Young’s Inequality for the Twisted Convolution

Abstract

We characterise the triples (p, q, r) for which the bi-linear $\lambda$-twisted convolution map $B_\lambda : (f,g) \rightarrow f \times _\lambda \, g$ is bounded from $L^p(\mathbb C^n) \times L^q(\mathbb C^n) \rightarrow L^r(\mathbb C^n)$ for $1 \le p,q,r \le \infty$. This gives the analogue of Young’s inequality for the twisted convolution.

1 Introduction

The generalised Young’s inequality says that if $f \in L^p(\mathbb R^n), g \in L^q(\mathbb R^n)$, then $f *g \in L^r(\mathbb R^n)$ and the inequality $\Vert f *g \Vert _r \le \Vert f\Vert _p \Vert g\Vert _q$ holds for $1 \le p,q,r \le \infty$ if and only if $\frac{1}{p} + \frac{1}{q} = 1+ \frac{1}{r}$, see [2]. It is a natural question to ask for a characterisation of triples (p, q, r) for the similar inequality with the twisted convolution. The $\lambda$-twisted convolution of two functions f and g on $\mathbb C^n$ is defined, for $\lambda \in \mathbb R{\setminus } \{0\}$, by

$$\begin{aligned} f \times _\lambda g(z) = \int \limits _{\mathbb {C}^n}f(z-w) \, g(w)\, e^{i \frac{\lambda }{2}\text {im} (z. \overline{w})} \, dw, \end{aligned}$$

(1.1)

whenever the integral converges. The twisted convolution was originally introduced by Segal, in the abstract set up of locally compact Abelian groups in [15].

The $\lambda$-twisted convolution arises naturally in the analysis on the Heisenberg group $\mathbb {H}^n$, which is topologically $\mathbb C^n \times \mathbb {R}$ see [3, 17], with the group law $(z,t) \cdot (w,s) = \left( z+w, t+s+ \, \frac{1}{2} \text {im}(z \cdot \bar{w}) \right) .$ The convolution of two functions f and g on $\mathbb {H}^n$ is given by

$$\begin{aligned} f *g(z,t):= & {} \int _{\mathbb C^n \times \mathbb {R}} f\Bigg (z-w, t-s- \frac{1}{2} \text {im}(z \cdot \bar{w}) \Bigg ) g(w,s)dw ds. \end{aligned}$$

Taking the Fourier transform in the t-variable at $\lambda$, the above convolution reduces to the $\lambda$-twisted convolution:

$$\begin{aligned} (f *g)^\lambda (z)= & {} (2\pi )^{1/2}\, f^\lambda \times _\lambda g^\lambda (z) \end{aligned}$$

(1.2)

where $f^\lambda (z) = \frac{1}{\sqrt{2 \pi }} \int _\mathbb Rf(z,t) \, e^{i \lambda t} \, dt$, for $\lambda \in \mathbb R$.

On the other hand, if we consider the reduced Heisenberg group $\mathbb {H}^n_0$, which is topologically $\mathbb C^n \times \mathbb {T}^1$, with group law

$$\begin{aligned} (z,t) \cdot (w,s) = \left( z+w, \left[ t+s+ \, \frac{1}{2} \text {im}(z \cdot \bar{w}) \right] (\text {mod} \, 2 \pi ) \right) , \end{aligned}$$

then any function f on $\mathbb C^n$ can be lifted to the function F on $\mathbb {H}^n_0$ by setting

$$\begin{aligned} F(z,t) = (2 \pi )^{-1} f(z) \, e^{-i\lambda t }, ~ \lambda \in \mathbb {N}. \end{aligned}$$

(1.3)

The group convolution of two such functions is again of the same form, and is given in terms of the $\lambda$-twisted convolution:

$$\begin{aligned} F *G(z,t) = (2\pi )^{-1}\, e^{-i\lambda t } f \times _\lambda g(z), ~ \text {for}~ \lambda \in \mathbb {N}. \end{aligned}$$

(1.4)

Note that the integral given by (1.1) is absolutely convergent for nice functions, say when $f \in L^\infty (\mathbb C^n)$ and $g\in L^1(\mathbb C^n)$, and in this case $f \times _\lambda g$ defines a bounded continuous function on $\mathbb {C}^n$, in view of the continuity of the map $z \rightarrow g(z-\cdot )\, e^{-i\frac{\lambda }{2}\text {im}(z. \overline{\, \cdot \, })}$ from $\mathbb C^n \rightarrow L^1(\mathbb C^n)$.

Since $|f \times _\lambda g| \le |f| * |g|$, the $\lambda$-twisted convolution satisfies all the inequalities in terms of |f| and |g| satisfied by the usual convolution. In particular we have

$$\begin{aligned} \Vert f \times _\lambda g\Vert _{L^{\infty }(\mathbb {C}^n)} \le \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^{p'}(\mathbb {C}^n)} \end{aligned}$$

(1.5)

by Hölder’s inequality, for $f\in L^p (\mathbb {C}^n)$ and $g \in L^{p^{\prime }}(\mathbb {C}^n),~ \frac{1}{p} + \frac{1}{p'}=1,$ $1 \le p \le \infty$, and also the Young’s inequality

$$\begin{aligned} \Vert f \times _\lambda g\Vert _{L^p(\mathbb {C}^n)} \le \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^1(\mathbb {C}^n)} \end{aligned}$$

(1.6)

for $f \in L^p(\mathbb {C}^n),g \in L^1(\mathbb {C}^n)$. But unlike the usual convolution, we also have

$$\begin{aligned} \Vert f\times _\lambda g\Vert _{L^2(\mathbb {C}^n)} \le (2\pi /|\lambda |)^{n/2}\, \Vert f\Vert _{L^2 (\mathbb {C}^n)}\Vert g\Vert _{L^2(\mathbb {C}^n) } \end{aligned}$$

(1.7)

for $f, g\in L^2(\mathbb {C}^n)$, see [3] p. 27, or [19] p. 14.

Note that the $\lambda$-twisted convolution is commutative only for $\lambda = 0$, the case in which it coincides with the usual convolution. However, we have $f \times _\lambda g = g \times _\lambda f$ when f and g are poly-radial functions: In fact when f and g are radial functions in $L^2(\mathbb C^n)$, this follows by their special Hermite expansions, in view of the identity (2.4.2) in [18]. Since a poly-radial function on $\mathbb C^n$ is radial in each of the n variables $z_i \in \mathbb C$, successive application of the one dimensional result lead to the same conclusion for poly-radial functions as well.

The boundedness properties of twisted convolution operators, i.e., operators of the form $f \rightarrow K \times f$, between Lebesgue spaces has been of interest, and addressed by several authors. For instance, in [7] Karapetyants and de Arellano, studied the $L^p \rightarrow L^q$ boundedness of twisted convolution with a singular kernel. A work somewhat similar in spirit, concerning potential operators in the twisted convolution setting has been carried out by Novak and Stempak in [13]. In [11], Mantero showed that there exists a class of distribution kernels K, for which the associated twisted convolution operator is bounded in $L^p$ for every $p \in (1,2]$ but not for any $p>2$. This is in contrast to the behaviour of the usual convolution operator. An amusing result in this context is given by Michael Cowling in [1], which says that an operator defined by twisted convolution with a distribution kernel K of compact support, is bounded on $L^p(\mathbb C^n)$ if and only if the operator defined by the usual convolution with K is bounded on $L^p(\mathbb C^n)$. This result has been extended by Leptin to the context of general connected Lie groups in [8], and a sharp version was obtained jointly with Müller in [9]. In a recent article [14], O’Neil has obtained some boundedness result for twisted convolution operators on $L^p(\mathbb C^n)$ for $1 \le p \le 2$. Some partial results for the boundedness of the twisted convolution operators from $L^p(\mathbb C^n) \times L^q (\mathbb C^n)\rightarrow L^r(\mathbb C^n)$ have been obtained in [10, 20]. We also refer the reader to [5, 12], for the work related to the Hardy spaces associated the twisted convolution, and the boundedness of local Riesz transform on them. We would also like to point out an interesting blog article by Terence Tao [16], where he illustrates the use of a similar twisted convolution defined on $l^2 (\mathbb {Z}_2^n)$, in the proof of the sensitivity conjecture in theoretical computer science.

