \Vert f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')} \Vert _{L^2_x}\,\text {d}t' \biggr \Vert _{L^2_t}. \end{aligned}$$

(10.1)

Recalling (8.9), the right hand side of (10.1) satisfies

$$\begin{aligned}&\Bigl \Vert \int \limits _0^t\Vert f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')} \Vert _{L^2_x}\,\text {d}t' \Bigr \Vert _{L^4_t} \\ {}&\quad \leqslant C\Bigl \Vert \int \limits _0^t \phi _s(t')^{1/4}\Vert f_{\phi _s(t')} \Vert _2\,\phi _s(t-t')^{1/4}\Vert g_{\phi _s(t-t')} \Vert _2\,\text {d}t' \Bigr \Vert _{L^2_t}\\ {}&\quad \leqslant C\int \limits _0^\infty \phi _s(t')^{1/4}\Vert f_{\phi _s(t')} \Vert _2\Bigl \Vert \mathbb {1}_{\{t\geqslant t'\}}(t')\phi _s(t-t')^{1/4}\Vert g_{\phi _s(t-t')} \Vert _2 \Bigr \Vert _{L^2_t}\,\text {d}t'\\ {}&\quad \leqslant C\Bigl \Vert \phi _s(t)^{1/4}\Vert g_{\phi _s(t)} \Vert _2 \Bigr \Vert _{L^2_t}\int \limits _{\psi _s(2^{k}s)}^{\psi _s(2^{k+1}s)}\phi _s(t')^{1/4}\Vert f_{ \phi _s(t') } \Vert _2\,\text {d}t', \end{aligned}$$

where in the last line we used the support condition for f. Recalling the support condition for g

$$\begin{aligned} \Bigl \Vert \phi _s(t)^{1/4}\Vert g_{\phi _s(t)} \Vert _{L^2({\mathbb {S}}^2)} \Bigr \Vert _{L^2_t}^2&=\int \limits _{\psi _s(2^{k'}s)}^{\psi _s(2^{k'+1}s)}\phi _s(t)^{1/2} \Vert g_{\phi _s(t) } \Vert _{L^2({\mathbb {S}}^2)} ^2 \,\textrm{d}t\\&\leqslant \bigl (\phi _s(\psi _s(2^{k'}s))\bigr )^{-1/2}\int \limits _0^\infty \phi _s(t)\Vert g_{\phi _s(t)} \Vert _{L^2({\mathbb {S}}^2)}^2 \,\textrm{d}t\\&=(2^{k'}s)^{-1/2}\Vert g \Vert _{L^2({\mathcal {H}}^3_s)}^2, \end{aligned}$$

where in the last line we used Lemma 8.4. Similarly

$$\begin{aligned} \int \limits _{\psi _s(2^{k}s)}^{\psi _s(2^{k+1}s)}\phi _s(t')^{1/4}\Vert f_{\phi _s(t') } \Vert _2\,\textrm{d}t'&\leqslant \Bigl (\int \limits _{\psi _s(2^{k}s)}^{\psi _s(2^{k+1}s)}\phi _s(t')^{1/2}\Vert f_{\phi _s(t') } \Vert _2^2\,\textrm{d}t'\Bigr )^{1/2}\\&\quad \times \Bigl (\int \limits _{\psi _s(2^{k}s)}^{\psi _s(2^{k+1}s)} 1\,\textrm{d}t'\Bigr )^{1/2}\\&\leqslant (2^ks)^{-1/4}(\psi _s(2^{k+1}s)-\psi _s(2^{k}s))^{1/2}\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}\\&\asymp (2^ks)^{-1/4}(2^{k}s)^{1/2}\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}\\&=(2^{k}s)^{1/4}\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}. \end{aligned}$$

We conclude that

$$\begin{aligned} \Vert f\mu _s *g\mu _s \Vert _{L^2_{x,t}}&\leqslant \Bigl \Vert \int \limits _0^t\Vert f\sigma _{\phi _s(t')} *g\sigma _{\phi _s(t-t')} \Vert _{L^2_x}\,\text {d}t' \Bigr \Vert _{L^4_t} \lesssim 2^{k/4}\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}2^{-k'/4}\Vert g \Vert _{L^2({\mathcal {H}}^3_s)}\\ {}&=2^{-\frac{1}{4}\vert k'-k\vert }\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}\Vert g \Vert _{L^2({\mathcal {H}}^3_s)}.\end{aligned}$$

$\square$

Proposition 10.2

Let $f,g\in L^2({\mathcal {H}}^3)$ and suppose that their supports are separated in the sense that there exist $k,k'\in \mathbb {N}$, $k\leqslant k'$, such that ${\text {supp}}(f)\subseteq \{\vert y\vert \leqslant 2^{k}\}$ and ${\text {supp}}(g)\subseteq \{\vert y\vert \geqslant 2^{k'}\}$. Then

$$\begin{aligned}\Vert Tf\cdot Tg \Vert _{L^2(\mathbb {R}^4)}\leqslant C2^{-\frac{1}{4}\vert k-k'\vert }\Vert f \Vert _{L^2(\mathcal H^3)}\Vert g \Vert _{L^2({\mathcal {H}}^3)}.\end{aligned}$$

Similarly, if there exist $k,R,R'\in \mathbb {N}$, $R\leqslant R'$, such that ${\text {supp}}(f)\subseteq \{2^{k-R}\leqslant \vert y\vert \leqslant 2^{k+R}\}$ and ${\text {supp}}(g)\subseteq \{\vert y\vert \leqslant 2^{k-R'} \}\cup \{\vert y\vert \geqslant 2^{k+R'}\}$, then

$$\begin{aligned}\Vert Tf\cdot Tg \Vert _{L^2(\mathbb {R}^4)}\leqslant C2^{-\frac{1}{4}\vert R-R'\vert }\Vert f \Vert _{L^2(\mathcal H^3)}\Vert g \Vert _{L^2({\mathcal {H}}^3)}.\end{aligned}$$

Proof

We decompose $f=\sum _{m\in \mathbb {N}}f_m$ and $g=\sum _{m'\in \mathbb {N}}g_{m'}$ where $f_m,g_m$ are supported where $2^{m}\leqslant \vert y\vert \leqslant 2^{m+1}$, $m\geqslant 0$. Then

$$\begin{aligned} \Vert Tf\cdot Tg \Vert _{L^2(\mathbb {R}^4)}&= \Bigl \Vert \sum _{m,m'}Tf_m\cdot Tg_{m'} \Bigr \Vert _{L^2}\leqslant \sum _{m,m'}\Vert Tf_m\cdot Tg_{m'} \Vert _{L^2}\\&\leqslant \sum _{m,m'} 2^{-\frac{1}{4}\vert m-m'\vert }\Vert f_m \Vert _{L^2}\Vert g_{m'} \Vert _{L^2}\\&=2^{-\frac{1}{4}\vert k'-k+1\vert }\sum _{m\leqslant 0,m'\geqslant 0} 2^{-\frac{1}{4}\vert m-m'\vert }\Vert f_{m+k-1} \Vert _{L^2}\Vert g_{m'+k'} \Vert _{L^2}\\&\leqslant C 2^{-\frac{1}{4}\vert k'-k\vert }\Vert f \Vert _{L^2({\mathcal {H}}^3)}\Vert g \Vert _{L^2({\mathcal {H}}^3)}. \end{aligned}$$

The second part of the proposition follows from the first and the triangle inequality. $\square$

Decomposing a function $f\in L^2({\overline{{\mathcal {H}}}}^3)$ as the sum of a function $f_+\in L^2({\mathcal {H}}^3)$ and $f_-\in L^2(-{\mathcal {H}}^3)$, $f=f_++f_-$, using that ${\overline{T}}f(\cdot ,\cdot )=Tf_+(\cdot ,\cdot )+Tf_-(\cdot ,-\cdot )$ and the triangle inequality we can obtain a statement analogous to the previous proposition for functions on the full one-sheeted hyperboloid ${\overline{{\mathcal {H}}}}^3$: if f, g belong to $L^2({\overline{{\mathcal {H}}}}^3)$ and satisfy for some $k,R,R'\in \mathbb {N}$, $R\leqslant R'$:

$$\begin{aligned}{\text {supp}}(f)\subseteq \{2^{k-R}\leqslant \vert y\vert \leqslant 2^{k+R}\},\,{\text {supp}}(g)\subseteq \{\vert y\vert \leqslant 2^{k-R'} \}\cup \{\vert y\vert \geqslant 2^{k+R'}\},\end{aligned}$$

then

$$\begin{aligned} \Vert {\overline{T}}f\cdot {\overline{T}}g \Vert _{L^2(\mathbb {R}^4)}\leqslant C2^{-\frac{1}{4}\vert R-R'\vert }\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3)}\Vert g \Vert _{L^2({\overline{{\mathcal {H}}}}^3)}. \end{aligned}$$

(10.2)

Proposition 10.3

An extremizing sequence for the adjoint Fourier restriction inequality (1.10) on $\overline{{\mathcal {H}}}^{\,3}$ does not satisfy dichotomy.

Proof

Let us argue by contradiction. Let $\{f_n\}_n$ be an extremizing sequence such that $\{\vert f_n\vert ^2\}_n$ satisfies condition (iii), dichotomy, in Lemma 9.1. Let $\varepsilon >0$ be given and $f_{n,1},f_{n,2}$, $n_0$ be as in the conclusion of the dichotomy condition. Then, for $n\geqslant n_0$

$$\begin{aligned}\Vert {\overline{T}}f_n-{\overline{T}}f_{n,1}-{\overline{T}}f_{n,2} \Vert _{L^4}\leqslant \overline{{{\textbf {H}} }}_4\Vert f_n-(f_{n,1}+f_{n,2}) \Vert _{L^2}\leqslant \overline{{{\textbf {H}} }}_4\,\varepsilon ,\end{aligned}$$

therefore

$$\begin{aligned} \Vert {\overline{T}}f_n \Vert _{L^4}\leqslant \overline{{{\textbf {H}} }}_4\,\varepsilon +\Vert {\overline{T}}(f_{n,1}+f_{n,2}) \Vert _{L^4}. \end{aligned}$$

(10.3)

Expanding, using Proposition 10.2 (or the comment thereafter) and the support condition for $f_{n,1}$ and $f_{n,2}$ as in (9.1)–(9.4), there exists $C<\infty$ independent of $\varepsilon$ such that for all n large enough

$$\begin{aligned} \Vert {\overline{T}}(f_{n,1}+f_{n,2}) \Vert _{L^4}^4&=\Vert ({\overline{T}}(f_{n,1}+f_{n,2}))^2 \Vert _{L^2}^2=\Vert (\overline{T}f_{n,1})^2+2\overline{T}f_{n,1}\cdot {\overline{T}}f_{n,2}+({\overline{T}}f_{n,2})^2 \Vert _{L^2}^2\\&=\Vert {\overline{T}}f_{n,1} \Vert _{L^4}^4+\Vert {\overline{T}}f_{n,2} \Vert _{L^4}^4+2\langle ({\overline{T}}f_{n,1})^2,(\overline{T}f_{n,2})^2\rangle \\&\quad +4\langle (\overline{T}f_{n,1})^2,{\overline{T}}f_{n,1}\cdot {\overline{T}}f_{n,2}\rangle +4\langle (\overline{T}f_{n,2})^2,{\overline{T}}f_{n,1}\cdot {\overline{T}}f_{n,2}\rangle \\&\leqslant \Vert {\overline{T}}f_{n,1} \Vert _{L^4}^4+\Vert {\overline{T}}f_{n,2} \Vert _{L^4}^4+\varepsilon \\&\leqslant \overline{{{\textbf {H}} }}_4^4\Vert f_{n,1} \Vert _2^4+\overline{{{\textbf {H}} }}_4^4\Vert f_{n,2} \Vert _2^4+\varepsilon \\&\leqslant \overline{{{\textbf {H}} }}_4^4(\alpha ^2+(1-\alpha )^2)+C\varepsilon , \end{aligned}$$

so that using (10.3) and taking $n\rightarrow \infty$ we find that for any $\varepsilon >0$

$$\begin{aligned} \overline{{{\textbf {H}} }}_4^4\leqslant \overline{{{\textbf {H}} }}_4^4(\alpha ^2+(1-\alpha )^2)+C\varepsilon , \end{aligned}$$

for some constant $C<\infty$ independent of $\varepsilon$.

