Log-Sobolev inequalities for semi-direct product operators and applications

Abstract

In this paper, we prove some type of logarithmic Sobolev inequalities (with parameters) for operators in semi-direct product forms (see Sect. 1 for a precise definition). This generalizes the tensorization procedure for this type of inequalities and allows to deal with some operators with varying coefficients. We provide many examples of applications and obtain ultracontractive bounds for some of these operators by using appropriate Hardy’s type inequalities necessary for our method. This theory is developed in the setting of carré du champ with diffusion property.

Appendix: Hardy type lemmas

This “Appendix” is devoted to the proof of the lemmas stated and used in Sect. 5.

1.1 Proof of Lemma 5.2

Since

$$\begin{aligned} \left\| \frac{\partial f}{\partial x} \right\| _{L^{2}\left( \mathbb {R}^{2}\right) }\le \Vert \nabla _\mathcal{L} f\Vert _{L^{2}\left( \mathbb {R}^{2}\right) }, \end{aligned}$$

we see that it is sufficient to prove the one dimensional estimate

$$\begin{aligned} \int _{\mathbb {R}} \left( -\ln \vert x\vert \right) f^2(x) dx \le t \left\| f^{\prime }\right\| _{2}^{2} + \frac{1}{2} \ln \left( 2et^{-1}\right) \vert \vert f\vert \vert ^2_2,\quad t>0. \end{aligned}$$

(6.40)

where the norms are now in $L^{2}(\mathbb {R})$. The proof of (6.40) will descend from the following proposition:

Proposition 5.4

For all $0<\delta \le 1$

$$\begin{aligned} \int _{0}^{1}- \ln x\cdot |f|^{2}dx\le |\ln \delta |\cdot \Vert f\Vert ^{2}_{L^{2}(0,1)} +\Vert f\Vert ^{2}_{L^{2}(0,\delta )} +\frac{2 \delta }{e}\Vert f\Vert _{L^{2}(0,\delta )}\Vert f^{\prime }\Vert _{L^{2}(0,\delta )}. \end{aligned}$$

(6.41)

Proof

It is sufficient to prove the estimate for a smooth function f. We have the identity

$$\begin{aligned} \int _{0}^{1}(-\ln x)|f|^{2}dx =\int _{0}^{1}(-\ln x)\frac{d}{dx}\int _{0}^{x}|f|^{2} =\int _{0}^{1}\frac{1}{x}\int _{0}^{x}|f|^{2}+ \left. (-\ln x)\int _{0}^{x}|f|^{2}\right| _{0}^{1} \end{aligned}$$

and we notice that the last boundary term vanishes. We estimate the remaining integral at the r.h.s. as follows. The piece of the integral with $\delta \le x\le 1$ is bounded by

$$\begin{aligned} \int _{\delta }^{1}\frac{1}{x}\int _{0}^{x}|f|^{2}d\xi dx\le \Vert f\Vert ^{2}_{L^{2}(0,1)}\int _{\delta }^{1}\frac{1}{x} dx= \Vert f\Vert ^{2}_{L^{2}(0,1)}\cdot |\ln \delta |. \end{aligned}$$

(6.42)

On the other hand, an integration by parts gives:

$$\begin{aligned} \begin{aligned} \int _{0}^{\delta }\frac{1}{x} \int _{0}^{x}|f|^{2}d\xi dx&= \int _{0}^{\delta }\frac{1}{x} \int _{0}^{x}\xi ^{\prime }|f|^{2}d\xi dx= \int _{0}^{\delta }\frac{1}{x} \left[ -\int _{0}^{x}2\xi ff^{\prime } d\xi + \left. \xi |f|^{2}\right| _{0}^{x} \right] dx\\&=\int _{0}^{\delta }\int _{0}^{x} \left( -\frac{2\xi }{x}\right) ff^{\prime }d\xi dx +\int _{0}^{\delta }|f|^{2}dx. \end{aligned} \end{aligned}$$

(6.43)

Now we notice that

$$\begin{aligned} \begin{aligned} \int _{0}^{\delta }\int _{0}^{x} \left( -\frac{2\xi }{x}\right) ff^{\prime }d\xi dx&= \int _{0}^{\delta }\int _{\xi }^{\delta } \left( -\frac{2\xi }{x}\right) ff^{\prime } dx d\xi \\&= - 2\int _{0}^{\delta } ff^{\prime } \left( \xi \ln \frac{\delta }{\xi }\right) d\xi . \end{aligned} \end{aligned}$$

