1 Introduction

The theory of weak turbulence is a physical theory which describes the transfer of energy between different wavelengths in a large class of wave systems. This theory can be applied to homogeneous problems that can be approximated to leading order by a system of linear waves that interact by means of weak nonlinearities. The basic mathematical model in the theory of weak turbulence is a kinetic equation that describes the transfer of energy between different wavelengths. Contrary to the starting wave equations, the kinetic equations arising in weak turbulence theory exhibit irreversible behaviour. Examples of applications of the theory of weak turbulence to specific physical systems can be found in [3, 4, 8, 9, 13, 17, 22] and [23].

One of the most extensively studied systems in the setting of weak turbulence theory is the one associated to the nonlinear Schrödinger equation

$$\begin{aligned} iu_t=-\Delta u+\varepsilon |u|^2u, \end{aligned}$$
(1.1)

with \(\varepsilon >0\) small (cf. [4, 13, 24]). Denoting \(F(t,k)=|\hat{u}(t,k)|^2\), where \(\hat{u}\) is the space Fourier transform of the solution of (1.1), then restricting to isotropic solutions one obtains up to rescaling the following equation for \(f(t,\omega ):=F(t,k)\) with \(\omega :=|k|^2\):

$$\begin{aligned} \partial _t f_1=\frac{1}{2}\iint _{[0,\infty )^2}W\big [(f_1+f_2)f_3f_4-(f_3+f_4)f_1f_2\big ]\mathrm {d}\omega _3\mathrm {d}\omega _4, \end{aligned}$$
(1.2)

where \(f_i=f(t,\omega _i)\) for each \(i\in \{1,2,3,4\}\), where \(\omega _2=(\omega _3+\omega _4-\omega _1)_+\), and where \(W=\min _i\{\sqrt{\omega _i}\}/\sqrt{\omega _1}\). The mathematical theory of this equation has been studied in detail in [6] where several properties of the solutions of (1.2) were obtained. As it is more convenient to study the evolution of the mass density function \(g(t,\omega )=\sqrt{\omega }f(t,\omega )\), we reformulate (1.2) as

$$\begin{aligned} \begin{aligned} \partial _t g_1=\frac{1}{2}\iint _{[0,\infty )^2}\tilde{W}\left[ \left( \frac{g_1}{\sqrt{\omega _1}}+\frac{g_2}{\sqrt{\omega _2}}\right) \frac{g_3g_4}{\sqrt{\omega _3\omega _4}}-\left( \frac{g_3}{\sqrt{\omega _3}}+\frac{g_4}{\sqrt{\omega _4}}\right) \frac{g_1g_2}{\sqrt{\omega _1\omega _2}}\right] \mathrm {d}\omega _3\mathrm {d}\omega _4, \end{aligned}\nonumber \\ \end{aligned}$$
(1.3)

where now \(g_i=g(t,\omega _i)\) for each \(i\in \{1,2,3,4\}\), where \(\omega _2=(\omega _3+\omega _4-\omega _1)_+\) and where \(\tilde{W}=\min _i\{\sqrt{\omega _i}\}\). Formally this equation has two conserved quantities, namely the mass, which is the integral of g, and the energy, which is the first moment of g.

For a class of weak solutions to (1.3) with finite mass, it was proved in [6] that all solutions converge, as \(t\rightarrow \infty \), to a Dirac mass supported at a well-defined point \(a\ge 0\), which depends only on the support of the initial distribution. It turns out that unless the initial distribution is contained in a periodic lattice, there holds \(a=0\). In this last case, it is possible to formally derive an equation that describes the behaviour of the fraction of mass that is not supported near the origin, which we denote by G. Formally this equation reads as

$$\begin{aligned}&\partial _tG(\omega )=\frac{1}{2}\int _0^\omega \frac{G(\omega -\xi )G(\xi )\mathrm {d}\xi }{\sqrt{(\omega -\xi )\xi }}-\frac{G(\omega )}{\sqrt{\omega }}\int _0^\infty \frac{G(\xi )\mathrm {d}\xi }{\sqrt{\xi }}\nonumber \\&\quad -\frac{1}{2}\frac{G(\omega )}{\sqrt{\omega }}\int _0^\omega \left[ \frac{G(\omega -\xi )}{\sqrt{\omega -\xi }} +\frac{G(\xi )}{\sqrt{\xi }}\right] \mathrm {d}\xi +\int _0^\infty \frac{G(\omega +\xi )}{\sqrt{\omega +\xi }}\left[ \frac{G(\omega )}{\sqrt{\omega }}+\frac{G(\xi )}{\sqrt{\xi }}\right] \mathrm {d}\xi . \end{aligned}$$
(1.4)

in which one may recognize a coagulation-fragmentation equation with nonlinear fragmentation. Note that many terms in (1.4) are singular and the meaning of this equation has to be precised. A more elaborate discussion on the sense in which G describes the asymptotic behaviour of g can be found in [6] and [10]. It is known that solutions to (1.3) can contain Dirac masses at the origin. If that is the case, then (1.4) can be obtained by only considering those collisions which are mediated by an interaction with the condensate.

Notice that if we assume that \(g=M\delta _0+G\), then the energy of g is contained in G. Therefore the analysis of the long time behaviour of G is relevant, even though the mass of G becomes negligible compared to the mass supported at the origin as \(t\rightarrow \infty \). As conjectured in [6], we expect a self-similar distribution of the energy among the different wavelengths as \(t\rightarrow \infty \), provided that g has initially finite energy. In [10] we have proved the existence of a family of self-similar solutions G to (1.4) with finite energy. These solutions are the natural candidates for describing the long time asymptotics of solutions g to (1.3) with finite mass and energy. Note however, that stability of these self-similar solutions is an open problem, even at the linearised level.

Of course, the assumption of finite energy is not really needed to have self-similar behaviour for the solutions g of (1.3). Long time self-similarity can be expected if the initial data has a power law tail as \(\omega \rightarrow \infty \). This gives a natural scaling law relating the energy density and \(\omega \). Actually, long time self-similarity can only be expected if either the initial distribution has a power law tail, or if the energy is finite. This is because otherwise the behaviour of the solutions is not stable for large values of \(\omega \) under the evolution equations (1.3) and (1.4). This is reminiscent of the situation for the coagulation equation with constant kernel, where in order to have self-similarity the power law behaviour for the initial data is needed (cf. [12]).

Let us briefly discuss the expected self-similar behaviour of a solution g to (1.3) if we assume the initial distribution to behave like \(\omega ^{-\rho }\) for large values of \(\omega \), where \(\rho >0\). Given that for \(\rho \le \frac{1}{2}\) it is not clear whether the collision terms in (1.3) can be given a meaning, we restrict ourselves to the case \(\rho >\frac{1}{2}\). A particularly relevant exponent is \(\rho =\frac{2}{3}\), which is the so called Kolmogorov–Zakharov exponent for (1.3). The interpretation of this exponent is the existence of a constant flux of particles from large values of \(\omega \) to smaller ones in the space of frequencies (cf. [22]).

If we suppose that \(\rho >1\), then g has finite mass and the heuristic derivation of (1.4) is valid (cf. [6, 10]). As discussed before, we thus expect the self-similar solutions G of (1.4) to describe the long time asymptotics of solutions g to (1.3). One of the main results of this paper will be the proof of existence of self-similar solutions to (1.4) with tail behaviour \(\omega ^{-\rho }\) for \(1<\rho <2\). If \(\rho >2\), then the solutions have finite energy. Existence of self-similar solutions with finite energy has been proved in [10]. In this paper we prove that these solutions decay exponentially as \(\omega \rightarrow \infty \).

The case \(\frac{1}{2}<\rho <1\) is different, since the mass of g is infinite. We therefore expect the amount of mass located at the origin to grow without limit. Dimensional analysis suggests that the long time asymptotics of solutions to (1.3) then cannot be approximated by solutions to a simpler quadratic equation, similar to (1.4), where all the collisions are mediated by interaction with one particle placed at the origin. More precisely, if we suppose that \(g=M\delta _0+G\), where \(M=M(t)\) is the amount of mass located at the origin, then there are terms that are cubic in G that cannot be ignored, and we expect self similar solutions to be of the form

$$\begin{aligned} G(t,\omega )=\frac{1}{t^{\frac{\rho }{2\rho -1}}}\Phi \left( \frac{\omega }{t^{\frac{1}{2\rho -1}}}\right) . \end{aligned}$$

We further note that dimensional analysis alone is insufficient to determine the exact scaling law for M. However, it suggests that \(M\sim t^\alpha \) for \(\alpha \le \frac{1-\rho }{2\rho -1}\).

Seemingly the first paper to consider the asymptotics of solutions of (1.3) in connection with solutions to the nonlinear Schrödinger equation is [18]. In particular, that paper describes the scaling properties of solutions g to (1.3) in the cases where either the energy is finite, or where g behaves for large frequencies according to the Kolmogorov–Zakharov power law. These two cases correspond to assuming that the spectral distribution \(F(t,k)=|\hat{u}(t,k)|^2\) has either finite energy or decays according to the Kolmogorov–Zakharov exponent for large |k|. However, in the case of infinite energy there is no particular reason for the exponent of the power law to be restricted to this one. Hence, it makes sense to study solutions g to (1.3) where g initially has arbitrary power law behaviour at infinity, at least from a mathematical point of view.

This paper is a continuation of the study of self-similar solutions to (1.4), which was initiated in [10]. We refer to that paper for a more extensive discussion of the connection of (1.2) to particle models, as well as other equations in mathematical physics such as the Boltzmann–Nordheim equation.

The structure of the paper is as follows. In Sect. 2 we introduce our notation, and we give the statements of the main results. Section 3 contains the proof of existence of self-similar profiles, while in Sect. 4 regularity is proven. In Sect. 5 we then demonstrate unique power law behaviour of the self-similar profiles in the case of infinite energy. Lastly, in the case of finite energy we prove a pointwise exponential upper bound and an exponential lower bound in averaged sense in Sect. 6.

2 Notation and Results

We start with some definitions and notations that we use throughout the paper.

Definition 2.1

We write \(\mathcal {M}([0,\infty ))\), \(\mathcal {M}_+([0,\infty ))\), and \(\mathcal {M}_+([0,\infty ])\) for the spaces of signed, nonnegative, and finite nonnegative Radon measures respectively.

Remark 2.2

Note that the notation for measure spaces as introduced in Definition 2.1 differs from the one in [10]. In that paper \(\mathcal {M}_+([0,\infty ))\) was used to denote the space of finite nonnegative Radon measures \(\mu \) on \([0,\infty ]\) for which \(\mu (\{\infty \})=0\).

Remark 2.3

For an integral with respect to a measure \(\mu \) we will always use the notation \(\mu (x)\mathrm {d}x\), even if \(\mu \) is not absolutely continuous with respect to Lebesgue measure.

Definition 2.4

Given an interval \(I\subset [-\infty ,\infty ]\), we write \(C(I)=C^0(I)\) for the set of functions that are continuous on I. Given further \(k\in \mathbb {N}\), we write \(C^k(I)\) for the subset of these functions in C(I) whose derivatives of order up to k exist and are in C(I), and \(C_c^k(I)\), or \(C_c(I)\), for the set of functions in \(C^k(I)\), or C(I), supported in a compact \(K\subset I\). Given finally \(k\in \mathbb {N}_0=\{0,1,\ldots \}\) and \(\alpha \in (0,1)\), we write \(C^{k,\alpha }(I)\) for the set of functions in \(C^k(I)\) for which the k-th derivative is \(\alpha \)-Hölder continuous on any compact \(K\subset I\).

Remark 2.5

Given \(\varphi \in C(I)\), we write \(\Vert \varphi \Vert _\infty =\Vert \varphi \Vert _{C(I)}\).

Remark 2.6

Note that if \(f\in C^k([0,\infty ])\), then not only are the functions \(f^{(\ell )}\), with \(\ell =0,1,\ldots ,k\), bounded on the interval \([0,\infty ]\), but also \(\lim _{x\rightarrow \infty }f(x)\) exists, and \(\lim _{x\rightarrow \infty }f^{(\ell )}(x)=0\) for \(\ell =1,\ldots ,k\).

Definition 2.7

Given \(k\in \mathbb {N}_0\), we denote by \(\mathcal {B}_k\) the subspace of those functions \(\vartheta \in C^k([0,\infty ])\) for which

$$\begin{aligned} \Vert (1+x)\vartheta (x)\Vert _{C^k([0,\infty ])}=:\Vert \vartheta \Vert _{\mathcal {B}_k}<\infty . \end{aligned}$$

Remark 2.8

Since \(\mathcal {B}_0\) is a separable Banach space, the unit ball in the dual space \(\mathcal {B}_0'\) endowed with the weak-\(*\) topology is metrizable (cf. [1, Theorem 3.28]). Consequently, the properties of the weak-\(*\) topology restricted to the unit ball in \(\mathcal {B}_0'\) can be characterized by means convergence of sequences. We recall that a sequence \(\{\mu _n\}\) in \(\mathcal {B}_0'\) converges to \(\mu \in \mathcal {B}_0'\) with respect to the weak-\(*\) topology (denoted \(\mu _n{{\mathrm{\rightharpoonup {^*}}}}\mu \) in \(\mathcal {B}_0'\)) if and only if

$$\begin{aligned} \int _{[0,\infty )}\vartheta (x)\mu _n(x)\mathrm {d}x\rightarrow \int _{[0,\infty )}\vartheta (x)\mu (x)\mathrm {d}x\text { for all }\vartheta \in \mathcal {B}_0. \end{aligned}$$

Remark 2.9

We use the notations \(a\vee b=\max \{a,b\}\) and \(a\wedge b=\min \{a,b\}\).

A robust characterization of the power law behaviour of measures \(\mu \) near infinity will be achieved by means of the functionals

$$\begin{aligned} R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) |\mu (x)|\mathrm {d}x. \end{aligned}$$
(2.1)

More precisely, we will make extensive use of the following normed spaces.

Definition 2.10

Given \(\rho \in (1,2]\), we define \(\mathcal {X}_\rho \) to be the subset of those nonnegative Radon measures \(\mu \in \mathcal {M}_+([0,\infty ))\) for which

$$\begin{aligned} \sup _{R>0}\left\{ R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) |\mu (x)|\mathrm {d}x\right\} =:\Vert \mu \Vert _\rho <\infty . \end{aligned}$$
(2.2)

Remark 2.11

Even though the space \(\mathcal {X}_\rho \) only contains nonnegative Radon measures, the norm \(\Vert \cdot \Vert _\rho \) is defined for arbitrary signed Radon measures by (2.2).

Also, since \(\Vert \mu \Vert _2=\int _{[0,\infty )}\mu (x)\mathrm {d}x\) we can identify any \(\mu \in \mathcal {X}_2\) with a unique element in \(\mathcal {M}_+([0,\infty ])\cap \{\mu (\{\infty \})=0\}\), and we will henceforth use the abbreviated notation \(\mathcal {X}_2=\mathcal {M}_+([0,\infty ])\cap \{\mu (\{\infty \})=0\}\).

Note lastly that if \(\rho \in (1,2)\), then for all \(\mu \in \mathcal {X}_\rho \) there holds \(\mu (\{0\})=0\), since \(0\le \mu (\{0\})\le R^{\rho -2}\int _{[0,\infty ]}(1\wedge \frac{R}{x})\mu (x)\mathrm {d}x\times R^{2-\rho }\le \Vert \mu \Vert _\rho R^{2-\rho }\) for all \(R>0\).

Lemma 2.12

Given \(\rho \in (1,2)\), there holds \(\mathcal {X}_\rho \subset \mathcal {B}_0'\), and \(\{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \) is weakly-\(*\) closed in \(\mathcal {B}_0'\).

Proof

Since for \(\vartheta \in \mathcal {B}_0\) with \(\Vert \vartheta \Vert _{\mathcal {B}_0}=1\) there holds \(|\vartheta (x)|\le \frac{1}{1+x}\le 1\wedge \frac{1}{x}\) for \(x>0\), we find for \(\mu \in \mathcal {X}_\rho \), which are nonnegative, that

$$\begin{aligned} \Vert \mu \Vert _{\mathcal {B}_0'}\overset{\mathrm{def}}{=}\sup _{\Vert \vartheta \Vert _{\mathcal {B}_0}=1}\int _{[0,\infty )}\vartheta (x)\mu (x)\mathrm {d}x \le \int _{[0,\infty )}\left( 1\wedge \tfrac{1}{x}\right) \mu (x)\mathrm {d}x\le \Vert \mu \Vert _\rho , \end{aligned}$$
(2.3)

which proves the inclusion. Given further a sequence \(\{\mu _n\}\) in \(\{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \) such that \(\mu _n{{\mathrm{\rightharpoonup {^*}}}}\mu \) in \(\mathcal {B}_0'\), then clearly \(\mu \ge 0\). Furthermore, for all \(R>0\) there holds by definition of weak-\(*\) convergence that

$$\begin{aligned} \zeta _n(R)=R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) \mu _n(x)\mathrm {d}x\rightarrow R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) \mu (x)\mathrm {d}x, \end{aligned}$$

and since for all \(R>0\) the sequence \(\{\zeta _n(R)\}\) is bounded by one, we conclude that \(\Vert \mu \Vert _\rho \le 1\). \(\square \)

Definition 2.13

By the weak-\(*\) topology on \(\mathcal {X}_\rho \) we mean the restriction of the weak-\(*\) topology of \(\mathcal {B}_0'\) to \(\mathcal {X}_\rho \).

Lemma 2.14

Given \(\rho \in (1,2)\), the unit ball in \(\mathcal {X}_\rho \) is weakly-\(*\) compact.

Proof

Using (2.3), there holds \(\{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \subset \{\Vert \mu \Vert _{\mathcal {B}_0'}\le 1\}\), so by weak-\(*\) closedness it suffices to check that \(\{\Vert \mu \Vert _{\mathcal {B}_0'}\le 1\}\) is weakly-\(*\) compact, which follows by Banach–Alaoglu (cf. [1, Theorem 3.16]). \(\square \)

Remark 2.15

Notice that for any function \(\varphi \in C([0,\infty ])\) there exists a unique \(\vartheta \in \mathcal {B}_0\) such that \(\varphi (x)=(1+x)\vartheta (x)\), and vice versa. Therefore

$$\begin{aligned} \Vert \mu \Vert _{\mathcal {B}_0'}= & {} \sup _{\Vert \vartheta \Vert _{\mathcal {B}_0}=1}\int _{[0,\infty )}\vartheta (x)\mu (x)\mathrm {d}x\nonumber \\= & {} \sup _{\Vert \varphi \Vert _{C([0,\infty ])}=1}\int _{[0,\infty )}\varphi (x)\tfrac{\mu (x)}{1+x}\mathrm {d}x=\left\| \tfrac{\mu (x)}{1+x}\right\| _{(C([0,\infty ]))'}, \end{aligned}$$

and \(\mathcal {B}_0'\) and \(\mathcal {M}([0,\infty ];\frac{\mathrm {d}x}{1+x})\) are isomorphic.

Definition 2.16

For \(\rho \in (1,2)\), we define for any \(R_0>0\) the subset \(\mathcal {Y}_\rho =\mathcal {Y}_\rho (R_0)\) to contain those elements \(\mu \in \mathcal {X}_\rho \) that satisfy both \(\Vert \mu \Vert _\rho \le 1\) and

$$\begin{aligned} \int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) \mu (x)\mathrm {d}x\ge R^{2-\rho }\lambda _\rho \big (\tfrac{R}{R_0}\big )\text { for all }R>0, \end{aligned}$$
(2.4)

with \(\lambda _\rho (x)=(1-|x|^{-(2-\rho )/2})_+\).

Remark 2.17

For any \(R_0>0\), the set \(\mathcal {Y}_\rho (R_0)\) is a nonempty (\((2-\rho )(\rho -1)x^{1-\rho } \mathrm {d}x\in \mathcal {Y}_\rho \)), convex and weakly-\(*\) compact subset of the unit sphere \(\{\Vert \mu \Vert _\rho =1\}\).

We now state the notion of weak solution to (1.4), which is analogous to the one that was introduced in [10].

Definition 2.18

A function \(G\in C([0,\infty ):\mathcal {X}_2)\) that for all \(t\in [0,\infty )\) and all \(\varphi \in C^1([0,\infty ):C_c^1([0,\infty )))\) satisfies

$$\begin{aligned}&\int _{[0,\infty )}\varphi (t,x)G(t,x)\mathrm {d}x-\int _{[0,\infty )}\varphi (0,x)G(0,x)\mathrm {d}x\nonumber \\&\quad =\int _0^t\left[ \int _{[0,\infty )}\varphi _s(x)G(x)\mathrm {d}x+\frac{1}{2}\iint _{[0,\infty )^2}\frac{G(x)G(y)}{\sqrt{xy}}\mathcal {D}_2[\varphi ](x,y)\mathrm {d}x\mathrm {d}y\right] \mathrm {d}s,\quad \end{aligned}$$
(2.5)

where \(\mathcal {D}_2\) for \(\varphi \in C([0,\infty ))\) is defined by

$$\begin{aligned}\nonumber \mathcal {D}_2[\varphi ](x,y)=\varphi (x+y)+\varphi (|x-y|)-2\varphi (x\vee y), \end{aligned}$$

will be called a weak solution to (1.4).

Remark 2.19

The use of the space \(\mathcal {X}_2=\mathcal {M}_+([0,\infty ])\cap \{\mu (\{\infty \})=0\}\) might seem artificial. We only impose the restriction to \(\{\mu (\{\infty \})=0\}\) to avoid trivial nonuniqueness due to the fact that (2.5) does not give any information about \(G(\cdot ,\{\infty \})\), which could be an arbitrary function since we are using test functions that are compactly supported in \([0,\infty )\).

Remark 2.20

We frequently use the following notation for the second difference:

$$\begin{aligned} \Delta _y^2f(x)=f(x+y)+f(x-y)-2f(x) \end{aligned}$$

Also, for notational convenience we introduce

$$\begin{aligned} \mathcal {D}_2^*[\vartheta ](x,y)= & {} \mathcal {D}_2[\varphi ](x,y)\text { with }\varphi (z)=z\vartheta (z)\nonumber \\= & {} (x+y)\vartheta (x+y)+|x-y|\vartheta (|x-y|)-2(x\vee y)\vartheta (x\vee y). \end{aligned}$$
(2.6)

Lemma 2.21

For \(f\in C^2(\mathbb {R})\) and \(x\in \mathbb {R}\) there hold

$$\begin{aligned} \Delta _y^2f(x)&=\int _\mathbb {R}(|y|-|w-x|)_+f''(w)\mathrm {d}w&\text {for }y&\in \mathbb {R},\end{aligned}$$
(2.7)
$$\begin{aligned} \partial _y\left[ \Delta _y^2f(x)\right]&=\int _{x-y}^{x+y}f''(w)\mathrm {d}w&\text {for }y&\ge 0. \end{aligned}$$
(2.8)

Proof

By the fundamental theorem of calculus we observe that

$$\begin{aligned} \Delta _y^2f(x)= & {} \int _x^{x+|y|}f'(z)\mathrm {d}z-\int _{x-|y|}^xf'(z)\mathrm {d}z\\= & {} \int _x^{x+|y|}\int _x^zf''(w)\mathrm {d}w\mathrm {d}z+\int _{x-|y|}^x\int _z^xf''(w)\mathrm {d}w\mathrm {d}z. \end{aligned}$$

Applying Fubini to the right hand side and rearranging terms, we find (2.7). The proof of (2.8) is similar. \(\square \)

Remark 2.22

For any \(f,g\in C([0,\infty ))\), we write \(f(x)\sim g(x)\) as \(x\rightarrow \infty \) if there holds \(\lim _{x\rightarrow \infty }\frac{f(x)}{g(x)}=1\).

2.1 Statement of Main Results

In this section we state the main results of this paper. The first result gives a sufficient condition for existence of a self-similar solution.

Proposition 2.23

Given \(\rho \in (1,2]\), if \(\Phi _\rho \in L^1(0,\infty )\) is a nonnegative function that for all \(\varphi \in C_c^1([0,\infty ))\) satisfies

$$\begin{aligned} \frac{1}{\rho }\int _{(0,\infty )}\big (x\varphi '(x)-(\rho -1)(\varphi (x)- & {} \varphi (0))\big )\Phi _\rho (x)\mathrm {d}x\nonumber \\&=\iint _{\{x>y>0\}}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\Delta _y^2\varphi (x)\mathrm {d}x\mathrm {d}y, \end{aligned}$$
(2.9)

then the function \(G\in C([0,\infty ):\mathcal {X}_2)\) that is given by

$$\begin{aligned} G(t,x)\mathrm {d}x=\left( M-\frac{\Vert \Phi _\rho \Vert _{L^1(0,\infty )}}{(t+t_0)^{(\rho -1)/\rho }}\right) \delta _0(x)\mathrm {d}x+\Phi _\rho \left( \frac{x}{(t+t_0)^{1/\rho }}\right) \frac{\mathrm {d}x}{t+t_0}, \end{aligned}$$

with \(t_0\in (0,\infty )\) and \(M\in [0,\infty )\) such that \(Mt_0^{(\rho -1)/\rho }\ge \Vert \Phi _\rho \Vert _{L^1(0,\infty )}\), is a weak solution to (1.4) in the sense of Definition 2.18.