We conclude $1\leqslant \alpha ^2+(1-\alpha )^2$. We reach a contradiction since $\alpha \in (0,1)$ and the numerical inequality $\alpha ^2+(1-\alpha )^2<1$ holds. $\square$

The proof we just gave to discard dichotomy can be seen in the context of the strict superaditivity condition as proposed by Lions [30, Sect. I.2]; see for instance the comment at the end of Appendix A in [35].

11 Dyadic Refinements and Discarding Vanishing

In this section we prove a dyadic improvement of the $L^2\rightarrow L^4$ inequality (1.4) that will imply that extremizing sequences for ${\overline{T}}$ do not satisfy the vanishing condition (ii) of Lemma 9.1. We start with the following proposition.

Proposition 11.1

There exists a constant $C<\infty$ with the following property. Let $f\in L^2({{\mathcal {H}}}^3)$ and for $k\in \mathbb {N}$ let $f_k(y)=f(y)\mathbb {1}_{\{2^{k}\leqslant \vert y\vert < 2^{k+1}\}}$. Then

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}\leqslant C\biggl (\sum _{k\geqslant 0}\Vert f_k \Vert _{L^2(\overline{{\mathcal {H}}}^3)}^3\biggr )^{1/3}. \end{aligned}$$

(11.1)

Proof

We follow [38, Proof of Prop. 3.4]. We have

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}^{3}= & {} \Vert Tf\cdot Tf\cdot Tf \Vert _{L^{4/3}}=\Bigl \Vert \sum _{k,l,m}Tf_k\cdot Tf_l\cdot Tf_m \Bigr \Vert _{L^{4/3}} \\\leqslant & {} \sum _{k,l,m}\Vert Tf_k\cdot Tf_l\cdot Tf_m \Vert _{L^{4/3}}. \end{aligned}$$

Fix a triplet (k, l, m). We can assume without loss of generality that $\vert k-l\vert =\max \{\vert k-l\vert ,\vert k-m\vert ,\vert l-m\vert \}$ so that the use of Hölder’s inequality and Proposition 10.1 give

$$\begin{aligned} \Vert Tf_k\cdot Tf_l\cdot Tf_m \Vert _{L^{4/3}}&\leqslant \Vert Tf_k\cdot Tf_l \Vert _{L^2}\Vert Tf_m \Vert _{L^4}\\&\lesssim 2^{-\frac{1}{4}\vert k-l\vert }\Vert f_k \Vert _{L^2}\Vert f_l \Vert _{L^2}\Vert f_m \Vert _{L^2}\\&\leqslant 2^{-\vert k-l\vert /12}2^{-\vert k-m\vert /12}2^{-\vert l-m\vert /12}\Vert f_k \Vert _{L^2}\Vert f_l \Vert _{L^2}\Vert f_m \Vert _{L^2}. \end{aligned}$$

We conclude that

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}^{3}\lesssim \sum _{k,l,m}2^{-\vert k-l\vert /12}2^{-\vert k-m\vert /12}2^{-\vert l-m\vert /12}\Vert f_k \Vert _{L^2}\Vert f_l \Vert _{L^2}\Vert f_m \Vert _{L^2}. \end{aligned}$$

Applying Hölder’s inequality to the last estimate we obtain

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}^{3}\lesssim \sum _{k,l,m}2^{-\vert k-l\vert /12}2^{-\vert k-m\vert /12}2^{-\vert l-m\vert /12}\Vert f_k \Vert _{L^2}^3\lesssim \sum _{k}\Vert f_k \Vert _{L^2}^3.\end{aligned}$$

$\square$

As an application we have the following corollary.

Corollary 11.2

There exists a constant $C<\infty$ with the following property. Let $f\in L^2({{\mathcal {H}}}^3)$ and for $k\in \mathbb {N}$ let $f_k(y)=f(y)\mathbb {1}_{\{2^{k}\leqslant \vert y\vert < 2^{k+1}\}}$. Then

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}\leqslant C\sup _{k\in \mathbb {N}}\Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}^{1/3}\Vert f \Vert _{L^2({{\mathcal {H}}}^3)}^{2/3}. \end{aligned}$$

(11.2)

Proof

From Proposition 11.1 we obtain

$$\begin{aligned} \Vert Tf \Vert _{L^4(\mathbb {R}^4)}&\leqslant C\Bigl (\sum _{k\geqslant 0}\Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}^3\Bigr )^{1/3}=C\Bigl (\sum _{k\geqslant 0}\Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}\cdot \Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}^2\Bigr )^{1/3}\\&\leqslant C\sup _{k\in \mathbb {N}}\Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}^{1/3} \Bigl (\sum _{k\geqslant 0}\Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}^2\Bigr )^{1/3}\\&=C\sup _{k\in \mathbb {N}}\Vert f_k \Vert _{L^2({{\mathcal {H}}}^3)}^{1/3} \Vert f \Vert _{L^2({{\mathcal {H}}}^3)}^{2/3}. \end{aligned}$$

$\square$

The same previous argument and (10.2) give

$$\begin{aligned} \Vert {{\overline{T}}}f \Vert _{L^4(\mathbb {R}^4)}\lesssim \sup _{k\in \mathbb {N}}\Vert f_k \Vert _{L^2({\overline{{\mathcal {H}}}}^3)}^{1/3}\Vert f \Vert _{L^2({\overline{{\mathcal {H}}}}^3)}^{2/3}, \end{aligned}$$

(11.3)

and thus it is immediate that for an extremizing sequence for ${\overline{T}}$ the vanishing alternative does not hold.

Proposition 11.3

Extremizing sequences for the adjoint Fourier restriction inequality (1.10) on $\overline{{\mathcal {H}}}^{\,3}$ do not satisfy vanishing.

12 Convergence to the Cone

The content of this section is important in the study of the compactness alternative of Lemma 9.1, in the case in which, in addition, the extremizing sequences concentrate at infinity.

Formally, we can write $\Gamma ^3={\mathcal {H}}^3_0$, $\sigma _c=\mu _0$ and $T_c=T_0$. It is natural then to study relationships between the adjoint Fourier restriction operator on the cone $(\Gamma ^3,\sigma _c)$ and on each member of the family $\{({\mathcal {H}}^3_s,\mu _s)\}_{s>0}$, in the limit $s\rightarrow 0^+$, and this is the content of this section (see also [29, Lemma 2.9] for related results for the case of the two-sheeted hyperboloid).

Note that if $0\leqslant t\leqslant s$ and $\vert y\vert \geqslant s$, then the inequality $\sqrt{\vert y\vert ^2-s^2}\leqslant \sqrt{\vert y\vert ^2-t^2}$ implies that for $f\in L^2(\mu _s)$

$$\begin{aligned} \Vert f\mathbb {1}_{\{\vert y\vert \geqslant s\}} \Vert _{L^2(\sigma _c)}\leqslant \Vert f\mathbb {1}_{\{\vert y\vert \geqslant s\}} \Vert _{L^2(\mu _t)}\leqslant \Vert f \Vert _{L^2(\mu _s)}, \end{aligned}$$

and for $f\in L^2(\mu _s)$, extended to be zero in the region where $\vert y\vert \leqslant s$,

$$\begin{aligned} \lim _{t\rightarrow 0^+}\Vert f \Vert _{L^2(\mu _t)}=\Vert f \Vert _{L^2(\sigma _c)}. \end{aligned}$$

Throughout this section we will commonly encounter the situation of having $f\in L^2({\mathcal {H}}^3_s)$ and regard it as a function in $L^2({\mathcal {H}}^3_t)$, $0\leqslant t\leqslant s$, via the orthogonal projection to $\mathbb {R}^3\times \{0\}$. In this case, it will be understood that f is extended by zero in the region where⁸$\vert y\vert \leqslant s$.

Let us consider the following situation. Let $a>0$, $\{s_n\}_n\subset \mathbb {R}$ satisfying $s_n\rightarrow 0$ as $n\rightarrow \infty$. Let $\{f_n\}_n$ be a family of functions with $f_n\in L^2({\mathcal {H}}_{s_n}^3)$, supported where $\vert y\vert \geqslant a$ and satisfying ${\sup _n\Vert f_n \Vert _{L^2(\mu _{s_n})}<\infty }$. As already noted, $\Vert f_n \Vert _{L^2(\mu _{s_n})}\geqslant \Vert f_n \Vert _{L^2(\sigma _c)}$, therefore $\{f_n\mathbb {1}_{\{\vert y\vert \geqslant s_n\}}\}_n$ is a bounded sequence in $L^2(\sigma _c)$. We can assume, possibly after passing to a subsequence, that $f_n\rightharpoonup f$ in $L^2(\sigma _c)$. The aim of this section is to compare $\Vert f\sigma _c *f\sigma _c \Vert _2$ and the limiting behavior of $\Vert f_n\mu _{s_n} *f_n\mu _{s_n} \Vert _2$, as $n\rightarrow \infty$, in the case when $f\ne 0$. We have some preliminary results.

Lemma 12.1

Let $a>0$ and $f\in L^2({\mathcal {H}}_s^3)$ for all small $s> 0$ and supported where $\vert y\vert \geqslant a$, then

$$\begin{aligned} \Vert T_{s}f-T_cf \Vert _{L^4(\mathbb {R}^4)}\rightarrow 0 \text { as } s\rightarrow 0^+. \end{aligned}$$

Proof

From the uniform in s bound $\Vert T_{s} \Vert =\Vert T \Vert$ and density arguments, it suffices to consider the case when $f\in C_c^\infty (\mathbb {R}^3)$. Let $b\in (a,\infty )$ be such that the support of f is contained in the region where $a\leqslant \vert y\vert \leqslant b$.

By Plancherel’s theorem, to show $T_sf\rightarrow Tf$ in $L^4(\mathbb {R}^4)$, as $s\rightarrow 0^+$, it suffices to show that $f\mu _{s}*f\mu _{s}\rightarrow f\sigma _c*f\sigma _c$ and $f\mu _{s}*f\sigma _c\rightarrow f\sigma _c*f\sigma _c$ in $L^2(\mathbb {R}^4)$, as $s\rightarrow 0^+$.

First, we claim that there is pointwise convergence $f\mu _{s}*f\mu _{s}(\xi ,\tau )\rightarrow f\sigma _c*f\sigma _c(\xi ,\tau )$ and $f\mu _{s}*f\sigma _c(\xi ,\tau )\rightarrow f\sigma _c*f\sigma _c(\xi ,\tau )$, a.e. $(\xi ,\tau )\in \mathbb {R}^4$, as $s\rightarrow 0^+$. Indeed, as in the proof of the explicit formula for $\mu _s*\mu _s$ in Sect. 3, we can write integral formulas for $f\mu _s*f\mu _s$, $f\mu _s*f\sigma _c$ and $f\sigma _c*f\sigma _c$ for any $s\geqslant 0$ as in (3.10)–(3.12). Given that ${\widetilde{R}}_s(\xi )$ and ${\widetilde{Q}}_s(\xi )$ are explicit, we can spell out (3.10) and (3.11) from where it becomes clear that there is a.e. pointwise convergence to $f\sigma _c*f\sigma _c$ as $s\rightarrow 0^+$. Note that for each fixed $\xi \ne 0$, $\mathbb {1}_{\widetilde{R}_s(\xi )}(u,v)\rightarrow \mathbb {1}_{\{\vert u\vert \leqslant \vert \xi \vert \leqslant v\}}(u,v)$ and $\mathbb {1}_{ \widetilde{Q}_s(\xi )}(u,v)\rightarrow \mathbb {1}_{\{\vert u\vert \leqslant \vert \xi \vert \leqslant v\}}(u,v)$ a.e. pointwise as $s\rightarrow 0^+$.