The function $\xi \ln (\delta /\xi )$ is non negative on $[0,\delta ]$ and vanishes at the boundary; its maximum is at $\xi =\delta /e$ so that

$$\begin{aligned} \left| \xi \ln \frac{\delta }{\xi }\right| \le \frac{\delta }{e}. \end{aligned}$$

This implies

$$\begin{aligned} \left| \int _{0}^{\delta }\int _{0}^{x} \left( -\frac{2\xi }{x}\right) f f^{\prime } d\xi dx\right| \le \frac{2 \delta }{e} \Vert f\Vert _{L^{2}(0,\delta )}\Vert f^{\prime }\Vert _{L^{2}(0,\delta )} \end{aligned}$$

and by (6.43)

$$\begin{aligned} \int _{0}^{\delta }\frac{1}{x} \int _{0}^{x}|f|^{2}d\xi dx\le \frac{2 \delta }{e} \Vert f\Vert _{L^{2}(0,\delta )}\Vert f^{\prime }\Vert _{L^{2}(0,\delta )} +\Vert f\Vert ^{2}_{L^{2}(0,\delta )}. \end{aligned}$$

Putting this estimate together with (6.42) we obtain (6.41). $\square$

Now, we are in position to prove Lemma 5.2. By changing f(x) by $f(-x)$, we get from (6.41)

$$\begin{aligned} \int _{-1}^{0}- \ln \vert x\vert \cdot |f|^{2}dx\le |\ln \delta |\cdot \Vert f\Vert ^{2}_{L^{2}(-1,0)} +\Vert f\Vert ^{2}_{L^{2}(-\delta ,0)} +\frac{2 \delta }{e}\Vert f\Vert _{L^{2}(-\delta ,0)}\Vert f^{\prime }\Vert _{L^{2}(-\delta ,0 )}. \end{aligned}$$

(6.44)

By the inequality $2ab\le \frac{1}{s} a^2+sb^2$ valid for any $s>0$, we deduce

$$\begin{aligned} 2\Vert f\Vert _{L^{2}(-\delta ,0)}\Vert f^{\prime }\Vert _{L^{2}(-\delta ,0 )} + 2\Vert f\Vert _{L^{2}(0,\delta )}\Vert f^{\prime }\Vert _{L^{2}(0,\delta )} \le \frac{1}{s} \Vert f'\Vert ^{2}_{L^{2}(-\delta ,\delta )} +s\Vert f\Vert ^{2}_{L^{2}(-\delta ,\delta )}. \end{aligned}$$

By summing up with (6.41), for any $s>0$ and $\delta \in (0,1]$, we have

$$\begin{aligned} \int _{\mathbb R}- \ln \vert x\vert \cdot |f|^{2}dx\le \int _{-1}^{1}- \ln \vert x\vert \cdot |f|^{2}dx\le \frac{ \delta }{e s} \Vert f^{\prime }\Vert ^{2}_2 + \left( |\ln \delta |+ \frac{ s\delta }{e}+1\right) \Vert f\Vert ^{2}_2. \end{aligned}$$

We have obtained an inequality of the form

$$\begin{aligned} \int _{\mathbb R}- \ln \vert x\vert \cdot |f|^{2}dx\le c_1 \Vert f^{\prime }\Vert ^{2}_2 + c_2 \Vert f\Vert ^{2}_2 \end{aligned}$$

with $c_i>0$. Here, we use a dilation argument by applying this inequality to the rescaled function

$$\begin{aligned} f(x)=g\left( x \sqrt{t/c_1}\right) ,\quad t>0, \end{aligned}$$

and we obtain

$$\begin{aligned} \int _{\mathbb R}- \ln \vert x\vert \cdot |f|^{2}dx\le t\Vert f^{\prime }\Vert ^{2}_2 + \ln \left( \frac{ \sqrt{c_1}e^{ c_2}}{\sqrt{t}}\right) \Vert f\Vert ^{2}_2. \end{aligned}$$

Taking $c_1= \frac{ \delta }{e s}$ and $c_2= |\ln \delta |+ \frac{ s \delta }{e}+1$,

$$\begin{aligned} c(s\delta ):=\sqrt{c_1}e^{ c_2} = \sqrt{\frac{e}{s\delta } } e^{\frac{s\delta }{e}}. \end{aligned}$$

We minimize $c(s\delta )$ over $s\delta >0$ and get $\inf _{s\delta >0} c(s\delta )=\inf _{u>0} \sqrt{\frac{1}{u}}e^u=\sqrt{2e}$, which implies (6.40) as claimed.