Proof

Making the necessary changes, this is identical to the proof of [10, Proposition 4.1], in which the case \(\rho =2\) was considered. \(\square \)

The rest of this paper is devoted to the proof of the following results.

Theorem 2.24

(Existence) Given \(\rho \in (1,2]\), there exists at least one \(\Phi _\rho \in \mathcal {X}_2\) that for all \(\varphi \in C_c^1([0,\infty ))\) satisfies

(2.10)

Proposition 2.25

(Regularity) Given \(\rho \in (1,2]\), if \(\Phi _\rho \in \mathcal {X}_2\) satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), then \(\Phi _\rho \) is absolutely continuous with respect to Lebesgue measure, and its Radon–Nykodim derivative is smooth on \((0,\infty )\) and satisfies

$$\begin{aligned} -\tfrac{1}{\rho }x\Phi _\rho '(x)-\Phi _\rho (x)= & {} \int _0^{x/2}\frac{\Phi _\rho (y)}{\sqrt{y}}\left[ \frac{\Phi _\rho (x+y)}{\sqrt{x+y}}+\frac{\Phi _\rho (x-y)}{\sqrt{x-y}}-2\frac{\Phi _\rho (x)}{\sqrt{x}}\right] \mathrm {d}y\nonumber \\&+\,\int _{x/2}^\infty \frac{\Phi _\rho (y)\Phi _\rho (x+y)}{\sqrt{y(x+y)}}\mathrm {d}y-2\frac{\Phi _\rho (x)}{\sqrt{x}}\int _{x/2}^x\frac{\Phi _\rho (y)}{\sqrt{y}}\mathrm {d}y. \end{aligned}$$
(2.11)

Actually, \(\Phi _\rho \) thus satisfies (2.9) for all \(\varphi \in C_c^1([0,\infty ))\), and furthermore, \(\Phi _\rho \) is either strictly positive or identically zero on \((0,\infty )\).

Proposition 2.26

Given \(\rho \in (1,2]\), if \(\Phi _\rho \in \mathcal {X}_2\) satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), then \(x\Phi _\rho (x)\in \mathcal {X}_\rho \). Furthermore, for any \(c>0\) the rescaled measure \(\Phi _*(x)\mathrm {d}x=\Phi _\rho (cx)\mathrm {d}x\) also satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), and there holds \(\Vert x\Phi _*(x)\Vert _\rho =c^{-\rho }\Vert x\Phi _\rho (x)\Vert _\rho \).

Remark 2.27

The statements in Theorem 2.24 and Proposition 2.26 have already been proven for the case \(\rho =2\) in [10, Section 4]. In the proofs in this paper we will thus restrict ourselves to the case \(\rho \in (1,2)\).

Theorem 2.28

(Power law asymptotics) Given \(\rho \in (1,2)\), if \(\Phi _\rho \in \mathcal {X}_2\) satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), and if furthermore \(\Vert x\Phi _\rho (x)\Vert _\rho =1\), then

$$\begin{aligned} \Phi _\rho (r)\sim (2-\rho )(\rho -1)r^{-\rho }\text { as }r\rightarrow \infty . \end{aligned}$$

Theorem 2.29

(Exponential bounds) If \(\Phi _2\in \mathcal {X}_2\) satisfies (2.10) with \(\rho =2\) for all \(\varphi \in C_c^1([0,\infty ))\), and if \(\Phi _2\) is not identically zero on \((0,\infty )\), then there exist constants \(a,c\in (0,1)\) such that

$$\begin{aligned} \Phi _2(r)\le \frac{e^{-ar}}{c}\text { for all }r\ge 1\text {, and }\int _{(R,R+1)}\Phi _2(x)\mathrm {d}x\ge ce^{-\frac{R}{a}}\text { for all }R\ge 0. \end{aligned}$$

3 Existence of Self-Similar Profiles

The proof of Theorem 2.24 for the case \(\rho =2\) was already given in [10], and the obtained profiles \(\Phi _2\) turned out to have finite energy. Due to this finiteness of the energy the existence result for self-similar solutions to (1.4) in [10] can be seen as the analogue to the existence result for self-similar solutions with finite mass to the coagulation equation obtained in [5] and [7].

For the coagulation equation, self-similar solutions with infinite mass, i.e. with fat tailed behaviour at infinity, have been obtained in [14] for locally bounded kernels, and in [16] for a class of singular kernels, which in particular includes the classical Smoluchowski kernel for Brownian coagulation.

In this paper we construct self-similar solutions with fat tailed behaviour at infinity to (1.4), adapting the methods of [14]. The main idea in the construction made in that paper, is that for fat tailed solutions the linear terms in the equation for the self-similar profile are dominant for large values of x. The effect of the nonlinear collision kernels can be seen as a nonlocal diffusive effect for large particles, which gives a lower order correction. Due to the fact that in coagulation equations the size of the particles is always increasing, the resulting diffusive effect is directed towards larger values. Conversely, in our case the collision kernel can transport particles to both larger or smaller values, and the resulting nonlocal diffusive effect is no longer directed. However, the operator describing this diffusive effect is more symmetric than in the case of coagulation. This has two main consequences. Firstly, the natural test functions required to study the transport of particles are those that are either convex or concave, while in the case of the coagulation equation the natural test functions were the monotone ones. Secondly, due to the symmetry of our collision kernel the singular terms in (1.4) have a weaker effect, and many of the technicalities that had to be introduced in [16] can be avoided.

We would like to mention that large parts of our construction below also work in the case \(\rho =2\). However, technicalities aside, it is not a priori clear that this construction yields a nontrivial solution where not all the mass is concentrated in the origin.

We now restrict ourselves to \(\rho \in (1,2)\). Introducing as in [10] the notations \(\Psi _\rho (x)=x\Phi _\rho (x)\) and \(\vartheta (x)=\frac{1}{x}(\varphi (x)-\varphi (0))\), we can rewrite (2.10) as

$$\begin{aligned} \frac{1}{\rho }\int _{[0,\infty )}\left( x\vartheta '(x)\!+\!(2\!-\!\rho ) \vartheta (x)\right) \Psi _\rho (x)\mathrm {d}x \!=\!\frac{1}{2}\iint _{[0,\infty )^2}\frac{\Psi _\rho (x)\Psi _\rho (y)}{(xy)^{3/2}}\mathcal {D}_2^*[\vartheta ](x,y)\mathrm {d}x\mathrm {d}y,\nonumber \\ \end{aligned}$$
(3.1)

where we recall the notation (2.6). Now, we would like to prove existence of a solution to (3.1) by showing that there exists a stationary solution to

$$\begin{aligned}&\int _{[0,\infty )}\vartheta (t,x)\Psi _\rho (t,x)\mathrm {d}x-\int _{[0,\infty )}\vartheta (0,x)\Psi _\rho (0,x)\mathrm {d}x\nonumber \\&\qquad =\int _0^t\Bigg [\int _{[0,\infty )}\left( \vartheta _s(s,x)-\tfrac{1}{\rho }\left( x\vartheta _x(s,x)+(2-\rho )\vartheta (s,x)\right) \right) \Psi _\rho (s,x)\mathrm {d}x\nonumber \\&\qquad \quad +\,\frac{1}{2}\iint _{[0,\infty )^2}\frac{\Psi _\rho (s,x)\Psi _\rho (s,y)}{(xy)^{3/2}}\mathcal {D}_2^*[\vartheta (s,\cdot )](x,y)\mathrm {d}x\mathrm {d}y\Bigg ]\mathrm {d}s. \end{aligned}$$
(3.2)

In order to avoid technical difficulties due to the singularity of the kernel, we will consider a regularized version of (3.2). We then prove existence of stationary solutions to that equation by a Schauder type fixed point theorem, and finally show by a compactness result that by removing the regularization we obtain a solution to (3.1).

Assumption 3.1

Let \(\rho \in (1,2)\) and \(\varepsilon >0\) be fixed arbitrarily, let \(a\in (0,\frac{\varepsilon }{2})\) be arbitrary, and let \(\phi \in C_c^\infty ((-1,1))\) be a fixed even function such that \(\phi \ge 0\) and \(\Vert \phi \Vert _{L^1(\mathbb {R})}=1\). For any \(b>0\) we define \(\phi _{b}(x)=\frac{1}{b}\phi (\frac{x}{b})\) for all \(x\in \mathbb {R}\).

Proposition 3.2

Under Assumption 3.1, there exist \(R_0>0\) and a weakly-\(*\) continuous semigroup \((S_a(t))_{t\ge 0}\) on \(\mathcal {Y}_\rho =\mathcal {Y}_\rho (R_0)\) such that if given \(\Psi _0\in \mathcal {Y}_\rho \), then \(\Psi _a(t,\cdot )= S_a(t)\Psi _0\in \mathcal {Y}_\rho \) satisfies

$$\begin{aligned}&\int _{[0,\infty )}\vartheta (t,x)\Psi _a(t,x)\mathrm {d}x-\int _{[0,\infty )}\vartheta (0,x)\Psi _0(x)\mathrm {d}x\nonumber \\&\qquad =\int _0^t\left[ \int _{[0,\infty )}\left( \vartheta _s(s,x)-\tfrac{1}{\rho }\left( x\vartheta _x(s,x)+(2-\rho )\vartheta (s,x)\right) \right) \Psi _a(s,x)\mathrm {d}x\right. \nonumber \\&\left. \qquad \quad +\iint _{\{x>y>0\}}\frac{\Psi _a(s,x)(\phi _a*\Psi _a(s,\cdot ))(y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[\vartheta (s,\cdot )](x,y)\mathrm {d}x\mathrm {d}y\right] \mathrm {d}s \end{aligned}$$
(3.3)

for all \(t\ge 0\) and all \(\vartheta \in C^1([0,\infty ):\mathcal {B}_1)\).

3.1 Construction of the Semigroup

To prove existence of an evolution semigroup for (3.3), it is useful to consider a reformulation where the transport term is removed. Introducing the variables

$$\begin{aligned} H_a(s,X)&=\Psi _a(s,x),&\psi (s,X)&=e^{-s/\rho }\vartheta (s,x),&X&=xe^{s/\rho }, \end{aligned}$$
(3.4)

we can write (3.3) as

$$\begin{aligned} \int _{[0,\infty )}\psi (t,X)&H_a(t,X)\mathrm {d}X-\int _{[0,\infty )}\psi (0,X)H_a(0,X)\mathrm {d}X\nonumber \\&=\int _0^t\left[ \int _{[0,\infty )}\left( \psi _s(s,X)+\tfrac{\rho -1}{\rho }\psi (s,X)\right) H_a(s,X)\mathrm {d}X\right. \nonumber \\&\left. \quad +\iint _{\{X>Y>0\}}\frac{e^{s/\rho }H_a(s,X)(\phi _{ae^{s/\rho }}*H_a(s,\cdot ))(Y)}{((X+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\right. \nonumber \\&\quad \left. \times \,\mathcal {D}_2^*[\psi (s,\cdot )](X,Y)\mathrm {d}X\mathrm {d}Y\right] \mathrm {d}s. \end{aligned}$$
(3.5)

To construct the evolution semigroup for (3.3) we thus construct a solution to (3.5). To this end we prove existence and uniqueness for suitable mild solutions, which turn out to be weak solutions in the sense of (3.5).

Lemma 3.3

Under Assumption 3.1, then given \(H_0\in \mathcal {X}_\rho \), there exist \(T>0\), depending on \(\varepsilon \) and \(\Vert H_0\Vert _\rho \), and a unique function \(H_a\in C([0,T]:\mathcal {X}_\rho )\) that is a fixed point for the operator \(\mathcal {T}_a\), from \(C([0,T]:\mathcal {X}_\rho )\) to itself, defined by

$$\begin{aligned} \mathcal {T}_a[H](t,X)= & {} H_0(X)e^{-\int _0^tA_a(s)[H(s,\cdot )](X)\mathrm {d}s}\nonumber \\&+\int _0^te^{-\int _s^tA_a(\sigma )[H(\sigma ,\cdot )](X)\mathrm {d}\sigma }B_a(s)[H(s,\cdot )](X)\mathrm {d}s, \end{aligned}$$
(3.6)

where \(A_a(s):\mathcal {X}_\rho \rightarrow C([0,\infty ])\), for \(s\in [0,T]\), is given by

$$\begin{aligned} A_a(s)[H](X)=\frac{2Xe^{s/\rho }}{(X+\varepsilon e^{s/\rho })^{3/2}}\int _0^X\frac{(\phi _{ae^{s/\rho }}*H)(Y)}{(Y+\varepsilon e^{s/\rho })^{3/2}}\mathrm {d}Y-\frac{\rho -1}{\rho }, \end{aligned}$$
(3.7)

and where \(B_a(s):\mathcal {X}_\rho \rightarrow \mathcal {X}_\rho \), again for \(s\in [0,T]\), is such that for all \(\psi \in \mathcal {B}_0\) we have

$$\begin{aligned} \int _{[0,\infty )}\psi (X)B_a(s)[H](X)\mathrm {d}X= & {} \iint _{\{X>Y>0\}}\frac{e^{s/\rho } H(X)(\phi _{ae^{s/\rho }}*H)(Y)}{((X+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\\&\times \left[ (X+Y)\psi (X+Y)+(X-Y)\psi (X-Y)\right] \mathrm {d}X\mathrm {d}Y. \end{aligned}$$

Moreover, for initial data in \(\{\Vert \mu \Vert _\rho \le E_0\}\cap \mathcal {X}_\rho \), the constant \(T>0\) depends only on \(\varepsilon \) and \(E_0\).

Lemma 3.4

The fixed point \(H_a\in C([0,T]:\mathcal {X}_\rho )\), obtained in Lemma 3.3, satisfies (3.5) for all \(t\in [0,T]\) and \(\psi \in C^1([0,T]:\mathcal {B}_0)\).

Proof of Lemma 3.3

To check that \(\mathcal {T}_a\) is well-defined from \(C([0,T]:\mathcal {X}_\rho )\) to itself, it suffices to check that \(B_a(s)\) maps \(\mathcal {X}_\rho \) into itself. To that end we note that

$$\begin{aligned} \Vert B_a(s)[H]\Vert _\rho= & {} \sup _{R>0}R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) B_a(s)[H](X)\mathrm {d}X\\\le & {} \frac{2}{\varepsilon }\int _{(0,\infty )}\left( \sup _{R>0}R^{\rho -2}\int _{(Y,\infty )}\left( 1\wedge \tfrac{R}{X}\right) H(X)\mathrm {d}X\right) \frac{(\phi _{ae^{s/\rho }}*H)(Y)}{Y+\varepsilon e^{s/\rho }}\mathrm {d}Y\\\le & {} \frac{2}{\varepsilon }\Vert H\Vert _\rho \int _{[0,\infty )}\left( \int _{(0,\infty )}\frac{\phi _{ae^{s/\rho }}(Y-Z)}{Y+\varepsilon e^{s/\rho }}\mathrm {d}Y\right) H(Z)\mathrm {d}Z, \end{aligned}$$

so using further that \(|Y-Z|\le ae^{s/\rho }<\frac{1}{2}\varepsilon e^{s/\rho }\) for all \(Y-Z\in \mathrm{supp}(\phi _{ae^{s/\rho }})\), we have

$$\begin{aligned} \Vert B_a(s)[H]\Vert _\rho\le & {} \frac{2}{\varepsilon }\Vert H\Vert _\rho \int _{[0,\infty )}\left( \int _{(0,\infty )}\frac{\phi _{ae^{s/\rho }}(Y-Z)}{Z+(\varepsilon -a)e^{s/\rho }}\mathrm {d}Y\right) H(Z)\mathrm {d}Z\nonumber \\\le & {} \frac{2}{\varepsilon }\Vert H\Vert _\rho \int _{[0,\infty )}\tfrac{1}{\frac{1}{2}\varepsilon e^{s/\rho }}\left( 1\wedge \tfrac{\frac{1}{2}\varepsilon e^{s/\rho }}{Z}\right) H(Z)\mathrm {d}Z \le \tfrac{2^\rho }{\varepsilon ^\rho }e^{-s(\rho -1)/\rho }\Vert H\Vert _\rho ^2.\nonumber \\ \end{aligned}$$
(3.8)

Using this estimate and exploiting the nonnegativity of the first term on the right hand side of (3.7), we find for any \(t\in [0,T]\) that

$$\begin{aligned} \Vert \mathcal {T}_a[H](t,\cdot )\Vert _\rho\le & {} e^{t(\rho -1)/\rho }\Vert H_0\Vert _\rho +\int _0^t e^{(t-s)(\rho -1)/\rho }\Vert B_a(s)[H(s,\cdot )]\Vert _\rho \mathrm {d}s\\\le & {} e^{t(\rho -1)/\rho }\left( \Vert H_0\Vert _\rho +\frac{2^\rho }{\varepsilon ^\rho }\int _0^t e^{-2s(\rho -1)/\rho }\mathrm {d}s\times \sup _{s\in [0,t]}\Vert H(s,\cdot )\Vert _\rho ^2\right) , \end{aligned}$$

implying that \(\mathcal {T}_a\) maps the subset

$$\begin{aligned} \mathcal {S}:=\left\{ H\in C([0,T]:\mathcal {X}_\rho )\,\Big |\,\sup _{t\in [0,T]}\Vert H(t,\cdot )\Vert _\rho =:\Vert H\Vert _{T,\rho }\le 2\Vert H_0\Vert _\rho \right\} \end{aligned}$$

into itself, provided that \(T>0\) is small enough. Note that for \(\varepsilon >0\) fixed, \(T>0\) can be chosen uniformly for \(H_0\in \{\Vert \mu \Vert _\rho \le E_0\}\cap \mathcal {X}_\rho \).

To check that the operator is actually strongly contractive on \(\mathcal {S}\) for sufficiently small \(T>0\), and thereby proving the lemma, we now first observe for \(H_1^*,H_2^*\in \mathcal {X}_\rho \) and \(\sigma \in [0,T]\) that

$$\begin{aligned}&\left\| A_a(\sigma )[H_1^*](\cdot )-A_a(\sigma )[H_2^*](\cdot )\right\| _\infty \\&\quad \le \sup _{X>0}\frac{2Xe^{\sigma /\rho }}{(X+\varepsilon e^{\sigma /\rho })^{3/2}}\left| \int _0^X\frac{(\phi _{ae^{\sigma /\rho }}*(H_1^*-H_2^*))(Y)}{(Y+\varepsilon e^{\sigma /\rho })^{3/2}}\mathrm {d}Y\right| \\&\quad \le \frac{2}{\varepsilon }\int _{[0,\infty )}\left( \int _0^\infty \frac{\phi _{ae^{\sigma /\rho }}(Y-Z)}{Y+\varepsilon e^{\sigma /\rho }}\mathrm {d}Y\right) |H_1^*-H_2^*|(Z)\mathrm {d}Z\\&\quad \le \tfrac{2^\rho }{\varepsilon ^\rho }e^{-\sigma (\rho -1)/\rho }\Vert H_1^*-H_2^*\Vert _\rho \end{aligned}$$

hence for \(H_1,H_2\in \mathcal {S}\) and \(0\le s\le t\le T\) we have

$$\begin{aligned}&\left\| e^{-\int _s^tA_a(\sigma )[H_1(\sigma ,\cdot )](\cdot )\mathrm {d}\sigma }-e^{-\int _s^tA_a(\sigma )[H_2(\sigma ,\cdot )](\cdot )\mathrm {d}\sigma }\right\| _\infty \nonumber \\&\quad \le e^{(t-s)(\rho -1)/\rho }\int _s^t\left\| A_a(\sigma )[H_1(\sigma ,\cdot )](\cdot )-A_a(\sigma )[H_2(\sigma ,\cdot )](\cdot )\right\| _\infty \mathrm {d}\sigma \nonumber \\&\quad \le \tfrac{2^\rho }{\varepsilon ^\rho }(t-s)e^{(t-s)(\rho -1)/\rho }\Vert H_1-H_2\Vert _{T,\rho }. \end{aligned}$$
(3.9)

Again for \(H_1^*,H_2^*\in \mathcal {X}_\rho \) we next note that

$$\begin{aligned}&\int _{(0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) |B_a(s)[H_1^*]-B_a(s)[H_2^*]|(X)\mathrm {d}X\\&\quad \le \iint _{\{X>Y>0\}}\frac{e^{s/\rho }|H_1^*(X)(\phi _{ae^{s/\rho }}*H_1^*)(Y)-H_2^*(X)(\phi _{ae^{s/\rho }}*H_2^*)(Y)|}{((X+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\\&\quad \quad \times \left( (X+Y)\wedge R+(X-Y)\wedge R\right) \mathrm {d}X\mathrm {d}Y, \end{aligned}$$

so analogous arguments as used to obtain (3.8) give us that

$$\begin{aligned}&\Vert B_a(s)[H_1^*]-B_a(s)[H_2^*]\Vert _\rho \nonumber \\&\quad \le \tfrac{2^\rho }{\varepsilon ^\rho }e^{-s(\rho -1)/\rho }\left( \Vert H_1^*\Vert _\rho +\Vert H_2^*\Vert _\rho \right) \Vert H_1^*-H_2^*\Vert _\rho . \end{aligned}$$
(3.10)

Combining finally (3.8), (3.9) and (3.10), we find for \(H_1,H_2\in \mathcal {S}\) the estimate

$$\begin{aligned} \Vert \mathcal {T}_a[H_1]-\mathcal {T}_a[H_2]\Vert _{T,\rho }\le K(T)\Vert H_1-H_2\Vert _{T,\rho }, \end{aligned}$$

with

$$\begin{aligned} K(T)=\tfrac{2^\rho }{\varepsilon ^2}Te^{T(\rho -1)/\rho }\Vert H_0\Vert _\rho \left( 1+4\left( \tfrac{2^\rho }{\varepsilon ^\rho }T\Vert H_0\Vert _\rho +e^{-T(\rho -1)/\rho }\right) \right) \xrightarrow {T\rightarrow 0}0, \end{aligned}$$

and noting again that for \(\varepsilon >0\) fixed we can again choose \(T>0\) uniformly for \(H_0\in \{\Vert \mu \Vert _\rho \le E_0\}\cap \mathcal {X}_\rho \), the proof is complete. \(\square \)

Proof of Lemma 3.4

By construction there holds \(H_a=\mathcal {T}_a[H_a]\), so multiplying this identity by \(\varphi \in C^1([0,T]:\mathcal {B}_0)\) and integrating with respect to X, we obtain for all \(t\in [0,T]\) that

$$\begin{aligned}&\int _{[0,\infty )}\varphi (t,X)H_a(t,X)\mathrm {d}X\nonumber \\&\quad =\int _{[0,\infty )}\varphi (t,X)H_0(X)e^{-\int _0^tA_a(s)[H_a(s,\cdot )](X)\mathrm {d}s}\mathrm {d}X\nonumber \\&\qquad +\int _0^t\int _{[0,\infty )}\varphi (t,X)e^{-\int _s^tA_a(\sigma )[H_a(\sigma ,\cdot )](X)\mathrm {d}\sigma }B_a(s)[H_a(s,\cdot )](X)\mathrm {d}X\mathrm {d}s. \end{aligned}$$
(3.11)

If now \(\psi \in C^1([0,T]:\mathcal {B}_0)\) is arbitrary, then taking the time derivative of (3.11) with \(\varphi \) replaced by \(\psi \), we get

$$\begin{aligned}&\partial _t\left[ \int _{[0,\infty )}\psi (t,X)H_a(t,X)\mathrm {d}X\right] \nonumber \\&\quad =\int _{[0,\infty )}\psi (t,X)B_a(t)[H_a(t,\cdot )](X)\mathrm {d}X\nonumber \\&\qquad +\int _{[0,\infty )}\big (\psi _t(t,X)-\psi (t,X)A_a(t)[H_a(t,\cdot )](X)\big )H_a(t,X)\mathrm {d}X, \end{aligned}$$
(3.12)

where the last term on the right hand side is obtained by combining two terms, using the identity obtained from (3.11) with \(\varphi (t,X)=\psi (t,X)A_a(t)[H_a(t,\cdot )](X)\). Integrating (3.12), we then obtain (3.5). \(\square \)

We are now able to show local in time existence of solutions to (3.3) by construction, as well as an estimate of the norm \(\Vert \cdot \Vert _\rho \) for these solutions.

Proposition 3.5

Under Assumption 3.1, suppose for \(\Psi _0\in \mathcal {X}_\rho \) that \(T>0\) and \(H_a\in C([0,T]:\mathcal {X}_\rho )\) are as obtained in Lemma 3.3 with \(H_0=\Psi _0\). Then the function \(\Psi _a\in C([0,T]:\mathcal {X}_\rho )\), defined via

$$\begin{aligned} \Psi _a(t,x)=H_a(t,X)\quad \text {and}\quad x=Xe^{-t/\rho }, \end{aligned}$$
(3.13)

satisfies (3.3) for all \(\vartheta \in C^1([0,T]:\mathcal {B}_1)\) and all \(t\in [0,T]\), and for all \(t\in [0,T]\) there holds \(\Vert \Psi _a(t,\cdot )\Vert _\rho \le \Vert \Psi _0\Vert _\rho \).