By the Dominated Convergence Theorem, to finish it suffices to show that there exists $F\in L^2(\mathbb {R}^4)$ such that $\vert f\mu _s*f\mu _s(\xi ,\tau )\vert \leqslant F(\xi ,\tau )$ and $\vert f\mu _s*f\sigma _c(\xi ,\tau )\vert \leqslant F(\xi ,\tau )$, for a.e. $(\xi ,\tau )\in \mathbb {R}^4$. We use the inequalities

$$\begin{aligned} \vert f\mu _{s}*f\mu _{s}(\xi ,\tau )\vert ^2&\leqslant \Vert f \Vert _{L^\infty }^4\,\bigl (\mu _{s}*\mu _{s}\bigr )^2(\xi ,\tau ),\\ \vert f\mu _{s}*f\sigma _c(\xi ,\tau )\vert ^2&\leqslant \Vert f \Vert _{L^\infty }^4\,\bigl (\mu _{s}*\sigma _c\bigr )^2(\xi ,\tau ). \end{aligned}$$

On the supports of $f\mu _{s}*f\mu _{s}$ and $f\mu _{s}*f\sigma _c$, the functions $\mu _{s}*\mu _{s}$ and $\mu _{s}*\sigma _c$ are uniformly bounded in $s\in (0,1)$, as can be seen from Lemma 3.2 and formula (3.9). It follows that we can take

$$\begin{aligned}F(\xi ,\tau )=4\pi \Vert f \Vert _{L^\infty }^2\bigl (1+a^{-1}\bigr )\mathbb {1}_{\{a\leqslant \tau \leqslant 2b\}}\mathbb {1}_{\{\vert \xi \vert \leqslant 2b\}}(\xi ,\tau ).\end{aligned}$$

$\square$

Remark 12.2

Another possible way to prove Lemma 12.1, which does not rely on the exponent being an even integer, can be to follow the outline in the proof of [29, Lemma 2.9 (d)] which makes use of the analysis of oscillatory integrals through the method of stationary phase. More in detail, we could proceed as follows. As in the proof above, we can restrict attention to the case when $f\in C_c^\infty (\mathbb {R}^3)$, supported in the region where $a\leqslant \vert y\vert \leqslant b$, for some $b<\infty$.

We first consider the pointwise convergence $T_sf(x,t)\rightarrow T_cf(x,t)$, as $s\rightarrow 0$ for a.e. $(x,t)\in \mathbb {R}^3\times \mathbb {R}$. Recall the definitions of $T_sf(x,t)$ and $T_cf(x,t)$ in (1.13) and (1.14) and note that there is pointwise convergence of their integrands, that is

$$\begin{aligned} e^{ix\cdot y} e^{it\sqrt{\vert y\vert ^2-s^2}} f(y)\frac{\mathbb {1}_{\{\vert y\vert>s\}}}{\sqrt{\vert y\vert ^2-s^2}}\rightarrow e^{ix\cdot y} e^{it\vert y\vert } f(y)\frac{\mathbb {1}_{\{\vert y\vert >0\}}}{\vert y\vert }, \text { as }s\rightarrow 0,\end{aligned}$$

for all $(x,t)\in \mathbb {R}^4, y\in \mathbb {R}^3$. On the other hand, as the support of f is contained in the region where $\vert y\vert \geqslant a$, for all $s\in (0,a/2)$ we have

$$\begin{aligned} \biggl \vert e^{ix\cdot y} e^{it\sqrt{\vert y\vert ^2-s^2}} f(y)\frac{\mathbb {1}_{\{\vert y\vert >s\}}}{\sqrt{\vert y\vert ^2-s^2}} \biggr \vert \leqslant \vert f(y)\vert \frac{1}{\sqrt{\vert a\vert ^2-s^2}}\leqslant \frac{2}{\sqrt{3}a}\vert f(y)\vert , \end{aligned}$$

so that as $\vert f\vert \in L^2(\mathbb {R}^3)$, we can use the dominated convergence theorem to conclude that $T_sf\rightarrow T_cf$ pointwise in $\mathbb {R}^4$.

Let us take $M\in [1,\infty )$ and $s\leqslant a/2$. We have the identity

$$\begin{aligned} \vert x\vert ^2\,T_sf(x,t)=-\int \limits _{\mathbb {R}^3} e^{ix\cdot y} \Delta _y\biggl (\frac{e^{it\sqrt{\vert y\vert ^2-s^2}}f(y)}{\sqrt{\vert y\vert ^2-s^2}}\biggr )\,\textrm{d}y, \end{aligned}$$

as can be seen by integration by parts, so that if $\vert t\vert \leqslant M$, we obtain

$$\begin{aligned} \vert T_sf(x,t)\vert \lesssim _{a,b} \frac{1+M^2}{1+\vert x\vert ^2}\Vert f \Vert _{H^2(\mathbb {R}^3)}\in L^4(\mathbb {R}^3\times [-M,M]), \end{aligned}$$

where $H^2(\mathbb {R}^3)$ denotes the inhomogeneous Sobolev space with norm $\Vert f \Vert _{H^2(\mathbb {R}^3)}^2=\int \limits _{\mathbb {R}^3}\vert \hat{f}(x)\vert ^2(1+\vert x\vert ^2)^2\,\textrm{d}x$. By the dominated convergence theorem we conclude that $T_sf\rightarrow T_cf$ in $L^4(\mathbb {R}^3\times [-M,M])$, as $s\rightarrow 0^+$, for each $M<\infty$.

To treat the region where $\vert t\vert \geqslant M$, recall the dispersive estimates

$$\begin{aligned} \Vert T_sg(\cdot ,t) \Vert _{L^\infty _x(\mathbb {R}^3)}\lesssim _{a,b}\frac{1}{t^{3/2}}\Vert g \Vert _{L^1(\mathbb {R}^3)}\, \text { and }\,\Vert T_cg(\cdot ,t) \Vert _{L^2_x(\mathbb {R}^3)}\lesssim _{a,b} \frac{1}{t}\Vert g \Vert _{L^1(\mathbb {R}^3)},\end{aligned}$$

valid for any $g\in L^1(\mathbb {R}^3)$ supported where $a\leqslant \vert y\vert \leqslant b$. They can be proved using the method of stationary phase or by studying the fundamental solutions of the respective underlying classical partial differential equation as mentioned in the Introduction. Since we also have the $L^2$-norm conservation $\Vert T_sg \Vert _{L^2_x(\mathbb {R}^3)}=\Vert T_cg \Vert _{L^2_x(\mathbb {R}^3)}=\Vert g \Vert _{L^2(\mathbb {R}^3)}$ we obtain the interpolated estimates

$$\begin{aligned} \Vert T_sg \Vert _{L^4_x(\mathbb {R}^3)}\lesssim _{a,b}\frac{1}{t^{3/4}}\Vert g \Vert _{L^{4/3}(\mathbb {R}^3)}\,\text { and }\,\Vert T_cg \Vert _{L^4_x(\mathbb {R}^3)}\lesssim _{a,b}\frac{1}{t^{1/2}}\Vert g \Vert _{L^{4/3}(\mathbb {R}^3)}. \end{aligned}$$

In this way

$$\begin{aligned} \Vert T_sf-T_cf \Vert _{L^4_{x,t}(\vert t\vert \geqslant M)}\leqslant \Vert T_sf \Vert _{L^4_{x,t}(\vert t\vert \geqslant M)}+\Vert T_cf \Vert _{L^4_{x,t}(\vert t\vert \geqslant M)}\lesssim _{a,b}\frac{1}{M^{1/4}}\Vert f \Vert _{L^{4/3}(\mathbb {R}^3)}. \end{aligned}$$

The previous estimate in the region $\{(x,t)\in \mathbb {R}^3\times \mathbb {R}:\vert t\vert \geqslant M\}$ and the $L^4$ convergence in the region $\mathbb {R}^3\times [-M,M]$, valid for any $M\in [1,\infty )$, imply the desired result.

Recall the Fourier multiplier notation and the $\dot{H}^{1/2}(\mathbb {R}^3)$ homogeneous Sobolev norm and inner product from (5.1) and (5.2). We have the following lemma.

Lemma 12.3

Let $a>0$, then for each fixed $t\in \mathbb {R}$

$$\begin{aligned}\lim \limits _{s\rightarrow 0}\sup _{\begin{array}{c} u\in \dot{H}^{1/2}(\mathbb {R}^3)\\ {\text {supp}}({\hat{u}})\subseteq \{\xi \in \mathbb {R}^3:\vert \xi \vert \geqslant a\} \end{array}}\frac{\Vert e^{it\sqrt{-\Delta -s^2}}u-e^{it\sqrt{-\Delta }}u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}}{\Vert u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}}=0.\end{aligned}$$

Proof

For any $s\geqslant 0$ we have $\Vert e^{it\sqrt{-\Delta -s^2}}u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}=\Vert u \Vert _{\dot{H}^{1/2}}$. Now

$$\begin{aligned}e^{it\sqrt{\vert y\vert ^2-s^2}}-e^{it\vert y\vert }=\int \limits _0^s \frac{\,\textrm{d}}{\,\textrm{d}r}e^{it\sqrt{\vert y\vert ^2-r^2}} \,\textrm{d}r=-it\int \limits _0^s e^{it\sqrt{\vert y\vert ^2-r^2}}\frac{r}{\sqrt{\vert y\vert ^2-r^2}}\,\textrm{d}r.\end{aligned}$$

Then,

$$\begin{aligned} \Vert (e^{it\sqrt{-\Delta -s^2}}-e^{it\sqrt{-\Delta }})u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}&\leqslant \vert t\vert \int \limits _0^s \Bigl \Vert e^{it\sqrt{-\Delta -r^2}}\frac{r}{\sqrt{-\Delta -r^2}}u \Bigr \Vert _{\dot{H}^{1/2}(\mathbb {R}^4)}\,\textrm{d}r\\&=\vert t\vert \int \limits _0^s\Bigl \Vert \frac{r}{\sqrt{-\Delta -r^2}}u \Bigr \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}\,\textrm{d}r. \end{aligned}$$

If $0\leqslant s<a$ and ${\text {supp}}({\hat{u}})\subseteq \{\vert \xi \vert \geqslant a\}$, then

$$\begin{aligned}\Bigl \Vert \frac{r}{\sqrt{-\Delta -r^2}}u \Bigr \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}\leqslant \frac{r}{\sqrt{a^2-r^2}}\Vert u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)},\end{aligned}$$

so that

$$\begin{aligned} \Vert (e^{it\sqrt{-\Delta -s^2}}-e^{-it\sqrt{-\Delta }})u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}\leqslant \vert t\vert (a-\sqrt{a^2-s^2})\Vert u \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}, \end{aligned}$$

and the conclusion follows. $\square$

We now address the pointwise convergence of $T_{s_n}f_n$ to $T_cf$.