In this article, we are interested in the boundedness of the bi-linear operator $B_\lambda : (f,g) \mapsto f \times _\lambda g$ from $L^{p} (\mathbb C^n) \times L^{q} (\mathbb C^n) \rightarrow L^{r}(\mathbb C^n)$. Our main result is the following geometric characterisation of the triples (p, q, r) for the inequality

$$\begin{aligned} \Vert B_\lambda (f,g)\Vert _{L^r(\mathbb C^n)} \le C_\lambda \, \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

(1.8)

to hold.

Theorem 1.1

Let $\lambda \in \mathbb R\setminus \{0\}$. The inequality

$$\begin{aligned} \Vert f \times _\lambda g\Vert _{L^r(\mathbb {C}^n)} \le C_\lambda \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

holds with a constant $C_\lambda$ independent of $f \in L^p(\mathbb C^n)$ and $g \in L^q(\mathbb C^n), 1\le p,q\le \infty$, if and only if $0 \le \frac{1}{p}+ \frac{1}{q}-1 \le \frac{1}{r} \le \text {min}\{ \frac{1}{p}, \frac{1}{q}\}$.

Actually for (p, q, r) satisfying the above condition, we prove the estimate with $C_\lambda =(2\pi / |\lambda | )^{n( 1+ \frac{1}{r}- \frac{1}{p}- \frac{1}{q} ) }$. Fixing $K \in L^{p_0}(\mathbb C^n)$, the above theorem also give conditions for the $L^p(\mathbb C^n) \rightarrow L^q(\mathbb C^n)$ boundedness of the twisted convolution operator $T_\lambda : f \rightarrow K \times _\lambda f$.

2 Some Auxiliary Results

Here we prove some elementary results useful in the next section. For $z\in \mathbb C^n$, the $\lambda$-twisted translation $\tau _z^\lambda$ of a function g on $\mathbb C^n$ is defined by

$$\begin{aligned} \tau _z^\lambda g \, (w) = e^{i\frac{\lambda }{2} \text {im}(w \cdot \bar{z})}g(w-z). \end{aligned}$$

(2.1)

Note that $f \times _\lambda g$ given by (1.1) is nothing but the convolution associated with the $\lambda$-twisted translation:

$$\begin{aligned} f \times _\lambda g(z)= \int _{\mathbb C^n} f(w) \, \tau _z^\lambda \tilde{g} (w) \, dw \end{aligned}$$

(2.2)

where $\tilde{g}(w)= g(-w)$. We can also define the twisted convolution of a tempered distribution K and a Schwartz class function f, by replacing the above integral by the pairing $\langle \,, \rangle$ of tempered distributions and Schwartz class functions: $K \times _\lambda f (z):= \langle K, \tau _z^\lambda \, \tilde{g} \rangle$ which defines a continuous function on $\mathbb C^n$. An interesting case is when K is the constant function, in which case $K \times _\lambda f$ is a constant multiple of the symplectic Fourier transform of f defined by

$$\begin{aligned} \mathcal {F}_\lambda f(z) =(2 \pi )^{-n} \int _{\mathbb C^n} f(w) \, e^{i\frac{\lambda }{2} \text {im}(z \cdot \bar{w})}\, dw, ~ f \in \mathcal {S}(\mathbb C^n) \end{aligned}$$

(2.3)

with the inverse transform $\mathcal {F}_\lambda ^{-1} = (\lambda /2)^{2n} \mathcal {F}_{-\lambda }$

In fact, we can also define the twisted convolutions of two tempered distributions $K_1$ and $K_2$, by setting $\langle K_1 \times _\lambda K_2, \varphi \rangle =\langle K_1, \tilde{K}_2 \times _\lambda \varphi \rangle$, for $\varphi \in \mathcal {S} (\mathbb C^n)$ whenever $\tilde{K}_2 \times \varphi \in \mathcal {S} (\mathbb C^n)$. This is the case for instance when $K_2$ is of compact support, as in the usual convolution. But the compactness assumption on the support of at least one of the distributions, is not necessary in the context of twisted convolution. For instance taking $K_1=K_2 \equiv 1$, the distribution given by the constant function 1, we have the identity $1 \times _\lambda 1 = (4\pi / \lambda )^{2n} \delta$ as tempered distributions, where $\delta$ denotes the Dirac delta distribution supported at $0 \in \mathbb C^n$. In fact, for $\varphi \in \mathcal {S} (\mathbb C^n)$ we have

$$\begin{aligned} \hspace{21.68121pt}\langle 1 \times _\lambda 1, \varphi \rangle= & {} \langle 1, 1 \times _\lambda \varphi \rangle = (2\pi )^n \langle 1, \mathcal {F}_\lambda (\varphi ) \rangle \nonumber \\= & {} (2\pi )^n \int _{\mathbb C^n} \mathcal {F}_\lambda (\varphi ) (w) dw. \end{aligned}$$

(2.4)

By Fourier inversion formula, the last term is $(4\pi / \lambda )^{2n} \varphi (0) = (4\pi / \lambda )^{2n} \, \langle \delta , \varphi \rangle$.

Remark 2.1

Even though the $\lambda$-twisted convolution is not commutative in general as mentioned above, the inequality (1.8) is still symmetric with respect to f and g. i.e., the inequality

$$\begin{aligned} \Vert f \times _\lambda g\Vert _{L^r(\mathbb {C}^n)} \le C_\lambda \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

holds for a triple $(p,q,r), ~ 1 \le p,q,r \le \infty$, if and only if the inequality

$$\begin{aligned} \Vert g \times _\lambda f\Vert _{L^r(\mathbb {C}^n)} \le C_\lambda \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)}. \end{aligned}$$

holds. This follows from the relation $( f \times _\lambda g)^\sharp = g^\sharp \times _\lambda f^\sharp$ where $f^\sharp (z):= \overline{ f(-z)}$ and the fact that $\Vert h^\sharp \Vert _\rho = \Vert h\Vert _\rho$ for $h \in L^\rho (\mathbb C^n)$, $1 \le \rho \le \infty$.

Lemma 2.2

Let K be a tempered distribution on $\mathbb C^n$. Then the operator $T: f \rightarrow K \times _\lambda \, f$ from $\mathcal {S}(\mathbb C^n) \rightarrow \mathcal {S}'(\mathbb C^n)$ is invariant under the $\lambda$-twisted translations. That is $T \circ \tau _{z_0}^\lambda = \tau _{z_0}^\lambda \circ T$ on $\mathcal {S}(\mathbb C^n)$, for all $z_0 \in \mathbb C^n$.

Moreover, if X is any Banach space of functions on $\mathbb C^n$ in which $\mathcal {S}(\mathbb C^n)$ is dense, then the same invariance holds for the continuous extension $\tilde{T}: X \rightarrow Y$, into any Banach space Y, provided the translation operators $\tau _z^\lambda$ are continuous in both X and Y.