Proof

To check that \(\Psi _a\) satisfies (3.3) is an elementary computation [use (3.4)], so we restrict ourselves to proving the estimate of the norm. We observe that

$$\begin{aligned} \Vert \Psi _a(t,\cdot )\Vert _\rho =e^{-t(\rho -1)/\rho }\Vert H_a(t,\cdot )\Vert _\rho , \end{aligned}$$

so it suffices to check that

$$\begin{aligned} \Vert H_a(t,\cdot )\Vert _\rho \le e^{t(\rho -1)/\rho }\Vert H_0\Vert _\rho \text { for all }t\in [0,T]. \end{aligned}$$
(3.14)

By Lemma 3.4, now \(H_a\) satisfies (3.5) for all \(t\in [0,T]\) and \(\psi \in C^1([0,T]:\mathcal {B}_0)\), and we note for any \(R>0\) that \(\psi (X)=1\wedge \frac{R}{X}\) satisfies \(\psi \in \mathcal {B}_0\). Moreover, the mapping \(X\mapsto X\psi (X)\) is concave, so \(\mathcal {D}_2^*[\psi ]\le 0\), and there thus holds

$$\begin{aligned} \int _{[0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) H_a(t,X)\mathrm {d}X\le & {} \int _{[0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) H_0(X)\mathrm {d}X\\&+\,\frac{\rho -1}{\rho }\int _0^t\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) H_a(s,X)\mathrm {d}X\mathrm {d}s. \end{aligned}$$

By Gronwall’s lemma and multiplying by \(R^{\rho -2}\) we then get

$$\begin{aligned} R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) H_a(t,X)\mathrm {d}X \le e^{t(\rho -1)/\rho }R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{X}\right) H_0(X)\mathrm {d}X, \end{aligned}$$

and (3.14) follows by taking the supremum over all \(R>0\). \(\square \)

We are now able to construct a family of operators on the unit ball of \(\mathcal {X}_\rho \), which will turn out to be the semigroup required in Proposition 3.2.

Definition 3.6

Under Assumption 3.1, we define the family \((S_a(t))_{t\ge 0}\) of operators from the unit ball in \(\mathcal {X}_\rho \) into itself as follows. Let \(T>0\) be as obtained in Lemma 3.3, depending only on the parameters \(\varepsilon >0\) and \(E_0=1\). Then for all \(\Psi _0\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \) there exists a unique function \(H_a\in C([0,T]:\mathcal {X}_\rho )\) that is a fixed point for the operator \(\mathcal {T}_a\), given by (3.6) with \(H_0=\Psi _0\). For \(t\ge 0\) we now set \(S_a(t)\Psi _0=\Psi _a(t,\cdot )\) if \(t\le T\), where \(\Psi _a\in C([0,T]:\mathcal {X}_\rho )\) is defined via (3.13), and

$$\begin{aligned} S_a(t)\Psi _0= S_a(t-nT)\left( S_a(T)\right) ^n\Psi _0\text { if}\, t\in (nT,(n+1)T] \text {for}\, n\in \mathbb {N}, \end{aligned}$$
(3.15)

which is possible since \(S(T)\Psi _*\) is in the unit ball for all \(\Psi _*\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \) (cf. Proposition 3.5).

Proposition 3.7

Under Assumption 3.1, the family of operators \((S_a(t))_{t\ge 0}\), as defined in Definition 3.6, has the semigroup property, i.e.

$$\begin{aligned} S_a(t_1+t_2)=S_a(t_1)S_a(t_2)\text { for all }t_1,t_2\ge 0. \end{aligned}$$
(3.16)

Moreover, given \(\Psi _0\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \), then the function defined as \(\Psi _a(t,x)=S_a(t)\Psi _0(x)\) satisfies (3.3) for all \(t\ge 0\) and all \(\vartheta \in C^1([0,\infty ):\mathcal {B}_1)\).

Proof

For any \(\Psi _0\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \), let \(H_a\in C([0,T]:\mathcal {X}_\rho )\) be the unique fixed point to \(\mathcal {T}_a\) with \(H_0=\Psi _0\). Using then (3.13), we find by careful computation that

$$\begin{aligned} A_a(s)[H_a(s,\cdot )](X)= & {} \frac{2Xe^{s/\rho }}{(X+\varepsilon e^{s/\rho })^{3/2}}\int _0^X\frac{(\phi _{ae^{s/\rho }}*H_a(s,\cdot ))(Y)}{(Y+\varepsilon e^{s/\rho })^{3/2}}\mathrm {d}Y-\frac{\rho -1}{\rho }\nonumber \\= & {} \,\frac{2Xe^{-s/\rho }}{(Xe^{-s/\rho }+\varepsilon )^{3/2}}\int _0^{Xe^{-s/\rho }}\frac{(\phi _a*\Psi _a(s,\cdot ))(y)}{(y+\varepsilon )^{3/2}}\mathrm {d}y-\frac{\rho -1}{\rho }\nonumber \\= & {} \,A_a(0)[\Psi _a(s,\cdot )](Xe^{-s/\rho }), \end{aligned}$$
(3.17)

and similarly we can check that

$$\begin{aligned} B_a(s)[H_a(s,\cdot )](X)=B_a(0)[\Psi _a(s,\cdot )](Xe^{-s/\rho }). \end{aligned}$$
(3.18)

Using now (3.17) and (3.18), it follows by the definition of \(H_a\) as the fixed point of \(\mathcal {T}_a\) that for \(t\in [0,T]\) and \(X\ge 0\) there holds

$$\begin{aligned} H_a(t,X)= & {} \Psi _0(X)e^{-\int _0^tA_a(0)[\Psi _a(s,\cdot )](Xe^{-s/\rho })\mathrm {d}s}\\&+\int _0^te^{-\int _s^tA_a(0)[\Psi _a(\sigma ,\cdot )](Xe^{-\sigma /\rho })\mathrm {d}\sigma }B_a(0)[\Psi _a(s,\cdot )](Xe^{-s/\rho })\mathrm {d}s, \end{aligned}$$

hence by again (3.13) for \(t\in [0,T]\) and \(x\ge 0\) there holds

$$\begin{aligned} \Psi _a(t,x)= & {} \Psi _0(xe^{t/\rho })e^{-\int _0^tA_a(0)[\Psi _a(s,\cdot )](xe^{(t-s)/\rho })\mathrm {d}s}\\&+\int _0^te^{-\int _s^tA_a(0)[\Psi _a(\sigma ,\cdot )](xe^{(t-\sigma )/\rho })\mathrm {d}\sigma }B_a(0)[\Psi _a(s,\cdot )](xe^{(t-s)/\rho })\mathrm {d}s. \end{aligned}$$

For any \(t_1,t_2\ge 0\) with \(t_1+t_2\le T\) we then use the following decomposition

$$\begin{aligned}&\Psi _0(\cdot )e^{\int _0^{t_1+t_2}[\cdots ]\mathrm {d}s}+\int _0^{t_1+t_2}e^{\int _s^{t_1+t_2}[\cdots ]\mathrm {d}\sigma }[\cdots ]\mathrm {d}s\\&\qquad =\,\left( \Psi _0(\cdot )e^{\int _0^{t_2}[\cdots ]\mathrm {d}s}+\int _0^{t_2}e^{\int _s^{t_2}[\cdots ]\mathrm {d}\sigma }[\cdots ]\mathrm {d}s\right) e^{\int _{t_2}^{t_1+t_2}[\cdots ]\mathrm {d}s}\\&\qquad \quad +\int _{t_2}^{t_1+t_2}e^{\int _s^{t_1+t_2}[\cdots ]\mathrm {d}\sigma }[\cdots ]\mathrm {d}s \end{aligned}$$

and after performing the changes of variables \(s\rightarrow t_2+s\) and \(\sigma \rightarrow t_2+\sigma \) in the integrals on the right hand side we obtain

$$\begin{aligned} \Psi _a(t_1+t_2,x)= & {} \Psi _a(t_2,xe^{t_1/\rho })e^{-\int _0^{t_1}A_a(0)[\Psi _a(t_2+s,\cdot )](xe^{(t_1-s)/\rho })\mathrm {d}s}\\&+\int _0^{t_1}e^{-\int _{s}^{t_1}A_a(0)[\Psi _a(t_2+\sigma ,\cdot )](xe^{(t_1-\sigma )/\rho })\mathrm {d}\sigma }B_a(0)[\Psi _a(t_2\!+\!s,\cdot )](xe^{(t_1-s)/\rho })\mathrm {d}s. \end{aligned}$$

We now see that \(H_*(s,xe^{s/\rho })=\Psi _*(s,x):=\Psi _a(t_2+s,x)\) is a fixed point for the operator \(\mathcal {T}_a\) with \(H_0=\Psi _a(t_2,\cdot )\), and by the short time uniqueness of fixed points, obtained in Lemma 3.3, we thus find that

$$\begin{aligned} S_a(t_1+t_2)\Psi _0=\Psi _a(t_1+t_2,\cdot )= S_a(t_1)\Psi _a(t_2,\cdot )= S_a(t_1)S_a(t_2)\Psi _0, \end{aligned}$$

which proves the semigroup property for \(t_1,t_2\ge 0\) with \(t_1+t_2\le T\).

Next we use the local semigroup property as derived above to observe for \(t_1,t_2\in [0,T]\) with \(t_1+t_2>T\) that

$$\begin{aligned} S_a(t_1+t_2-T)S_a(T)=S_a(t_1+t_2-T)S_a(T-t_2)S_a(t_2)=S_a(t_1)S_a(t_2), \end{aligned}$$

so since the left hand side equals \(S_a(t_1+t_2)\) by definition [cf. (3.15)], we have

$$\begin{aligned} S_a(t_1+t_2)=S_a(t_1)S_a(t_2)=S_a(t_2)S_a(t_1)\text { for all }t_1,t_2\in [0,T]. \end{aligned}$$
(3.19)

Using lastly (3.15) and (3.19) for arbitrary \(t_1,t_2\ge 0\), and writing \(n_i\) for the integer part of \(\frac{t_i}{T}\), we find that

$$\begin{aligned} S_a(t_1)S_a(t_2)= & {} S_a(t_1-n_1T)(S_a(T))^{n_1}S_a(t_2-n_2T)(S_a(T))^{n_2}\nonumber \\= & {} S_a(t_1-n_1T)S_a(t_2-n_2T)(S_a(T))^{n_1+n_2}\nonumber \\= & {} S_a(t_1+t_2-(n_1+n_2)T)(S_a(T))^{n_1+n_2}. \end{aligned}$$
(3.20)

If \(t_1+t_2<(n_1+n_2+1)T\), then the right hand side of (3.20) equals \(S_a(t_1+t_2)\) by definition. On the other hand, if \(t_1+t_2\ge (n_1+n_2+1)T\), then by again (3.19) we have

$$\begin{aligned}&S_a(t_1+t_2-(n_1+n_2)T)(S_a(T))^{n_1+n_2}\\&\qquad =S_a(t_1+t_2-(n_1+n_2+1)T)(S_a(T))^{n_1+n_2+1}, \end{aligned}$$

and here the right hand side equals \(S_a(t_1+t_2)\) by definition. We therefore conclude that (3.16) holds. \(\square \)

3.2 Two Useful Lemmas

In this subsection we give two lemmas that will be useful for obtaining the lower bound in our proof of existence of a set \(\mathcal {Y}_\rho =\mathcal {Y}_\rho (R_0)\) that is invariant under the previously defined evolution. These results will also be used in the final section of this paper.

Lemma 3.8

For any \(\alpha \in (0,2)\) the fundamental solution \(u^\alpha \) to the integro-differential equation

$$\begin{aligned} u_t(t,x)=\int _{\mathbb {R}_+}y^{-\alpha -1}\Delta _y^2[u(t,\cdot )](x)\mathrm {d}y, \end{aligned}$$
(3.21)

i.e. the solution to (3.21) with initial datum \(u(0,\cdot )=\delta _0\), is given by \(u^\alpha (t,x)=t^{-1/\alpha }v_\alpha (xt^{-1/\alpha })\), where \(v_\alpha \in C^\infty (\mathbb {R})\) is the probability density function that has characteristic function \(\exp (-c_\alpha |k|^\alpha )\), with \(c_\alpha =-2\Gamma (-\alpha )\cos (\frac{\alpha \pi }{2})\) if \(\alpha \ne 1\) and \(c_1=\pi \). In particular, \(v_\alpha \) is positive, symmetric, nonincreasing on \(\mathbb {R}_+\), and it satisfies \(\lim _{|z|\rightarrow \infty }|z|^{\alpha +1}v_\alpha (z)=1\).

Proof

Taking the Fourier transform of (3.21) gives us

$$\begin{aligned} \hat{u}_t(t,k)=-c_\alpha |k|^\alpha \hat{u}(t,k), \end{aligned}$$

hence \(u^\alpha \) is the inverse Fourier transform of \(\exp (-c_\alpha |k|^\alpha t)\):

$$\begin{aligned}&u^\alpha (t,x)=\frac{1}{2\pi }\int _\mathbb {R}e^{ikx}e^{-c_\alpha |k|^\alpha t}\mathrm {d}k=\frac{1}{t^{1/\alpha }}v_\alpha \left( \frac{x}{t^{1/\alpha }}\right) \\&\quad \text {with }v_\alpha (z)=\frac{1}{2\pi }\int _\mathbb {R}e^{ikz}e^{-c_\alpha |k|^\alpha }\mathrm {d}k \end{aligned}$$

Smoothness and symmetry of \(v_\alpha \) are immediate, while for the remaining properties of \(v_\alpha \) we note that \(\exp (-c_\alpha |k|^\alpha )\) is the characteristic function of a symmetric stable probability distribution (cf. [11, Theorem 5.7.3]). Now, [11, Theorem 5.10.1] states that all stable distributions are unimodal, so since by symmetry the maximum of \(v_\alpha \) is located at zero we have that \(V_\alpha (x)=\int _{-\infty }^xv_\alpha (z)\mathrm {d}z\) is concave for \(x\ge 0\). Therefore \(v_\alpha '\le 0\) on \(\mathbb {R}_+\) and it is shown that \(v_\alpha \) is nonincreasing on \(\mathbb {R}_+\). The asymptotics of \(v_\alpha \) follow by a standard contour deformation argument, and strict positivity follows from combining the decay behaviour of \(v_\alpha \) with the monotonicity result. \(\square \)

Since we will frequently use solutions to (3.21) with odd initial data, we give the following lemma.

Lemma 3.9

For \(\alpha \in (0,2)\), let \(u^\alpha \) be the fundamental solution to (3.21), let \(u_0\in C(\mathbb {R})\cap L^1(\mathbb {R};|x|^{-\alpha -1}\mathrm {d}x)\) be odd, and for \(t>0\) let \(u(t,\cdot ):=[u_0*u^\alpha (t,\cdot )](\cdot )\). Then the following hold.

  • For all \(t>0\), \(u(t,\cdot )\) is odd and smooth.

  • Maximum principle. If \(u_0\ge 0\,[\,\le 0\,]\) on \(\mathbb {R}_+\), then \(u(t,\cdot )\ge 0\,[\,\le 0\,]\) on \(\mathbb {R}_+\) for all \(t>0\).

  • If \(u_0\) is concave [convex] on \(\mathbb {R}_+\), then \(u(t,\cdot )\) is concave [convex] on \(\mathbb {R}_+\) for all \(t>0\), and in particular

    $$\begin{aligned} \Delta _y^2[u(t,\cdot )](x)\le 0\,[\,\ge 0\,]\text { for all }x\ge 0\text {, }y\in \mathbb {R}\text { and }t\ge 0. \end{aligned}$$
    (3.22)

Proof

For all \(t>0\), \(u_0(t,\cdot )\) is odd since it is the convolution of an odd and an even function, while smoothness follows from the fact that \(u^\alpha (t,\cdot )\) is smooth for all \(t>0\). Suppose now that \(u_0\ge 0\,[\,\le 0\,]\) on \(\mathbb {R}_+\). For \(x\ge 0\) we then find, by the facts that \(u_0\) is odd and that \(u^\alpha (t,\cdot )\) is even for all \(t>0\), that we can write

$$\begin{aligned} u(t,x)=\int _{\mathbb {R}_+} u_0(y)\left( u^\alpha (t,x-y)-u^\alpha (t,x+y)\right) \mathrm {d}y, \end{aligned}$$

and it follows that \(u(t,\cdot )\ge 0\,[\,\le 0\,]\) on \(\mathbb {R}_+\) for all \(t>0\), since \(u^\alpha (t,\cdot )\) is even and monotonically decreasing on \(\mathbb {R}_+\) (\(u^\alpha (t,x-z)-u^\alpha (t,x+z)\ge 0\) for \(x,z\ge 0\)). We next restrict ourselves to the case where \(u_0\) is concave on \(\mathbb {R}_+\), since the other case is similar. Then, for all \(y\in \mathbb {R}\) there holds \(\Delta _y^2u_0\le 0\) on \(\mathbb {R}_+\). For \(|y|\le x\) this follows immediately from the definition of concavity, while for \(|y|>x>0\) we note, using that \(u_0\) is odd, that

$$\begin{aligned} \Delta _y^2u_0(x)=\Delta _x^2u_0(|y|)+2\big (u_0(|y|)-u_0(|y|-x)-u_0(x)\big )\le 0. \end{aligned}$$

Here the first term on the right hand side of the equality is nonpositive by the previous argument, and the remaining terms are nonpositive by

$$\begin{aligned} u_0(|y|-x)+u_0(x)= & {} u_0\left( \tfrac{x}{|y|}\times 0+\tfrac{|y|-x}{|y|}|y|\right) +u_0\left( \tfrac{|y|-x}{|y|}\times 0+\tfrac{x}{|y|}|y|\right) \\\ge & {} \left( \tfrac{x}{|y|}+\tfrac{|y|-x}{|y|}\right) u_0(0)+\left( \tfrac{|y|-x}{|y|}+\tfrac{x}{|y|}\right) u_0(|y|)=u_0(|y|), \end{aligned}$$

where we have used that \(u_0(0)=0\) since \(u_0\) is odd. Next, since the second difference operator is linear, it commutes with the integral operator on the right hand side of (3.21), and since further the second difference of an odd function is odd, we find by the maximum principle proven above that (3.22) holds. Additionally we then find that \(u(t,\cdot )\) is concave on \(\mathbb {R}_+\) since for all \(x\ge 0\) we have \(u_{xx}(t,x)=\lim _{y\rightarrow 0}\frac{1}{y^2}\Delta _y^2[u(t,\cdot )](x)\le 0\), where we recall that we may take two derivatives by smoothness of \(u^\alpha (t,\cdot )\). \(\square \)

3.3 Invariance of \(\mathcal {Y}_\rho (R_0)\)

Our goal in this subsection is to show that, for some suitable \(R_0>0\), the semigroup introduced in Definition 3.6 maps the set \(\mathcal {Y}_\rho (R_0)\) (cf. Definition 2.16) into itself. The proof of invariance of the lower bound [cf. (2.4)] shows strong similarities with the approach in [14]. As mentioned before, the main difference here compared with the approach in that paper is that because of the form of the nonlocal diffusion operator in our collision kernel [cf. (2.5)] it is convenient to use test functions that are concave, while in [14] it was natural to use monotone test functions. A consequence of this is that in order to measure the size of \(\Psi _a\) it is now natural to use the functionals given by (2.1), as opposed to the functionals \(\int _0^Rh(x)\mathrm {d}x\) which were used in [14].

We first derive the following estimate.

Lemma 3.10

Under Assumption 3.1, let \((S_a(t))_{t\ge 0}\) be the semigroup on the unit ball of \(\mathcal {X}_\rho \) as defined in Definition 3.6. Let further \(\vartheta \in C(\mathbb {R})\) be such that the mapping \(z\mapsto z\vartheta (z)\) is odd, bounded, and concave on \(\mathbb {R}_+\). Then for all \(t\ge 0\) and all \(\Psi _0\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \) it holds that

$$\begin{aligned}&\int _{[0,\infty )}\vartheta (x)S_a(t)\Psi _0(x)\mathrm {d}x\nonumber \\&\quad \ge e^{t(\rho -1)/\rho }\int _{[0,\infty )}\left( \int _\mathbb {R}\frac{y\vartheta (y)}{(\rho t)^{1/\rho }}v_\rho \left( \frac{xe^{-t/\rho }-y}{(\rho t)^{1/\rho }}\right) \mathrm {d}y\right) \tfrac{1}{x}\Psi _0(x)\mathrm {d}x, \end{aligned}$$
(3.23)

where \(v_\rho \) is the self-similar profile associated to the fundamental solution of (3.21) with \(\alpha =\rho \), which was obtained in Lemma 3.8.

Proof

Throughout this proof we let \(t\ge 0\) and \(\Psi _0\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \) be fixed, and for all \(s\in [0,t]\) we write \(\Psi _a(s,x)=S_a(s)\Psi _0(x)\).

For \(\vartheta \in C(\mathbb {R})\) fixed as in the statement of the lemma, we define

$$\begin{aligned}&u(s,x)=e^{(t-s)(\rho -1)/\rho }[\varphi *u^\rho (\rho (t-s),\cdot )](xe^{(s-t)/\rho })\times \tfrac{1}{x},\nonumber \\&\quad \text { for}\, s\in [0,t] \text { and}\, x\in \mathbb {R}, \text { and with}\, \varphi (y):=y\vartheta (y), \end{aligned}$$
(3.24)

where \(u^\rho \) is the fundamental solution of (3.21) with \(\alpha =\rho \) as obtained in Lemma 3.8. We then note that \(u(t,\cdot )=\vartheta \), and that u(0, x) is equal to the integral with respect to y in the right hand side of (3.23). Therefore (3.23) can be written as

$$\begin{aligned} \int _{[0,\infty )}u(t,x)\Psi _a(t,x)\mathrm {d}x\ge \int _{[0,\infty )}u(0,x)\Psi _0(x)\mathrm {d}x, \end{aligned}$$

so, using u as a test function in (3.3), we see that (3.23) is equivalent to

$$\begin{aligned}&\int _0^t\Bigg [\int _{[0,\infty )}\left( u_s(s,x)-\tfrac{1}{\rho }\left( x u_x(s,x)+(2-\rho )u(s,x)\right) \right) \Psi _a(s,x)\mathrm {d}x\nonumber \\&\quad \,\,\,+\int _{[0,\infty )}\left( \int _0^x\frac{(\phi _a*\Psi _a(s,\cdot ))(y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[u(s,\cdot )](x,y)\mathrm {d}y\right) \Psi _a(s,x)\mathrm {d}x\Bigg ]\mathrm {d}s\ge 0,\qquad \qquad \end{aligned}$$
(3.25)

where we recall that \(\mathcal {D}_2^*\) is defined in (2.6).