Lemma 12.4

Let $a>0$ and $\{s_n\}_n$ be a sequence of positive real numbers converging to zero. Let $f\in L^2(\Gamma ^3)$ and $\{f_n\}_{n}$ be a sequence satisfying $f_n\in L^2({\mathcal {H}}_{s_n}^3)$, $\sup _n\Vert f_n \Vert _{L^2(\mu _{s_n})}<\infty$ and supported where $\vert y\vert \geqslant a$, for all n. Suppose that $f_n\rightharpoonup f$ in $L^2(\Gamma ^3)$, as $n\rightarrow \infty$. Then, as $n\rightarrow \infty$

$$\begin{aligned} T_{s_n}f_{n}(x,t)\rightarrow T_cf(x,t) \text { for a.e. } (x,t)\in \mathbb {R}^4. \end{aligned}$$

Proof

Following the argument in the proof of Proposition 5.1, we start by defining $v_n$ and v by their Fourier transforms

$$\begin{aligned}{\hat{v}}_n(y)=\frac{f_n(y)}{\sqrt{\vert y\vert ^2-s_n^2}},\quad {\hat{v}}(y)=\frac{f(y)}{\vert y\vert }.\end{aligned}$$

Since $\sup _n\Vert f_n \Vert _{L^2(\Gamma ^3)}\leqslant \sup _n\Vert f_n \Vert _{L^2(\mu _{s_n})}<\infty$ and the functions are supported where $\vert y\vert \geqslant a>0$ we see that

$$\begin{aligned}\sup _n\Vert v_n \Vert _{\dot{H}^{1/2}(\mathbb {R}^3)}^2=\sup _n\int \limits _{\mathbb {R}^3}\vert {\hat{v}}_n(y)\vert ^2\vert y\vert \,\textrm{d}y\leqslant \sup _n\frac{a}{\sqrt{a^2-s_n^2}}\Vert f_n \Vert _{L^2(\mu _{s_n})}^2<\infty ,\end{aligned}$$

and

$$\begin{aligned}\sup _n\Vert v_n \Vert _{ L^{2}(\mathbb {R}^3)}^2=(2\pi )^{-3}\sup _n\int \limits _{\mathbb {R}^3}\vert {\hat{v}}_n(y)\vert ^2\,\textrm{d}y\leqslant (2\pi )^{-3}\sup _n\frac{1}{\sqrt{a^2-s_n^2}}\Vert f_n \Vert _{L^2(\mu _{s_n})}^2<\infty .\end{aligned}$$

If $\varphi \in \dot{H}^{1/2}(\mathbb {R}^3)$, then ${{\hat{\varphi }}}(\cdot )\vert \cdot \vert \in L^2(\Gamma ^3)$, from where we can deduce that $v_n\rightharpoonup v$ in $\dot{H}^{1/2}(\mathbb {R}^3)$, as $n\rightarrow \infty$. The operator $T_{s_n}$ applied to $f_n$ equals $(2\pi )^3e^{it\sqrt{-\Delta -s_n^2}}v_n$. Fix $t\in \mathbb {R}$. From Lemma 12.3 we know $\Vert (e^{it\sqrt{-\Delta -s_n^2}}-e^{it\sqrt{-\Delta }})\mathbb {1}_{\{\sqrt{-\Delta }\geqslant a\}} \Vert \rightarrow 0$ as $n\rightarrow \infty$, the norm being as operators on $\dot{H}^{1/2}(\mathbb {R}^3)$. This, added to the continuity of $e^{it\sqrt{-\Delta }}$ in $\dot{H}^{1/2}(\mathbb {R}^3)$ implies

$$\begin{aligned}e^{it\sqrt{-\Delta -s_n^2}}v_n\rightharpoonup e^{it\sqrt{-\Delta }}v,\end{aligned}$$

weakly in $\dot{H}^{1/2}(\mathbb {R}^3)$, as $n\rightarrow \infty$. Then, by the Rellich–Kondrashov Theorem, for any $R>0$

$$\begin{aligned}e^{it\sqrt{-\Delta -s_n^2}}v_n\rightarrow e^{it\sqrt{-\Delta }}v,\end{aligned}$$

strongly in $L^2(B(0,R))$, as $n\rightarrow \infty$. Denote by

$$\begin{aligned}F_n(t):=\int \limits _{\vert x\vert <R}\biggl \vert e^{it\sqrt{-\Delta -s_n^2}}v_n-e^{it\sqrt{-\Delta }}v \biggr \vert ^2\,\textrm{d}x=\Vert e^{it\sqrt{-\Delta -s_n^2}}v_n-e^{ it\sqrt{-\Delta }}v \Vert _{ L^2(B(0,R))}^2.\end{aligned}$$

Since we have $\Vert {\hat{v}}_n \Vert _{L^2(\mathbb {R}^3)}\lesssim _a \Vert f_n \Vert _{L^2(\mu _{s_n})}$ and $\Vert {\hat{v}} \Vert _{L^2(\mathbb {R}^3)}\lesssim _a \Vert f \Vert _{L^2(\sigma _c)}$, we obtain

$$\begin{aligned} F_n(t)&\leqslant \Vert e^{it\sqrt{-\Delta -s_n^2}}v_n-e^{ it\sqrt{-\Delta }}v \Vert _{L^2(\mathbb {R}^3)}^2 \leqslant (\Vert v_n \Vert _{L^2(\mathbb {R}^3)}+\Vert v \Vert _{L^2(\mathbb {R}^3)})^2\\&\lesssim \Vert f_n \Vert _{L^2(\mu _s)}^2+\Vert f \Vert _{L^2(\sigma _c)}^2. \end{aligned}$$

We can now finish exactly as in the proof of Proposition 5.1 and conclude that there exists a subsequence $\{n_k\}_k$ such that

$$\begin{aligned}T_{s_{n_k}}f_{n_k}-T_cf\rightarrow 0\quad \text {a.e. in }\mathbb {R}^4.\end{aligned}$$

$\square$

Finally, we prove that the existence of an extremizing sequence that concentrates at infinity with a nonzero weak limit, after appropriate rescaling, implies that the operator norm of T is upper bounded by that of $T_c$ (which in the end we will pair with Proposition 4.1 to rule out this scenario).

Lemma 12.5

Let $\{s_n\}_n$ be a sequence of positive real numbers converging to zero. Let $f\in L^2(\Gamma ^3)$ be a nonzero function and $\{f_n\}_{n}$ be a sequence satisfying $f_n\in L^2({\mathcal {H}}_{s_n}^3)$, for all n. Suppose that:

• (i) $\Vert f_n \Vert _{L^2(\mu _{s_n})}=1$,

• (ii) $\Vert T_{s_n}f_n \Vert _{L^4}\rightarrow \Vert T \Vert \,(=\Vert T_1 \Vert )$,

• (iii) $f_n\rightharpoonup f\ne 0$ in $L^2(\Gamma ^3)$,

If there exists $a>0$ such that

• (iv) ${\text {supp}}(f),{\text {supp}}(f_n)\subseteq \{y\in \mathbb {R}^3:\vert y\vert \geqslant a\}$, for all n,

then

$$\begin{aligned}\Vert T \Vert \leqslant \Vert T_c \Vert .\end{aligned}$$

If condition (iv) is relaxed to

• (v) $\sup _{n\in \mathbb {N}}\Vert f_n\mathbb {1}_{\{\vert y\vert \leqslant a\}} \Vert _{L^2(\mu _{s_n})}\leqslant \varepsilon$, for some $\varepsilon >0$,

then

$$\begin{aligned} \Vert T \Vert ^2\Vert f\mathbb {1}_{\{\vert y\vert \geqslant a\}} \Vert _{L^2(\sigma _c)}^2\leqslant \Vert T_c \Vert ^2\Vert f\mathbb {1}_{\{\vert y\vert \geqslant a\}} \Vert _{L^2(\sigma _c)}^2+C\varepsilon , \end{aligned}$$

for some universal constant C. In particular, if we have $\sup _{n\in \mathbb {N}}\Vert f_n\mathbb {1}_{\{\vert y\vert \leqslant a\}} \Vert _{L^2(\mu _{s_n})}\rightarrow 0$ as $a\rightarrow 0^+$, then $\Vert T \Vert \leqslant \Vert T_c \Vert$.

An analog statement applies if we change T and $T_c$ by ${\overline{T}}$ and ${\overline{T}}_c$, respectively, the proof being identical.

Proof

We argue as in [20]. By the weak convergence condition (iii),

$$\begin{aligned} \Vert f_n-f \Vert _{L^2(\sigma _c)}^2=\Vert f_n \Vert _{L^2(\sigma _c)}^2-\Vert f \Vert _{L^2(\sigma _c)}^2+o(1). \end{aligned}$$

(12.1)

Now consider that (iv) holds. By (i) and (iv), $\Vert f_n \Vert _{L^2(\sigma _c)}^2 -\Vert f_n \Vert _{L^2(\mu _{s_n})}^2\rightarrow 0$. Indeed,

$$\begin{aligned} \begin{aligned} 0&\leqslant \Vert f_n \Vert _{L^2(\mu _{s_n})}^2-\Vert f_n \Vert _{L^2(\sigma _c)}^2=\int \limits _{\vert y\vert \geqslant a} \vert f_n(y)\vert ^2\biggl \vert \frac{1}{\sqrt{\vert y\vert ^2-s_n^2}}-\frac{1}{\vert y\vert } \biggr \vert \,\textrm{d}y\\&\leqslant \Vert f_n \Vert _{L^2(\mu _{s_n})}^2 \biggl \Vert \frac{\vert y\vert -\sqrt{\vert y\vert ^2-s_n^2}}{\vert y\vert }\mathbb {1}_{\{\vert y\vert \geqslant a\}} \biggr \Vert _{L_y^\infty (\mathbb {R}^3)}\\&=\Bigl (1-\sqrt{1-s_n^2a^{-2}}\Bigr )\rightarrow 0, \end{aligned} \end{aligned}$$

(12.2)

as $n\rightarrow \infty$. Then, (12.1) implies

$$\begin{aligned} \Vert f_n-f \Vert _{L^2(\sigma _c)}^2=\Vert f_n \Vert _{L^2(\mu _{s_n})}^2-\Vert f \Vert _{L^2(\sigma _c)}^2+o(1). \end{aligned}$$

(12.3)

Because of conditions (iii) and (iv) and Lemma 12.4, $T_{s_n}f_n\rightarrow T_cf$ a.e. pointwise in $\mathbb {R}^4$, as $n\rightarrow \infty$, and we can apply the Brézis–Lieb lemma to the sequence $\{T_{s_n}f_n\}_n\subset L^4(\mathbb {R}^4)$ to obtain

$$\begin{aligned}\Vert T_{s_n}f_n-T_cf \Vert _{L^4(\mathbb {R}^4)}^4=\Vert T_{s_n}f_n \Vert _{L^4(\mathbb {R}^4)}^4-\Vert T_cf \Vert _{L^4(\mathbb {R}^4)}^4+o(1).\end{aligned}$$

Because by scaling the norm of the operator $T_{s_n}$ is independent of n (see Sect. 14) and by (ii) $\Vert T_{s_n}f_n \Vert _{L^4(\mathbb {R}^4)}\rightarrow \Vert T \Vert$ as $n\rightarrow \infty$, we obtain

$$\begin{aligned} \Vert T_{s_n} \Vert ^2&=\Vert T \Vert ^2=\frac{\Vert T_{s_n}f_n \Vert _{L^4(\mathbb {R}^4)}^2}{\Vert f_n \Vert _{L^2(\mu _{s_n})}^2}+o(1)\nonumber \\&=\frac{(\Vert T_{s_n}f_n-T_cf \Vert _{L^4}^4+\Vert T_cf \Vert _{L^4}^4+o(1))^{1/2}}{\Vert f_n-f \Vert _{L^2(\sigma _c)}^2+\Vert f \Vert _{L^2(\sigma _c)}^2+o(1)}+o(1)\nonumber \\&\leqslant \frac{\Vert T_{s_n}f_n-T_cf \Vert _{L^4}^2+\Vert T_cf \Vert _{L^4}^2+o(1)}{\Vert f_n-f \Vert _{L^2(\sigma _c)}^2+\Vert f \Vert _{L^2(\sigma _c)}^2+o(1)}+o(1)\nonumber \\&\leqslant \frac{\Vert T_{s_n}f_n-T_{s_n}f \Vert _{L^4}^2+\Vert T_cf \Vert _{L^4}^2+o(1)}{\Vert f_n-f \Vert _{L^2(\sigma _c)}^2+\Vert f \Vert _{L^2(\sigma _c)}^2+o(1)}+o(1), \end{aligned}$$