Proof

First we assume that $K \in \mathcal {S}(\mathbb C^n)$, so that

$$\begin{aligned} T \big (\tau _{z_0}^\lambda f\big ) (z) = \int _{\mathbb C^n} K(w) \, \tau _z^\lambda \big ( \tau _{z_0}^\lambda \tilde{f}\big ) (w) \, dw = \int _{\mathbb C^n} K(w) \, \big (\tau _z^\lambda \tau _{-z_0}^\lambda \tilde{f}\big ) (w) \, dw. \end{aligned}$$

Since $\tau _z^\lambda \tau _{-z_0}^\lambda =e^{i\frac{\lambda }{2} \text {im}(z \cdot \bar{z_0})} \tau _{z-z_0}^\lambda$ the last expression is same as

$$\begin{aligned} e^{i\frac{\lambda }{2} \text {im}(z \cdot \bar{z_0})} K \times _\lambda f (z-z_0) =: \tau _{z_0}^\lambda (K \times _\lambda f) (z)= \tau _{z_0}^\lambda Tf(z) \end{aligned}$$

for all $z_0 \in \mathbb C^n$.

The same idea works for $K \in \mathcal {S}' (\mathbb C^n)$ as well, by replacing the integrals defining $K \times _\lambda f (z)$ by the pairing $\langle K, \tau _{z}^\lambda \tilde{f} \rangle$, which makes sense for $f \in \mathcal {S} (\mathbb C^n)$:

$$\begin{aligned} T \big (\tau _{z_0}^\lambda f\big ) (z)= & {} \big \langle K, \tau _z^\lambda \big ( \tau _{z_0}^\lambda \tilde{f}\big ) \big \rangle = \big \langle K, \tau _z^\lambda \tau _{-z_0}^\lambda \tilde{f} \big \rangle \\= & {} e^{i\frac{\lambda }{2} \text {im}(z \cdot \bar{z_0})} \, \big \langle K, \tau _{z-z_0}^\lambda \tilde{f} \big \rangle = \tau _{z_0} ^\lambda (T f) (z). \end{aligned}$$

The invariance for the extension $T: X\rightarrow Y$ follows by approximating f in the Banach space X by a sequence $f_n \in \mathcal {S} (\mathbb C^n)$, and using the continuity of $T: X\rightarrow Y$ and the continuity of the twisted translations in X and Y. $\square$

Remark 2.3

For $\lambda \in \mathbb Z$, the first part of Lemma 2.2 can also be deduced from the translation invariance of the convolution on the reduced Heisenberg group by considering the translations by $(z_0,0) \in \mathbb {H}^n_0$, using the lifting given by (1.3) in view of the identity (1.4).

Lemma 2.4

Let $1 \le p,q,r \le \infty$, $0 \ne \lambda \in \mathbb R$ and $f\in L^p(\mathbb C^n), g\in L^q(\mathbb C^n).$ Then the inequality

$$\begin{aligned} \Vert f \times _\lambda g\Vert _{L^r (\mathbb {C}^n)} \le C_\lambda \, \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

holds for a triple (p, q, r) with a constant $C_\lambda$, if and only if the inequality $\Vert f \times g\Vert _{L^r (\mathbb {C}^n)} \le C_1 \, \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)}$ holds. More over $C_\lambda =C_1 \, |\lambda |^{n(\frac{1}{p} + \frac{1}{q}- \frac{1}{r}-1 )}$.

Proof

For $f \in L^p(\mathbb C^n)$ and $g \in L^q(\mathbb C^n)$, $1 \le p,q,r \le \infty$, set $f_\lambda (z) = f\big (\frac{z}{\sqrt{|\lambda |}}\big )$ and $g_\lambda (z) = g\big (\frac{z}{\sqrt{|\lambda |}}\big )$ for $\lambda \ne 0$. By a simple change of variable, we see that

$$\begin{aligned} \hspace{36.135pt}f_\lambda \times g_\lambda (z)= |\lambda | ^ n (f \times _{|\lambda |} g)\Bigg (\frac{z}{\sqrt{|\lambda |}}\Bigg ) = |\lambda | ^ n (g \times _{-|\lambda |} f)\Bigg (\frac{z}{\sqrt{|\lambda |}}\Bigg ). \end{aligned}$$

(2.5)

Since $\Vert h_\lambda \Vert _\rho = |\lambda |^{n/\rho } \Vert h \Vert _\rho$, the conclusion follows for $\lambda >0$, in view of the first equality above, with $C_\lambda =C_1 \, |\lambda |^{n( \frac{1}{p} + \frac{1}{q} - \frac{1}{r}-1 )}$. Note that when $\lambda <0$, we have $\lambda = -|\lambda |$, hence this case follows using the second expression for $f_\lambda \times g_\lambda$, in view of Remark 2.1. $\square$

Definition 2.5

We call a triple $(p,q,r), ~ 1\le p,q,r \le \infty$ admissible (for $B_\lambda$), if the inequality $\Vert f \times _\lambda g\Vert _{L^r (\mathbb {C}^n)} \le C_\lambda \, \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)}$ holds with a constant $C_\lambda$ valid for all $f \in L^p(\mathbb {C}^n)$ and $g \in L^q(\mathbb {C}^n)$.

By Lemma 2.4, the admissibility does not depend on the particular $\lambda \in \mathbb R\setminus \{0\}$. Thus to study the boundedness of $B_\lambda : L^p\times L^q \rightarrow L^r$ for $\lambda \ne 0$, it is enough to study the case $\lambda =1$. We write B(f, g) and $f \times g$ dropping the subscript when $\lambda =1$.

We next observe a symmetry property of admissible triples, a crucial step in the proof of Theorem 1.1.

Lemma 2.6

A triple (p, q, r) with $1 \le p,q,r \le \infty$ is admissible if and only if any of the triples $(q,r',p')$, $(r',p,q')$, (q, p, r), $(r', q, p')$, $(p, r', q')$ are admissible, where $\frac{1}{s} + \frac{1}{s'} =1$ for $s=p,q,r$.

Proof

Note that admissibility of a triple (p, q, r) is equivalent to the quantity

$$\begin{aligned} \Vert B\Vert _{L^p \times L^q \rightarrow L^r}:= \sup _{ \{ f,g \, \in \, \mathcal {S}(\mathbb C^n) \,: \, \Vert f \Vert _{ L^{p} (\mathbb {C}^n) } =1= \Vert g \Vert _{ L^{q} (\mathbb {C}^n) \} } } \Vert f \times g \Vert _{L^r(\mathbb C^n)} \end{aligned}$$

(2.6)

being finite. Since $\Vert B(f,g) \Vert _{L^r(\mathbb C^n)} = sup_{ \{ \Vert \varphi \Vert _{ L^{r'} (\mathbb {C}^n) } =1 \} } | \langle f \times g, \varphi \rangle |$ for $1 \le r \le \infty$, (2.6) can be expressed as

$$\begin{aligned} \hspace{36.135pt}\Vert B\Vert _{L^p \times L^q \rightarrow L^r}:= \sup _{\{ f,g \, \in \, \mathcal {S}(\mathbb C^n) \,: \, \Vert f \Vert _{ L^{p} (\mathbb {C}^n) } = \Vert g \Vert _{ L^{q} (\mathbb {C}^n) } = 1= \Vert \varphi \Vert _{ L^{r'} (\mathbb {C}^n) } \} } | \langle f \times g, \varphi \rangle |\nonumber \\ \end{aligned}$$

(2.7)