Next, for \(x\ge y\ge 0\) we note that \((x+\varepsilon )^{-3/2}\le \frac{1}{x}(y+\varepsilon )^{-1/2}\), and, defining \(U(s,x):=xu(s,x)\), that \(\mathcal {D}_2^*[u(s,\cdot )](x,y)=\Delta _y^2[U(s,\cdot )](x)\). Further, since \(U(s,\cdot )\) is odd, bounded and concave on \(\mathbb {R}_+\) for all \(s\ge 0\) (cf. Lemma 3.9), by (3.22) it holds for all \(x,y\ge 0\) that \(\Delta _y^2[U(s,\cdot )](x)\le 0\), which together yields the estimate

$$\begin{aligned}&\int _0^x\frac{(\phi _a*\Psi _a(s,\cdot ))(y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[u(s,\cdot )](x,y)\mathrm {d}y\nonumber \\&\quad \,\, \ge \frac{1}{x}\int _{\mathbb {R}_+}\frac{(\phi _a*\Psi _a(s,\cdot ))(y)}{(y+\varepsilon )^2}\Delta _y^2[U(s,\cdot )](x)\mathrm {d}y. \end{aligned}$$
(3.26)

By an integration by parts, and using (2.8) for \(\partial _y[\Delta _y^2[U(s,\cdot )](x)]\), we can now write the right hand side of (3.26) as

$$\begin{aligned} \frac{1}{x}\int _{\mathbb {R}_+}\left( \int _y^\infty \frac{(\phi _a*\Psi _a(s,\cdot ))(z)}{(z+\varepsilon )^2}\mathrm {d}z\right) \left( \int _{x-y}^{x+y}U_{ww}(s,w)\mathrm {d}w\right) \mathrm {d}y, \end{aligned}$$
(3.27)

where we note that the integral with respect to w on the right hand side is nonpositive for \(x,y\ge 0\), since \(U(s,\cdot )\) is odd and concave on \(\mathbb {R}_+\) (note that \(\int _{-a}^aU''(w)\mathrm {d}w=0\) for \(a\ge 0\)). To find a lower bound for (3.27) we thus need an upper bound for the integral with respect to z. We thereto note for \(z\ge y\) that \((z+\varepsilon )^{-2}\le \frac{1}{y^2}(1\wedge \frac{y}{z+\varepsilon })\), and by expanding the domain of integration we find

$$\begin{aligned} \int _y^\infty \frac{(\phi _a*\Psi _a(s,\cdot ))(z)}{(z+\varepsilon )^2}\mathrm {d}z\le & {} \frac{1}{y^2}\int _{[0,\infty )}\left( 1\wedge \tfrac{y}{z+\varepsilon }\right) (\phi _a*\Psi _a(s,\cdot ))(z)\mathrm {d}z\nonumber \\= & {} \frac{1}{y^2}\int _{[0,\infty )}\left( \int _{[0,\infty )}\left( 1\wedge \tfrac{y}{z+\varepsilon }\right) \phi _a(z-x)\mathrm {d}z\right) \Psi _a(s,x)\mathrm {d}x,\nonumber \\ \end{aligned}$$
(3.28)

where the equality holds by Fubini. Now, since for all \(z-x\in \mathrm{supp}(\phi _a)\) there holds \(|z-x|\le a<\frac{\varepsilon }{2}\), we have for all \(x\ge 0\) that

$$\begin{aligned} \int _{[0,\infty )}\left( 1\wedge \tfrac{y}{x+(z-x+\varepsilon )}\right) \phi _a(z-x)\mathrm {d}z \le \left( 1\wedge \tfrac{y}{x+\frac{\varepsilon }{2}}\right) \int _{\mathbb {R}}\phi _a(z-x)\mathrm {d}z \le \left( 1\wedge \tfrac{y}{x}\right) , \end{aligned}$$

which, using the definition of the norm, we can use to estimate the right hand side of (3.28) by

$$\begin{aligned} \frac{1}{y^2}\int _{[0,\infty )}\left( 1\wedge \tfrac{y}{x}\right) \Psi _a(s,x)\mathrm {d}x\le \tfrac{1}{y^2}\Vert \Psi _a(s,\cdot )\Vert _\rho \,y^{2-\rho }\le y^{-\rho }. \end{aligned}$$

Combining then the previous estimates, and recalling the nonpositivity of the integral with respect to w in (3.27), we obtain

$$\begin{aligned}&\int _{[0,\infty )}\left( \int _0^x\frac{(\phi _a*\Psi _a(s,\cdot ))(y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[u(s,\cdot )](x,y)\mathrm {d}y\right) \Psi _a(s,x)\mathrm {d}x\nonumber \\&\quad \ge \int _{[0,\infty )}\left( \int _{\mathbb {R}_+}y^{-\rho }\times \int _{x-y}^{x+y}U_{ww}(s,w)\mathrm {d}w\ \mathrm {d}y\right) \tfrac{1}{x}\Psi _a(s,x)\mathrm {d}x, \end{aligned}$$
(3.29)

where, by an integration by parts in the integral with respect to y, the right hand side of (3.29) can be rewritten as

$$\begin{aligned} \int _{[0,\infty )}\left( \rho \int _{\mathbb {R}_+}y^{-\rho -1}\Delta _y^2[U(s,\cdot )](x)\mathrm {d}y\right) \tfrac{1}{x}\Psi _a(s,x)\mathrm {d}x. \end{aligned}$$
(3.30)

Also, since \(u^\rho \) is the fundamental solution of (3.21), we note that \(U(s,x)=xu(s,x)\) by construction satisfies [cf. (3.24)]

$$\begin{aligned} U_s(s,x)+\tfrac{\rho -1}{\rho }U(s,x)-\tfrac{1}{\rho }xU_x(s,x)=-\rho e^{t-s}\int _{\mathbb {R}_+}y^{-\rho -1}\Delta _y^2[U(s,\cdot )](x)\mathrm {d}y. \end{aligned}$$
(3.31)

Checking then that the left hand side of (3.31) can be rewritten as

$$\begin{aligned} x\left( u_s(s,x)-\tfrac{1}{\rho }\left( x u_x(s,x)+(2-\rho )u(s,x)\right) \right) , \end{aligned}$$

we find that the first integral between square brackets on the left hand side of (3.25) equals

$$\begin{aligned} \int _{[0,\infty )}\left( -\rho e^{t-s}\int _{\mathbb {R}_+}y^{-\rho -1}\Delta _y^2[U(s,\cdot )](x)\mathrm {d}y\right) \tfrac{1}{x}\Psi _a(s,x)\mathrm {d}x. \end{aligned}$$
(3.32)

Concluding, since the first and second integral between square brackets on the left hand side of (3.25) can be estimated from below by (3.32) and (3.30) respectively, the left hand side of (3.25) can be bounded from below by

$$\begin{aligned} \int _0^t\left( \int _{[0,\infty )}\left( \rho (1-e^{t-s})\int _{\mathbb {R}_+}y^{-\rho -1}\Delta _y^2[U(s,\cdot )](x)\mathrm {d}y\right) \tfrac{1}{x}\Psi _a(s,x)\mathrm {d}x\right) \mathrm {d}s, \end{aligned}$$

which is nonnegative since both \(1-e^{t-s}\) and \(\Delta _y^2[U(s,\cdot )](x)\) are nonpositive on the domain of integration, while all other terms are nonnegative. This proves that (3.25) holds, and since (3.23) and (3.25) are equivalent the proof is complete. \(\square \)

The following two lemmas will be useful in the actual proof of invariance of (2.4) under the evolution (3.3), where we will use a suitable function \(\vartheta \) in (3.23).

Lemma 3.11

For \(\rho \in (1,2)\) and \(\Psi \in \mathcal {X}_\rho \), it holds for all odd \(\Theta \in C^2([-\infty ,\infty ])\) that satisfy \(\lim _{x\rightarrow \infty }\Theta '(x)x^{2-\rho }=0\) that

$$\begin{aligned} \int _{[0,\infty )}\Theta (x)\cdot \tfrac{1}{x}\Psi (x)\mathrm {d}x=-\int _{[0,\infty )}\Theta ''(x)\left( \int _{[0,\infty )}\left( 1\wedge \tfrac{x}{z}\right) \Psi (z)\mathrm {d}z\right) \mathrm {d}x. \end{aligned}$$
(3.33)

Proof

Observing that

$$\begin{aligned} \int _{[0,\infty )}\left( 1\wedge \tfrac{x}{z}\right) \Psi (z)\mathrm {d}z=\int _0^x\int _y^\infty \tfrac{1}{z}\Psi (z)\mathrm {d}z\mathrm {d}y, \end{aligned}$$

it is clear that (3.33) follows by integrating by parts twice, provided that the boundary values vanish. Since \(\Theta \) is odd with bounded first derivative, we have \(\Theta (x)=\Theta '(0)x+o(x^2)\) as \(x\rightarrow 0\), so

$$\begin{aligned} \left| \Theta (x)\int _x^\infty \tfrac{1}{z}\Psi (z)\mathrm {d}z\right|\le & {} 2|\Theta '(0)|\int _{[0,\infty )}\left( 1\wedge \tfrac{x}{z}\right) \Psi (z)\mathrm {d}z\\\le & {} 2|\Theta '(0)|\Vert \Psi \Vert _\rho \cdot x^{2-\rho }\rightarrow 0\text { as }x\rightarrow 0, \end{aligned}$$

where the second inequality holds by definition of the norm. Notice further that by our choice of \(\Theta \), and using again the definition of the norm, we have

$$\begin{aligned} \left| \Theta '(x)\int _{[0,\infty )}\left( 1\wedge \tfrac{x}{z}\right) \Psi (z)\mathrm {d}z\right| \le \Vert \Psi \Vert _\rho \cdot |\Theta '(x)|x^{2-\rho }\rightarrow 0\text { as }x\rightarrow \infty . \end{aligned}$$

The claim then follows as the remaining boundary values vanish trivially. \(\square \)

Lemma 3.12

For \(\rho \in (1,2)\), let \(v_\rho \) be the self-similar profile associated to the fundamental solution of (3.21) with \(\alpha =\rho \). Then for all \(\theta _1,\theta _2>0\) the function

$$\begin{aligned} \Theta (x)=\int _\mathbb {R}y\left( 1\wedge \left| \tfrac{\theta _1}{y}\right| \right) v_\rho \left( \tfrac{x-y}{\theta _2}\right) \tfrac{\mathrm {d}y}{\theta _2} \end{aligned}$$
(3.34)

is odd, smooth, and satisfies \(\lim _{x\rightarrow \infty }\Theta '(x)x^{2-\rho }=0\) and

$$\begin{aligned} -\Theta ''(x)=\left( v_\rho \left( \tfrac{x-\theta _1}{\theta _2}\right) -v_\rho \left( \tfrac{x+\theta _1}{\theta _2}\right) \right) \tfrac{1}{\theta _2}\ge 0\text { for }x\ge 0. \end{aligned}$$
(3.35)

Proof

That \(\Theta \) is odd follows from the fact that it is the convolution of an odd and an even function, while smoothness follows since \(v_\rho \in C^\infty (\mathbb {R})\). We now note that \(\partial _x f(x-y)=-\partial _y f(x-y)\), so differentiating (3.34) we obtain

$$\begin{aligned} \Theta '(x) =-\int _\mathbb {R}y\left( 1\wedge \left| \tfrac{\theta _1}{y}\right| \right) \left[ v_\rho \left( \tfrac{x-y}{\theta _2}\right) \right] _y\tfrac{\mathrm {d}y}{\theta _2}=\int _{-\theta _1}^{\theta _1}v_\rho \left( \tfrac{x-y}{\theta _2}\right) \tfrac{\mathrm {d}y}{\theta _2}, \end{aligned}$$
(3.36)

where the second equality follows by integration by parts. Differentiating (3.36) once more, we then obtain the equality in (3.35), while the nonnegativity follows from the symmetry and monotonicity properties of \(v_\rho \). Finally, by symmetry and using the tail behaviour of \(v_\rho \) (cf. Lemma 3.8), we find that

$$\begin{aligned} \Theta '(x)=\int _{(x-\theta _1)/\theta _2}^{(x+\theta _1)/\theta _2}v_\rho (z)\mathrm {d}z \le \int _{(x-\theta _1)/\theta _2}^\infty v_\rho (z)\mathrm {d}z\sim \frac{1}{\rho }\frac{\theta _2^\rho }{x^\rho }\text { as }x\rightarrow \infty , \end{aligned}$$

hence \(\Theta '(x)x^{2-\rho }\le \frac{1}{\rho }\theta _2^\rho \cdot x^{2(1-\rho )}\rightarrow 0\) as \(x\rightarrow \infty \). \(\square \)

We are now able to prove the following.

Proposition 3.13

Under Assumption 3.1, there exists some \(R_0>0\), independent of a and \(\varepsilon \), such that the set \(\mathcal {Y}_\rho =\mathcal {Y}_\rho (R_0)\) is invariant under the evolution of the semigroup \((S_a(t))_{t\ge 0}\) as defined in Definition 3.6.

Proof

Since the semigroup \((S_a(t))_{t\ge 0}\) maps the unit ball of \(\mathcal {X}_\rho \) into itself, we only need to prove preservation of the lower bound [cf. (2.4)] for some \(R_0>0\). Let thus \(R_0>0\) be arbitrary for now, and fix any \(\Psi _0\in \mathcal {Y}_\rho =\mathcal {Y}_\rho (R_0)\). We will then write \(\Psi _a(t,x)=S(t)\Psi _0(x)\) for all \(t\ge 0\).

Now, for any \(R>0\) the function \(\vartheta (z)=1\wedge |\frac{R}{z}|\) satisfies the assumptions from Lemma 3.10, so by (3.23) and the change of variables \(y\rightarrow ye^{-t/\rho }\) we find for all \(t\in [0,T]\) and all \(R>0\) that

$$\begin{aligned}&\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) \Psi _a(t,x)\mathrm {d}x\nonumber \\&\qquad \ge e^{t(\rho -2)/\rho }\int _{[0,\infty )}\left( \int _\mathbb {R}\frac{y\left( 1\wedge |\frac{Re^{t/\rho }}{y}|\right) }{(\rho te^t)^{1/\rho }}v_\rho \left( \frac{x-y}{(\rho te^t)^{1/\rho }}\right) \mathrm {d}y\right) \tfrac{1}{x}\Psi _0(x)\mathrm {d}x,\qquad \end{aligned}$$
(3.37)

where we recall that \(v_\rho \) is the self-similar profile associated to the fundamental solution of (3.21) with \(\alpha =\rho \) (cf. Lemma 3.8). Multiplying then (3.37) by \(R^{\rho -2}\), and using Lemmas 3.11 and 3.12, we obtain

$$\begin{aligned}&R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) \Psi _a(t,x)\mathrm {d}x\nonumber \\&\quad \ge \left( Re^{t/\rho }\right) ^{\rho -2}\int _{[0,\infty )}\left( v_\rho \left( \tfrac{x-Re^{t/\rho }}{(\rho te^t)^{1/\rho }}\right) -v_\rho \left( \tfrac{x+Re^{t/\rho }}{(\rho te^t)^{1/\rho }}\right) \right) \nonumber \\&\qquad \times \left( \int _{[0,\infty )}\left( 1\wedge \tfrac{x}{z}\right) \Psi _0(z)\mathrm {d}z\right) \tfrac{\mathrm {d}x}{(\rho te^t)^{1/\rho }}, \end{aligned}$$
(3.38)

which, by the inequality in (3.35) and using the lower bound on \(\Psi _0\in \mathcal {Y}_\rho \) (cf. Definition 2.16), we can bound from below by

$$\begin{aligned}&\left( Re^{t/\rho }\right) ^{\rho -2}\int _{[0,\infty )}\left( v_\rho \left( \tfrac{x-Re^{t/\rho }}{(\rho te^t)^{1/\rho }}\right) -v_\rho \left( \tfrac{x+Re^{t/\rho }}{(\rho te^t)^{1/\rho }}\right) \right) x^{2-\rho }\lambda _\rho \big (\tfrac{x}{R_0}\big )\tfrac{\mathrm {d}x}{(\rho te^t)^{1/\rho }}\nonumber \\&\qquad =\,\left( Re^{t/\rho }\right) ^{\rho -2}\int _{\mathbb {R}}v_\rho \left( \tfrac{x-Re^{t/\rho }}{(\rho te^t)^{1/\rho }}\right) x|x|^{1-\rho }\lambda _\rho \big (\tfrac{x}{R_0}\big )\tfrac{\mathrm {d}x}{(\rho te^t)^{1/\rho }}\nonumber \\&\qquad =\left( \tfrac{Re^{t/\rho }}{R_0}\right) ^{\rho -2}u\left( \tfrac{\rho te^t}{R_0^\rho },\tfrac{Re^{t/\rho }}{R_0}\right) , \end{aligned}$$
(3.39)

where u is the solution to (3.21) with \(\alpha =\rho \) and \(u(0,x)=x|x|^{1-\rho }\lambda _\rho (x)\). In view of the fact that the left hand side of (3.38) is nonnegative for all \(R>0\), and since the right hand side of (3.39) does not depend on \(\Psi _0\) any more, it now only remains to show that, for all \(R>R_0\), with \(R_0>0\) chosen appropriately, the left hand side of (3.39) is bounded from below by \(\lambda _\rho (\frac{R}{R_0})\) as \(t\rightarrow 0\), or equivalently that

$$\begin{aligned} \left( \xi e^{t/\rho }\right) ^{\rho -2}u\left( \tfrac{\rho te^t}{R_0^\rho },\xi e^{t/\rho }\right) \ge \lambda _\rho (\xi )\text { as }t\rightarrow 0,\quad \text {for all }\xi >1. \end{aligned}$$
(3.40)

Let now \(u^*\) be the solution to (3.21) with \(\alpha =\rho \) and

$$\begin{aligned} u^*(0,x)=x|x|^{1-\rho }\left( 1-|x|^{-(2-\rho )/2}\right) \in C(\mathbb {R})\cap L^1(\mathbb {R};|x|^{-\rho -1}\mathrm {d}x). \end{aligned}$$

We then observe that \(u^*(0,\cdot )\le u(0,\cdot )\) on \(\mathbb {R}_+\), so by the maximum principle in Lemma 3.9 we have that \(u^*(\tau ,\cdot )\le u(\tau ,\cdot )\) on \(\mathbb {R}_+\) for all \(\tau >0\), and in particular

$$\begin{aligned} \left( \xi e^{t/\rho }\right) ^{\rho -2}u\left( \tfrac{\rho te^t}{R_0^\rho },\xi e^{t/\rho }\right) \ge \left( \xi e^{t/\rho }\right) ^{\rho -2}u^*\left( \tfrac{\rho te^t}{R_0^\rho },\xi e^{t/\rho }\right) \end{aligned}$$
(3.41)

for \(\xi >1\) and \(t>0\). Now, for \(x\ge 0\) and \(\tau >0\), we expand the convolution of \(u^*(0,\cdot )\) with the fundamental solution to (3.21) to obtain

$$\begin{aligned} x^{\rho -2}u^*(\tau ,x)= & {} x^{\rho -2}\int _\mathbb {R}(x-y)\left| x-y\right| ^{1-\rho }\left( 1-\left| x-y\right| ^{-(2-\rho )/2}\right) v_\rho \left( \tfrac{y}{\tau ^{1/\rho }}\right) \tfrac{\mathrm {d}y}{\tau ^{1/\rho }}\\= & {} \int _\mathbb {R}\left( 1-\tfrac{y}{x}\right) \left| 1-\tfrac{y}{x}\right| ^{1-\rho }\left( 1-x^{-(2-\rho )/2}\left| 1-\tfrac{y}{x}\right| ^{-(2-\rho )/2}\right) v_\rho \left( \tfrac{y}{\tau ^{1/\rho }}\right) \tfrac{\mathrm {d}y}{\tau ^{1/\rho }}, \end{aligned}$$

which, using the notation

$$\begin{aligned} W(x,\zeta ):=\left( 1-\zeta \right) \left| 1-\zeta \right| ^{1-\rho }\left( 1-x^{-(2-\rho )/2}\left| 1-\zeta \right| ^{-(2-\rho )/2}\right) , \end{aligned}$$

and the facts that \(\int _\mathbb {R}v_\rho (z)\mathrm {d}z=1\) and \(\int _\mathbb {R}zv_\rho (z)\mathrm {d}z=0\), yields

$$\begin{aligned} x^{\rho -2}u^*(\tau ,x)= & {} W(x,0)\nonumber \\&+\int _\mathbb {R}\left( W\left( x,\tfrac{y}{x}\right) -W(x,0)-\tfrac{y}{x}W_\zeta (x,0)\right) v_\rho \left( \tfrac{y}{\tau ^{1/\rho }}\right) \tfrac{\mathrm {d}y}{\tau ^{1/\rho }}, \end{aligned}$$
(3.42)

Next, we estimate \(v_\rho \) by its tail behaviour (cf. Lemma 3.8), to find that we can bound the absolute value of the second term on the right hand side of (3.42) by

$$\begin{aligned}&\int _\mathbb {R}\left| W\left( x,\tfrac{y}{x}\right) -W(x,0)-\tfrac{y}{x}W_\zeta (x,0)\right| \left| \tfrac{y}{\tau ^{1/\rho }}\right| ^{-1-\rho }\tfrac{\mathrm {d}y}{\tau ^{1/\rho }}\\&\quad \le \frac{\tau }{x^\rho }\int _\mathbb {R}\left| W(x,\zeta )-W(x,0)-\zeta W_\zeta (x,0)\right| \left| \zeta \right| ^{-1-\rho }\mathrm {d}\zeta =:\frac{\tau }{x^\rho }K(x), \end{aligned}$$

so noticing then that K(x) can be uniformly bounded for \(x\ge 1\) by a constant \(\kappa >0\), we obtain that

$$\begin{aligned} x^{\rho -2}u^*(\tau ,x)\ge \left( 1-x^{-(2-\rho )/2}\right) -\frac{\tau \kappa }{x^\rho }\text { for all }x\ge 1\text { and all }\tau >0. \end{aligned}$$
(3.43)

Combining thus (3.41) and (3.43) we find for \(\xi >1\) and \(t>0\) that

$$\begin{aligned} \left( \xi e^{t/\rho }\right) ^{\rho -2}u\left( \tfrac{\rho te^t}{R_0^\rho },\xi e^{t/\rho }\right)\ge & {} \left( 1-\left( \xi e^{t/\rho }\right) ^{-(2-\rho )/2}\right) -\frac{\frac{\rho te^t}{R_0^\rho }\cdot \kappa }{\left( \xi e^{t/\rho }\right) ^\rho }\nonumber \\= & {} \left( 1-\xi ^{-(2-\rho )/2}\right) \!+\!\left( 1-e^{-t(2-\rho )/(2\rho )}\right) \xi ^{-(2-\rho )/2}-t\tfrac{\rho \kappa }{R_0^\rho }\xi ^{-\rho },\nonumber \\ \end{aligned}$$
(3.44)

where we note that the first term on the right hand side equals \(\lambda _\rho (\xi )\) for \(\xi >1\) (since it is positive). Expanding then finally the exponential in the second term, we can bound the right hand side of (3.44), for \(t\rightarrow 0\), from below by

$$\begin{aligned} \lambda _\rho (\xi )+t\left( \tfrac{2-\rho }{4\rho }\xi ^{-(2-\rho )/2}-\tfrac{\rho \kappa }{R_0^\rho }\xi ^{-\rho }\right) , \end{aligned}$$

where, provided that \(R_0>0\) is such that \(R_0^\rho \ge \frac{4\rho ^2\kappa }{2-\rho }\), the second term is nonnegative for \(\xi >1\), which implies that (3.40) holds, and thus proves the claim. \(\square \)

3.4 Proof of Proposition 3.2

Proof of Proposition 3.2

Let \((S_a(t))_{t\ge 0}\) be the semigroup on the unit ball in \(\mathcal {X}_\rho \) that was defined in Definition 3.6. Following Proposition 3.13 there then exists some \(R_0>0\) such that the set \(\mathcal {Y}_\rho =\mathcal {Y}_\rho (R_0)\) (cf. Definition 2.16) is invariant under the evolution of \((S_a(t))_{t\ge 0}\). It thus remains to check, for any fixed \(t>0\), that the mapping \(\Psi _0\mapsto \Psi _a(t,\cdot ):=S_a(t)\Psi _0\) is weakly-\(*\) continuous, which by continuity of the change of variables (3.13) is equivalent to checking weak-\(*\) continuity of \(\Psi _0\mapsto H_a(t,\cdot )\).