(12.4)

where in the last inequality we used the triangle inequality and that $\Vert T_{s_n}f-T_cf \Vert _{L^4}\rightarrow 0$ as $n\rightarrow \infty$, from Lemma 12.1. Then

$$\begin{aligned}\Vert T_{s_n} \Vert ^2\leqslant \frac{\Vert T_{s_n} \Vert ^2\Vert f_n-f \Vert _{L^2(\mu _{s_n})}^2+\Vert T_cf \Vert _{L^4}^2+o(1)}{\Vert f_n-f \Vert _{L^2(\sigma _c)}^2+\Vert f \Vert _{L^2(\sigma _c)}^2+o(1)}+o(1),\end{aligned}$$

and hence

$$\begin{aligned}\Vert T_{s_n} \Vert ^2(\Vert f_n-f \Vert _{L^2(\sigma _c)}^2+\Vert f \Vert _{L^2(\sigma _c)}^2+o(1))\leqslant \Vert T_{s_n} \Vert ^2\Vert f_n-f \Vert _{L^2(\mu _{s_n})}^2+\Vert T_cf \Vert _{L^4}^2+o(1),\end{aligned}$$

which is equivalent to

$$\begin{aligned}\Vert T_{s_n} \Vert ^2\Vert f \Vert _{L^2(\sigma _c)}^2\leqslant \Vert T_cf \Vert _{L^4}^2+\Vert T_{s_n} \Vert ^2(\Vert f_n-f \Vert _{L^2(\mu _{s_n})}^2-\Vert f_n-f \Vert _{L^2(\sigma _c)}^2)+o(1). \end{aligned}$$

Arguing as in (12.2) we obtain $\Vert f_n-f \Vert _{L^2(\mu _{s_n})}^2-\Vert f_n-f \Vert _{L^2(\sigma _c)}^2\rightarrow 0$, and therefore,

$$\begin{aligned}\Vert T \Vert =\Vert T_{s_n} \Vert \leqslant \frac{\Vert T_c f \Vert _{L^4}}{\Vert f \Vert _{L^2(\sigma _c)}}\leqslant \Vert T_c \Vert .\end{aligned}$$

Finally, if we relax the support condition (iv) to the $\varepsilon$-small norm condition (v), it will be enough if in (12.4) we use

$$\begin{aligned} \frac{\Vert T_{s_n}f_n \Vert _{L^4(\mathbb {R}^4)}^2}{\Vert f_n \Vert _{L^2(\mu _{s_n})}^2}\leqslant \frac{\Vert T_{s_n}(f_n\mathbb {1}_{\{\vert y\vert \geqslant a\}}) \Vert _{L^4(\mathbb {R}^4)}^2}{\Vert f_n\mathbb {1}_{\{\vert y\vert \geqslant a\}} \Vert _{L^2(\mu _{s_n})}^2}+C\varepsilon , \end{aligned}$$

where $C<\infty$ is independent of n and a, together with $f_n\mathbb {1}_{\{\vert y\vert \geqslant a\}}\rightharpoonup f\mathbb {1}_{\{\vert y\vert \geqslant a\}}$ in $L^2(\Gamma ^3)$ and $T_{s_n}(f_n\mathbb {1}_{\{\vert y\vert \geqslant a\}})\rightarrow T_c(f\mathbb {1}_{\{\vert y\vert \geqslant a\}})$ a.e. in $\mathbb {R}^4$, as $n\rightarrow \infty$, the latter property being a consequence of the former and Lemma 12.4. $\square$

13 Proof of Theorem 1.3

In previous Sects. 10 and 11, we proved that if $\{f_n\}_n$ is an extremizing sequence for ${\overline{T}}$, then subsequences of $\{\vert f_n\vert ^2\}_n$ can not satisfy vanishing nor dichotomy of Lemma 9.1, which as we saw, were a consequence of bilinear estimates for ${\overline{T}}$. In this section we prove that, as a consequence of the compactness alternative and Lemma 12.5 of the previous section, extremizing sequences posses convergent subsequences and, as a result, extremizers exist.

Proof of Theorem 1.3

Let $\{f_n\}_{n}\subset L^2(\overline{{\mathcal {H}}}^3)$ be an $L^2$ normalized complex valued extremizing sequence for ${\overline{T}}$. After passing to a subsequence if necessary we can assume that alternative (i) in Lemma 9.1 holds for $\{\vert f_n\vert ^2\}_{n}$, that is, there exists $\{\ell _n\}_n\subset \mathbb {N}$ with the property that for all $\varepsilon >0$ there exists $R_\varepsilon <\infty$ such that for all $R\geqslant R_\varepsilon$ and $n\in \mathbb {N}$

$$\begin{aligned} \int \limits _{\{2^{\ell _n-R}\leqslant \vert y\vert \leqslant 2^{\ell _n+R}\}}\vert f_n(y)\vert ^2\,\textrm{d}{{\bar{\mu }}}(y)\geqslant 1-\varepsilon . \end{aligned}$$

(13.1)

If there exists a subsequence $\{n_k\}_{k}\subset \mathbb {N}$ such that $\{\ell _{n_k}\}_{k}$ is bounded above, then we can apply the same method provided in the proof of Proposition 5.2 for the upper half of the hyperboloid, ${\mathcal {H}}^3$, to conclude that there exists a subsequence $\{f_{n_k}\}_k$ that satisfies the conclusion of the theorem with all $L_{n_k}$’s equal to the identity matrix. Therefore, in what follows we will assume that $\ell _{n}\rightarrow \infty$ as $n\rightarrow \infty$.

Passing to a subsequence if necessary we can assume then that $\{f_n\}_{n}$ satisfies the following: $\Vert f_n \Vert _{L^2}=1$, $\Vert {\overline{T}}f_n \Vert _{L^4}\rightarrow \overline{{{\textbf {H}} }}_4$ and there exists a sequence $\{\ell _n\}_{n\in \mathbb {N}}\subset \mathbb {N}$ such that $\ell _n\rightarrow \infty$ as $n\rightarrow \infty$ and for any $\varepsilon >0$ there exists $R_\varepsilon <\infty$ such that for all $R\geqslant R_\varepsilon$ and all $n\in \mathbb {N}$ equation (13.1) holds. Therefore, with $R_\varepsilon$ as before, we have that for all $R\geqslant R_\varepsilon$ there exists $N_n\in [\ell _n-R,\ell _n+R]\cap \mathbb {N}$ such that for all $n\in \mathbb {N}$

$$\begin{aligned} \int \limits _{\{2^{N_n}\leqslant \vert y\vert \leqslant 2^{N_n+1}\}}\vert f_n(y)\vert ^2\,\textrm{d}{{\bar{\mu }}}(y)\geqslant \frac{1-\varepsilon }{2R}. \end{aligned}$$

Denote $P_N$ the dyadic cut off at scale $2^N$, that is, $P_Nf(y):=f(y)\mathbb {1}_{\{2^N\leqslant \vert y\vert < 2^{N+1}\}}$. Using the continuity of ${\overline{T}}$ and the triangle inequality we obtain

$$\begin{aligned} \Vert {\overline{T}}(P_{N_n}f_n) \Vert _{L^4}&\geqslant \Vert {\overline{T}}f_n \Vert _{L^4}-\overline{{{\textbf {H}} }}_4\Vert f_n-P_{N_n}f_n \Vert _{L^2({\bar{\mu }})}\geqslant \Vert {\overline{T}}f_n \Vert _{L^4}-\overline{{{\textbf {H}} }}_4\Bigl (1-\frac{1-\varepsilon }{2R}\Bigr )^{1/2}\\&=\overline{{{\textbf {H}} }}_4-\overline{{{\textbf {H}} }}_4\Bigl (1-\frac{1-\varepsilon }{2R}\Bigr )^{1/2}+o_n(1). \end{aligned}$$

Choosing $\varepsilon =\varepsilon _0$ close to 0 and $R=R_{\varepsilon _0}+1$, we obtain a sequence $\{N_n\}_{n}\subset \mathbb {N}$, with $\vert N_n-\ell _n\vert \leqslant R_{\varepsilon _0}+1$, so that $N_n\rightarrow \infty$ as $n\rightarrow \infty$, and a constant $c>0$ such that for all n large enough⁹

$$\begin{aligned}\Vert P_{N_n}f_n \Vert _{L^2({\bar{\mu }})}> c,\quad \Vert {\overline{T}}(P_{N_n}f_n) \Vert _{L^4}> c.\end{aligned}$$

We rescale $f_n$ defining $g_n$ by $g_n(y)=2^{N_n}f(2^{N_n}y)$. Letting $s_n=2^{-N_n}$ we have $s_n\rightarrow 0$ as $n\rightarrow \infty$, $g_n\in L^2(\overline{{\mathcal {H}}}^3_{s_n})$,

$$\begin{aligned} \Vert g_n \Vert _{L^2({{\bar{\mu }}}_{s_n})}&=\Vert f_n \Vert _{L^2({{\bar{\mu }}})}=1,\nonumber \\ \Vert {\overline{T}}_{s_n}g_n \Vert _{L^4}&=\Vert {\overline{T}}f_n \Vert _{L^4}\rightarrow \overline{{{\textbf {H}} }}_4 \text { as }n\rightarrow \infty ,\nonumber \\ \Vert P_1g_n \Vert _{L^2({{\bar{\mu }}}_{s_n})}&=\Vert P_{N_n}f_n \Vert _{L^2({{\bar{\mu }}})}>c\text { and} \end{aligned}$$

(13.2)

$$\begin{aligned} \Vert {\overline{T}}_{s_n}(P_1g_n) \Vert _{L^4}&=\Vert {\overline{T}}(P_{N_n}f_n) \Vert _{L^4}>c, \end{aligned}$$

(13.3)

as obtained by simple scaling (see Sect. 14). Moreover, from (13.1) for any small $\varepsilon >0$, $R>2R_\varepsilon$ and $n\in \mathbb {N}$

$$\begin{aligned} \int \limits _{\{2^{-R}\leqslant \vert y\vert \leqslant 2^{R}\}}\vert g_n(y)\vert ^2 \,\textrm{d}{{\bar{\mu }}}_{s_n}(y)\geqslant 1-\varepsilon , \end{aligned}$$

(13.4)

so that the $g_n$’s are “localized at scale 1”. Using Lemma 8.1 applied to ${\overline{T}}_{s_n}$ and $P_1g_n$, which is possible given (13.2) and (13.3), we obtain that for all $n\in \mathbb {N}$ there exist caps ${\mathcal {C}}_n\subset {\overline{{\mathcal {H}}}}^3_{s_n}$, which we may consider all to be contained in the upper half, ${\mathcal {H}}^3_{s_n}$, possibly after passing to a subsequence,¹⁰, for some spherical caps , such that

$$\begin{aligned} \int \limits _{{\mathcal {C}}_n}\vert g_n(y)\vert \,\textrm{d}{{\bar{\mu }}}_{s_n}(y)=\int \limits _{{\mathcal {C}}_n}\vert P_1g_n(y)\vert \,\textrm{d}{{\bar{\mu }}}_{s_n}(y)\geqslant c{{\bar{\mu }}}_{s_n}({\mathcal {C}}_n)^{1/2}\Vert P_1g_n \Vert _{L^2({{\bar{\mu }}}_{s_n})}\gtrsim {{\bar{\mu }}}_{s_n}({\mathcal {C}}_n)^{1/2}, \end{aligned}$$

as a consequence of (8.6). Equivalently

$$\begin{aligned} \int \limits _{2^{N_n}{\mathcal {C}}_n}\vert f_n(y)\vert \,\textrm{d}{{\bar{\mu }}}(y)\gtrsim {{\bar{\mu }}}(2^{N_n}{\mathcal {C}}_n)^{1/2}. \end{aligned}$$

(13.5)

Let $\alpha =\limsup _{n\rightarrow \infty } {{\bar{\mu }}}_{s_n}({\mathcal {C}}_n)$. Two cases arise.