1.2 Proof of Lemma 5.3

Let $0<\alpha <1,\delta >0$ and $f\in C_0^{\infty }(\mathbb {R})$. We write

$$\begin{aligned} I_{\alpha }(f)=\int _0^{\infty }\frac{1}{\vert x\vert ^{\alpha }} f^2(x)dx=J_{\alpha }+K_{\alpha } \end{aligned}$$

with $J_{\alpha }=\int _0^{\delta }\frac{1}{\vert x\vert ^{\alpha }} f^2(x)dx$ and $K_{\alpha }=\int _{\delta }^{\infty }\frac{1}{\vert x\vert ^{\alpha }} f^2(x)dx$. Obviously, $K_{\alpha } \le {\delta }^{-\alpha }\vert \vert f\vert \vert _2^2$. By integration by parts,

$$\begin{aligned} J_{\alpha }= & {} \left[ \frac{x^{1-\alpha }}{1-\alpha }f^2(x)\right] _0^{\delta } - \frac{2}{1-\alpha }\int _0^{\delta }{ x^{1-\alpha }} f f^{\prime }dx\\\le & {} \frac{\delta ^{1-\alpha }}{1-\alpha }f^2(\delta )+\frac{1}{1-\alpha } {\delta ^{1-\alpha }} \left( \frac{1}{\delta }\vert \vert f\vert \vert ^2_2+{\delta }\vert \vert f^{\prime }\vert \vert ^2_2\right) . \end{aligned}$$

This last inequality comes from:

$$\begin{aligned} 2\vert f f^{\prime }\vert \le \frac{1}{\delta } f^2+{\delta }\left( f^{\prime }\right) ^2. \end{aligned}$$

We now prove

$$\begin{aligned} \vert f(\delta )\vert \le \frac{1}{\sqrt{\delta }}\vert \vert f\vert \vert _2 + \sqrt{\delta } \vert \vert f^{\prime }\vert \vert _2. \end{aligned}$$

(6.45)

Let $x_0\in [0,\delta ]$ such that $\vert f(x_0)\vert =\inf _{ [0,\delta ]}\vert f(x)\vert$. Then

$$\begin{aligned} \vert f(\delta )\vert \le \vert f(\delta )-f(x_0)\vert +\vert f(x_0)\vert \le \int _0^{\delta }\vert f^{\prime }\vert + \frac{1}{\sqrt{\delta }} \vert \vert f\vert \vert _2 \le {\sqrt{\delta }} \vert \vert f^{\prime } \vert \vert _2 + \frac{1}{\sqrt{\delta }} \vert \vert f\vert \vert _2 \end{aligned}$$

by Hölder inequality. We deduce

$$\begin{aligned} \left| f(\delta )\right| ^2 \le 2{\delta } \vert \vert f^{\prime } \vert \vert _2^2 + \frac{2}{\delta }\vert \vert f\vert \vert _2^2. \end{aligned}$$

Therefore,

$$\begin{aligned} J_{\alpha }\le \frac{3}{1-\alpha }\left( {\delta }^{2-\alpha }\vert \vert f^{\prime } \vert \vert _2^2 + {\delta }^{-\alpha }\vert \vert f \vert \vert _2^2 \right) . \end{aligned}$$

From this bound and the bound on $K_{\alpha }$, we get

$$\begin{aligned} I_{\alpha }(f)\le \frac{3}{1-\alpha } {\delta }^{2-\alpha }\vert \vert f^{\prime } \vert \vert _2^2 +\frac{4-\alpha }{1-\alpha } {\delta }^{-\alpha }\vert \vert f \vert \vert _2^2. \end{aligned}$$

(6.46)

This inequality is stable by dilation. Indeed, changing f(x) by $f_{\lambda }(x)=f({\lambda }x)$, we obtain

$$\begin{aligned} I_{\alpha }(f)\le \frac{3}{1-\alpha } \left( {\delta {\lambda }}\right) ^{2-\alpha }\vert \vert f^{\prime } \vert \vert _2^2 +\frac{4-\alpha }{1-\alpha } \left( {\delta {\lambda }}\right) ^{-\alpha }\vert \vert f \vert \vert _2^2. \end{aligned}$$

This reduces to (6.46) by setting $s={\delta {\lambda }}$.