Since $\langle f \times g, \varphi \rangle = (f \times g) \times \tilde{\varphi } \, (0)$ with $\tilde{\varphi }(x)= \varphi (-x)$, by the associativity of the twisted convolution and the relation $(\varphi \times \psi \tilde{)} = \tilde{\varphi } \times \tilde{\psi }$, we see that

$$\begin{aligned} \langle B(f, g), \varphi \rangle = \langle f \times g, \varphi \rangle = \langle f, \tilde{g} \times \varphi \rangle = \langle f, B(\tilde{g}, \varphi ) \rangle . \end{aligned}$$

(2.8)

Now since $\tilde{g} \in L^q(\mathbb C^n)$ is as arbitrary as $g \in L^q(\mathbb C^n)$, we see from (2.7) and (2.8) that

$$\begin{aligned} \Vert B\Vert _{L^p \times L^q \rightarrow L^r} = \Vert B\Vert _{L^{q} \times L^{r'} \rightarrow L^{p'}}. \end{aligned}$$

(2.9)

From this identity we also conclude that $\Vert B\Vert _{L^{q} \times L^{r'} \rightarrow L^{p'}} = \Vert B\Vert _{L^{r'} \times L^{p} \rightarrow L^{q'}}.$ Hence the equivalence of admissibility of the triples $(q,r',p')$ and $(r',p,q')$ with (p, q, r) also follows,

Again, in view of Remark 2.1, if we flip the first two indices in an admissible triple, the resulting triple is also admissible. This leads to the fact that the admissibility of (p, q, r) is equivalent to the admissibility of the remaining three triples: (q, p, r), $(r', q, p')$ and $(p, r', q')$. This completes the proof. $\square$

Using Lemma 2.6, we can easily see the non admissibility of certain triples, which will be useful in the proof of Theorem 3.1:

Corollary 2.7

The triples $(p,\infty , 1)$, $(\infty , q, 1)$ and $(\infty , \infty ,r)$ are not admissible for $1 \le p,q,r \le \infty .$

Proof

The fact that $(\infty , \infty ,r)$ is not admissible is obvious from (2.4), which says that $f \times f$ is a constant multiple of the Dirac measure for the $L^\infty$ function $f \equiv 1$. Since $\delta$ is not given by any locally integrable function, it follows that B cannot map $L^\infty (\mathbb C^n) \times L^\infty (\mathbb C^n)$ into $L^r(\mathbb C^n)$ for any $r\in [1,\infty ]$, hence in particular $(\infty , \infty , r)$ cannot be admissible. In view of Lemma 2.6, it also follows from this that the triples $(\infty , q, 1)$ and $(p,\infty , 1)$ are not admissible for any $p,q \in [1,\infty ]$. $\square$

As we saw before, the twisted convolution operator shares many properties of the usual convolution operators. Because of the presence of $e^{-i\frac{\lambda }{2}\text {im} (z. \overline{w})}$, it also behaves like the Fourier transform, as we already saw when f is constant. The proof of the following auxiliary necessary condition is motivated by this fact.

Proposition 2.8

Let $1\le r \le \infty$. If the inequality

$$\begin{aligned} \Vert f \times g\Vert _{L^r (\mathbb {C}^n)} \le C \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

holds with a constant C valid for all $f \in L^p (\mathbb {C}^n)$ and $g \in L^q (\mathbb {C}^n)$, $1\le p, q \le \infty ,$ then $\frac{1}{p}+ \frac{1}{q} + \frac{1}{r} \ge 1$.

Proof

Note that for $r=1$, the condition $\frac{1}{p}+ \frac{1}{q}+ \frac{1}{r} \ge 1$ clearly holds. So we assume $1<r \le \infty$, and start with dimension $n=1$. Taking $\varphi = \chi _{[-1,1]}$, $\psi = \chi _{[-\frac{1}{2},\frac{1}{2}]}$, for $w= u+iv \in \mathbb C$ and $t>0$, we define

$$\begin{aligned} f_t(w)= & {} (4t)^{-1/p} \varphi \left( \frac{u}{2t}\right) \psi \left( \frac{v}{2}\right) \nonumber \\ g_t(w)= & {} (t)^{-1/q} \varphi \left( \frac{u}{t}\right) \psi (v). \end{aligned}$$

(2.10)

Then $\Vert f_t\Vert _{L^p(\mathbb C)} = \Vert f_1\Vert _{L^p(\mathbb C)}$ and $\Vert g_t\Vert _{L^q(\mathbb C)} = \Vert g_1\Vert _{L^q(\mathbb C)}$, for all $t>0$ with $\text {supp} (f_t) \subset [-2t,2t] \times [-1,1]$ and $\text {supp} (g_t) \subset [-t,t] \times [-\frac{1}{2},\frac{1}{2}]$. Now setting $\alpha (t)= 4^{-1/p} \, t^{-(1/p+ 1/q)}$ and $z =x+iy \sim (x,y) \in \mathbb R^2$, we have

$$\begin{aligned} f_t(z-w) g_t(w)= & {} \alpha (t) \, \varphi \left( \frac{x-u}{2t}\right) \psi \left( \frac{y-v}{2}\right) \, \varphi \left( \frac{u}{t}\right) \psi (v), ~ w = u + iv \\= & {} \alpha (t) \, \chi _{[x-2t, x+2t]} (u) \chi _{[y-1, y+1]} (v) \, \chi _{[-t, t]} (u) \chi _{\big [-\frac{1}{2},\frac{1}{2}\big ]} (v) \\= & {} \alpha (t) \, \chi _{A_t(x)} (u) \, \chi _{ B(y)} (v) \\= & {} \alpha (t) \, \chi _{A_t(x) \times B(y) } (u,v) \end{aligned}$$

where $A_t(x) = [x-2t, x+2t] \cap [-t, t],$ and $B(y) = A_{1/2}.$ More explicitly,

$$\begin{aligned}{} & {} A_t (x) = \left\{ \begin{array}{lr} \emptyset &{} \text {if}~ |x|> 3t\\ \left[ -t, x+2t\right] &{} \text {if}~ -3t \leqslant x \leqslant -t \\ \left[ -t,t \right] &{} \text {if} -t \leqslant x \leqslant t \\ \left[ x-2t,t \right] ~&{} \text {if}~ t \leqslant x \leqslant 3t \end{array} \right. \\{} & {} B (y) = \left\{ \begin{array}{lr} \emptyset &{} \text {if}~ |y| > 3/2\\ \left[ -1/2, y+1\right] &{} \text {if}~ -3/2 \leqslant y \leqslant -1/2 \\ \left[ -1/2,1/2\right] &{} \text {if} -1/2 \leqslant y \leqslant 1/2 \\ \left[ y-1,1/2\right] &{} \text {if}~ 1/2 \leqslant y \leqslant 3/2. \end{array} \right. \end{aligned}$$

Writing $\text {im} (z \cdot \overline{w})= uy-vx$ for $z=x+iy, w=u+iv$, we see that

$$\begin{aligned} f_t \times g_t (z)= & {} \int _\mathbb Cf_t(z-w) \,g_t(w) \,e^{\frac{i}{2} (uy-vx)} \, du\, dv\\= & {} \alpha (t) \, \int _{u\in \mathbb R} \chi _{A_t(x)} (u) \, e^{\frac{i}{2} uy} \, du \int _{v\in \mathbb R} \chi _{B(y)} (v) \,e^{-\frac{i}{2} vx} \, dv. \end{aligned}$$