Case 1: $\alpha >0$. Passing to a subsequence if necessary, we can assume that there exists a constant $c>0$ such that for all n

$$\begin{aligned} \int \limits _{{\mathcal {C}}_n}\vert g_n(y)\vert \,\textrm{d}{{\bar{\mu }}}_{s_n}(y)\geqslant c>0. \end{aligned}$$

We can view $g_n$ as a function on the double cone via the usual identification using the orthogonal projection onto $\mathbb {R}^3$, where we extend it to be zero in the region where $\vert y\vert \leqslant s_n$. Since $\Vert g_n \Vert _{L^2({{\bar{\sigma }}}_c)}\leqslant \Vert g_n \Vert _{L^2({{\bar{\mu }}}_{s_n})}=1$ and

$$\begin{aligned} 0<c<\int \limits _{{\mathcal {C}}_n}\vert g_n(y)\vert \,\textrm{d}{{\bar{\mu }}}_{s_n}(y)\lesssim \int \limits _{{\mathcal {C}}_n}\vert g_n(y)\vert \,\textrm{d}{{\bar{\sigma }}}_c(y), \end{aligned}$$

(13.6)

for all n large enough (as a consequence of (13.4)), there is weak convergence of $\{\vert g_n\vert \}_n$ in $L^2({{\bar{\sigma }}}_c)$ after the possible extraction of a subsequence, $\vert g_n\vert \rightharpoonup g$, for some $g\in L^2({{\bar{\sigma }}}_c)$ which satisfies $g\ne 0$ by (13.6). Inequality (13.4) implies that

$$\begin{aligned}\lim _{a\rightarrow 0^+}\sup _{n\in \mathbb {N}}\Vert g_n\mathbb {1}_{\{\vert y\vert \leqslant a\}} \Vert _{L^2({\bar{\mu }}_{s_n})}=0.\end{aligned}$$

Because $\Vert {\overline{T}}_{s_n}(g_n) \Vert _{L^4}\leqslant \Vert {\overline{T}}_{s_n}(\vert g_n\vert ) \Vert _{L^4}$, it is then also the case that $\Vert {\overline{T}}_{s_n}(\vert g_n\vert ) \Vert _{L^4}\rightarrow \overline{{{\textbf {H}} }}_4$, so that we can use part (v) of Lemma 12.5 applied to $\{\vert g_n\vert \}_n$ to conclude

$$\begin{aligned}\Vert {\overline{T}} \Vert \leqslant \Vert {\overline{T}}_c \Vert ,\end{aligned}$$

which is in contradiction with Proposition 4.3. Therefore, this case does not arise.

Case 2: $\alpha =0$. Let $\{\gamma _n\}_n\subset [0,\pi ]$ and $\{R_n\}_n\subset SO(3)$ be such that

$$\begin{aligned} R_n^{-1}({\mathcal {C}}_n)= & {} \{(r\omega ,\sqrt{r^2-s_n^2}):1\leqslant r\leqslant 2,\\{} & {} \omega =(\cos \varphi ,\cos \theta \sin \varphi ,\sin \theta \sin \varphi ),\theta \in [0,2\pi ],\,\varphi \in [0,\gamma _n] \}. \end{aligned}$$

The condition $\alpha =0$ implies $\gamma _n\rightarrow 0$ as $n\rightarrow \infty$. Let $\beta =\limsup _{n\rightarrow \infty }{\bar{\mu }}(2^{N_n}{\mathcal {C}}_n)=\limsup _{n\rightarrow \infty }2^{2N_n}{\bar{\mu }}_{s_n}({\mathcal {C}}_n)$. Two subcases arise.

Subcase 2a: $\beta <\infty$. This implies that the sequence $\{{\bar{\mu }}(2^{N_n}{\mathcal {C}}_n)\}_n$ is bounded. We may assume that the angles $\gamma _n$ are less that $\pi /2$ as $\{\gamma _n\}_n$ tends to zero. From Lemma 2.3 with $s=1$, there exists $\{t_n\}_n\subset [0,1)$ such that the caps $\{L^{-t_n}R_n^{-1}(2^{N_n}{\mathcal {C}}_n):n\in \mathbb {N}\}$ are contained in a fixed bounded ball of $\mathbb {R}^4$. It therefore follows from (13.5) and the Cauchy–Schwarz inequality that $\{(R_nL^{t_n})^*f_n\}_n\subset L^2({\overline{{\mathcal {H}}}}^3)$ is an extremizing sequence with $L^2$ norm uniformly bounded below by a constant $c>0$ in a fixed ball. We can then conclude the precompactness modulo characters of the sequence $\{(R_nL^{t_n})^*f_n\}_n$ using the argument in the proof of Proposition 5.2.

Subcase 2b: $\beta =\infty$. From (2.9) in Lemma 2.3 with $s=1$, after passing to a subsequence if necessary, $\lim _{n\rightarrow \infty }2^{2N_n}\sin ^2(\gamma _n)=\infty$. Set $t_n=\cos \gamma _n$, so that $t_n\rightarrow 1$ as $n\rightarrow \infty$. From Lemma 2.4 with $s=s_n$, the set ${{\widetilde{{\mathcal {C}}}}}_n:=L_{t_n}^{-1}R_n^{-1}({\mathcal {C}}_n)\subset {\overline{{\mathcal {H}}}}^3_{s_n(1-t_n^2)^{-1/2}}$ satisfies, for all n large enough for which $2^{2N_n}\sin ^2(\gamma _n)\geqslant 8$ and $\gamma _n\leqslant \pi /3$,

$$\begin{aligned} {{\bar{\mu }}}_{\frac{s_n}{\sqrt{1-t_n^2}}}({{\widetilde{{\mathcal {C}}}}}_n)\geqslant \frac{\pi }{2}\quad \text { and }\quad {{\widetilde{{\mathcal {C}}}}}_n\subseteq [\tfrac{7}{16},\tfrac{33}{16}]\times {\mathbb {S}}^2.\end{aligned}$$

Set $a_n=s_n(1-t_n^2)^{-1/2}=(2^{N_n}\sin \gamma _n)^{-1}\rightarrow 0$, as $n\rightarrow \infty$. Let $\tilde{f}_n=(R_nL^{t_n})^*f_n$ so that $\{{\tilde{f}}_n\}_n\subset L^2({\overline{{\mathcal {H}}}}^3)$ is also an $L^2$-normalized extremizing sequence which satisfies, for some constant $c>0$,

$$\begin{aligned} \int \limits _{a_n^{-1}{{\widetilde{{\mathcal {C}}}}}_n}\vert {\tilde{f}}_n(y)\vert \,\textrm{d}{{\bar{\mu }}}(y)\geqslant c{{\bar{\mu }}}(a_n^{-1}\widetilde{\mathcal {C}}_n)^{1/2},\quad \int \limits _{a_n^{-1}{{\widetilde{{\mathcal {C}}}}}_n}\vert {\tilde{f}}_n(y)\vert ^2 \,\textrm{d}{{\bar{\mu }}}(y)\geqslant c^2, \end{aligned}$$

and $a_n^{-1}{{\widetilde{{\mathcal {C}}}}}_n\subseteq [\frac{7}{16a_n},\frac{33}{16a_n}]\times {\mathbb {S}}^2$.

Define the rescale ${\tilde{g}}_n(\cdot ):=a_n^{-1}{\tilde{f}}_n(a_n^{-1}\,\cdot )$, so that for each n we have ${\tilde{g}}_n\in L^2({\overline{{\mathcal {H}}}}^3_{a_n})$, $\Vert {\tilde{g}}_n \Vert _{L^2({\overline{{\mathcal {H}}}}^3_{a_n})}=1$ and there is a constant $c'>0$ such that

$$\begin{aligned} \int \limits _{{{\widetilde{{\mathcal {C}}}}}_n}\vert {\tilde{g}}_n(y)\vert \,\textrm{d}{{\bar{\mu }}}_{a_n}(y)\geqslant c{{\bar{\mu }}}_{a_n}(\widetilde{\mathcal {C}}_n)^{1/2}>c'>0. \end{aligned}$$

We are almost in the same situation as in Case 1, but we need the analog of (13.4) for the sequence $\{{\tilde{g}}_n\}_n$. After passing to a subsequence if necessary, $\{{\tilde{f}}_n \}_n$ satisfies the compactness alternative in Lemma 9.1. Denoting $\{{\tilde{\ell }}_n\}_n$ the corresponding sequence associated to $\{{\tilde{f}}_n\}_n$ as in (13.1) we then necessarily have that $\{{{\tilde{\ell }}_n}-\log _2(a_n^{-1})\}_n$ is bounded. This implies the desired analog of (13.4) for $\{{\tilde{g}}_n\}_n$. Therefore the argument in Case 1 applies showing that this subcase does not arise.

As a result, only Subcase 2a of Case 2 is possible, proving the theorem. $\square$

14 Scaling

Here we record scaling properties of the family of operators $\{T_s\}_{s>0}$. Recall from Sect. 3 that for $s>0$, $\mathcal H^3_s=\{(y,\sqrt{\vert y\vert ^2-s^2}):y\in \mathbb {R}^3\}$, equipped with the measure $\mu _s$ with density $\,\textrm{d}\mu _s(y,t)=\mathbb {1}_{\{\vert y\vert >s\}}\delta (t-\sqrt{\vert y\vert ^2-s^2})\frac{\,\textrm{d}y\,\textrm{d}t}{\sqrt{\vert y\vert ^2-s^2}}$.

The operator $T_s$, defined on ${\mathcal {S}}(\mathbb {R}^3)$, is given by

$$\begin{aligned}T_sf(x,t)=\widehat{f\mu _s}(-x,-t)=\int \limits _{\{y\in \mathbb {R}^3,\vert y\vert \geqslant s\}} e^{ix\cdot y}e^{it\sqrt{\vert y\vert ^2-s^2}}f(y)\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-s^2}}.\end{aligned}$$

We want to study the scaling of the quantity ${{\textbf {H}} }_{p,s}$ defined by

$$\begin{aligned} {{\textbf {H}} }_{p,s}:=\sup _{0\ne f\in L^2({\mathcal {H}}^3_s)}\frac{\Vert T_sf \Vert _{L^p(\mathbb {R}^4)}}{\Vert f \Vert _{L^2({\mathcal {H}}^3_s)}}. \end{aligned}$$

Changing variables $y\rightsquigarrow sy$ in the expression defining $Tf(x,t)=T_1f(x,t)$ we obtain

$$\begin{aligned} Tf(x,t)&=\int \limits _{\{y\in \mathbb {R}^3,\vert y\vert \geqslant 1\}} e^{ix\cdot y}e^{it\sqrt{\vert y\vert ^2-1}}f(y)\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}}\\&=s^{-1}\int \limits _{\{y\in \mathbb {R}^3,\vert y\vert \geqslant s\}} e^{is^{-1}x\cdot y}e^{is^{-1}t\sqrt{\vert y\vert ^2-s^2}}s^{-1}f(s^{-1}y)\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-s^2}}, \end{aligned}$$

from where $sTf(sx,st)=T_s(s^{-1}f(s^{-1}\cdot ))(x,t)$ and it follows that

$$\begin{aligned}s^{1-4/p}\Vert Tf \Vert _{L^p(\mathbb {R}^{4})}=\Vert T_ss^{-1}f(s^{-1}\cdot ) \Vert _{L^p(\mathbb {R}^{4})}.\end{aligned}$$

On the other hand

$$\begin{aligned} \int \limits _{\{y\in \mathbb {R}^3,\,\vert y\vert \geqslant 1\}} \vert f(y)\vert ^q\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-1}} =\int \limits _{\{y\in \mathbb {R}^3,\,\vert y\vert \geqslant s\}} \vert s^{-2/q}f(s^{-1}y)\vert ^q\frac{\,\textrm{d}y}{\sqrt{\vert y\vert ^2-s^2}}, \end{aligned}$$

that is $\Vert f \Vert _{L^q(\mu )}=\Vert s^{-2/q}f(s^{-1}\cdot ) \Vert _{L^q(\mu _s)}$. Thus

$$\begin{aligned} s^{1-4/p}\Vert Tf \Vert _{L^p(\mathbb {R}^{4})}\Vert f \Vert _{L^2(\mu )}^{-1}&=\Vert T_ss^{-1}f(s^{-1}\cdot ) \Vert _{L^p(\mathbb {R}^{4})} \Vert s^{-1}f(s^{-1}\cdot ) \Vert _{L^2(\mu _s)}^{-1}, \end{aligned}$$

and it follows that for all $s>0$

$$\begin{aligned} {{\textbf {H}} }_{p,s}=s^{1-4/p}{{\textbf {H}} }_{p}. \end{aligned}$$

In particular, if $p=4$,

$$\begin{aligned} {{\textbf {H}} }_{4,s}={{\textbf {H}} }_{4}, \end{aligned}$$

for all $s>0$.