Now, for any two measures \(\Psi _1,\Psi _2\in \{\Vert \mu \Vert _\rho \le 1\}\cap \mathcal {X}_\rho \), we write \(\Psi _a^i(s,\cdot ):=S_a(s)\Psi _i\), \(i\in \{1,2\}\), for all \(s\in [0,t]\), and we let \(H_a^i\in C([0,t]:\mathcal {X}_\rho )\) be defined via the change of variables (3.13), which are functions that satisfy (3.5) for all \(\psi \in C^1([0,t]:\mathcal {B}_0)\). What we need to show is that for any \(\psi ^*\in \mathcal {B}_0\) and any \(\delta >0\) small, there exists a weakly-\(*\) open set \(\mathcal {U}=\mathcal {U}(\psi ^*,\delta )\) such that \(\Psi _1-\Psi _2\in \mathcal {U}\) implies

$$\begin{aligned} \left| \int _{[0,\infty )}\psi ^*(X)H_a^1(t,X)\mathrm {d}X-\int _{[0,\infty )}\psi ^*(X)H_a^2(t,X)\mathrm {d}X\right| <\delta . \end{aligned}$$
(3.45)

By a density argument we may restrict ourselves to \(\psi ^*\in \mathcal {B}_1\), and by (3.5) we know that

$$\begin{aligned}&\int _{[0,\infty )}\psi ^*(X)\left( H_a^1(t,X)-H_a^2(t,X)\right) \mathrm {d}X\nonumber \\&\quad =\int _{[0,\infty )}\psi (0,X)\left( \Psi _1(X)-\Psi _2(X)\right) \mathrm {d}X, \end{aligned}$$
(3.46)

if \(\psi \in C^1([0,t]:\mathcal {B}_1)\) satisfies \(\psi (t,\cdot )=\psi ^*\) on \([0,\infty ]\), and, for all \(s\in (0,t)\),

$$\begin{aligned} 0= & {} \int _{[0,\infty )}\left( \psi _s(s,X)+\tfrac{\rho -1}{\rho }\psi (s,X)\right) \left( H_a^1(s,X)-H_a^2(s,X)\right) \mathrm {d}X\\&+\,q(H_a^1,\psi ,s)-q(H_a^2,\psi ,s) \end{aligned}$$

with

$$\begin{aligned} q(H_a^i,\psi ,s)= & {} \iint _{\{X>Y>0\}}\frac{e^{s/\rho }H_a^i(s,X)(\phi _{ae^{s/\rho }}*H_a^i(s,\cdot ))(Y)}{((X+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\\&\times \mathcal {D}_2^*[\psi (s,\cdot )](X,Y)\mathrm {d}X\mathrm {d}Y. \end{aligned}$$

Using the identity \(h_1h_1^*-h_2h_2^*=\frac{1}{2}(h_1-h_2)(h_1^*+h_2^*)+\frac{1}{2}(h_1+h_2)(h_1^*-h_2^*)\), we now rewrite the difference \(q(H_a^1,\psi ,s)-q(H_a^2,\psi ,s)\) as

$$\begin{aligned}&\frac{1}{2}\iint _{\{X>Y>0\}}\Big [(H_a^1(s,X)-H_a^2(s,X))(\phi _{ae^{s/\rho }}*(H_a^1(s,\cdot )+H_a^2(s,\cdot )))(Y)\\&\qquad +\,(H_a^1(s,X)+H_a^2(s,X))(\phi _{ae^{s/\rho }}*(H_a^1(s,\cdot )-H_a^2(s,\cdot )))(Y)\Big ]\\&\qquad \times \frac{e^{s/\rho }\mathcal {D}_2^*[\psi (s,\cdot )](X,Y)}{((X+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\mathrm {d}X\mathrm {d}Y, \end{aligned}$$

which in turn equals

$$\begin{aligned}&\int _{[0,\infty )}\left( \ell (H_a^1,\psi ;s,X)+\ell (H_a^2,\psi ;s,X)+\ell ^*(H_a^1,\psi ;s,X)+\ell ^*(H_a^2,\psi ;s,X)\right) \\&\quad \times \left( H_a^1(s,X)-H_a^2(s,X)\right) \mathrm {d}X, \end{aligned}$$

with

$$\begin{aligned} \ell (H_a^i,\psi ;s,X)=\frac{1}{2}\int _0^X\frac{e^{s/\rho }(\phi _{ae^{s/\rho }}*H_a^i(s,\cdot ))(Y)}{((X+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\mathcal {D}_2^*[\psi (s,\cdot )](X,Y)\mathrm {d}Y, \end{aligned}$$

and

$$\begin{aligned} \ell ^*(H_a^i,\psi ;s,X)= & {} \frac{1}{2}\int _{[0,\infty )}\phi _{ae^{s/\rho }}(Y-X)\int _{[Y,\infty ]}\frac{e^{s/\rho }H_a^i(s,X')}{((X'+\varepsilon e^{s/\rho })(Y+\varepsilon e^{s/\rho }))^{3/2}}\\&\times \,\mathcal {D}_2^*[\psi (s,\cdot )](X',Y)\mathrm {d}X'\mathrm {d}Y. \end{aligned}$$

We thus obtain the linear backward in time boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _s=-\tfrac{\rho -1}{\rho }\psi -\left( \ell (H_a^1,\psi )+\ell (H_a^2,\psi )+\ell ^*(H_a^1,\psi )+\ell ^*(H_a^2,\psi )\right) ,\\ \psi (t,\cdot )=\psi ^*, \end{array}\right. } \end{aligned}$$

which can be uniquely solved in \(C^1([0,t],\mathcal {B}_1)\) by a standard fixed point argument, since \(\ell (H_a^i,\cdot )\) and \(\ell ^*(H_a^i,\cdot )\) are bounded linear operators from \(\mathcal {B}_1\) to itself. Moreover, by estimate (3.14) we find that there exists a constant \(C>0\), independent of \(\psi ^*\), such that \(\Vert \psi (0,\cdot )\Vert _{\mathcal {B}_1}\le C\Vert \psi ^*\Vert _{\mathcal {B}_1}\). Now, by compactness of \(\Vert \cdot \Vert _{\mathcal {B}_1}\)-bounded sets in \(\mathcal {B}_0\), we can select finitely many \(\omega _1,\ldots ,\omega _n\in \mathcal {B}_0\) such that \(\min _i\Vert \psi (0,\cdot )-\omega _i\Vert _{\mathcal {B}_0}<\tfrac{1}{3}\delta \). For any \(i\in \{1,\ldots ,n\}\) we then write the right hand side of (3.46) as

$$\begin{aligned}&\int _{[0,\infty )}\left( \psi (0,X)-\omega _i(X)\right) \left( \Psi _1(X)-\Psi _2(X)\right) \mathrm {d}X,\\&\qquad +\int _{[0,\infty )}\omega _i(X)\left( \Psi _1(X)-\Psi _2(X)\right) \mathrm {d}X, \end{aligned}$$

so defining finally

$$\begin{aligned} \mathcal {U}=\left\{ \mu \in \mathcal {M}([0,\infty ))\,:\,\Vert \mu \Vert _\rho <\infty \text { and }\max _i\left| \int _{[0,\infty )}\omega _i(X)\mu (X)\mathrm {d}X\right| <\tfrac{1}{3}\delta \right\} , \end{aligned}$$

it follows, by choosing \(i\in \{1,\ldots ,n\}\) such that \(\Vert \psi (0,\cdot )-\omega _i\Vert _{\mathcal {B}_0}\) is minimal, that if \(\Psi _1-\Psi _2\in \mathcal {U}\), then (3.45) holds, which completes the proof. \(\square \)

3.5 Proof of Theorem 2.24

Proof of Theorem 2.24

In view of Remark 2.27 we restrict ourselves to the case \(\rho \in (1,2)\). For any \(\varepsilon >0\) fixed and \(a\in (0,\frac{\varepsilon }{2})\) arbitrary, let \(R_0>0\) and \((S_a(t))_{t\ge 0}\) be as obtained in Proposition 3.2. Then, by a variant of Tychonoff’s theorem (cf. [5, Theorem 1.2]) there exists some \(\Psi _a^\varepsilon \in \mathcal {Y}_\rho \) for which for all \(t\ge 0\) there holds \(S(t)\Psi _a^\varepsilon =\Psi _a^\varepsilon \), and which for all \(\vartheta \in \mathcal {B}_1\) satisfies

$$\begin{aligned}&\frac{1}{\rho }\int _{[0,\infty )}\left( x\vartheta '(x)+(2-\rho )\vartheta (x)\right) \Psi _a^\varepsilon (x)\mathrm {d}x\nonumber \\&\quad =\iint _{\{x>y>0\}}\frac{\Psi _a^\varepsilon (x)(\phi _a*\Psi _a^\varepsilon )(y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[\vartheta ](x,y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$
(3.47)

Further, since \(\mathcal {Y}_\rho \) is compact and independent of a, there exist \(a_n\rightarrow 0\) and \(\Psi ^\varepsilon \in \mathcal {Y}_\rho \) such that \(\Psi _{a_n}^\varepsilon {{\mathrm{\rightharpoonup {^*}}}}\Psi ^\varepsilon \) in \(\mathcal {X}_\rho \), and we will see that \(\Psi ^\varepsilon \) satisfies

$$\begin{aligned}&\frac{1}{\rho }\int _{[0,\infty )}\left( x\vartheta '(x)+(2-\rho )\vartheta (x)\right) \Psi ^\varepsilon (x)\mathrm {d}x\nonumber \\&\quad =\frac{1}{2}\iint _{[0,\infty )^2}\frac{\Psi ^\varepsilon (x)\Psi ^\varepsilon (y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[\vartheta ](x,y)\mathrm {d}x\mathrm {d}y, \end{aligned}$$
(3.48)

for all \(\vartheta \in \mathcal {B}_1\). Indeed, writing a for \(a_n\) and \(a\rightarrow 0\) for \(a_n\rightarrow 0\), for any \(\vartheta \in \mathcal {B}_1\) fixed it follows from the definition of weak-\(*\) convergence that the left hand side of (3.47) converges to the left hand side of (3.48). Convergence of the right hand side is more tricky since in the limit there might be a nontrivial contribution along the diagonal \(\{x=y\ge 0\}\). Expanding the convolution and using Fubini, we first of all rewrite the right hand side of (3.47) as

$$\begin{aligned}&\int _{[0,\infty )}\frac{\Psi _a^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\left( \int _0^x\left( \int _{[0,\infty )}\phi _a(y-z)\Psi _a^\varepsilon (z)\mathrm {d}z\right) \frac{\mathcal {D}_2^*[\vartheta ](x,y)}{(y+\varepsilon )^{3/2}}\mathrm {d}y\right) \mathrm {d}x\\&\quad =\iint _{[0,\infty )^2}\frac{\Psi _a^\varepsilon (x)\Psi _a^\varepsilon (z)}{((x+\varepsilon )(z+\varepsilon ))^{3/2}}\left( \int _0^x\frac{\phi _a(y-z)\mathcal {D}_2^*[\vartheta ](x,y)}{((z+\varepsilon )^{-1}(y+\varepsilon ))^{3/2}}\mathrm {d}y\right) \mathrm {d}x\mathrm {d}z, \end{aligned}$$

which by symmetrization equals

$$\begin{aligned} \frac{1}{2}\iint _{[0,\infty )^2}\frac{\Psi _a^\varepsilon (x)\Psi _a^\varepsilon (z)}{((x+\varepsilon )(z+\varepsilon ))^{3/2}}\mathcal {E}_a(x,z)\mathrm {d}x\mathrm {d}z, \end{aligned}$$
(3.49)

with

$$\begin{aligned} \mathcal {E}_a(x,z) =\int _{-\infty }^x\frac{\phi _a(y-z)\mathcal {D}_2^*[\vartheta ](x,y)}{((z+\varepsilon )^{-1}(y+\varepsilon ))^{3/2}}\mathrm {d}y +\int _{-\infty }^z\frac{\phi _a(y-x)\mathcal {D}_2^*[\vartheta ](z,y)}{((x+\varepsilon )^{-1}(y+\varepsilon ))^{3/2}}\mathrm {d}y, \end{aligned}$$

where the extension of the domains of integration until \(-\infty \) is possible if we extend \(\mathcal {D}_2^*[\vartheta ]\) to a continuous function on \(\mathbb {R}^2\) by setting \(\mathcal {D}_2^*[\vartheta ]=0\) on \(\mathbb {R}^2\setminus \mathbb {R}_+^2\). Observing next that

$$\begin{aligned}&\int _{-\infty }^x\phi _a(y-z)\mathrm {d}y+\int _{-\infty }^z\phi _a(y-x)\mathrm {d}y =\int _{-\infty }^{x-z}\phi _a(y)\mathrm {d}y+\int _{-\infty }^{z-x}\phi _a(y)\mathrm {d}y\\&\qquad =\int _{-\infty }^{x-z}\phi _a(y)\mathrm {d}y+\int _{x-z}^\infty \phi _a(-y)\mathrm {d}y=1\text { for all }x,z\in \mathbb {R}, \end{aligned}$$

where the last equality holds since \(\phi _a\) is even, we then find by continuity and symmetry of \(\mathcal {D}_2^*[\vartheta ]\) that for all \(\delta >0\) there exists some \(a_\delta >0\) such that

$$\begin{aligned} \left| \mathcal {E}_a(x,z)-\mathcal {D}_2^*[\vartheta ](x,z)\right| <\delta \text { for all }a\in (0,a_\delta )\text { and all }x,z\ge 0. \end{aligned}$$

Using then the fact that \((x+\varepsilon )^{-3/2}\le \varepsilon ^{-3/2}(1+\frac{x}{\varepsilon })^{-1}\le \varepsilon ^{-3/2}(1\wedge \frac{\varepsilon }{x})\) for \(x\ge 0\), the definition of the norm \(\Vert \cdot \Vert _\rho \), and the fact that \(\Vert \Psi _a^\varepsilon \Vert _\rho =1\) for all \(\Psi _a^\varepsilon \in \mathcal {Y}_\rho \), we obtain for \(a\in (0,a_\delta )\) that

$$\begin{aligned}&\left| \frac{1}{2}\iint _{[0,\infty )^2}\frac{\Psi _a^\varepsilon (x)\Psi _a^\varepsilon (z)}{((x+\varepsilon )(z+\varepsilon ))^{3/2}}\left( \mathcal {E}_a(x,z)-\mathcal {D}_2^*[\vartheta ](x,z)\right) \mathrm {d}x\mathrm {d}z\right| \\&\quad \le \frac{\delta }{2}\left( \varepsilon ^{-3/2}\int _{[0,\infty )}\left( 1\wedge \tfrac{\varepsilon }{x}\right) \Psi _a^\varepsilon (x)\mathrm {d}x\right) ^2 \le \tfrac{1}{2}\varepsilon ^{1-2\rho }\times \delta . \end{aligned}$$

Further, as \((x+\varepsilon )^{-3/2}\le r^{-1/2}\varepsilon ^{-1}(1\wedge \frac{\varepsilon }{x})\) for \(x\ge r>0\), we similarly get that

$$\begin{aligned}&\left| \frac{1}{2}\iint _{[0,\infty )^2\setminus [0,r]^2}\frac{\Psi _a^\varepsilon (x)\Psi _a^\varepsilon (z)}{((x+\varepsilon )(z+\varepsilon ))^{3/2}}\mathcal {D}_2^*[\vartheta ](x,z)\mathrm {d}x\mathrm {d}z\right| \\&\quad \le 4\Vert \vartheta \Vert _{\mathcal {B}_0}\left( \int _{[0,\infty )}\frac{\Psi _a^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x\right) \left( \int _{(r,\infty )}\frac{\Psi _a^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x\right) \\&\quad \le 4\Vert \vartheta \Vert _{\mathcal {B}_0}\varepsilon ^{(3-4\rho )/2}\times r^{-1/2}, \end{aligned}$$

so recalling that on compact squares the finite sums of products of single variable functions are dense in the uniform topology in the continuous functions, it follows that in the limit \(a\rightarrow 0\), (3.49) becomes

$$\begin{aligned} \frac{1}{2}\iint _{[0,r]^2}\frac{\Psi ^\varepsilon (x)\Psi ^\varepsilon (y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[\vartheta ](x,y)\mathrm {d}x\mathrm {d}y+O(\delta )+O(r^{-1/2}) \end{aligned}$$

as \(\delta \rightarrow 0\) and \(r\rightarrow \infty \), and taking these limits we obtain that \(\Psi ^\varepsilon \) indeed satisfies (3.48) for all \(\vartheta \in \mathcal {B}_1\).

Before we take the limit \(\varepsilon \rightarrow 0\) we will need a further estimate: We show that there is a constant \(K>0\), independent of \(\varepsilon \), such that

$$\begin{aligned} \int _{(0,z]}\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x\le K\cdot z^{1-\rho /2}\text { for all }z>0. \end{aligned}$$
(3.50)

To that end, we note that \(x\vartheta '(x)+(2-\rho )\vartheta (x)=[x\vartheta (x)]_x-(\rho -1)\vartheta (x)\), and choosing \(\vartheta \in \mathcal {B}_1\) such that the mapping \(z\mapsto z\vartheta (z)\) is nondecreasing and concave we obtain from (3.48) that

$$\begin{aligned}&\tfrac{\rho -1}{\rho }\int _{[0,\infty )}\vartheta (x)\Psi ^\varepsilon (x)\mathrm {d}x\nonumber \\&\quad \ge -\frac{1}{2}\iint _{[0,\infty )^2}\frac{\Psi ^\varepsilon (x)\Psi ^\varepsilon (y)}{((x+\varepsilon )(y+\varepsilon ))^{3/2}}\mathcal {D}_2^*[\vartheta ](x,y)\mathrm {d}x\mathrm {d}y, \end{aligned}$$
(3.51)

where \(\mathcal {D}_2^*[\vartheta ]\le 0\). Now, by an approximation argument (3.51) also holds with \(\vartheta _r(x)=(1\wedge \frac{r}{x})\), for arbitrary \(r>0\). We then use the definition of the norm \(\Vert \cdot \Vert _\rho \), and the fact that \(\Vert \Psi ^\varepsilon \Vert _\rho =1\), to estimate the left hand side of (3.51) by \(\frac{\rho -1}{\rho }r^{2-\rho }\). Noting furthermore that the second difference of an affine function is zero, we find for \(x,y\ge 0\) that

$$\begin{aligned} \mathcal {D}_2^*[\vartheta _r](x,y)= & {} \mathcal {D}_2[z\vartheta _r(z)-r](x,y)=-\mathcal {D}_2[(r-z)_+](x,y)\\= & {} -\big [(r-(x+y))_++(r-|x-y|)_+-2(r-(x\vee y))_+\big ]\\= & {} -\big [(x+y-r)_+\wedge (r-|x-y|)_+\big ], \end{aligned}$$

and it follows from (3.51) that

$$\begin{aligned} \tfrac{\rho -1}{\rho }r^{2-\rho }\ge \frac{1}{2}\iint _{[0,\infty )^2}\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\frac{\sqrt{y}\Psi ^\varepsilon (y)}{(y+\varepsilon )^{3/2}}\tfrac{(x+y-r)_+\wedge (r-|x-y|)_+}{\sqrt{xy}}\mathrm {d}x\mathrm {d}y. \end{aligned}$$
(3.52)

Set now \(\alpha =\frac{1}{3}(\sqrt{7}+1)>1\), which solves \(1-(\alpha ^2-\alpha )=\frac{1}{2}\alpha ^2\), and notice that \((xy)^{-1/2}(r-|x-y|)\ge (\alpha ^2r)^{-1}(r-(\alpha ^2-\alpha )r)=\frac{1}{2}\) for \(x,y\in (\alpha r,\alpha ^2 r]\). We then restrict the domain of integration on the right hand side of (3.52) to \((\alpha r,\alpha ^2 r]^2\) to obtain

$$\begin{aligned} \frac{1}{4}\left( \int _{(\alpha r,\alpha ^2 r]}\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x\right) ^2\le \tfrac{\rho -1}{\rho }r^{2-\rho }, \end{aligned}$$

hence there holds

$$\begin{aligned} \int _{(\alpha ^{-1}r,r]}\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x\le \left( 2\alpha ^{\rho -2}\sqrt{\tfrac{\rho -1}{\rho }}\right) r^{1-\rho /2}\text { for all }r>0, \end{aligned}$$

and using the decomposition \((0,z]=\bigcup _{j=0}^\infty (\alpha ^{-j-1}z,\alpha ^{-j}z]\) we obtain (3.50).

Now, as \(\mathcal {Y}_\rho \) is independent of \(\varepsilon \), there also exist \(\varepsilon _n\rightarrow 0\) and \(\Psi _\rho \in \mathcal {Y}_\rho \) such that \(\Psi ^{\varepsilon _n}{{\mathrm{\rightharpoonup {^*}}}}\Psi _\rho \) in \(\mathcal {X}_\rho \). Writing then \(\varepsilon \) for \(\varepsilon _n\) and \(\varepsilon \rightarrow 0\) for \(\varepsilon _n\rightarrow 0\), the left hand side of (3.48), by definition of weak-\(*\) convergence, converges to

$$\begin{aligned}&\frac{1}{\rho }\int _{[0,\infty )}\left( x\vartheta '(x)+(2-\rho )\vartheta (x)\right) \Psi _\rho (x)\mathrm {d}x\nonumber \\&\quad =\frac{1}{\rho }\int _{[0,\infty )}\big (x[x\vartheta (x)]_x-(\rho -1)x\vartheta (x)\big )\tfrac{1}{x}\Psi _\rho (x)\mathrm {d}x. \end{aligned}$$
(3.53)

We next check that (3.50) carries over to the limit. Let thereto \(\eta _\delta \in C(\mathbb {R})\), for \(\delta >0\), be nondecreasing with \(\eta _\delta =0\) on \((-\infty ,\delta )\) and \(\eta _\delta =1\) on \((2\delta ,\infty )\). Using then (3.50), for all \(z>0\) we obtain

$$\begin{aligned} \int _{(0,z]}\tfrac{1}{x}\Psi _\rho (x)\mathrm {d}x= & {} \lim _{\delta \rightarrow 0}\int _{(0,z+2\delta ]}\tfrac{1}{x}\Psi _\rho (x)(\eta _\delta (x)-\eta _\delta (x-z))\mathrm {d}x\nonumber \\= & {} \lim _{\delta \rightarrow 0}\lim _{\varepsilon \rightarrow 0}\int _{(0,z+2\delta ]}\frac{\sqrt{x}\Psi _\rho (x)}{(x+\varepsilon )^{3/2}}(\eta _\delta (x)-\eta _\delta (x-z))\mathrm {d}x \le K\cdot z^{1-\rho /2}.\nonumber \\ \end{aligned}$$
(3.54)

Now, with \(\eta _\delta \) as above and using (3.50), we find for any \(\psi \in C([0,\infty ])\) that

$$\begin{aligned}&\left| \int _{[0,\infty )}\psi (x)\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x-\int _{[0,\infty )}\frac{\eta _\delta (x)\psi (x)}{(1+\frac{\varepsilon }{x})^{3/2}}\frac{1}{x}\Psi ^\varepsilon (x)\mathrm {d}x\right| \\&\quad \le \int _{(0,2\delta ]}\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\psi (x)\mathrm {d}x\le 2^{1-\rho /2}K\Vert \psi \Vert _{C([0,\infty ])}\times \delta ^{1-\rho /2}, \end{aligned}$$

so similarly using (3.54), and since \(\Psi _\rho \in \mathcal {X}_\rho \) implies \(\Psi _\rho (\{0\})=0\), we obtain

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\int _{[0,\infty )}\psi (x)\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x =\int _{[0,\infty )}\psi (x)\tfrac{1}{x}\Psi _\rho (x)\mathrm {d}x. \end{aligned}$$

We thus find that the finite measures \(\frac{\sqrt{x}\Psi ^\varepsilon (x)}{(x+\varepsilon )^{3/2}}\mathrm {d}x\) converge with respect to the weak-\(*\) topology in \((C([0,\infty ]))'\) to some \(\Phi _\rho \in \mathcal {X}_2\) that satisfies \(\Phi _\rho (x)=\frac{1}{x}\Psi _\rho (x)\) for \(x>0\). Recall now that \(|\mathcal {D}_2^*[\vartheta ](x,y)|\le 4\Vert \vartheta \Vert _{\mathcal {B}_0}\) for all \(x,y\ge 0\), and also \(|\mathcal {D}_2^*[\vartheta ](x,y)|\le 2\Vert \vartheta \Vert _{\mathcal {B}_1}(x\wedge y)\). Then \((xy)^{-1/2}\mathcal {D}_2^*[\vartheta ](x,y)\) is bounded and continuous on \([0,\infty ]^2\), and vanishes uniformly on the boundary, and we thus obtain that the right hand side of (3.48) converges to the right hand side of

$$\begin{aligned}&\frac{1}{\rho }\int _{[0,\infty )}\big (x[x\vartheta (x)]_x-(\rho -1)x\vartheta (x)\big )\Phi _\rho (x)\mathrm {d}x\nonumber \\&\quad =\iint _{[0,\infty )^2}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\mathcal {D}_2^*[\vartheta ](x,y)\mathrm {d}x\mathrm {d}y, \end{aligned}$$
(3.55)

which \(\Phi _\rho \) satisfies for all \(\vartheta \in \mathcal {B}_1\) [cf. (3.53)]. To conclude, we note that by appropriate approximation it can be shown that \(\Phi _\rho \) satisfies (3.55) for all \(\vartheta \) for which the mapping \(z\mapsto z\vartheta (z)\) is differentiable and constant from a certain point onwards towards infinity. The proof is then completed by the observation that any \(\varphi \in C_c^1([0,\infty ))\) defines such a function via \(\varphi (x)-\varphi (0)= x\vartheta (x)\). \(\square \)

4 Regularity, and a Decay Result

We start this section with a useful integrability estimate, which can be seen as an improvement on (3.50).

Lemma 4.1

Given \(\rho \in (1,2]\), if \(\Phi _\rho \in \mathcal {X}_2\) satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), then there exists a constant \(C>0\) such that

$$\begin{aligned} \int _{(0,R]}\Phi _\rho (x)\mathrm {d}x\le C\cdot \sqrt{R}\text { for all }R>0. \end{aligned}$$
(4.1)

Proof

Given any nonincreasing convex function \(\varphi \in C_c^1([0,\infty ))\), we obtain from (2.10) that there holds

$$\begin{aligned}&\tfrac{\rho -1}{\rho }\int _{[0,\infty )}\left( \varphi (0)-\varphi (x)\right) \Phi _\rho (x)\mathrm {d}x\nonumber \\&\quad \ge \frac{1}{2}\iint _{[0,\infty )^2}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\mathcal {D}_2[\varphi ](x,y)\mathrm {d}x\mathrm {d}y, \end{aligned}$$
(4.2)

so using appropriate arguments to approximate \(\varphi (x)=(r-x)_+\), with \(r>0\) arbitrary, and arguing as in the proof of (3.50), we find from (4.2) that

$$\begin{aligned} \tfrac{\rho -1}{\rho }\Vert \Phi _\rho \Vert _2\cdot r\ge \tfrac{1}{2}\mathcal {I}[\Phi _\rho ](r), \end{aligned}$$
(4.3)

with

$$\begin{aligned} \mathcal {I}[\Phi _\rho ](r):=\iint _{\mathbb {R}_+^2}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}(x+y-r)_+\wedge (r-|x-y|)_+\mathrm {d}x\mathrm {d}y. \end{aligned}$$
(4.4)

Also as in the proof of (3.50), we then restrict the domain of integration on the right hand side of (4.3) to \((\alpha r,\alpha ^2 r]^2\), with \(\alpha =\frac{1}{3}(\sqrt{7}+1)\), to obtain

$$\begin{aligned} \frac{1}{4}\left( \int _{(\alpha r,\alpha ^2r]}\Phi _\rho (x)\mathrm {d}x\right) ^2\le \tfrac{\rho -1}{\rho }\Vert \Phi _\rho \Vert _2\cdot r, \end{aligned}$$

hence there holds

$$\begin{aligned} \int _{(\alpha ^{-1}r,r]}\Phi _\rho (x)\mathrm {d}x\le \tfrac{2}{\alpha }\sqrt{\tfrac{\rho -1}{\rho }\Vert \Phi _\rho \Vert _2}\cdot \sqrt{r}\text { for all }r>0, \end{aligned}$$

and (4.1) follows by the decomposition \((0,R]=\bigcup _{j=0}^\infty (\alpha ^{-j-1}R,\alpha ^{-j}R]\). \(\square \)

We now first show that self-similar profiles are Hölder continuous.