Note that $A_t(x) \times B(y) = \emptyset$ when $|x|>3t$ or $|y|>1$, and hence $f_t \times g_t$ is supported in the rectangle $[-3t,3t] \times [-1,1]$. Also, $A_t(x) = [-t,t]$ for all $x \in [-t,t]$ and $B(y) = [-\frac{1}{2},\frac{1}{2}]$ for all $y \in [-\frac{1}{2},\frac{1}{2}]$. Hence for all $z \sim (x,y) \in [-t,t] \times [-\frac{1}{2},\frac{1}{2}]$, we have

$$\begin{aligned} f_t \times g_t (z)= & {} \alpha (t) \, \int _{-t}^t e^{\frac{i}{2} uy} du ~ \int _{-1/2}^{1/2} e^{-\frac{i}{2} vx} dv\nonumber \\= & {} \alpha (t) \, \frac{4 \sin (yt/2)}{y} \, \frac{4 \sin (x/4)}{x}. \end{aligned}$$

(2.11)

It follows that,

$$\begin{aligned} \Vert f_t \times g_t \Vert _{L^\infty (\mathbb C^n)}\ge & {} C_p \, t^{-(1/p+ 1/q)} \, \sup _{(x,y) \in [-t,t] \times [-\frac{1}{2},\frac{1}{2}]} \left| \frac{4 \sin (yt/2)}{y} \, \frac{4 \sin (x/4)}{x}\right| \nonumber \\\ge & {} C \, t^{1-(1/p+ 1/q)} \end{aligned}$$

(2.12)

as the continuous function $(x,y) \rightarrow \frac{4 \sin (yt/2)}{y} \, \frac{4 \sin (x/4)}{x}$ attains the supremum 2t at $z \sim (x, y)=(0,0)$. Similarly for $r < \infty$

$$\begin{aligned} \Vert f_t \times g_t \Vert _r^r\ge & {} \int _{ [-t,t] \times \big [-\frac{1}{2},\frac{1}{2}\big ] } | f_t \times g_t (z)|^r \, dx \, dy \nonumber \\= & {} ( \alpha (t))^r \, \int _{-t}^t \left| \frac{4 \sin (x/4)}{x}\right| ^r dx \int _{-\frac{1}{2}}^{\frac{1}{2}} \left| \frac{4 \sin (yt/2)}{y}\right| ^r dy \nonumber \\= & {} 4^{1-r/p} 2^{1+r} t^{(r-1-r/p \, - \, r/q)} \left( \int _{-t/4}^{t/4} \left| \frac{ \sin (\theta )}{\theta }\right| ^r d\theta \right) ^2. \end{aligned}$$

(2.13)

Note that the last integral converges to some finite positive number as $t \rightarrow \infty$, for $1<r < \infty$. From (2.12) and (2.13) we see that for large t, the inequality

$$\begin{aligned} \Vert f_t \times g_t \Vert _r \ge C \, t^{1-1/r-1/p - 1/q} \end{aligned}$$

(2.14)

holds for $1 < r \le \infty$.

If the inequality in the proposition holds for $1<r \le \infty$, then the left hand side of (2.14) is bounded by $C \Vert f_1\Vert _p \Vert g\Vert _q$, by the choice of $f_t$ and $g_t$ in (2.10). Thus since C is independent of t, letting $t \rightarrow \infty$, (2.14) will lead to a contradiction unless $1-1/r- 1/p - 1/q \le 0$. This proves the necessity of the condition $\frac{1}{p}+ \frac{1}{q}+ \frac{1}{r} \ge 1$ for $1< r \le \infty$ as well, in dimension $n=1$.

The higher dimensional case also follows by similar arguments working with $F_t(z) = \Pi _{j=1}^n f_t(z_j)$ and $G_t(z) = \Pi _{j=1}^n g_t(z_j)$ for $z=(z_1,...,z_n) \in \mathbb C^n$, $f_t,g_t$ as in (2.10), leading to the following n-dimensional analogue of (2.11):

$$\begin{aligned} F_t \times G_t (z)= & {} \big (\alpha _{p,q}(t)\big )^n \, \Pi _{j=1}^n \frac{4 \sin (y_jt/2)}{y_j} \, \frac{4 \sin (x_j/4)}{x_j}. \end{aligned}$$

(2.15)

valid for all $z= (z_1,\dots ,z_n),$ with $z_j =(x_j + iy) \sim (x_j, y_j) \in [-t,t] \times [-\frac{1}{2},\frac{1}{2}]$. Using this identity, the above arguments leads to the same conclusion in $\mathbb C^n$ for $n > 1$. $\square$

We end this section with the following multi-linear interpolation theorem from Zygmund [21](Theorem 3.3), specialised to $L^p$ spaces on Euclidean spaces with respect to the Lebesgue measure.

Theorem 2.9

Let $T: (f_1, f_2, \dots ,f_k) \rightarrow T (f_1,f_2, \dots ,f_k)$ be a multilinear map defined for k-tuples $(f_1, f_2,\dots , f_k)$ of simple functions on $\mathbb R^n, ~k \in \mathbb {N}$, and taking values in the space of measurable functions on $\mathbb R^n$. Suppose that T satisfies the inequalities

$$\begin{aligned} \Vert T(f_1,f_2, \dots ,f_k)\Vert _{L^{q_1} (\mathbb R^n) }\le & {} M_1 \Vert f_1\Vert _{ L^{p_1^1} (\mathbb R^n) } \dots \Vert f_k\Vert _{ L^{p_k^1} (\mathbb R^n) } \\ \Vert T(f_1,f_2, \dots ,f_k)\Vert _{L^{q_2} (\mathbb R^n) }\le & {} M_2 \Vert f_1\Vert _{ L^{p_1^2} (\mathbb R^n) } \dots \Vert f_k\Vert _{ L^{p_k^2} (\mathbb R^n) } \end{aligned}$$

where $1 \le q_i, p_i^j \le \infty$ for $i=1,2, \dots ,k, \, j =1,2$. Then T also satisfies the inequality

$$\begin{aligned} \Vert T(f_1,f_2, \dots ,f_k)\Vert _{L^{q} (\mathbb R^n) }\le & {} M_1^{1-t}M_2^t \, \Vert f_1\Vert _{L^{p_1} (\mathbb R^n) } \cdots \Vert f_k\Vert _{L^{p_n} (\mathbb R^n) } \end{aligned}$$

for $\frac{1}{q} = \frac{1-t}{q_1} + \frac{t}{q_2}$ and $\frac{1}{p_i} = \frac{(1-t) }{p_i^1} + \frac{t}{p_i^2}$ for simple functions $f_i \in L^{p_i}(\mathbb R^n), \, i= 1,\dots ,k$. More over, if $p_i < \infty$ for $i=1,\dots , k$, then T extends as a continuous operator from $L^{p_1} (\mathbb R^n) \times \dots \times L^{p_k} (\mathbb R^n) \rightarrow L^q(\mathbb R^n)$.

3 Main Results

We start with proving a necessary condition for the boundedness of the bi-linear map $(f,g) \rightarrow f \times g$ from $L^p(\mathbb C^n) \times L^q(\mathbb C^n) \rightarrow L^r(\mathbb C^n)$.

Theorem 3.1

Let $B(f,g) = f \times g$. If the inequality

$$\begin{aligned} \Vert B(f,g)\Vert _{L^r(\mathbb C^n)} \le C \Vert f \Vert _{L^p(\mathbb {C}^n)} \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

holds for a triple $(p, q,r), ~ 1\le p, q, r \le \infty$ with a constant C valid for all $f \in L^p(\mathbb {C}^n)$ and $g \in L^q (\mathbb {C}^n)$, then $1 - \frac{1}{r} \le \frac{1}{p}+ \frac{1}{q} \le 1+ \frac{1}{r}$ and $r \ge \text {max}\{ p.q\}$.