Appendix A: Computation of a Limit

Let

$$\begin{aligned} I(a)&=16\pi ^3 \int \limits _0^{\infty }e^{-a\tau }\Bigl (\tau ^2\sqrt{\tau ^2+4}-\frac{2}{3}(\tau ^2+4)\sqrt{\tau ^2+1} +\frac{8}{3}\\ {}&\quad +2\tau \log (\tau +\sqrt{\tau ^2+1})\Bigr )\,\text {d}\tau , \end{aligned}$$

and

$$\begin{aligned} II(a)=16\pi ^2\biggl (\int \limits _0^\infty e^{-a\tau }\sqrt{\tau ^2+1}\,\textrm{d}\tau \biggr )^2. \end{aligned}$$

Fig. 1

Graph of the ratio I(a)/II(a) and the constant $2\pi$ for $0<a<0.25$, illustrating the content of Lemma A.1

The ratio I(a)/II(a) appeared in the proof of Proposition 4.1 while establishing that the best constant for the hyperboloid ${\mathcal {H}}^3$ is strictly greater than the best constant for the cone $\Gamma ^3$ in their respective $L^2\rightarrow L^4(\mathbb {R}^4)$ adjoint Fourier restriction inequalities. The purpose of this appendix is to prove the following lemma (Fig. 1).

Lemma A.1

$$\begin{aligned}&\lim _{a\rightarrow 0^+}\frac{I(a)}{II(a)}=2\pi ,\quad \lim _{a\rightarrow 0^+}\frac{\,\textrm{d}}{\,\textrm{d}a}\frac{I(a)}{II(a)}=0,\quad \lim _{a\rightarrow 0^+}\frac{\,\textrm{d}^2}{\,\textrm{d}a^2}\frac{I(a)}{II(a)}=0, \end{aligned}$$

and

$$\begin{aligned}&\lim _{a\rightarrow 0^+}\frac{\,\textrm{d}^3}{\,\textrm{d}a^3}\frac{I(a)}{II(a)}=8\pi . \end{aligned}$$

Therefore there exists $a_0>0$ such that

$$\begin{aligned} \frac{I(a)}{II(a)}>2\pi , \end{aligned}$$

for all $0<a<a_0$.

Throughout this section we use the asymptotic notation $o_a(f(a))$ and $O_a(f(a))$ as $a\rightarrow 0^+$ in the usual way, namely $g(a)=o_a(f(a))$ if $g(a)/f(a)\rightarrow 0$ as $a\rightarrow 0^+$, and $g(a)=O_a(f(a))$ if there exists a constant C, independent of a, such that $\vert g(a)\vert \leqslant C\vert f(a)\vert$ for all $a>0$ small enough.

Changing variable $u=a\tau$ we obtain

$$\begin{aligned} I(a)&=\frac{16\pi ^3}{a^4}\biggl ( \int \limits _0^{\infty }e^{-u}\Bigl (u^2\sqrt{u^2+4a^2}-\frac{2}{3}(u^2+4a^2)\sqrt{u^2+a^2}\\&\quad + 2a^2u\, \log (u+\sqrt{u^2+a^2})\Bigr )\,\textrm{d}\tau +\frac{8a^3}{3}-2a^2\log (a)\biggr ), \end{aligned}$$

and

$$\begin{aligned} II(a)&=\frac{16\pi ^2}{a^4}\biggl (\int \limits _0^\infty e^{-u}\sqrt{u^2+a^2}\,\textrm{d}u\biggr )^2. \end{aligned}$$

Using the Dominated Convergence Theorem it is direct to check that

$$\begin{aligned} \lim _{a\rightarrow 0^+}a^4I(a)=32\pi ^3\,\text { and }\,\lim _{a\rightarrow 0^+}a^4II(a)=16\pi ^2, \end{aligned}$$

so that

$$\begin{aligned} \lim _{a\rightarrow 0^+}\frac{I(a)}{II(a)}=2\pi . \end{aligned}$$

To address the limit of the derivatives of the ratio I(a)/II(a) it will be convenient to introduce a rescaling. Let

$$\begin{aligned} N(a)&:=a^{4/3}I(a^{1/3})=16\pi ^3\biggl ( \int \limits _0^{\infty }e^{-u}\Bigl (u^2\sqrt{u^2+4a^{2/3}}-\frac{2}{3}(u^2+4a^{2/3})\sqrt{u^2+a^{2/3}}\\&\quad + 2a^{2/3}u\, \log (u+\sqrt{u^2+a^{2/3}})\Bigr )\,\textrm{d}\tau +\frac{8a}{3}-\frac{2}{3}a^{2/3}\log (a)\biggr ), \end{aligned}$$

and

$$\begin{aligned} D(a)&:=a^{4/3}II(a^{1/3})=16\pi ^2\biggl (\int \limits _0^\infty e^{-u}\sqrt{u^2+a^{2/3}}\,\textrm{d}u\biggr )^2. \end{aligned}$$

As we already know, and can readily check, $N(a)\rightarrow 32\pi ^3,\, D(a)\rightarrow 16\pi ^2$ and $N(a)/D(a)\rightarrow 2\pi$ as $a\rightarrow 0^+$. The remaining properties of the derivatives of I(a)/II(a) in Lemma A.1 will follow if we show that $\frac{\,\textrm{d}}{\,\textrm{d}a}(N(a)/D(a))\rightarrow \frac{4\pi }{3}$ as $a\rightarrow 0^+$.

In what follows we write $(\cdot )'$ as a short for the derivative with respect to a. Given that both $N'(a)$ and $D'(a)$ diverge to $+\infty$ as $a\rightarrow 0^+$ it will be convenient to write the derivative of N(a)/D(a) in the following way

$$\begin{aligned} \begin{aligned} \frac{\,\textrm{d}}{\,\textrm{d}a}\frac{N(a)}{D(a)}&=\frac{16\pi ^2N'(a)-32\pi ^3 D'(a)}{D(a)^2}\\&\quad +\frac{(D(a)-16\pi ^2)\,N'(a)-(N(a)-32\pi ^3)\,D'(a)}{D(a)^2}. \end{aligned} \end{aligned}$$

(A.1)

We have the following lemma.

Lemma A.2

• (i) $\displaystyle \lim _{a\rightarrow 0^+}\frac{\,\textrm{d}}{\,\textrm{d}a}\frac{N(a)}{D(a)}=\frac{4\pi }{3}$.

• (ii) As $a\rightarrow 0^+$,

  $$\begin{aligned} N'(a)=O_a\Bigl (\frac{\log a}{a^{1/3}}\Bigr )\,\text { and }\, D'(a)=O_a\Bigl (\frac{\log a}{a^{1/3}}\Bigr ). \end{aligned}$$

• (iii) $\displaystyle \lim _{a\rightarrow 0^+}(N(a)-32\pi ^3)\, D'(a)=0\,$ and $\,\displaystyle \lim _{a\rightarrow 0^+}(D(a)-16\pi ^2)\,N'(a)=0.$

Proof

In the course of the proof of this lemma we will make repeated use of the asymptotic behavior of some integrals as contained in Lemma A.3 below. We start with property (ii). For $a>0$ the derivative of N is as follows,

$$\begin{aligned} N'(a)&=16\pi ^3\biggl ( \int \limits _0^{\infty }e^{-u}\Bigl (u^2\frac{4}{3a^{1/3}\sqrt{u^2+4a^{2/3}}}-\frac{16}{9a^{1/3}}\sqrt{u^2+a^{2/3}}\nonumber \\&\quad -\frac{2}{9}(u^2+4a^{2/3})\frac{1}{a^{1/3}\sqrt{u^2+a^{2/3}}}+ \frac{4}{3a^{1/3}}u\, \log (u+\sqrt{u^2+a^{2/3}})\nonumber \\&\quad +\frac{2}{3}a^{1/3}u\,\frac{1}{(u+\sqrt{u^2+a^{2/3}})\sqrt{u^2+a^{2/3}}}\Bigr ) \,\textrm{d}u+\frac{8}{3}-\frac{4}{9a^{1/3}}\log (a)-\frac{2}{3a^{1/3}}\biggr )\nonumber \\&=16\pi ^3\biggl ( \frac{8}{3}-\frac{4}{3a^{1/3}}-\frac{4}{9a^{1/3}}\log (a)+ \frac{4}{3a^{1/3}}\int \limits _0^\infty e^{-u}u\log (u+\sqrt{u^2+a^{2/3}})\,\textrm{d}u\biggr )\nonumber \\&\quad +o_a(1)\nonumber \\&=O_a\Bigl (\frac{\log a}{a^{1/3}}\Bigr ), \end{aligned}$$

(A.2)

where we used (A.5), (A.8), (A.7), (A.10) and (A.11). The derivative of the function D is as follows

$$\begin{aligned} D'(a)=\frac{32\pi ^2}{3}\int \limits _{0}^{\infty } e^{-u}\sqrt{u^2+a^{2/3}}\,\textrm{d}u\cdot \int \limits _{0}^{\infty }e^{-u}\frac{1}{a^{1/3}\sqrt{u^2+a^{2/3}}}\,\textrm{d}u, \end{aligned}$$

so that (A.4) and (A.5) imply

$$\begin{aligned} D'(a)=O_a\Bigl (\frac{1}{a^{1/3}}\Bigr )\,O_a(\log a)=O_a\Bigl (\frac{\log a}{a^{1/3}}\Bigr ), \end{aligned}$$

and more explicitly using (A.13), as we will need later,

$$\begin{aligned} D'(a)=\frac{32\pi ^2}{3a^{1/3}}\biggl (\int \limits _{0}^{\infty }e^{-u}\log (u+\sqrt{u^2+a^{2/3}})\,\textrm{d}u-\frac{1}{3}\log a\biggr ) +o_a(1). \end{aligned}$$

(A.3)

We now turn to the proof of part (iii). Using that $\int \limits _0^\infty e^{-u}u^3\,\textrm{d}u=6$ we can write

$$\begin{aligned} N(a)-32\pi ^3&=16\pi ^3\biggl ( \int \limits _0^{\infty }e^{-u}\Bigl (u^2(\sqrt{u^2+4a^{2/3}}-u)-\frac{2}{3}u^2(\sqrt{u^2+a^{2/3}}-u)\\&\quad -\frac{8}{3}a^{2/3}\sqrt{u^2+a^{2/3}}+ 2a^{2/3}u\, \log (u+\sqrt{u^2+a^{2/3}})\Bigr )\,\textrm{d}u\\&\quad +\frac{8a}{3}-\frac{2}{3}a^{2/3}\log (a)\biggr )\\&=16\pi ^3a^{1/3}\biggl ( \int \limits _0^{\infty }e^{-u}\Bigl (u^2\frac{4a^{1/3}}{\sqrt{u^2+4a^{2/3}}+u} -\frac{2}{3}u^2\frac{a^{1/3}}{\sqrt{u^2+a^{2/3}}+u}\\&\quad -\frac{8}{3}a^{1/3}\sqrt{u^2+a^{2/3}}+ 2a^{2/3}u\, \log (u+\sqrt{u^2+a^{2/3}})\Bigr )\,\textrm{d}u\\&\quad +\frac{8a^{2/3}}{3}-\frac{2}{3}a^{1/3}\log (a)\biggr )\\&=O_a(a^{2/3}\log a). \end{aligned}$$

Then

$$\begin{aligned} (N(a)-32\pi ^3)\cdot D'(a)=O_a(a^{2/3}\log a)\,O_a\Bigl (\frac{\log a}{a^{1/3}}\Bigr ) =O_a(a^{1/3}\log ^2 a)=o_a(1). \end{aligned}$$