Lemma 4.2

Given \(\rho \in (1,2]\), if \(\Phi _\rho \in \mathcal {X}_2\) satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), then it is absolutely continuous with respect to Lebesgue measure, its Radon–Nykodim derivative is locally \(\alpha \)-Hölder continuous on \((0,\infty )\) for any \(\alpha <\frac{1}{2}\), and it actually satisfies (2.9) for all \(\varphi \in C_c^1([0,\infty ))\).

Proof

Given \(\chi \in C_c^\infty ((0,\infty ))\), we set \(\varphi (x)=-\int _x^\infty \frac{1}{z}\chi (z)\mathrm {d}z\) and use this function in (2.10) to obtain

$$\begin{aligned}&\int _{(0,\infty )}\chi (x)\Phi _\rho (x)\mathrm {d}x-(\rho -1)\int _{(0,\infty )}\int _0^x\tfrac{1}{z}\chi (z)\mathrm {d}z\Phi _\rho (x)\mathrm {d}x\nonumber \\&\quad =\frac{\rho }{2}\iint _{[0,\infty )^2}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\left( \int _{x\vee y}^{x+y}\tfrac{1}{z}\chi (z)\mathrm {d}z-\int _{|x-y|}^{x\vee y}\tfrac{1}{z}\chi (z)\mathrm {d}z\right) \mathrm {d}x\mathrm {d}y. \end{aligned}$$
(4.5)

Writing then \(\Sigma _\chi =\mathrm{supp}(\chi )\) and \(\varsigma _\chi =\frac{1}{2}\min (\Sigma _\chi )\), we first of all note that

$$\begin{aligned} \left| \int _{(0,\infty )}\int _0^x\tfrac{1}{z}\chi (z)\mathrm {d}z\Phi _\rho (x)\mathrm {d}x\right|\le & {} \Phi _\rho ([2\varsigma _\chi ,\infty ))\times \int _{\Sigma _\chi }\tfrac{1}{z}|\chi (z)|\mathrm {d}z\\\le & {} \left( \Phi _\rho ([2\varsigma _\chi ,\infty ))\times \Vert \tfrac{1}{z}\Vert _{L^q(\Sigma _\chi )}\right) \Vert \chi \Vert _{L^p(\Sigma _\chi )}, \end{aligned}$$

with \(p\in [1,\infty )\) and \(q=\frac{p}{p-1}\), and similarly we find that

$$\begin{aligned}&\left| \iint _{[\varsigma _\chi ,\infty )^2}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\left( \int _{x\vee y}^{x+y}\tfrac{1}{z}\chi (z)\mathrm {d}z-\int _{|x-y|}^{x\vee y}\tfrac{1}{z}\chi (z)\mathrm {d}z\right) \mathrm {d}x\mathrm {d}y\right| \\&\quad \le \tfrac{1}{\varsigma _\chi }(\Phi _2([\varsigma _\chi ,\infty )))^2\times 2\int _{\Sigma _\chi }\tfrac{1}{z}|\chi (z)|\mathrm {d}z\\&\quad \le \left( \tfrac{2}{\varsigma _\chi }(\Phi _2([\varsigma _\chi ,\infty )))^2\times \Vert \tfrac{1}{z}\Vert _{L^q(\Sigma _\chi )}\right) \Vert \chi \Vert _{L^p(\Sigma _\chi )}. \end{aligned}$$

Noticing now that the term between brackets in the double integral on the right hand side of (4.5) vanishes on \(\{x+y\le 2\varsigma _\chi \}\), by symmetry it remains only to estimate the integral over \((x,y)\in [\varsigma _\chi ,\infty ]\times (0,\varsigma _\chi ]\). We thereto let \(p\in (1,\infty )\) be arbitrary, \(r\in (\frac{p}{p-1},\infty )\) large, and \(q\in (1,\infty ]\) such that \(1=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\), and we obtain that

$$\begin{aligned}&\left| \iint _{[\varsigma _\chi ,\infty )\times (0,\varsigma _\chi ]}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\left( \int _x^{x+y}\tfrac{1}{z}\chi (z)\mathrm {d}z-\int _{x-y}^x\tfrac{1}{z}\chi (z)\mathrm {d}z\right) \mathrm {d}(x,y)\right| \\&\quad \le \int _{[\varsigma _\chi ,\infty )}\frac{\Phi _\rho (x)}{\sqrt{x}}\mathrm {d}x\times \int _{(0,\varsigma _\chi ]}2\Vert \tfrac{1}{z}\chi (z)\Vert _{L^{\frac{r}{r-1}}(\Sigma _\chi )}y^{\frac{1}{r}-\frac{1}{2}}\Phi _\rho (y)\mathrm {d}y, \end{aligned}$$

which, using for the integral with respect to y a dyadic decomposition and Lemma 4.1, can be bounded by a constant times \(\Vert \chi \Vert _{L^p(\Sigma _\chi )}\). Note here that the dependence on \(\chi \) of the constants in the preceding estimates is limited to dependence on \(\varsigma _\chi \). Combining thus the preceding estimates and using a density argument, we then find for any \(p\in (1,\infty )\) and any \(K\subset (0,\infty ]\) compact that

$$\begin{aligned} \left| \int _{K}\chi (x)\Phi _\rho (x)\mathrm {d}x\right| \le C(\Phi _\rho ,\min (K),p,\rho )\Vert \chi \Vert _{L^p(K)}\text { for all }\chi \in L^p(K), \end{aligned}$$
(4.6)

with \(C(\Phi _\rho ,k,p,\rho )\le c(\Phi _\rho ,p,\rho )O(k^{-\frac{1}{p}}\Phi _2([k,\infty )))\) as \(k\rightarrow \infty \). By duality it now follows that \(\Phi _\rho \in \bigcap _{q\in (1,\infty )}L_\mathrm{loc}^q((0,\infty ])\), and since \(\Phi _\rho \in \mathcal {X}_2\) is finite we have \(\Phi _\rho \in L^1(0,\infty )\). Moreover, from the dependence on \(\min (K)\) of the constant C in (4.6) we find for all \(q\in [1,\infty )\) that \(\Vert \Phi _\rho \Vert _{L^q(r,\infty )}\le O\big (r^{\frac{1}{q}-1}\Vert \Phi _2\Vert _{L^1(r,\infty )}\big )\) as \(r\rightarrow \infty \). Note further that the contribution of integrals over lines to the double integral on the right hand side of (2.10) is zero for Lebesgue integrable functions, so \(\Phi _\rho \) actually satisfies (2.9) for all \(\varphi \in C_c^1([0,\infty ))\).

For the remaining continuity claim we fix \(\gamma \in (0,\frac{1}{2})\) and \([a,b]\subset (0,\infty )\) arbitrarily, and we start by showing that for any \(\varphi \in C_c^\infty (\mathbb {R})\) with support in [ab] we have

$$\begin{aligned} \left| \int _\mathbb {R}\varphi '(x)\,x\Phi _\rho (x)\mathrm {d}x\right| \le C(\Phi _\rho ,a,b,\gamma ,\rho )\Vert \varphi \Vert _{H^\gamma (\mathbb {R})}. \end{aligned}$$
(4.7)

Indeed, by (2.9) we immediately have

$$\begin{aligned}&\left| \int _{(0,\infty )}\varphi '(x)\,x\Phi _\rho (x)\mathrm {d}x\right| \le (\rho -1)\Vert \Phi _\rho \Vert _{L^2(a,b)}\Vert \varphi \Vert _{L^2(\mathbb {R})}\\&\quad +\rho \left| \iint _{S_1}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\Delta _y^2\varphi (x)\mathrm {d}x\mathrm {d}y\right| +\rho \left| \iint _{S_2}\Big [\cdots \Big ]\mathrm {d}x\mathrm {d}y\right| , \end{aligned}$$

where we have split the double integral over the domains

$$\begin{aligned} S_1=\{(y\vee (a-y))<x<y+b<2b\}\quad \text {and}\quad S_2=\left\{ x>y>\left( \tfrac{x}{2}\vee b\right) \right\} . \end{aligned}$$

For the integral over \(S_1\) we now first note for all \(y\in [0,b]\) that

$$\begin{aligned} \left| \int _{(y\vee (a-y),y+b)}\Phi _\rho (x)\Delta _y^2\varphi (x)\mathrm {d}x\right| \le 2\Vert \Phi _\rho \Vert _{L^2(\frac{a}{2},2b)}\Vert \Delta _y^2\varphi \Vert _{L^2(\mathbb {R})}, \end{aligned}$$

and since \(\Vert \Delta _y^2\varphi \Vert _{L^2(\mathbb {R})}\le C(\gamma )\Vert \varphi \Vert _{H^\gamma (\mathbb {R})}\,y^\gamma \) (cf. [1921]) we obtain

$$\begin{aligned} \left| \iint _{S_1}\Big [\cdots \Big ]\mathrm {d}x\mathrm {d}y\right| \le C(\Phi _\rho ,a,b)C(\gamma )\sqrt{\frac{2}{a}}\int _{(0,b)}y^{\gamma -\frac{1}{2}}\Phi _\rho (y)\mathrm {d}y\times \Vert \varphi \Vert _{H^\gamma (\mathbb {R})}, \end{aligned}$$

where the remaining integral with respect to y is bounded (use Lemma 4.1 and a dyadic decomposition of the interval (0, b)). For the integral over \(S_2\) we note that the second difference of \(\varphi \) is now completely given by \(\varphi (x-y)\). Applying thus Hölder’s and Young’s inequalities we obtain

$$\begin{aligned} \left| \iint _{S_2}\Big [\cdots \Big ]\mathrm {d}x\mathrm {d}y\right|&\le \frac{1}{b}\Vert \Phi _\rho \Vert _{L^2(b,\infty )}\left\| \int _{(b,\infty )}\Phi _\rho (y)|\varphi (z-y)|\mathrm {d}y\right\| _{L^2(b,\infty )}\\&\le C(\Phi _\rho ,b)\Vert \Phi _\rho \Vert _{L^1(0,\infty )}\Vert \varphi \Vert _{L^2(\mathbb {R})}, \end{aligned}$$

and it follows that (4.7) holds. We lastly fix any \(\zeta \in C_c^\infty ((a,b))\) with \(\zeta (x)=\frac{1}{x}\) for \(x\in I_{a,b}:=\big [\frac{2a+b}{3},\frac{a+2b}{3}\big ]\), and we set \(\Theta (x):=\zeta (x)x\Phi _\rho (x)\). Given then any \(\varphi \in C^\infty (\mathbb {R})\), we use (4.7) to get

$$\begin{aligned} \left| \int _\mathbb {R}\varphi '(x)\Theta (x)\mathrm {d}x\right|&\le \left| \int _\mathbb {R}(\zeta \varphi )'(x)\,x\Phi _\rho (x)\mathrm {d}x\right| +\left| \int _\mathbb {R}\zeta '(x)\varphi (x)\,x\Phi _\rho (x)\mathrm {d}x\right| \\&\le C(\Phi _\rho ,a,b,\gamma ,\rho )\Vert \zeta \varphi \Vert _{H^\gamma (\mathbb {R})}+\Vert z\zeta '(z)\Vert _\infty \Vert \Phi _\rho \Vert _{L^2(a,b)}\Vert \varphi \Vert _{L^2(\mathbb {R})}, \end{aligned}$$

and since \(C^\infty (\mathbb {R})\) is dense in \(H^\gamma (\mathbb {R})\) we obtain

$$\begin{aligned} \left| \int _\mathbb {R}\varphi '(x)\Theta (x)\mathrm {d}x\right| \le C(\Phi _\rho ,a,b,\gamma ,\zeta ,\rho )\Vert \varphi \Vert _{H^\gamma (\mathbb {R})}\text { for all }\varphi \in H^\gamma (\mathbb {R}), \end{aligned}$$

hence \(\Theta '\in H^{-\gamma }(\mathbb {R})=(H^\gamma (\mathbb {R}))'\). Therefore \(\Theta \in H^{1-\gamma }(\mathbb {R})\subset C^{0,\frac{1}{2}-\gamma }(\mathbb {R})\) (cf. [2, Theorem 19.6(b)] and [20, Section 2.7.1, Remark 2] resp.) and since \(\Theta =\Phi _\rho \) on \(I_{a,b}\) we have \(\Phi _\rho \in C^{0,\frac{1}{2}-\gamma }(I_{a,b})\). We then complete the proof by observing that \(\gamma \in (0,\frac{1}{2})\) was chosen arbitrarily, and that for any \(K\subset (0,\infty )\) compact there are \(a,b\in (0,\infty )\) such that \(K\subset I_{a,b}\). \(\square \)

We are now able to prove Propositions 2.25 and 2.26.

Proof of Proposition 2.25

By Lemma 4.2 we have that \(\Phi _\rho \in \bigcap _{\alpha <\frac{1}{2}}C^{0,\alpha }((0,\infty ))\), and \(\Phi _\rho \) satisfies (2.9) for all \(\varphi \in C_c^1([0,\infty ))\). We will prove smoothness via a bootstrap argument, for which we need to rewrite (2.9). For any \(\delta >0\), let \(\eta _\delta \in C([0,\infty ])\) be a nondecreasing function with \(\eta _\delta =0\) on \([0,\delta )\) and \(\eta _\delta =1\) on \((2\delta ,\infty ]\). For fixed \(\varphi \in C_c^\infty ((0,\infty ))\) it then holds by dominated convergence that

$$\begin{aligned}&\int _{(0,\infty )}\left( \tfrac{1}{\rho }[x\varphi (x)]_x-\varphi (x)\right) \Phi _\rho (x)\mathrm {d}x\nonumber \\&\quad =\lim _{\delta \rightarrow 0}\iint _{\{x>y>0\}}\frac{\eta _\delta (x)\Phi _\rho (x)\eta _\delta (y)\Phi _\rho (y)}{\sqrt{xy}}\Delta _y^2\varphi (x)\mathrm {d}x\mathrm {d}y, \end{aligned}$$
(4.8)

and if \(\delta <\frac{1}{4}\min (\mathrm{supp}(\varphi ))\), then

$$\begin{aligned}&\iint _{\{x>y>0\}}\frac{\eta _\delta (x)\Phi _\rho (x)\eta _\delta (y)\Phi _\rho (y)}{\sqrt{xy}}\Delta _y^2\varphi (x)\mathrm {d}x\mathrm {d}y\nonumber \\&\quad =\int _{(0,\infty )}\left( \int _0^{x/2}\frac{\eta _\delta (y)\Phi _\rho (y)}{\sqrt{y}}\left[ \frac{\Phi _\rho (x+y)}{\sqrt{x+y}}+\frac{\Phi _\rho (x-y)}{\sqrt{x-y}}-2\frac{\Phi _\rho (x)}{\sqrt{x}}\right] \mathrm {d}y\right. \nonumber \\&\quad \quad \left. +\int _{x/2}^\infty \frac{\Phi _\rho (y)\Phi _\rho (x+y)}{\sqrt{y(x+y)}}\mathrm {d}y-2\frac{\Phi _\rho (x)}{\sqrt{x}}\int _{x/2}^x\frac{\Phi _\rho (y)}{\sqrt{y}}\mathrm {d}y\right) \varphi (x)\mathrm {d}x. \end{aligned}$$
(4.9)

Using then the local Hölder regularity of \(\Phi _\rho \) and the integral estimate from Lemma 4.1, we find that \(\frac{\Phi _\rho (y)}{\sqrt{y}}\left[ \frac{\Phi _\rho (x+y)}{\sqrt{x+y}}+\frac{\Phi _\rho (x-y)}{\sqrt{x-y}}-2\frac{\Phi _\rho (x)}{\sqrt{x}}\right] \) is integrable with respect to y near zero, and we are able to take the limit \(\delta \rightarrow 0\) in the right hand side of (4.9). Combining (4.8) and (4.9), we thus obtain that \(\Phi _\rho \) satisfies (2.11), where the derivative on the left hand is still taken in the distributional sense.

Suppose now that \(\Phi _\rho \in C^{k,\alpha }((0,\infty ))\) for some \(k\in \mathbb {N}_0\) and \(\alpha \in (0,1)\). To show that \(\Phi _\rho \in C^{k+1,\alpha -\epsilon }((0,\infty ))\) for some arbitrarily small \(\epsilon >0\), it then suffices to check that the right hand side of (2.11) is in \(C^{k,\alpha -\epsilon }((0,\infty ))\), and since the second and third terms are actually even more regular it is enough to check this for the first. Moreover, writing \(f(x)=\frac{\Phi _\rho (x)}{\sqrt{x}}\) we observe that

$$\begin{aligned} f\big (\tfrac{1}{2}x\big )\left[ f^{(\ell )}\big (\tfrac{3}{2}x\big )+f^{(\ell )}\big (\tfrac{1}{2}x\big )-2f^{(\ell )}(x)\right] \in C^{k-\ell ,\alpha }((0,\infty ))\text { for }\ell =0,1,\ldots ,k, \end{aligned}$$

and we can restrict ourselves to proving that \(f\in C^{0,\alpha }((0,\infty ))\) implies

$$\begin{aligned} \int _0^{x/2}\frac{\Phi _\rho (y)}{\sqrt{y}}\big [f(x+y)+f(x-y)-2f(x)\big ]\mathrm {d}y=:F(x)\in C^{0,\alpha -\epsilon }((0,\infty )). \end{aligned}$$
(4.10)

Let thereto \(K\subset [k_1,k_2]\subset (0,\infty )\) be compact, and let \(\kappa >0\) be a constant such that \(|f(x)-f(y)|\le \kappa |x-y|^\alpha \) for all \(x,y\in [\frac{1}{2}k_1,2k_2]\). For \(x_1,x_2\in K\) with \(x_1\le x_2\) there then holds

$$\begin{aligned} \left| F(x_1)-F(x_2)\right|\le & {} \left| \int _{x_1/2}^{x_2/2}\frac{\Phi _\rho (y)}{\sqrt{y}}\Delta _y^2f(x_2)\mathrm {d}y\right| \\&+\int _0^{x_1/2}\frac{\Phi _\rho (y)}{\sqrt{y}}\left| \Delta _y^2f(x_1)-\Delta _y^2f(x_2)\right| \mathrm {d}y, \end{aligned}$$

where the first term on the right hand side is bounded by a constant times \(|x_1-x_2|\). Writing further \(\xi =\min \{\frac{1}{2}x_1,x_2-x_1\}\), we find that

$$\begin{aligned}&\int _0^{x_1/2}\frac{\Phi _\rho (y)}{\sqrt{y}}\left| \Delta _y^2f(x_1)-\Delta _y^2f(x_2)\right| \mathrm {d}y\\&\quad \le 4\kappa \left( \int _0^{\xi }\frac{\Phi _\rho (y)}{\sqrt{y}}y^\alpha \mathrm {d}y+|x_1-x_2|^\alpha \int _\xi ^{x_1/2}\frac{\Phi _\rho (y)}{\sqrt{y}}\mathrm {d}y\right) , \end{aligned}$$

which, using dyadic decompositions of the domains of integration and the estimate from Lemma 4.1, can be bounded by a constant times

$$\begin{aligned} |x_1-x_2|^\alpha (1+\log |x_1-x_2|)\le |x_1-x_2|^{\alpha -\epsilon }\text { as }|x_1-x_2|\rightarrow 0, \end{aligned}$$

with \(\epsilon >0\) arbitrarily small, and we have shown (4.10). By induction it then follows that \(\Phi _\rho \in C^\infty ((0,\infty ))\).

To lastly prove our positivity claim we suppose that there is some \(x\in (0,\infty )\) such that \(\Phi _\rho (x)=0\). As we have \(\Phi _\rho \ge 0\) on \((0,\infty )\) there then holds \(\Phi _\rho '(x)=0\), and it follows from (2.11) that \(\Phi _\rho (y)\Phi _\rho (x+y)=0\) for all \(y\in (0,\infty )\), and that \(\Phi _\rho (y)\Phi _\rho (x-y)=0\) for all \(y\in (0,x)\). As a consequence of the latter identity we find that \(\Phi _\rho (\frac{x}{2})=0\), hence \(\Phi _\rho (2^{-n}x)=0\) for all \(n\in \mathbb {N}\) by induction. From the iterated first identity we therefore have

$$\begin{aligned} \Phi _\rho (y)\Phi _\rho (2^{-n}x+y)=0\text { for all }y\in (0,\infty )\text { and all }n\in \mathbb {N}, \end{aligned}$$

so local uniform continuity implies \((\Phi _\rho (y))^2=0\) for all \(y\in K\) with \(K\subset (0,\infty )\) compact, hence \(\Phi _\rho =0\) on \((0,\infty )\), and we conclude that \(\Phi _\rho \) is indeed either strictly positive or identically zero on \((0,\infty )\). \(\square \)

Proof of Proposition 2.26

Since the rescaling statement is an easy exercise, we restrict ourselves to proving that \(x\Phi _\rho (x)\in \mathcal {X}_\rho \). Moreover, in view of Remark 2.27 we restrict ourselves to the case \(\rho \in (1,2)\), but see also Lemma 6.1.

Now, we first show that

$$\begin{aligned} R^{\rho -2}\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) x\Phi _\rho (x)\mathrm {d}x=\frac{\rho }{2}\int _0^R\mathcal {I}[\Phi _\rho ](r)r^{\rho -3}\mathrm {d}r\text { for all }R>0, \end{aligned}$$
(4.11)

with \(\mathcal {I}\) given by (4.4). To that end we note that by continuity of \(\Phi _\rho \) (cf. Proposition 2.25), after an approximation argument we can use \(\varphi (x)=(r-x)_+\), with \(r>0\) arbitrary, directly in (2.9) to obtain

$$\begin{aligned} \frac{1}{\rho }\left( (\rho -2)\int _0^rx\Phi _\rho (x)\mathrm {d}x+(\rho -1)r\int _r^\infty \Phi _\rho (x)\mathrm {d}x\right) =\tfrac{1}{2}\mathcal {I}[\Phi _\rho ](r). \end{aligned}$$
(4.12)

Using then the estimate from Lemma 4.1 in a dyadic decomposition for the first integral on the left hand side of (4.12), we find that for small \(r>0\) the second term is dominant and of order O(r) as \(r\rightarrow 0\). As a consequence the product of (4.12) and \(r^{\rho -3}\) is integrable near zero, and for any \(R>0\) we find that the right hand side of (4.11) equals

$$\begin{aligned} \int _0^R\left[ r^{\rho -2}\int _0^rx\Phi _\rho (x)\mathrm {d}x+r^{\rho -1}\int _r^\infty \Phi _\rho (x)\mathrm {d}x\right] _r\mathrm {d}r, \end{aligned}$$

hence (4.11) holds.

We next note that \(\mathcal {I}[\Phi _\rho ]\ge 0\), so the right hand side of (4.11) is nondecreasing as a function of R, hence the supremum over \(R>0\) is given by the limit \(R\rightarrow \infty \). Observing lastly that \((xy)^{-1/2}(x+y-r)_+\wedge (r-|x-y|)_+\le 1\) for all \(x,y\ge 0\) and \(r>0\), we obtain the uniform bound \(\mathcal {I}[\Phi _\rho ]\le \Vert \Phi _\rho \Vert _2^2\), hence the right hand side of (4.11) is bounded as a function of R, and we conclude that \(\Vert x\Phi _\rho (x)\Vert _\rho <\infty \). \(\square \)

5 Power Law Asymptotics: Theorem 2.28

The proof of Theorem 2.28 is given after the following useful result.

Lemma 5.1

Given \(\rho \in (1,2)\), if \(\Phi _\rho \in \mathcal {X}_2\) satisfies (2.10) for all \(\varphi \in C_c^1([0,\infty ))\), then the limits

$$\begin{aligned} \lim _{R\rightarrow \infty }\frac{R^{\rho -1}}{2-\rho }\int _{(R,\infty )}\Phi _\rho (x)\mathrm {d}x\quad \text {and}\quad \lim _{R\rightarrow \infty }\frac{R^{\rho -2}}{\rho -1}\int _{(0,R)}x\Phi _\rho (x)\mathrm {d}x \end{aligned}$$

exist, and both equal \(\Vert x\Phi _\rho (x)\Vert _\rho \).