Proof

The fact that the condition $1 - \frac{1}{r} \le \frac{1}{p}+ \frac{1}{q}$ is necessary is already shown in Proposition 2.8. The necessity of the condition $\frac{1}{p}+ \frac{1}{q} \le 1+ \frac{1}{r}$ follows by a simple dilation argument as in the case of the usual convolution, as observed in [6]. For the sake of completeness, we give a slightly different argument as in [10], where a very special case of Theorem 1.1 is obtained.

Taking dilations $f_t$ and $g_t$ as in the proof of Lemma 2.4 and using (2.5), we see that the above bi-linear twisted convolution inequality is same as

$$\begin{aligned} |t| ^{n(1+ 1/r -1/p - 1/q )} \Vert f \times _t g \Vert _r \le C \Vert f \Vert _p \, \Vert g \Vert _q \end{aligned}$$

(3.1)

with a constant C independent of $t>0$, valid for $f, g \in \mathcal {S}(\mathbb C^n)$. Since $f \times _t g \rightarrow f *g$ pointwise as $t \rightarrow 0$, and since $| f \times _t g| \le |f| *|g| \in L^\rho (\mathbb C^n)$ for any $1\le \rho \le \infty$, we see that $\lim _{t \rightarrow 0}\Vert f \times _t g\Vert _r$ exists by dominated convergence theorem, and is positive if we choose positive functions $f, g \in \mathcal {S}(\mathbb C^n)$, say $f(z)=e^{-|z|^2} =g(z)$. Thus letting $t \rightarrow 0$, we see that (3.1) is possible for all $t >0$ only if $1+ \frac{1}{r} \ge \frac{1}{p}+ \frac{1}{q}$, which proves the second inequality of the first condition.

Note that when $r=\infty$, the assertion $r \ge \text {max}\{ p,q\}$ clearly holds. Therefore we assume $r<\infty$. In view of Lemma 2.2, we can employ an idea of Hörmander in [4], to prove the assertion $r \ge \text {max}\{ p, q\}$, when $1 \le p,q, r < \infty$. Note that the inequality of Theorem 3.1 holds for B(f, g) for all $f \in L^p(\mathbb {C}^n)$ and $g \in L^q (\mathbb {C}^n)$ for some triple (p, q, r) with $1 \le p,q,r < \infty$, if and only if it holds for all smooth functions f and g with compact support, by density of $C_c^\infty (\mathbb C^n)$ in $L^\rho (\mathbb C^n)$ for any $\rho \in [1, \infty )$.

Let f and g be arbitrary smooth functions with compact support, and choose a sequence $\{z_j \} \subset \mathbb C^n$ such that $|z_j| \rightarrow \infty$ as $j \rightarrow \infty$. If the inequality in Theorem 3.1 holds, then applying that with f and $g_j = \tau _{z_j} g$ (the twisted translate of g given by 2.1) and taking C as the optimal constant, we get

$$\begin{aligned} \Vert (f \times g)+(f \times g_j)\Vert _r = \Vert f \times (g+g_j)\Vert _r \le C \Vert f \Vert _p \Vert g+ g_j\Vert _q. \end{aligned}$$

(3.2)

Since $\text {supp}( g_j)=\text {supp}( g) +z_j$ is disjoint from $\text {supp}( g)$ for large j, we have $\Vert g+ g_j\Vert _q = 2^ {1/q} \Vert g\Vert _q$ as $\Vert \tau _z g \Vert _q = \Vert g\Vert _q$. In the same way, since $f \times g_j$ is supported in $\text {supp}( f) + \text {supp}( g_j) =\text {supp}( f) + \text {supp}( g) + z_j$ which is compact, we see that when j is large, $\Vert (f \times g)+(f \times g_j)\Vert _r = 2^ {1/r}\Vert f \times g\Vert _r$. Using these observations in (3.2), we get for large j

$$\begin{aligned} \Vert f \times g \Vert _r \le C \, 2^{1/q -1/r} \Vert f \Vert _p \Vert g\Vert _q. \end{aligned}$$

Since $f,g \in C_c^\infty (\mathbb C^n)$ are arbitrary, this contradicts the optimality of C unless $r\ge q$, in view of the density of $C_c^\infty (\mathbb C^n)$ in $L^s(\mathbb C^n)$ for $s =p,q, 1 \le p,q < \infty )$.

Using the same arguments with $f_n = \tau _{z_n} f$ and g as above, we get the inequality

$$\begin{aligned} \Vert f \times g\Vert _r = \Vert g \times f\Vert _r \le C \, 2^{\big (\frac{1}{p}- \frac{1}{r}\big )} \Vert f \Vert _p \Vert g\Vert _q \end{aligned}$$

leading to the conclusion $r \ge p$, again by the optimality of the constant C. This verifies the assertion $r\ge \text {max}(p,q)$ when $1 \le p,q,r < \infty$.

To deal with the cases when at least one of p, q is $\infty$ and $1 \le r<\infty$, we first consider the case $1<r<\infty$: ie the cases $(\infty ,q,r)$ with $1 \le q < \infty$, and $(p, \infty ,r)$ with $1 \le p < \infty$ need to be addressed for $1<r<\infty$.

By Lemma 2.6, the triple $(\infty ,q,r)$ is admissible if and only if $(q,r',1)$ is admissible. Thus since $1<r<\infty$ and $1 \le q < \infty$, we are in the case $1\le q,r' <\infty$ for the triple $(q,r',1)$, which was settled above using Hörmander type arguments, and we conclude that $r' \le 1$, which is a contradiction as $r<\infty$. It follows that $(\infty ,q,r)$ cannot be admissible for $1\le q<\infty , 1<r <\infty$. From this it also follows by Lemma 2.6 that $(p,\infty ,r)$ is not admissible for any pair (p, r) such that $1\le p<\infty , 1< r <\infty$.

The cases remaining are: $r=1$ with at least one of p, q is $\infty$, and the case $1\le r < \infty , p=q=\infty$. That is, triples of the form $(p,\infty ,1)$ or $(\infty ,q, 1)$ with $1\le p,q < \infty$ and triples $(\infty , \infty , r)$ with $1\le r < \infty$. But by Corollary 2.7, none of these triples are admissible. Hence the condition $r \ge \text {max}\{ p,q\}$ vacuously holds in these cases. This completes the proof. $\square$

Remark 3.2

Note that the triple (p, q, r) given in the above theorem has a simple geometric description. Representing $p, q, r \in [1, \infty ]$ as the point $(x,y,z)= (\frac{1}{p}, \frac{1}{q}, \frac{1}{r}) \subset [0,1] ^3$, the condition of the above theorem says that the point (x, y, z) lies in the bounded region determined by the four planes given by the equations $x+y +z=1, x+y-z=1, x=z, y=z$, which is the convex hull of the points (1, 0, 0), (0, 1, 0), (1, 1, 1) and $\frac{1}{3} (1,1,1).$

We next show that the sufficiency holds in a subregion of the above region of necessity, with the extra condition $\frac{1}{p}+ \frac{1}{q}\ge 1$.

Theorem 3.3

Let $f \in L^p(\mathbb C^n), g \in L^q(\mathbb C^n), 1\le p,q\le \infty$. Then the inequality

$$\begin{aligned} \Vert f \times _\lambda g\Vert _{L^r(\mathbb {C}^n)} \le C_\lambda \, \Vert f \Vert _{L^p(\mathbb {C}^n)} \, \Vert g\Vert _{L^q(\mathbb {C}^n)} \end{aligned}$$

holds with $C_\lambda = (2\pi / |\lambda | )^{n( 1+ \frac{1}{r}- \frac{1}{p}- \frac{1}{q} ) }$ for $0 \ne \lambda \in \mathbb {R}$, for all $p,q,r \in [1, \infty ]$ satisfying $1 \le \frac{1}{p}+ \frac{1}{q} \le 1+ \frac{1}{r}$ and $r \ge \text {max}\{ p, q\}$.