On the other hand

$$\begin{aligned} D(a)-16\pi ^2&=16\pi ^2\biggl (\int \limits _0^\infty e^{-u}\sqrt{u^2+a^{2/3}}\,\textrm{d}u+1\biggr )\biggl (\int \limits _0^\infty e^{-u}\sqrt{u^2+a^{2/3}}\,\textrm{d}u-1\biggr )\\&=O_a(1)\biggl (\int \limits _0^\infty e^{-u}(\sqrt{u^2+a^{2/3}}-u)\,\textrm{d}u\biggr )\\&=O_a(1)\biggl (\int \limits _0^\infty e^{-u}\frac{a^{2/3}}{\sqrt{u^2+a^{2/3}}+u}\,\textrm{d}u\biggr )\\&=O_a(a^{2/3}\log a), \end{aligned}$$

where in the last line we used (A.9). Then

$$\begin{aligned} (D(a)-16\pi ^2)\cdot N'(a)=O_a(a^{2/3}\log a)\, O_a\Bigl (\frac{\log a}{a^{1/3}}\Bigr )=O_a(a^{1/3}\log ^2 a)=o_a(1). \end{aligned}$$

We now turn to the proof of (i). By (iii), the limit as $a\rightarrow 0^+$ of the second summand on the right hand side of (A.1) equals zero. We proceed to calculate the limit of the first summand. Combining (A.2) and (A.3) we obtain

$$\begin{aligned} 16\pi ^2 N'(a)&-32\pi ^3D'(a)\\&=\frac{8(16)^2\pi ^5}{3}-\frac{4(16)^2\pi ^5}{3a^{1/3}}\\&\quad +\frac{(32)^2\pi ^5}{3a^{1/3}} \int \limits _{0}^{\infty }e^{-u}(u-1)\log (u+\sqrt{u^2+a^{2/3}})\,\textrm{d}u+o_a(1)\\&=\frac{2(32)^2\pi ^5}{3}+\frac{(32)^2\pi ^5}{3a^{1/3}} \int \limits _{0}^{\infty }e^{-u}\Bigl ((u-1)\log (u+\sqrt{u^2+a^{2/3}})-1\Bigr )\,\textrm{d}u\\&\quad +o_a(1). \end{aligned}$$

Using (A.12) to treat the integral in the previous expression we obtain

$$\begin{aligned} \lim _{a\rightarrow 0^+}(16\pi ^2N'(a)-32\pi ^3D'(a))=\frac{(32)^2\pi ^5}{3}, \end{aligned}$$

therefore

$$\begin{aligned} \lim _{a\rightarrow 0^+}\frac{\,\textrm{d}}{\,\textrm{d}a}\frac{N(a)}{D(a)}=\frac{(32)^2\pi ^5}{3(16\pi ^2)^2}=\frac{4\pi }{3}. \end{aligned}$$

$\square$

Finally, we state the asymptotic behavior of the many integrals used during the proof of the previous lemma.

Lemma A.3

We have the following identities as $a\rightarrow 0^+$

$$\begin{aligned}&\int \limits _{0}^{\infty }e^{-u}\frac{1}{\sqrt{u^2+a^{2/3}}}\,\textrm{d}u=O_a(\log a), \end{aligned}$$

(A.4)

$$\begin{aligned}&\int \limits _0^{\infty }e^{-u}\frac{\sqrt{u^2+a^{2/3}}}{a^{1/3}}\,\textrm{d}u =\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a), \end{aligned}$$

(A.5)

$$\begin{aligned}&\int \limits _0^{\infty }e^{-u}\frac{u\sqrt{u^2+a^{2/3}}}{a^{1/3}}\,\textrm{d}u =\frac{2}{a^{1/3}}+O_a(a^{1/3}), \end{aligned}$$

(A.6)

$$\begin{aligned}&\int \limits _0^{\infty }e^{-u}\frac{u^2}{a^{1/3}\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u =\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a), \end{aligned}$$

(A.7)

$$\begin{aligned}&\int \limits _0^{\infty }e^{-u}\frac{u^2+4a^{2/3}}{a^{1/3}\sqrt{u^2+a^{2/3}}}\,\textrm{d}u =\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a), \end{aligned}$$

(A.8)

$$\begin{aligned}&\int \limits _0^\infty e^{-u}\frac{a^{2/3}}{u+\sqrt{u^2+a^{2/3}}}\,\textrm{d}u =O_a(a^{2/3}\log a), \end{aligned}$$

(A.9)

$$\begin{aligned}&\int \limits _0^{\infty }e^{-u}\,\frac{a^{1/3}u}{(u+\sqrt{u^2+4a^{2/3}})\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u =O_a(a^{1/3}\log a), \end{aligned}$$

(A.10)

$$\begin{aligned}&\int \limits _0^{\infty }e^{-u}\frac{u}{a^{1/3}}\, \log (u+\sqrt{u^2+a^{2/3}})\,\textrm{d}u =O_a\Bigl (\frac{1}{a^{1/3}}\Bigr ), \end{aligned}$$

(A.11)

$$\begin{aligned}&\int _{0}^{\infty }\frac{e^{-u}}{a^{1/3}}\big ((u-1)\log (u+\sqrt{u^2+a^{2/3}})-1\big )\,\textrm{d}u=-1+o_a(1). \end{aligned}$$

(A.12)

Proof

The identities are elementary but we choose to give details for the sake of completeness.

Verification of (A.4) and (A.5) Integration by parts shows that

$$\begin{aligned} \int \limits _{0}^{\infty }e^{-u}\frac{1}{\sqrt{u^2+a^{2/3}}}\,\textrm{d}u= & {} \int \limits _{0}^{\infty }e^{-u}\log (u+\sqrt{u^2+a^{2/3}})\,\textrm{d}u-\frac{1}{3}\log a\nonumber \\= & {} O_a(1)+O_a(\log a), \end{aligned}$$

(A.13)

and

$$\begin{aligned} \frac{1}{a^{1/3}}\int \limits _{0}^{\infty } e^{-u}\sqrt{u^2+a^{2/3}}\,\textrm{d}u&=\frac{1}{2a^{1/3}}\int \limits _{0}^{\infty } e^{-u}(a^{2/3}\log (u+\sqrt{u^2+a^{2/3}}) \\&\quad +u\sqrt{u^2+a^{2/3}}-\frac{1}{3}a^{2/3}\log a)\,\textrm{d}u\\&=O_a(a^{1/3})+O_a(a^{1/3}\log a)+\frac{1}{a^{1/3}}\\&\quad +\frac{1}{2}\int \limits _{0}^{\infty } e^{-u}\frac{u}{a^{1/3}}(\sqrt{u^2+a^{2/3}}-u)\,\textrm{d}u\\&=\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a)+\frac{a^{1/3}}{2}\int \limits _{0}^{\infty } e^{-u}\frac{u}{\sqrt{u^2+a^{2/3}}+u}\,\textrm{d}u\\&=\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a)+O_a(a^{1/3}). \end{aligned}$$

Verification of (A.6) Using that $\int _0^\infty e^{-u}u^2\,\textrm{d}u=2$ we have

$$\begin{aligned} \int \limits _0^{\infty }e^{-u}\frac{u\sqrt{u^2+a^{2/3}}}{a^{1/3}}\,\textrm{d}u&=\frac{2}{a^{1/3}}+\frac{1}{a^{1/3}}\int \limits _0^{\infty }e^{-u}u(\sqrt{u^2+a^{2/3}}-u)\,\textrm{d}u\\&=\frac{2}{a^{1/3}}+a^{1/3}\int \limits _0^{\infty }e^{-u}\frac{u}{\sqrt{u^2+a^{2/3}}+u}\,\textrm{d}u\\&=\frac{2}{a^{1/3}}+O_a(a^{1/3}). \end{aligned}$$

Verification of (A.7)

$$\begin{aligned} \int \limits _0^{\infty }e^{-u}\frac{u^2}{a^{1/3}\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u&=\int \limits _0^{\infty }e^{-u}\frac{\sqrt{u^2+4a^{2/3}}}{a^{1/3}}\,\textrm{d}u-4a^{1/3}\int \limits _0^{\infty }e^{-u}\frac{1}{\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u\\&=\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a)+a^{1/3}O_a(\log a), \end{aligned}$$

where we used (A.4) and (A.5).

Verification of (A.8)

$$\begin{aligned} \int \limits _0^{\infty }e^{-u}\frac{u^2+4a^{2/3}}{a^{1/3}\sqrt{u^2+a^{2/3}}}\,\textrm{d}u&=\frac{1}{a^{1/3}}\int \limits _0^{\infty }e^{-u}\sqrt{u^2+a^{2/3}}\,\textrm{d}u+3a^{1/3}\int \limits _0^{\infty }e^{-u}\frac{1}{\sqrt{u^2+a^{2/3}}}\,\textrm{d}u\\&=\frac{1}{a^{1/3}}+O_a(a^{1/3}\log a)+O_a(a^{1/3}), \end{aligned}$$

where in the last line we used (A.4) and (A.5).

Verification of (A.9)

$$\begin{aligned} \int \limits _0^\infty e^{-u}\frac{a^{2/3}}{u+\sqrt{u^2+a^{2/3}}}\,\textrm{d}u&= \int \limits _0^\infty e^{-u}(\sqrt{u^2+a^{2/3}}-u)\,\textrm{d}u\\&=1+a^{1/3}O_a(a^{1/3}\log a)-1\\&=O_a(a^{2/3}\log a), \end{aligned}$$

where we used (A.5).

Verification of (A.10)

$$\begin{aligned} \int \limits _0^{\infty }e^{-u}\,\frac{a^{1/3}u}{(u+\sqrt{u^2+4a^{2/3}})\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u&=\int \limits _0^{\infty }e^{-u}\,\frac{a^{1/3}}{\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u\\&\quad -\int \limits _0^{\infty }e^{-u}\,\frac{a^{1/3}}{u+\sqrt{u^2+4a^{2/3}}}\,\textrm{d}u\\&=O_a(a^{1/3}\log a), \end{aligned}$$

where we used (A.4) and (A.9).

Verification of (A.11) The identity is immediate since $e^{-u}u\log (u)\in L^p([0,\infty ))$ for all $p\in [1,\infty ]$.

Verification of (A.12) For $a>0$, integration by parts shows

$$\begin{aligned} \int \limits _{0}^{\infty }e^{-u}(u-1)\log (u+\sqrt{u^2+a^{2/3}})\,\textrm{d}u=\int \limits _0^\infty e^{-u}\frac{u}{\sqrt{u^2+a^{2/3}}}\,\textrm{d}u, \end{aligned}$$

so that to prove the last identity we need to show

$$\begin{aligned} \lim _{a\rightarrow 0^+}\frac{1}{a}\int \limits _{0}^{\infty }e^{-u}\biggl (1-\frac{u}{\sqrt{u^2+a^2}}\biggr )\,\textrm{d}u=1. \end{aligned}$$

Changing variable $u\rightsquigarrow au$ gives

$$\begin{aligned} \frac{1}{a}\int \limits _{0}^{\infty }e^{-u}\biggl (1-\frac{u}{\sqrt{u^2+a^2}}\biggr )\,\textrm{d}u&= \int \limits _{0}^{\infty }e^{-au}\biggl (1-\frac{u}{\sqrt{u^2+1}}\biggr )\,\textrm{d}u\\&=\int \limits _{0}^{\infty }e^{-au}\frac{1}{(u+\sqrt{u^2+1})\sqrt{u^2+1}}\,\textrm{d}u, \end{aligned}$$

hence

$$\begin{aligned} \lim _{a\rightarrow 0^+}\frac{1}{a}\int \limits _{0}^{\infty }e^{-u}\biggl (1-\frac{u}{\sqrt{u^2+a^2}}\biggr )\,\textrm{d}u=\int \limits _{0}^{\infty }\frac{1}{(u+\sqrt{u^2+1})\sqrt{u^2+1}}\,\textrm{d}u.\end{aligned}$$

Changing variable $u=\sinh t$ we obtain

$$\begin{aligned} \int \limits _{0}^{\infty }\frac{1}{(u+\sqrt{u^2+1})\sqrt{u^2+1}}\,\textrm{d}u=\int \limits _0^\infty \frac{1}{\sinh t+\cosh t}\,\textrm{d}t=\int \limits _0^\infty e^{-t} \,\textrm{d}t=1. \end{aligned}$$

$\square$