We set $c_1(\alpha )=\frac{3}{1-\alpha }$ and $c_2(\alpha ) =\frac{4-\alpha }{1-\alpha }$. Let $t>0$ and choose $\delta$ such that $t=c_1(\alpha ){\delta }^{2-\alpha }$. We set $\gamma =\frac{\alpha }{2-\alpha }$. The inequality (6.46) is equivalent to

$$\begin{aligned} I_{\alpha }(f)\le t\vert \vert f^{\prime } \vert \vert _2^2 +c_3 t^{-\gamma } \vert \vert f \vert \vert _2^2. \end{aligned}$$

with $c_3=c_2c_1^{\gamma }$. The $L^2$-norm are the norm on $L^2(\mathbb {R}^+)$. We easily deduce the result on $\mathbb {R}$,

$$\begin{aligned} \int _{-\infty }^{\infty }\frac{1}{\vert x\vert ^{\alpha }} f^2(x) dx = \int _0^{\infty }\frac{1}{\vert x\vert ^{\alpha }} f^2(x) dx+\int _0^{\infty }\frac{1}{\vert x\vert ^{\alpha }} f^2(-x) dx \le t\vert \vert f^{\prime } \vert \vert _2^2 +c_3 t^{-\gamma } \vert \vert f \vert \vert _2^2, \end{aligned}$$

where now, the $L^2$-norm are the norm on $L^2(\mathbb {R})$. To finish the proof of Lemma 5.3, for $g\in C_0^{\infty }(\mathbb {R}^2)$ and any $y\in \mathbb {R}$, we set $f(x)=g(x,y)$ in the inequality just above and integrate this inequality over $\mathbb {R}$ in y. We obtain

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{1}{\vert x\vert ^{\alpha }} g^2(x,y) dxdy \le t \vert \vert \frac{\partial g}{\partial x} \vert \vert _2^2 +c_3 t^{-\gamma } \vert \vert g \vert \vert _2^2. \end{aligned}$$

We conclude the lemma by the fact that

$$\begin{aligned} \left\| \frac{\partial g}{\partial x} \right\| ^2_{L^{2}\left( \mathbb {R}^{2}\right) }\le \left( \mathcal{L}g,g\right) . \end{aligned}$$

We take $b=\gamma$ to prove our inequality.

Proof of uniqueness of b. We use a dilation argument. Let $b^{\prime }>0$ such that, for any $t>0$,

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{1}{\vert x\vert ^{\alpha }} g^2(x,y) dxdy \le t \left( \mathcal{L}g,g\right) +c_3 t^{-b^{\prime }} \vert \vert g \vert \vert _2^2. \end{aligned}$$

Replace now g with $g \circ H_{\lambda }$ where $H_{\lambda }(x,y)=(\lambda x, \lambda ^{\beta } y)$, $\lambda >0$, for a fixed $\beta >1$; after a change of variables, we get for any $t>0$ and $\lambda >0$:

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{1}{\vert x\vert ^{\alpha }} g^2(x,y) dxdy \le t{\lambda }^{2-\alpha } (\mathcal{L}_{\lambda }g,g) +c_3 t^{-b^{\prime }}{\lambda }^{-\alpha } \vert \vert g \vert \vert _2^2 \end{aligned}$$

(6.47)

with

$$\begin{aligned} \mathcal{L}_{\lambda }=\mathcal{L}_{\lambda ,\beta }:= -\left( \frac{\partial }{\partial {x}}\right) ^2 - {\lambda }^{2\beta -2} \exp \left( -\frac{2\lambda ^{\alpha }}{\vert x\vert ^{\alpha }}\right) \left( \frac{\partial }{\partial {y}}\right) ^2. \end{aligned}$$

Let $s>0,{\lambda }>0$ and choose $t>0$ in (6.47) such that $s=t{\lambda }^{2-\alpha }$, then

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{1}{\vert x\vert ^{\alpha }} g^2(x,y) dxdy \le s \left( \mathcal{L}_{\lambda } g,g\right) +c_3 s^{-b^{\prime }}{\lambda }^{-b^{\prime }(\alpha -2)-\alpha } \vert \vert g \vert \vert _2^2. \end{aligned}$$

Assume $b^{\prime }>b$ and let $\lambda$ tend to 0 and s also (in that order), we get

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{1}{\vert x\vert ^{\alpha }} g^2(x,y) dxdy=0 \end{aligned}$$

for any function g: contradiction.

Now, let $s>0,{\lambda }>0$ and choose $t>0$ in (6.47) such that $s=t^{-b^{\prime }}{\lambda }^{-\alpha }$. Then

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{1}{\vert x\vert ^{\alpha }} g^2(x,y) dxdy \le s^{- \frac{1}{b^{\prime }}} {\lambda }^{-\frac{\alpha }{b^{\prime }}+2-\alpha } \left( \mathcal{L}_{\lambda }g,g\right) +c_3 s \vert \vert g \vert \vert _2^2. \end{aligned}$$

Assume $b>b^{\prime }$ and let $\lambda$ tend to $+\infty$ and s tend to 0 (in that order), we get the same contradiction. So $b^{\prime }=b$. The proof is completed.