Proof

Recalling from the proof of Proposition 2.26 that \(\Phi _\rho \) satisfies (4.12) for all \(r>0\), we find by a rearrangement of terms that

$$\begin{aligned}&R\int _{(R,\infty )}\Phi _\rho (x)\mathrm {d}x-(2-\rho )\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) x\Phi _\rho (x)\mathrm {d}x=\tfrac{\rho }{2}\,\mathcal {I}[\Phi _\rho ](R)\nonumber \\&\quad =(\rho -1)\int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) x\Phi _\rho (x)\mathrm {d}x-\int _{(0,R)}x\Phi _\rho (x)\mathrm {d}x\text { for all }R>0. \end{aligned}$$
(5.1)

Also from the proof of Proposition 2.26 [cf. (4.11)], we know that

$$\begin{aligned} \int _{[0,\infty )}\left( 1\wedge \tfrac{R}{x}\right) x\Phi _\rho (x)\mathrm {d}x=R^{2-\rho }\left( \Vert x\Phi _\rho (x)\Vert _\rho -\frac{\rho }{2}\int _R^\infty \mathcal {I}[\Phi _\rho ](r)r^{\rho -3}\mathrm {d}r\right) , \end{aligned}$$

so we can rewrite the first equality in (5.1) as

$$\begin{aligned}&\frac{R^{\rho -1}}{2-\rho }\int _{(R,\infty )}\Phi _\rho (x)\mathrm {d}x-\Vert x\Phi _\rho (x)\Vert _\rho \\&\quad =\frac{\rho }{2}\left( \frac{\mathcal {I}[\Phi _\rho ](R)R^{\rho -2}}{2-\rho }-\int _R^\infty \mathcal {I}[\Phi _\rho ](r)r^{\rho -3}\mathrm {d}r\right) , \end{aligned}$$

where the right hand side tends to zero as \(R\rightarrow \infty \). Doing the same for the second equality in (5.1) completes the proof. \(\square \)

Proof of Theorem 2.28

Recall that \(\Phi _\rho \in C^\infty ((0,\infty ))\) (cf. Proposition 2.25).

We first remark that for any \(r>0\) we have

$$\begin{aligned} \left| \frac{\Phi _\rho (r)}{(2-\rho )(\rho -1)r^{-\rho }}-1\right|\le & {} \left| \frac{r\Phi _\rho (r)}{(\rho -1)\int _{(r,\infty )}\Phi _\rho (x)\mathrm {d}x}-1\right| \nonumber \\&+\, \frac{r\Phi _\rho (r)}{(\rho -1)\int _{(r,\infty )}\Phi _\rho (x)\mathrm {d}x}\left| \frac{r^{\rho -1}}{2-\rho }\int _{(r,\infty )}\Phi _\rho (x)\mathrm {d}x-1\right| ,\nonumber \\ \end{aligned}$$
(5.2)

which by Lemma 5.1 reduces the problem to showing that the first term on the right hand side of (5.2) vanishes as \(r\rightarrow \infty \). Recall now from the proof of Proposition 2.26 that \(\Phi _\rho \) satisfies (4.12) for all \(r>0\), and note that we may differentiate this equation with respect to r to obtain

$$\begin{aligned} (\rho -1)\int _{(r,\infty )}\Phi _\rho (x)\mathrm {d}x-r\Phi _\rho (r)= & {} \tfrac{\rho }{2}\,\lim _{h\rightarrow 0}\tfrac{1}{h}\big (\mathcal {I}[\Phi _\rho ](r+h)-\mathcal {I}[\Phi _\rho ](r)\big )\nonumber \\= & {} \frac{\rho }{2}\Big (\iint _{\begin{array}{c} \{|x-y|<r,\\ (x\vee y)>r\} \end{array}}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\mathrm {d}x\mathrm {d}y\nonumber \\&-\iint _{\begin{array}{c} \{x+y>r,\\ (x\vee y)<r\} \end{array}}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\mathrm {d}x\mathrm {d}y\Big ), \end{aligned}$$
(5.3)

Then, since \(\int _{(r,\infty )}\Phi _\rho (x)\mathrm {d}x\sim (2-\rho )r^{1-\rho }\) as \(r\rightarrow \infty \) (cf. Lemma 5.1), we find by (5.3) that the first term on the right hand side of (5.2) vanishes as \(r\rightarrow \infty \) if

$$\begin{aligned} \iint _{\{x+y>r,|x-y|<r\}}\frac{\Phi _\rho (x)\Phi _\rho (y)}{\sqrt{xy}}\mathrm {d}x\mathrm {d}y=o(r^{1-\rho })\text { as }r\rightarrow \infty , \end{aligned}$$
(5.4)

so by proving (5.4) we prove the theorem. Now, the left hand side of (5.4) can be estimated by

$$\begin{aligned} \tfrac{4}{r}\Vert \Phi _\rho \Vert _{L^1\big (\frac{1}{4}r,\infty \big )}^2+2\int _{\frac{3}{4}r}^{\frac{5}{4}r}\frac{\Phi _\rho (x)}{\sqrt{x}}\left( \int _{|r-x|}^{\frac{1}{2}r}\frac{\Phi _\rho (y)}{\sqrt{y}}\mathrm {d}y\right) \mathrm {d}x, \end{aligned}$$
(5.5)

where the first term is \(O(r^{1-2\rho })\), which decays sufficiently fast. For the second term of (5.5) we use a dyadic decomposition of the interval \(\big (|r-x|,\frac{1}{2}r\big )\), and Lemma 4.1, to find that the term between brackets can be bounded up to a constant by \(\log \Big (\tfrac{r}{|r-x|}\Big )\). Hölder’s inequality further gives us the estimate

$$\begin{aligned} \int _{\frac{3}{4}r}^{\frac{5}{4}r}\frac{\Phi _\rho (x)}{\sqrt{x}}\log \big |\tfrac{r}{r-x}\big |\mathrm {d}x \le \frac{\Vert \Phi _\rho \Vert _{L^q\big (\frac{3}{4}r,\infty \big )}\big \Vert \log \big |\frac{r}{r-z}\big |\big \Vert _{L^p\big (\frac{3}{4}r,\frac{5}{4}r\big )}}{\sqrt{\tfrac{3}{4}r}}, \end{aligned}$$

so recalling for \(q\in (1,\infty )\) that \(\Vert \Phi _\rho \Vert _{L^q(r,\infty )}\le O\left( r^{\frac{1}{q}-1}\Vert \Phi _\rho \Vert _{L^1(r,\infty )}\right) =O\left( r^{\frac{1}{q}-\rho }\right) \) as \(r\rightarrow \infty \) (cf. proof of Lemma 4.2), we find that the second term in (5.5) is bounded by a term of order \(O(r^{\frac{1}{2}-\rho })\) as \(r\rightarrow \infty \), hence (5.4) holds. \(\square \)

6 Exponential Bounds: Theorem 2.29

6.1 A Pointwise Exponential Upper Bound

Our first result gives an explicit upper bound to the moments of self-similar profiles, and can be seen as an improvement on [10, Lemma 4.22].

Lemma 6.1

If \(\Phi _2\in \mathcal {X}_2\) satisfies (2.10) with \(\rho =2\) for all \(\varphi \in C_c^1([0,\infty ))\), then there exists a finite constant \(A>0\) such that

$$\begin{aligned} \int _{(0,\infty )}x^\gamma \Phi _2(x)\mathrm {d}x\le \gamma ^\gamma A^{\gamma +1}\text { for all }\gamma >0. \end{aligned}$$
(6.1)

Proof

Let \(r>0\) be fixed arbitrarily, and let \(m_\gamma =\int _{(0,r)}x^\gamma \Phi _2(x)\mathrm {d}x\) (for \(\gamma \ge 0\)). To prove the result it suffices to show that there exists a finite constant \(A>0\), independent of r, such that \(m_\gamma \le \gamma ^\gamma A^{\gamma +1}\) for all \(\gamma >0\).

We first recall from the proof of Proposition 2.26 that \(\Phi _2\) satisfies

$$\begin{aligned} \int _{(r,\infty )}\Phi _2(x)\mathrm {d}x=r^{-1}\mathcal {I}[\Phi _2](r), \end{aligned}$$
(6.2)

with \(\mathcal {I}\) as defined in (4.4). Similar to the derivation of (4.12), for any \(\gamma >1\) we now approximate \(\varphi _{r,\gamma }(x)=(r^\gamma -x^\gamma )_+\) by functions in \(C_c^1([0,\infty ))\) to obtain

$$\begin{aligned}&(1-\gamma )\int _{(0,r)}x^\gamma \Phi _2(x)\mathrm {d}x+r^\gamma \int _{(r,\infty )}\Phi _2(x)\mathrm {d}x\nonumber \\&\qquad =2\iint _{\{x>y>0\}}\frac{\Phi _2(x)\Phi _2(y)}{\sqrt{xy}}\Delta _y^2\varphi _{r,\gamma }(x)\mathrm {d}x\mathrm {d}y. \end{aligned}$$
(6.3)

Introducing the notation \(\phi _\gamma (x)=x^\gamma \), we note for \(x\ge y\ge 0\) that we have

$$\begin{aligned} \Delta _y^2\varphi _{r,\gamma }(x)= {\left\{ \begin{array}{ll} ((x+y)^\gamma -r^\gamma )_+-\Delta _y^2\phi _\gamma (x)&{}\text {if }x\le r,\\ (r^\gamma -(x-y)^\gamma )_+&{}\text {if }x>r. \end{array}\right. } \end{aligned}$$

Noticing further that

$$\begin{aligned} ((x+y)^\gamma -r^\gamma )_+&\ge r^{\gamma -1}((x+y)-r)_+,\\ \nonumber (r^\gamma -(x-y)^\gamma )_+&\ge r^{\gamma -1}(r-(x-y))_+, \end{aligned}$$

we find from (6.3), where we use (6.2) in the left hand side, that

$$\begin{aligned}&(1-\gamma )\int _{(0,r)}x^\gamma \Phi _2(x)\mathrm {d}x+r^{\gamma -1}\mathcal {I}[\Phi _2](r)\\&\quad \ge r^{\gamma -1}\mathcal {I}[\Phi _2](r)-2\iint _{\{0<y<x<r\}}\frac{\Phi _2(x)\Phi _2(y)}{\sqrt{xy}}\Delta _y^2\phi _\gamma (x)\mathrm {d}x\mathrm {d}y, \end{aligned}$$

hence

$$\begin{aligned} (\gamma -1)\int _{(0,r)}x^\gamma \Phi _2(x)\mathrm {d}x\le 2\iint _{\{0<y<x<r\}}\frac{\Phi _2(x)\Phi _2(y)}{\sqrt{xy}}\Delta _y^2\phi _\gamma (x)\mathrm {d}x\mathrm {d}y. \end{aligned}$$
(6.4)

Since for \(x\ge y\ge 0\) there holds \(\Delta _y^2\phi _2(x)=2y^2\le 2y\sqrt{xy}\), (6.4) now yields

$$\begin{aligned} m_2\le & {} 2\iint _{\{0<y<x<r\}}2y\,\Phi _2(x)\Phi _2(y)\mathrm {d}x\mathrm {d}y\nonumber \\= & {} 2\iint _{(0,r)^2}(x\wedge y)\Phi _2(x)\Phi _2(y)\mathrm {d}x\mathrm {d}y\le 2m_0m_1, \end{aligned}$$
(6.5)

so using Hölder’s inequality, then (6.5), and then again Hölder’s inequality, we obtain for all \(\gamma \in [0,2]\) that

$$\begin{aligned} m_\gamma \le m_0^{1-\frac{\gamma }{2}}m_2^{\frac{\gamma }{2}}= & {} m_0^{1-\frac{\gamma }{2}}\left( m_2\right) ^\gamma m_2^{-\frac{\gamma }{2}}\le m_0^{1-\frac{\gamma }{2}}\left( 2m_0m_1\right) ^{\gamma }m_2^{-\frac{\gamma }{2}}\\= & {} 2^\gamma m_0^{1+\frac{\gamma }{2}}\left( m_1\right) ^{\gamma }m_2^{-\frac{\gamma }{2}}\le 2^\gamma m_0^{1+\frac{\gamma }{2}}\left( m_0^{\frac{1}{2}}m_2^{\frac{1}{2}}\right) ^{\gamma }m_2^{-\frac{\gamma }{2}}=\tfrac{1}{2}\left( 2m_0\right) ^{\gamma +1}, \end{aligned}$$

hence \(m_\gamma \le \gamma ^\gamma A^{\gamma +1}\) for all \(\gamma \in (0,2]\) if \(A \ge 2\Vert \Phi _2\Vert _2 \ge 2m_0\) (since \(\gamma ^\gamma >\frac{1}{2}\)).

For \(n\in \mathbb {N}\cap (2,\infty )\) we use the binomial formula to note for \(x\ge y\ge 0\) that

$$\begin{aligned} \Delta _y^2\phi _n(x)= \sum _{j=2}^{n}(1+(-1)^j)\times {n\atopwithdelims (){j}}x^{n-j}y^j\le 2\sum _{j=2}^{n}{n\atopwithdelims (){j}}x^{n-j}y^{j-1}\times \sqrt{xy}, \end{aligned}$$

which we then use in (6.4) to obtain

$$\begin{aligned} m_n\le \frac{4}{n-1}\sum _{j=2}^{n}{n\atopwithdelims (){j}}m_{n-j}m_{j-1}. \end{aligned}$$
(6.6)

Supposing now that \(m_\gamma \le \gamma ^\gamma A^{\gamma +1}\) for all \(\gamma \in \mathbb {N}\cap (0,n)\) with some \(A\ge 2\Vert \Phi _2\Vert _2\), and since in particular \(m_0\le \Vert \Phi _2\Vert _2\le A\), we use (6.6) to find

$$\begin{aligned} m_n\le \left( \frac{4}{n-1}\sum _{j=2}^{n}{n\atopwithdelims (){j}}(n-j)^{n-j}(j-1)^{j-1}\right) A^{n+1}, \end{aligned}$$
(6.7)

where we suppose that \(0^0=1\). Also by the binomial formula, we note that

$$\begin{aligned} {n\atopwithdelims ()j}(n-j)^{n-j}j^j\le \sum _{l=0}^n{n\atopwithdelims ()l}(n-j)^{n-l}j^l= n^n\text { for }j\in \mathbb {N}\cap (0,n], \end{aligned}$$

hence

$$\begin{aligned} \frac{4}{n-1}\sum _{j=2}^{n}{n\atopwithdelims (){j}}(n-j)^{n-j}(j-1)^{j-1}\le n^n\times 4\left( \frac{1}{n-1}\sum _{j=2}^{n}\frac{(j-1)^{j-1}}{j^j}\right) . \end{aligned}$$
(6.8)

Noticing then that the term between brackets on the right hand side of (6.8) is actually the average of terms that are all bounded by \(\frac{1}{4}\), it follows that the right hand side of (6.7) is bounded by \(n^nA^{n+1}\), hence by induction we have that \(m_\gamma \le \gamma ^\gamma A^{\gamma +1}\) for all \(\gamma \in (0,2]\cup \mathbb {N}\) with any \(A\ge 2\Vert \Phi _2\Vert _2\).

Finally, suppose that \(\gamma \in (2,\infty )\setminus \mathbb {N}\) and let n be the smallest integer larger than \(\gamma \). By Hölder’s inequality we then have \(m_\gamma \le m_0^{1-\frac{\gamma }{n}}m_n^{\frac{\gamma }{n}}\), so with the above estimates on \(m_n\) for \(n\in \mathbb {N}\) we find that

$$\begin{aligned} m_\gamma \le A^{1-\frac{\gamma }{n}}\left( n^nA^{n+1}\right) ^{\frac{\gamma }{n}} =\gamma ^\gamma \big (\tfrac{n}{\gamma }\big )^\gamma A^{\gamma +1}\le \gamma ^\gamma \left( \tfrac{3}{2}A\right) ^{\gamma +1}\text { with any }A\ge 2\Vert \Phi _2\Vert _2, \end{aligned}$$

whereby \(m_\gamma \le \gamma ^\gamma (3\Vert \Phi _2\Vert _2)^{\gamma +1}\) for all \(\gamma >0\), so (6.1) holds with \(A\ge 3\Vert \Phi _2\Vert _2\). \(\square \)

Mimicking the proof in [15], we are now able to prove the pointwise exponential upper bound.

Proposition 6.2

If \(\Phi _2\in \mathcal {X}_2\) satisfies (2.10) with \(\rho =2\) for all \(\varphi \in C_c^1([0,\infty ))\), then there exists a constant \(a\in (0,1)\) such that \(\Vert e^{ar}\Phi _2(r)\Vert _{L^\infty (1,\infty )}<\infty \).

Proof

Recall first of all that \(\Phi _2\in C^\infty ((0,\infty ))\) (cf. Proposition 2.25).

Now, let \(A>\frac{1}{2e}\) be a constant such that (6.1) holds, which exists by the proof of Lemma 6.1. For any \(r>0\) there then holds

$$\begin{aligned} \int _{(r,\infty )}\Phi _2(x)\mathrm {d}x\le r^{-\gamma }\int _{(0,\infty )}x^\gamma \Phi _2(x)\mathrm {d}x\le A\exp \left( \gamma \log \left( \tfrac{\gamma A}{r}\right) \right) \text { for all }\gamma >0, \end{aligned}$$

where the right hand side is minimal if \(\gamma =\frac{r}{eA}\), so

$$\begin{aligned} \int _{(r,\infty )}\Phi _2(x)\mathrm {d}x\le A\exp \left( -\tfrac{r}{eA}\right) \text { for all }r>0. \end{aligned}$$
(6.9)

We next differentiate (4.12) with respect to r, i.e. we set \(\rho =2\) in (5.3), and we drop the double integral over \(\{|x-y|<r,(x\vee y)>r\}\) to find for \(r>0\) that

$$\begin{aligned} r\Phi _2(r)\le \int _{(r,\infty )}\Phi _2(x)\mathrm {d}x+\iint _{\{x+y>r,(x\vee y)<r\}}\frac{\Phi _2(x)\Phi _2(y)}{\sqrt{xy}}\mathrm {d}x\mathrm {d}y. \end{aligned}$$
(6.10)

Since the integrand in the double integral on the right hand side of (6.10) is symmetric, for \(r>1\) we can now estimate that term by

$$\begin{aligned} 2\iint _{\{x>\frac{r}{2},y>\frac{1}{2}\}}\frac{\Phi _2(x)\Phi _2(y)}{\sqrt{xy}}\mathrm {d}x\mathrm {d}y+2\int _{r-\frac{1}{2}}^r\frac{\Phi _2(x)}{\sqrt{x}}\left( \int _{r-x}^{\frac{1}{2}}\frac{\Phi _2(y)}{\sqrt{y}}\mathrm {d}y\right) \mathrm {d}x. \end{aligned}$$
(6.11)

For \(\xi \in (0,\frac{1}{2})\) we then let n be the smallest integer such that \(2^{-n-1}<\xi \), and we use Lemma 4.1 to obtain

$$\begin{aligned} \int _{\xi }^\frac{1}{2}\frac{\Phi _2(y)}{\sqrt{y}}\mathrm {d}y\le \sum _{j=1}^n\int _{2^{-j-1}}^{2^{-j}}\frac{\Phi _2(y)}{\sqrt{y}}\mathrm {d}y\le \sum _{j=1}^n\frac{C\cdot \sqrt{2^{-j}}}{\sqrt{2^{-j-1}}}\le \tfrac{C\sqrt{2}}{\log 2}|\log \xi |, \end{aligned}$$
(6.12)

so combining (6.10), (6.11) and (6.12), and using (6.9), we find for \(r>1\) that

$$\begin{aligned} r\Phi _2(r)\le e^{-\frac{1}{2}\frac{r}{eA}}\left( Ae^{-\frac{1}{2}\frac{r}{eA}}+\tfrac{4}{\sqrt{r}}A^2\right) +\frac{\frac{4C}{\log 2}}{\sqrt{r}}\int _{r-\frac{1}{2}}^r\Phi _2(x)|\log (r-x)|\mathrm {d}x. \end{aligned}$$
(6.13)

Multiplying (6.13) by \(\frac{1}{r}e^{\frac{1}{2}\frac{r}{eA}}\), and choosing \(R\gg 1\) sufficiently large, then yields

$$\begin{aligned} e^{\frac{1}{2}\frac{r}{eA}}\Phi _2(r)\le & {} \frac{1}{r}\left( 1+\frac{\frac{4C}{\log 2}}{\sqrt{r}}\int _0^{\frac{1}{2}}e^{\frac{1}{2}\frac{x}{eA}}|\log x|\mathrm {d}x\times \left\| e^{\frac{1}{2}\frac{z}{eA}}\Phi _2(z)\right\| _{L^\infty (r-\frac{1}{2},r)}\right) \nonumber \\\le & {} \frac{1}{r}\left( 1+\left\| e^{\frac{1}{2}\frac{z}{eA}}\Phi _2(z)\right\| _{L^\infty (1,r)}\right) \text { for all }r>R, \end{aligned}$$
(6.14)

so setting \(a=\frac{1}{2}\frac{1}{eA}\in (0,1)\), using (6.14), and iterating, we obtain

$$\begin{aligned} \Vert e^{az}\Phi _2(z)\Vert _{L^\infty (1,r)}\le & {} \Vert e^{az}\Phi _2(z)\Vert _{L^\infty (1,R)}+\tfrac{1}{R}\left( 1+\Vert e^{az}\Phi _2(z)\Vert _{L^\infty (1,r)}\right) \nonumber \\\le & {} \tfrac{R}{R-1}\left( \Vert e^{az}\Phi _2(z)\Vert _{L^\infty (1,R)}+\tfrac{1}{R}\right) \text { for all }r>R. \end{aligned}$$
(6.15)

The claim now follows since the right hand side of (6.15) is independent of r. \(\square \)

6.2 An Exponential Lower Bound in Integral Form

We will prove the following result, of which the lower bound is a corollary.

Proposition 6.3

If \(\Phi _2\in \mathcal {X}_2\) satisfies (2.10) with \(\rho =2\) for all \(\varphi \in C_c^1([0,\infty ))\), and if \(\Phi _2\) is not identically zero on \((0,\infty )\), then there exists a finite constant \(B>1\) such that

$$\begin{aligned} \inf _{R\ge 0}\left\{ \int _{(R,R+1)}e^{Bx}\Phi _2(x)\mathrm {d}x\right\} >0. \end{aligned}$$

Supposing that \(\Phi _2\in \mathcal {X}_2\) is as assumed in Proposition 6.3, then \(\Phi _2\) is smooth and strictly positive on \((0,\infty )\), and it satisfies (2.9) in particular for all \(\varphi \in C_c^1((0,\infty ))\), which we can rewrite as

$$\begin{aligned} \int _0^\infty \left[ \tfrac{1}{2}(x\varphi _x(x)-\varphi (x))-\int _0^x\frac{\Phi _2(y)}{\sqrt{xy}}\Delta _y^2\varphi (x)\mathrm {d}y\right] \Phi _2(x)\mathrm {d}x=0. \end{aligned}$$
(6.16)

Note that (6.16) also holds for \(\varphi \in W_0^{1,\infty }((0,\infty ))\), so given a function \(\varphi :(0,t)\rightarrow W_0^{1,\infty }((0,\infty ))\) with \(\psi _s(s,\cdot )\in L^\infty (0,\infty )\) that satisfies

$$\begin{aligned} \varphi _s(s,x)\le -\tfrac{1}{2}(x\varphi _x(s,x)-\varphi (s,x))+\int _0^x\frac{\Phi _2(y)}{\sqrt{xy}}\Delta _y^2[\varphi (s,\cdot )](x)\mathrm {d}y \end{aligned}$$
(6.17)

for almost all \(s\in (0,t)\) and \(x\in (0,\infty )\), there holds

$$\begin{aligned} \int _0^\infty \varphi (0,x)\Phi _2(x)\mathrm {d}x\ge \int _0^\infty \varphi (s,x)\Phi _2(x)\mathrm {d}x\text { for all }s\in [0,t]. \end{aligned}$$

In the following we will construct such a function.

From this point onwards we denote by u the solution to (3.21) with \(\alpha =\frac{1}{2}\) and initial data \(u(0,x)=\mathrm{sgn}(x)\). Now, this function has the following explicit self-similar form

$$\begin{aligned} u(s,x)=w(\tfrac{x}{s^2})\text { for }s>0\text { and }x\in \mathbb {R},\quad \text {with }w(\xi )=2\int _0^\xi v_{\frac{1}{2}}(z)\mathrm {d}z, \end{aligned}$$

where \(v_{\frac{1}{2}}\) is the self-similar profile associated to the fundamental solution of (3.21) with \(\alpha =\frac{1}{2}\) (cf. Lemma 3.8). Since \(v_\frac{1}{2}\) is odd, strictly positive, and nonincreasing on \(\mathbb {R}_+\), and since it has \(L^1\)-norm equal to one, it follows that w is a bijection from \(\mathbb {R}\) onto \((-1,1)\), that is furthermore concave on \(\mathbb {R}_+\). Consequently we can thus fix unique constants \(c_2>c_1>0\) such that \(w(c_1)=\frac{1}{2}\) and \(w(c_2)=\frac{3}{4}\). Moreover, for all \(s>0\) the concave mappings \(x\mapsto \frac{1}{2}\big (\frac{x}{c_1s^2}\wedge 1\big )\) and \(x\mapsto \frac{3}{4}\big (\frac{x}{c_2s^2}\wedge 1\big )\) then lie below \(u(s,\cdot )\) on \(\mathbb {R}_+\).