Proof

We give a rather a simple proof inspired by the geometric description of the region of necessity given in Remark 3.2. In view of Lemma 2.4, we can assume $\lambda =1$, and we use the following basic inequalities

$$\begin{aligned} \Vert f \times g\Vert _{L^\infty (\mathbb {C}^n)}\le & {} \Vert f \Vert _{L^1(\mathbb {C}^n)} \Vert g\Vert _{L^\infty (\mathbb {C}^n)} \end{aligned}$$

(3.3)

$$\begin{aligned} \Vert f \times g\Vert _{L^\infty (\mathbb {C}^n)}\le & {} \Vert f \Vert _{L^\infty (\mathbb {C}^n)} \Vert g\Vert _{L^1(\mathbb {C}^n)} \end{aligned}$$

(3.4)

$$\begin{aligned} \Vert f \times g\Vert _{L^1(\mathbb {C}^n)}\le & {} \Vert f \Vert _{L^1(\mathbb {C}^n)} \Vert g\Vert _{L^1(\mathbb {C}^n)} \end{aligned}$$

(3.5)

$$\begin{aligned} \Vert f \times g\Vert _{L^2(\mathbb {C}^n)}\le & {} (2 \pi )^{n/2} \Vert f \Vert _{L^2(\mathbb {C}^n)} \Vert g\Vert _{L^2(\mathbb {C}^n)} \end{aligned}$$

(3.6)

The inequalities (3.3) and (3.4) are straightforward and (3.5) follows by Fubini’s theorem as in the case of usual convolution. The inequality (3.6) is special to the twisted convolution, and essentially follows from the Plancherel theorem for the Weyl transform, see [3, 19].

Multi-linear interpolation between (3.3) and (3.4), using Theorem  2.9 gives

$$\begin{aligned} \Vert f \times g\Vert _{L^\infty (\mathbb {C}^n)}\le & {} \Vert f \Vert _{L^{p_1}(\mathbb {C}^n)} \Vert g\Vert _{L^{q_1} (\mathbb {C}^n)} \end{aligned}$$

(3.7)

where $\frac{1}{p_1}=\frac{\theta _1}{1} +\frac{1-\theta _1}{\infty }= \theta _1$ and $\frac{1}{q_1}= 1-\theta _1 = 1- \frac{1}{p_1}$, so that $q_1=p_1'$. Now a further multi-linear interpolation of (3.7) and (3.5) yields

$$\begin{aligned} \Vert f \times g\Vert _{L^{r_2}(\mathbb {C}^n)}\le & {} \Vert f \Vert _{L^{p_2}(\mathbb {C}^n)} \Vert g\Vert _{L^{q_2} (\mathbb {C}^n)} \end{aligned}$$

(3.8)

where $\frac{1}{r_2}=\frac{\theta _2}{1} +\frac{1-\theta _2}{\infty }= \theta _2, ~ \frac{1}{p_2}=\frac{\theta _2}{1} +\frac{1-\theta _2}{p_1} =\theta _2 +\theta _1(1-\theta _2)$ and $\frac{1}{q_2}=\frac{\theta _2}{1} +\frac{1-\theta _2}{q_1} =\theta _2 +(1-\theta _1)(1-\theta _2)$, $0 \le \theta _1, \theta _2 \le 1$. Since $\frac{1}{p_2}+ \frac{1}{q_2} =1+ \theta _2 = 1+ \frac{1}{r_2},$ we arrive at the condition $\frac{1}{p_2}+ \frac{1}{q_2} =1+ \frac{1}{r_2}$, (the same condition for the classical generalised Young’s inequality for the usual convolution). Using multi-linear interpolation once again between (3.8) and the inequality (3.6), which is specific to the twisted convolution, we get

$$\begin{aligned} \Vert f \times g\Vert _{L^{r}(\mathbb {C}^n)}\le & {} (2 \pi )^{\frac{n\theta _3}{2}} \, \Vert f \Vert _{L^{p}(\mathbb {C}^n)} \, \Vert g\Vert _{L^{q} (\mathbb {C}^n)} \end{aligned}$$

(3.9)

where $\frac{1}{r}=\frac{\theta _3}{2} +\frac{1-\theta _3}{r_2}, ~ \frac{1}{p}=\frac{\theta _3}{2} +\frac{1-\theta _3}{p_2}, \frac{1}{q}=\frac{\theta _3}{2} +\frac{1-\theta _3}{q_2}, 0 \le \theta _3 \le 1$. Expressing $p_2,q_2,r_2$ in terms of the parameters $\theta _1, \theta _2$, we arrive at the conditions

$$\begin{aligned} \frac{1}{p}= & {} \frac{\theta _3}{2} +(1-\theta _3) [ \theta _2 + (1-\theta _2) \theta _1 ],\\ \frac{1}{q}= & {} \frac{\theta _3}{2} +(1-\theta _3) [ \theta _2 + (1-\theta _2) (1-\theta _1) ], \\ \frac{1}{r}= & {} \frac{\theta _3}{2} + (1-\theta _3) \theta _2 \end{aligned}$$

$0 \le \theta _1, \theta _2,\theta _3 \le 1$. Note that the above identities give the parametric representation of the points $(\frac{1}{p}, \frac{1}{q}, \frac{1}{r})$ of a sub-region of the cube $[0,1]^3$, as $\theta _1, \theta _2, \theta _3$ vary independently over [0, 1]. This region is precisely the convex hull of the points (1, 1, 1), (1, 0, 0), (0, 1, 0) and $\frac{1}{2} (1,1,1)$. Since $1+ \frac{1}{r} - \frac{1}{p} - \frac{1}{q} = \frac{\theta _3}{2}$, (3.9) gives the required inequality for $\lambda =1$. The estimate for $\lambda \ne 0$ follows from this by Lemma  2.4. $\square$

Remark 3.4

We expect that the constant $C_\lambda$ in Theorem 3.3 is optimal. In fact $C_\lambda = C_1 |\lambda |^{n(\frac{1}{p} + \frac{1}{q}- \frac{1}{r}-1 )}$ in view of Lemma 2.4, and $C_1$ was obtained by interpolation with the inequalities (3.3) to (3.6), which have optimal constants:

Proposition 3.5

The constants in the inequalities (3.3) to (3.6) are optimal.

Proof

We first note that the equality holds in (3.6), for $f = g = (2 \pi )^{-n/2} \varphi$ where $\varphi (z)= e^{-|z|^2/4}$. In fact, since $\varphi \times \varphi = (2 \pi )^ n \varphi$ (see [18], Cor. 2.3.4, p. 59), and $\Vert f\Vert _{L^2(\mathbb C^n)} =1= \Vert g \Vert _{L^2(\mathbb C^n)}$, we see that

$$\begin{aligned} \Vert f \times g\Vert _{L^2(\mathbb C^n)}=(2 \pi )^{- n} \Vert \varphi \times \varphi \Vert _{L^2(\mathbb C^n)}= \Vert \varphi \Vert _{L^2(\mathbb C^n)} =(2 \pi )^{n/2}. \end{aligned}$$

To see that 1 is the optimal constant in (3.3), we choose $f (z) = (2 \pi )^{-n} e^{-|z|^2/2}$ and $g\equiv 1$, so that $\Vert f\Vert _{L^1(\mathbb C^n)} =1= \Vert g \Vert _{L^\infty (\mathbb C^n)}$. Since $f \times g (z) = \widehat{f}(z/2) =e^{-|z|^2/8}$, we see that $\Vert f \times g \Vert _{L^\infty (\mathbb C^n)} =1$ as desired. The optimality of the constant 1 in (3.4) also follows similarly.