For fixed \(R,b>1\) we first compare the functions \(f^1(s,x)=e^{bR}u(s,x-R)\) and \(f^2(s,x)=\frac{1}{4}e^{bx}\). In particular we are interested in the solutions x(s) to \(f^1(s,x(s))=f^2(s,x(s))\), which for small \(s>0\) are given by \(x(s)=R+\ell _i(s)\), \(i=1,2\), with \(\ell _2(s)\ge \ell _1(s)>0\) the solutions to \(u(s,\ell _i(s))=\frac{1}{4}e^{b\ell _i(s)}\). From Fig. 1 we then find that as long as \(u(s,c_1s^2)=\frac{1}{2}>\frac{1}{4}e^{bc_1s^2}\) and \(u(s,c_2s^2)=\frac{3}{4}>\frac{1}{4}e^{bc_2s^2}\) hold, i.e. as long as \(s^2<\frac{1}{b}\min \big \{\frac{1}{c_1}\log 2,\frac{1}{c_2}\log 3\big \}\), then there exist two different solutions \(\ell _i(s)\), and there holds \(\ell _2(s)-\ell _1(s)>(c_2-c_1)s^2>0\).

Fig. 1
figure 1

For \(s>0\) small and \(b>1\) large, we have \(\mathrm{sgn}(x)\), u(sx) (thick) and \(\frac{1}{4}e^{bx}\) (dashed), as well as \(\frac{1}{2}\big (\frac{x}{c_1s^2}\wedge 1\big )\) and \(\frac{3}{4}\big (\frac{x}{c_2s^2}\wedge 1\big )\)

We then compare \(f^3(s,x)=e^{b(R+1)}(u(s,R+1-x)+b(x-(R+1)))\) with \(f^2\). We are interested in the solutions x(s) to \(f^3(s,x(s))=f^2(s,x(s))\), which for small \(s>0\) are given by \(x(s)=R+r_i(s)\), \(i=1,2\), where \(r_2(s)\le r_1(s)<1\) are the solutions to \(e^b(u(s,1-r_i(s))+b(r_i(s)-1))=\frac{1}{4}e^{br_i(s)}\). Figure 2 now shows that as long as both \(e^b(u(s,c_1s^2)-bc_1s^2)=\frac{1}{2}e^b-be^bc_1s^2>\frac{1}{4}e^{bc_1s^2}\) and \(e^b(u(s,c_2s^2)-bc_2s^2)=\frac{3}{4}e^b-be^bc_2s^2>\frac{1}{4}e^{bc_2s^2}\) hold, i.e. as long as \(s^2<\frac{1}{b}\min \big \{\frac{1}{c_1}q(b,\frac{1}{2}),\frac{1}{c_2}q(b,\frac{3}{4})\big \}\) with \(q(b,\alpha )\) such that \(e^b(\alpha -q(b,\alpha ))=\frac{1}{4}e^{q(b,\alpha )}\), then there exist two different solutions \(r_i(s)\), and \(r_1(s)-r_2(s)>(c_2-c_1)s^2>0\). Note further that indeed \(r_2(s)\ge \frac{1}{b}q(b,1)=O\big (\frac{1}{b}\big )\) as \(b\rightarrow \infty \).

Fig. 2
figure 2

For \(s>0\) small and \(b>1\) large, we have \(e^b(\mathrm{sgn}(1-x)+b(x-1))\), \(e^b(u(s,1-x)+b(x-1))\) (thick), \(\frac{1}{4}e^{bx}\) (dashed) and \(e^{bx}\) (dotted), as well as \(e^b(\frac{1}{2}(\frac{1-x}{c_1s^2}\wedge 1)+b(x-1))\) and \(e^b(\frac{3}{4}(\frac{1-x}{c_2s^2}\wedge 1)+b(x-1))\)

For \(b\gg 1\) sufficiently large, we then define \(f:(0,\frac{1}{b})\rightarrow W_0^{1,\infty }((0,\infty ))\) by

$$\begin{aligned} f(s,\cdot )= {\left\{ \begin{array}{ll} f^1(s,\cdot )&{}\text {on }[R,R+\ell _2(s)),\\ f^2(s,\cdot )&{}\text {on }[R+\ell _2(s),R+r_2(s)],\\ f^3(s,\cdot )&{}\text {on }(R+r_2(s),R+1],\\ 0&{}\text {else.} \end{array}\right. } \end{aligned}$$
(6.18)

The following lemma will be useful.

Lemma 6.4

Let \(\Phi _2\in \mathcal {X}_2\) be as in the statement of Proposition 6.3. Then there exists a finite constant \(C>0\) such that

$$\begin{aligned} \int _z^\infty \frac{\Phi _2(e^{s/2}y)}{\sqrt{y}}\mathrm {d}y\le C\left( (1+|\log z|)\wedge \frac{1}{\sqrt{z}}\right) \text { for all }z>0\text { and all }s\ge 0. \end{aligned}$$

Proof

We first note the trivial estimate that

$$\begin{aligned} \int _z^\infty \frac{\Phi _2(e^{s/2}y)}{\sqrt{y}}\mathrm {d}y\le \frac{e^{-s/2}}{\sqrt{z}}\int _{ze^{s/2}}^\infty \Phi _2(y)\mathrm {d}y\le \frac{\Vert \Phi _2\Vert _2}{\sqrt{z}}\text { for} z>0 \text { and}\, s\ge 0. \end{aligned}$$
(6.19)

For \(z\in (0,\frac{1}{2})\), let now n be the smallest integer such that \(2^{-n-1}<z\), so that using Lemma 4.1 we find

$$\begin{aligned} \int _z^\frac{1}{2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{y}}\mathrm {d}y\le & {} \sum _{j=1}^n\int _{2^{-j-1}}^{2^{-j}}\frac{\Phi _2(e^{s/2}y)}{\sqrt{y}}\mathrm {d}y \le \sum _{j=1}^n\frac{e^{-s/2}\int _0^{e^{s/2}2^{-j}}\Phi _2(y)\mathrm {d}y}{\sqrt{2^{-j-1}}}\nonumber \\\le & {} \sum _{j=1}^n\frac{e^{-s/2}C\cdot \sqrt{e^{s/2}2^{-j}}}{\sqrt{2^{-j-1}}}\le C\sqrt{2}\,n \le \tfrac{C\sqrt{2}}{\log 2}|\log z|\text { for all }s\ge 0.\nonumber \\ \end{aligned}$$
(6.20)

The claim then follows by combining (6.19) and (6.20). \(\square \)

Lemma 6.5

Let \(\Phi _2\in \mathcal {X}_2\) be as in the statement of Proposition 6.3. Then there exist large constants \(R_0,b_0\gg 1\) such that if for \(R\ge R_0\) and \(b\ge b_0\) the function \(f:(0,\frac{1}{b})\rightarrow W_0^{1,\infty }((0,\infty ))\) is given by (6.18), then for all \(s\in (0,\frac{1}{b})\) the function \(\psi (s,x):=e^{-\log (b)s}f(s,x)\) satisfies

$$\begin{aligned} \psi _s(s,x)\le \int _0^x\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\Delta _y^2[\psi (s,\cdot )](x)\mathrm {d}y\text { for almost all }x\ge 0. \end{aligned}$$
(6.21)

Proof

Clearly, by nonnegativity of \(\Phi _2\) and \(\psi \), the right hand side of (6.21) is nonnegative if \(x\in [0,R)\cup (R+1,\infty )\), while the left hand side is identically equal to zero. For \(x\in [R,R+1]\) we now set \(\bar{c}=c_2-c_1>0\), and we estimate the right hand side of (6.21) from below by

$$\begin{aligned} \int _0^{\bar{c}/b^2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\Delta _y^2[\psi (s,\cdot )](x)\mathrm {d}y-2\psi (s,x)\int _{\bar{c}/b^2}^\infty \frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\mathrm {d}y, \end{aligned}$$
(6.22)

where, for sufficiently large \(R_0\gg 1\), the second term is bounded from below, uniformly for all \(R\ge R_0\), by \(-\log b\times \psi (s,x)\) (use Lemma 6.4 and \(x\ge R_0\)). By the smallness of the domain of integration in the first term of (6.22), we further find that we can bound this term by

$$\begin{aligned} e^{-\log (b)s}\int _0^{\bar{c}/b^2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\Delta _y^2[f^i(s,\cdot )](x)\mathrm {d}y, \end{aligned}$$
(6.23)

where \(i=1\) if \(x\in [R,R+\ell _2(s))\), \(i=2\) if \(x\in [R+\ell _2(s),R+r_2(s)]\), and \(i=3\) if \(x\in (R+r_2(s),R+1]\). By the semigroup property of the exponential function this then means that, for \(x\in (R+\ell _2(s),R+r_2(s))\), (6.22) can be bounded from below by

$$\begin{aligned}&\left( \int _0^{\bar{c}/b^2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\left( e^{by}+e^{-by}-2\right) \mathrm {d}y-\log b\right) e^{-\log (b)s}\tfrac{1}{4}e^{bx}\\&\quad \ge -\log b\times \psi (s,x)=\psi _s(s,x). \end{aligned}$$

Note now, for \(x\in (R,R+\ell _2(s))\), that \(\Delta _y^2[f^1(s,\cdot )](x)=e^{bR}\Delta _y^2[u(s,\cdot )](x-R)\le 0\) for all \(y\in \mathbb {R}\) (cf. Lemma 3.9), so we can estimate the integral in (6.23) with \(i=1\) from below by

$$\begin{aligned}&\frac{1}{\sqrt{R}}\int _{\mathbb {R}_+}\frac{\Phi _2(e^{s/2}y)}{\sqrt{y}}\Delta _y^2[f^1(s,\cdot )](x)\mathrm {d}y\nonumber \\&\qquad =\frac{1}{\sqrt{R}}\int _{\mathbb {R}_+}\left( \int _y^\infty \frac{\Phi _2(e^{s/2}z)}{\sqrt{z}}\mathrm {d}z\right) \left( \int _{x-y}^{x+y}f^1_{ww}(s,w)\mathrm {d}w\right) \mathrm {d}y, \end{aligned}$$
(6.24)

where the equality follows by integration by parts and (2.8) for \(\partial _y[\Delta _y^2[f^1(s,\cdot )](x)]\), and where the integral with respect to w in the right hand side of (6.24) is nonpositive for all \(y\in \mathbb {R}_+\) (cf. proof of Lemma 3.10). By Lemma 6.4, and choosing \(R_0\gg 1\) sufficiently large, we then bound the right hand side of (6.24) from below, uniformly for all \(R\ge R_0\), by

$$\begin{aligned} \int _{\mathbb {R}_+}\frac{2}{\sqrt{y}}\left( \int _{x-y}^{x+y}f^1_{ww}(s,w)\mathrm {d}w\right) \mathrm {d}y=e^{bR}\int _{\mathbb {R}_+}y^{-\frac{3}{2}}\Delta _y^2[u(s,\cdot )](x-R)\mathrm {d}y, \end{aligned}$$
(6.25)

where for the equality we have integrated by parts back again, and noting that the right hand side of (6.25) by construction equals \(e^{bR}u_s(s,x-R)=f_s^1(s,x)\) it follows that (6.22) can be estimated from below by

$$\begin{aligned} e^{-\log (b)s}f_s^1(s,x)-\log (b)e^{-\log (b)s}f^1(s,x)=\psi _s(s,x). \end{aligned}$$

Recalling lastly that the second difference of an affine function is zero, similar arguments show that the inequality in (6.21) also holds for \(x\in (R+r_2(s),R+1)\), which completes the proof. \(\square \)

Now, for \(R\gg 1\) sufficiently large, let \(\eta _R\in C^\infty (\mathbb {R})\) be such that \(\mathrm{supp}(\eta _R)=\big [\frac{3}{5}R+1,\frac{4}{5}R\big ]\), such that \(\eta _R=1\) on \(\big [\frac{3}{5}R+2,\frac{4}{5}R-1\big ]\), and such that \(\eta _R\) is increasing on \(\big (\frac{3}{5}R+1,\frac{3}{5}R+2\big )\), and decreasing on \(\big (\frac{4}{5}R-1,\frac{4}{5}R\big )\). For \(b\gg 1\) and \(t\in \big (0,\frac{1}{b}\big )\) we then define \(f^0\in C^\infty ((0,\infty ))\) by

$$\begin{aligned} f^0(x)=\eta _R(x)\times \frac{e^{bx}}{8R}\inf _{r\in [0,\frac{4}{5}R]}\left\{ \int _r^{r+1}e^{be^{-t/2}y}\Phi _2(y)\mathrm {d}y\right\} . \end{aligned}$$
(6.26)

Lemma 6.6

Let \(\Phi _2\in \mathcal {X}_2\) be as in the statement of Proposition 6.3. Then there exist large constants \(R_0,b_0\gg 1\) such that if for \(R\ge R_0\) and \(b\ge b_0\) the functions \(f:(0,t)\rightarrow W_0^{1,\infty }((0,\infty ))\) and \(f^0\in W_0^{1,\infty }((0,\infty ))\), with \(t\in (0,\frac{1}{b})\), are given by (6.18) and (6.26), then the function

$$\begin{aligned} \psi (s,x):=e^{-\log (b)s}f(s,x)+\tfrac{1}{2}se^{-\log (b)s}f^0(x) \end{aligned}$$
(6.27)

satisfies (6.21) for all \(s\in (0,t)\).

Proof

By Lemma 6.5 we can restrict ourselves to \(x\in (\frac{3}{5}R+1,\frac{4}{5}R)\) in checking that \(\psi \) satisfies (6.21) for all \(s\in (0,t)\). Similar to the first estimate in that lemma, we now note that

$$\begin{aligned}&\int _0^x\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\Delta _y^2[\psi (s,\cdot )](x)\mathrm {d}y \ge \int _{R+\ell _1(s)-x}^{R+r_1(s)-x}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\psi (s,x+y)\mathrm {d}y\nonumber \\&\qquad +\int _0^{\bar{c}/b^2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\Delta _y^2[\psi (s,\cdot )](x)\mathrm {d}y-2\psi (s,x)\int _{\bar{c}/b^2}^\infty \frac{\Phi _2(e^{t/2}y)}{\sqrt{xy}}\mathrm {d}y, \end{aligned}$$
(6.28)

where the last term on the right hand side is bounded from below by \(-\log b\times \psi (s,x)\) if \(R_0\gg 1\) is sufficiently large (cf. proof of Lemma 6.5). We then estimate the first term on the right hand side of (6.28) from below by

$$\begin{aligned}&e^{-\log (b)s}\int _{R+\ell _1(s)-x}^{R+r_1(s)-x}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\tfrac{1}{4}e^{b(x+y)}\mathrm {d}y\\&\qquad \ge e^{-\log (b)s}\times \frac{e^{bx}}{4R}\inf _{r\in [0,\frac{2}{5}R]}\left\{ \int _r^{r+(1-2c_1s^2)}e^{by}\Phi _2(e^{s/2}y)\mathrm {d}y\right\} \\&\qquad \ge e^{-\log (b)s}\times \frac{e^{bx}}{8R}\inf _{r\in [0,\frac{4}{5}R]}\left\{ \int _r^{r+1}e^{be^{t/2}y}\Phi _2(y)\mathrm {d}y\right\} , \end{aligned}$$

where we have used that \((1-2c_1s^2)e^{s/2}=1+\frac{1}{2}s+O(s^2)\ge 1\) as \(s\rightarrow 0\). Note next that the second term on the right hand side of (6.28) can be written as

$$\begin{aligned}&\tfrac{1}{2}se^{-\log (b)s}\int _0^{\bar{c}/b^2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{xy}}\left( \eta _R(x+y)e^{by}+\eta _R(x-y)e^{-by}-2\eta _R(x)\right) \mathrm {d}y\nonumber \\&\qquad \times \,\frac{e^{bx}}{8R}\inf _{r\in [0,\frac{4}{5}R]}\left\{ \int _r^{r+1}e^{be^{t/2}y}\Phi _2(y)\mathrm {d}y\right\} . \end{aligned}$$
(6.29)

Recalling now (2.7), the integral over the second difference in (6.29) can be bounded from below by

$$\begin{aligned} -\frac{1}{\sqrt{\frac{3}{5}R}}\int _0^{\bar{c}/b^2}\frac{\Phi _2(e^{s/2}y)}{\sqrt{y}}\left| \int _\mathbb {R}\left( y-|w|\right) _+\left[ \eta _R(x+w)e^{bw}\right] _{ww}\mathrm {d}w\right| \mathrm {d}y\\ \ge -\frac{1}{\sqrt{\frac{3}{5}R}}\int _0^{\bar{c}/b^2}y^{\frac{3}{2}}\Phi _2(e^{s/2}y)\mathrm {d}y\times \sup _{|w|<\frac{\bar{c}}{b^2}}\left| \left[ \eta _R(x+w)e^{bw}\right] _{ww}\right| , \end{aligned}$$

which is \(O\big (R^{-\frac{1}{2}}b^{-2}\big )\) as \(R,b\rightarrow \infty \) (use Lemma 4.1 and a dyadic decomposition), hence bounded from below by \(-1\) if \(R_0,b_0\gg 1\) are sufficiently large. Recalling lastly that \(1-\frac{1}{2}s\ge \frac{1}{2}\) for \(s\in (0,\frac{1}{b_0})\), we conclude with the above that the right hand side of (6.28) is bounded from below by

$$\begin{aligned} \left[ \tfrac{1}{2}se^{-\log (b)s}\right] _s\times \eta _R(x)\times \frac{e^{bx}}{8R}\inf _{r\in [0,\frac{4}{5}R]}\left\{ \int _r^{r+1}e^{be^{t/2}y}\Phi _2(y)\mathrm {d}y\right\} =\psi _s(s,x), \end{aligned}$$

and the proof is complete. \(\square \)

Lemma 6.7

Let \(\Phi _2\in \mathcal {X}_2\) be as in the statement of Proposition 6.3. Then there exist constants \(R_0,b_0\gg 1\) and \(c>0\) such that \(\log (I(b_0,R_0))\ge \log (b_0)+1\), and such that for all \(R\ge R_0\) and all \(b\ge b_0\) there holds

$$\begin{aligned} I(b,R)\ge t\left( I(be^{-t/2},\tfrac{4}{5}R)\right) ^2\text { for all }t\in (0,\tfrac{1}{b}), \end{aligned}$$
(6.30)

where

$$\begin{aligned} I(b,R):=c\times \inf _{r\in [0,R]}\left\{ \int _r^{r+1}e^{bx}\Phi _2(x)\mathrm {d}x\right\} . \end{aligned}$$
(6.31)

Proof

Let \(R_0,b_0\gg 1\) be as obtained in Lemma 6.6, and let \(R\ge R_0\), \(b\ge b_0\) and \(t\in \big (0,\frac{1}{b}\big )\) be fixed arbitrarily. We then define \(\varphi :(0,t)\rightarrow W_0^{1,\infty }((0,\infty ))\) as \(\varphi (s,x):=e^{s/2}\psi (s,xe^{-s/2})\), with \(\psi \) given by (6.27), and we note that \(\varphi \) satisfies (6.17) for almost all \((s,x)\in (0,t)\times (0,\infty )\) and \(\varphi (0,x)\le e^{bx}\mathbf {1}_{(R,R+1)}(x)\), hence

$$\begin{aligned} \int _R^{R+1}e^{bx}\Phi _2(x)\mathrm {d}x\ge \int _0^\infty \psi (t,xe^{-t/2})\Phi _2(x)\mathrm {d}x, \end{aligned}$$

where the right hand side can be bounded from below by

$$\begin{aligned}&\int _{\frac{2}{3}Re^{t/2}}^{\frac{11}{15}Re^{t/2}}e^{be^{-t/2}x}\Phi _2(x)\mathrm {d}x\times \frac{\frac{1}{2}t}{8R}\inf _{r\in [0,\frac{4}{5}R]}\left\{ \int _r^{r+1}e^{be^{-t/2}y}\Phi _2(y)\mathrm {d}y\right\} \\&\qquad \ge \frac{R}{16}\frac{t}{16R}\left( \inf _{r\in [0,\frac{4}{5}R]}\left\{ \int _r^{r+1}e^{be^{-t/2}y}\Phi _2(y)\mathrm {d}y\right\} \right) ^2. \end{aligned}$$

Taking then the infimum, (6.30) follows with \(c=\frac{1}{256}\). We lastly note that

$$\begin{aligned} I(b_0,R_0)\ge ce^{\frac{1}{2}b_0}\times \inf _{r\in [0,R_0]}\left\{ \int _{r+\frac{1}{2}}^{r+1}\Phi _2(x)\mathrm {d}x\right\} , \end{aligned}$$

where the logarithm of the right hand side is linear as a function of \(b_0\), and it follows that indeed \(\log (I(b_0,R_0))\ge \log (b_0)+1\) if \(b_0\gg 1\) is sufficiently large. \(\square \)

Proof of Proposition 6.3

Let \(R_0,b_0\gg 1\) and \(c>0\) be as obtained in Lemma 6.7, and let I be given by (6.31). Now, set \(B=b_0e^{\pi ^2/12}\), and for all \(n\in \mathbb {N}\) define \(t_n=n^{-2}\wedge B^{-1}\), \(b_n=b_{n-1}e^{t_n/2}\) and \(R_n=\frac{5}{4}R_{n-1}\). For every \(n\in \mathbb {N}\) there then holds \(b_n\le b_0\exp (\frac{1}{2}\sum _{j=1}^nj^{-2})<B\), hence \(t_n\in (0,\frac{1}{b_n})\), so it follows from (6.30) that \(I(b_n,R_n)\ge t_n(I(b_{n-1},R_{n-1}))^2\) for all \(n\in \mathbb {N}\). Taking the logarithm and multiplying by \(2^{-n}\), we now find for all \(n\in \mathbb {N}\) by iteration that

$$\begin{aligned} 2^{-n}\log (I(b_n,R_n))\ge & {} 2^{-(n-1)}\log (I(b_{n-1},R_{n-1}))+2^{-n}\log (t_n)\\\ge & {} \log (I(b_0,R_0))+\sum _{j=1}^n2^{-j}\log (t_j), \end{aligned}$$

from which we obtain that \(\log (I(b_n,R_n))>0\) for all \(n\in \mathbb {N}\), since \(\log (I(b_0,R_0))\ge \log (b_0)+1\) (cf. Lemma 6.7), and since

$$\begin{aligned} \sum _{j=1}^n2^{-j}\log (t_j)\ge & {} \sum _{j=1}^n2^{-j-1}\left( \log (j^{-2})+\log (B^{-1})\right) \\\ge & {} -\tfrac{1}{2}\left( \log (b_0)+\tfrac{\pi ^2}{12}\right) -\sum _{j=1}^\infty 2^{-j}\log (j)>-\log (b_0)-1. \end{aligned}$$

The proof is then completed by the observation that

$$\begin{aligned} \inf _{R\ge 0}\left\{ \int _{(R,R+1)}e^{Bx}\Phi _2(x)\mathrm {d}x\right\} =\frac{1}{c}\times \inf _{n\in \mathbb {N}}I(B,R_n)\ge \frac{1}{c}\times \inf _{n\in \mathbb {N}}I(b_n,R_n), \end{aligned}$$

where the right hand side is strictly positive, as the infimum is taken over terms that are strictly larger than 1. \(\square \)

Proof of Theorem 2.29

Corollary of Propositions 6.2 and 6.3.

Remark 6.8

To deduce a pointwise exponential lower bound on \(\Phi _2\), we seem to require existence of a solution \(\varphi \) on \((s,x)\in [0,T)\times [0,\infty )\) to

$$\begin{aligned} \left\{ \begin{array}{rl} \varphi _s(s,x)&{}=-\tfrac{1}{2}(x\varphi _x(s,x)-\varphi (s,x))+\frac{1}{\sqrt{x}}\int _0^x\frac{\Phi _2(y)}{\sqrt{y}}\Delta _y^2[\varphi (s,\cdot )](x)\mathrm {d}y\\ \varphi (0,x)&{}=\delta _0(x-r) \end{array}\right. \end{aligned}$$
(6.32)

for all \(r\ge R_0\) with \(R_0\gg 1\) finite. However, existence of such solutions is nontrivial due to the possibly divergent behaviour of \(\Phi _2\) near zero. A better understanding of well-posedness of (6.32) requires a more detailed analysis of the asymptotics of \(\Phi _2(z)\) as \(z\rightarrow 0\).