For the optimality in (3.5), we choose an even function $0 \le \varphi \in C_c^\infty (|z| \le 1)$ with $\int _{\mathbb C^n} \varphi (z) dz =1$, and set $\varphi _t(z) = t^{-2n } \varphi (z/t)$ for $t>0$. Taking $f =g= \varphi _t$, we see that $\Vert f\Vert _{L^1(\mathbb C^n)} =1= \Vert g \Vert _{L^1(\mathbb C^n)}.$ Now

$$\begin{aligned} \varphi _t \times \varphi _t (z) = t^{-4n} \int _{\mathbb C^n} \varphi \Big (\frac{z-w}{t}\Big ) \varphi \Big (\frac{w}{t}\Big ) e^{\frac{i}{2} \text {Im}(z \cdot \bar{w} ) } dw =A_t(z) + B_t(z) \end{aligned}$$

(3.10)

where

$$\begin{aligned} A_t(z)= & {} t^{-4n} \int _{\mathbb C^n} \varphi \Big (\frac{z-w}{t}\Big ) \varphi \Big (\frac{w}{t}\Big ) \, dw \end{aligned}$$

(3.11)

$$\begin{aligned} B_t(z)= & {} t^{-4n} \int _{\mathbb C^n} \varphi \Big (\frac{z-w}{t}\Big ) \varphi (\frac{w}{t}) \left( e^{\frac{i}{2} \text {Im}(z \cdot \bar{w} ) } -1 \right) dw \end{aligned}$$

(3.12)

Since $\varphi \ge 0$, we have

$$\begin{aligned} \Vert A_t \Vert _{L^1(\mathbb C^n)}= & {} t^{-4n} \int _{\mathbb C^n} \int _{\mathbb C^n} \varphi \Big (\frac{z-w}{t}\Big ) \varphi \Big (\frac{w}{t}\Big ) dz dw \nonumber \\= & {} \int _{\mathbb C^n} \int _{\mathbb C^n} \varphi (z-w) \varphi (w) dz dw = \Vert \varphi \Vert _{L^1(\mathbb C^n)} ^2 =1 \end{aligned}$$

(3.13)

for all $t>0$. Also by a change of variable, we have

$$\begin{aligned} B_t(z)= & {} t^{-2n} \int _{\mathbb C^n} \varphi \Big (\frac{z}{t} - w\Big ) \varphi (w) \left( e^{\frac{i}{2} \text {Im}(tz \cdot \bar{w} ) } -1 \right) dw. \end{aligned}$$

Hence

$$\begin{aligned} \Vert B_t\Vert _{L^1(\mathbb C^n)}\le & {} t^{-2n} \int _{\mathbb C^n} \int _{\mathbb C^n} \varphi \Big (\frac{z}{t} - w\Big ) \varphi (w) \left| e^{\frac{i}{2} \text {Im}(tz \cdot \bar{w} ) } -1 \right| dz \, dw \nonumber \\\le & {} \int _{\mathbb C^n} \varphi (w) \int _{\mathbb C^n} \varphi (z - w) \left| e^{\frac{i}{2} \text {Im}(t^2z \cdot \bar{w} ) } -1 \right| dz \, dw \nonumber \\= & {} \int _{\mathbb C^n} \varphi (w) \int _{\mathbb C^n} \varphi (z) \left| e^{\frac{i}{2} \text {Im}(t^2z \cdot \bar{w} ) } -1 \right| dz \, dw \rightarrow 0 \end{aligned}$$

(3.14)

as $t \rightarrow 0$, which can be seen by successive application of the dominated convergence theorem.

In view of (3.13) and (3.14), we conclude from (3.10) that $\Vert \varphi _t \times \varphi _t \Vert _{L^1(\mathbb C^n)} \rightarrow 1$ as $t\rightarrow 0$, asserting the optimality of the constant 1 in (3.4) as well. $\square$

Theorem 3.3 shows that the sufficiency holds in a subregion of the region of necessity given by Theorem 3.1, with the extra condition $\frac{1}{p}+ \frac{1}{q}\ge 1$, as pointed out before. It turns out that this condition is also necessary for the boundedness of $B_\lambda : L^p \times L^q \rightarrow L^r$. Since the above condition does not involve r, we believe that there is no counter example leading to this condition. The best lower bound for $\frac{1}{p}+ \frac{1}{q}$ that we could get by counterexample was $1-\frac{1}{r}$, as given in Theorem 3.1. However, by a simple symmetry argument given below, we are able to show that the condition $\frac{1}{p}+ \frac{1}{q} \ge 1$ is indeed necessary.

Theorems 3.1 and 3.3, together with the above observation yields the following proof of Theorem 1.1.

Proof of Theorem 1.1

By Theorem 3.1, the necessary condition on the triple (p, q, r) , for the inequality in Theorem 1.1 to hold is $1 - \frac{1}{r} \le \frac{1}{p}+ \frac{1}{q} \le 1+ \frac{1}{r}$ and $r \ge \text {max}\{ p,q\}$. Next we show that for $\frac{1}{p}+ \frac{1}{q} < 1$ the boundedness cannot hold. In fact, if (p, q, r) is admissible, then we have $(q, r', p')$ is also admissible by Lemma 2.6, and hence $q\le p'$ by Theorem 3.1. This is same as the condition $\frac{1}{p}+ \frac{1}{q} \ge 1$. Thus it follows that the condition $0 \le \frac{1}{p}+ \frac{1}{q}-1 \le \frac{1}{r} \le \text {min}\{ \frac{1}{p}, \frac{1}{q}\}$ is necessary.

Theorem 3.3 shows that the above condition is also sufficient, and we have $C_\lambda = (2\pi / |\lambda | )^{n( 1+ \frac{1}{r}- \frac{1}{p}- \frac{1}{q} ) }$ in view of Lemma 2.4. Thus we conclude that the condition $0 \le \frac{1}{p}+ \frac{1}{q}-1 \le \frac{1}{r} \le \text {min}\{ \frac{1}{p},\frac{1}{q}\}$ characterises the triples (p, q, r) for the boundedness of $B_\lambda : (f,g) \rightarrow f \times _\lambda g$ from $L^p(\mathbb {C}^n) \times L^q(\mathbb {C}^n) \rightarrow L^r(\mathbb C^n)$ for $1 \le p,q,r \le \infty$. $\square$

As an immediate corollary of Theorem 1.1 we get the following $L^p \rightarrow L^q$ boundedness of the twisted convolution operator with an $L^q$ kernel.

Theorem 3.6

Let $K \in L^{p_0}(\mathbb C^n)$. Then the operator $T_\lambda : f \rightarrow K \times _\lambda f$ is bounded from $L^p(\mathbb C^n) \rightarrow L^q(\mathbb C^n)$ for all pairs (p, q) such that $0 \le \frac{1}{p_0}+ \frac{1}{p}-1 \le \frac{1}{q} \le \text {min}\{ \frac{1}{p}, \frac{1}{p_0}\}$. In particular $T_\lambda : L^p(\mathbb C^n) \rightarrow L^p(\mathbb C^n)$ is bounded, for $p_0 \le p \le p_0'$.