A Generalization of the Logarithmic Gross–Sobolev Inequality

Abstract

A sharp integral inequality is proved that is used to derive a Sobolev interpolation inequality. A generalization of the logarithmic Sobolev inequality is proposed based on the Sobolev interpolation inequality.

1 SOBOLEV INTERPOLATION INEQUALITY

In this section, we prove a sharp integral inequality implying, due to the Hausdorff–Young inequality, a Sobolev interpolation inequality.

1.1 Integral Inequality

For convenience, we use the following notation:

$${\text{||}}U{\text{|}}{{{\text{|}}}_{p}} = {{\left\{ {\mathop \smallint \limits_{{{R}^{n}}}^{} {{{\left| {U\left( x \right)} \right|}}^{p}}dx} \right\}}^{{1/p}}},\quad p \geqslant 1,$$

is the norm in ${{L}_{p}}({{R}^{n}})$; the index p in $\parallel \cdot {{\parallel }_{p}}$ with p = 2 is omitted, that is, we write $\parallel \cdot \parallel$ in this case. Let k be any positive number. Let $\rho$ be a given positive number that is arbitrary if $n - k \leqslant 0$ and satisfies the inequality $\rho < \frac{{2k}}{{n - k}}$ if $n - k > 0$. We set $\alpha = \frac{{n\rho }}{{k(\rho + 2)}}$; for a given $\alpha ,$ we introduce the quantity $\chi = \sqrt {{{\alpha }^{\alpha }}{{{\left( {1 - \alpha } \right)}}^{{1 - \alpha }}}}$.

For any θ > 0, we define the Euler gamma function ${\Gamma }\left( \theta \right) = \mathop \smallint \limits_0^{ + \infty } {{e}^{{ - t}}}{{t}^{{\theta - 1}}}dt$; $B\left( {\beta ,\gamma } \right) = \mathop \smallint \limits_0^1 {{t}^{{\beta - 1}}}{{\left( {1 - t} \right)}^{{\gamma - 1}}}dt$ for all β > 0 and γ > 0 is the Euler beta function; ${{\sigma }_{n}} = \frac{{2{{\pi }^{{n/2}}}}}{{{\Gamma }\left( {\frac{n}{2}} \right)}}$;

$$\begin{gathered} {{K}_{g}}\left( \alpha \right) = \frac{1}{\chi }{{\left[ {\frac{{{{\sigma }_{n}}}}{k}~B\left( {\frac{n}{k},\,\,\frac{{n\left( {1 - \alpha } \right)}}{{k\alpha }}} \right)} \right]}^{{\alpha k/2n}}} \\ = \frac{1}{\chi }{{\left[ {\frac{{\frac{{{{\sigma }_{n}}}}{k}{\Gamma }\frac{n}{k}{\Gamma }\frac{{n(1 - \alpha )}}{{k\alpha }}}}{{{\Gamma }\left( {n{\text{/}}\left( {k\alpha } \right)} \right)}}} \right]}^{{\frac{{\alpha k}}{{2n}}}}}. \\ \end{gathered} $$

((1))

Lemma 1.Let$k,\rho ,$and α be the numbers defined above, $V(x) \in {{L}_{2}}({{R}^{n}}),~~{{r}^{{k/2}}}V(x) \in {{L}_{2}}({{R}^{n}})$, r = |x|. Then the following integral inequality holds:

$${\text{||}}V{\text{|}}{{{\text{|}}}_{{\left( {\rho + 2} \right)/\left( {\rho + 1} \right)}}} \leqslant {{K}_{g}}\left( \alpha \right){\text{||}}{{r}^{{k/2}}}V{\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}V{\text{|}}{{{\text{|}}}^{{1 - \alpha }}},$$

((2))

where${{K}_{g}}(\alpha )$is the constant defined by (1). The constant is sharp: inequality (2) turns into equality with

$$V\left( x \right) = {{V}_{0}}\left( r \right) = \frac{{{{\omega }_{1}}}}{{{{{({{\omega }_{2}} + {{\omega }_{2}}{{r}^{k}})}}^{{1 + 1/\rho }}}}},$$

where${{\omega }_{1}},{{\omega }_{2}},$and ω₃are arbitrary positive numbers.

1.2 Hausdorff–Young Inequality

Lemma 2.Let

$$\hat {U}\left( \xi \right) = \frac{1}{{{{{\left( {2\pi } \right)}}^{{n/2}}}}}\mathop \smallint \limits_{{{R}^{n}}}^{} {{e}^{{ - i\left( {x,\xi } \right)}}}U(x)dx,~~~~\xi \in {{R}^{n}},$$

be the Fourier transform of a function U(x), $\hat {U} \in {{L}_{p}}({{R}^{n}})$, $1 \leqslant p \leqslant 2$. Then the Hausdorff–Young inequality

$$\begin{gathered} {\text{||}}U{\text{|}}{{{\text{|}}}_{{p'}}} \leqslant {{K}_{B}}\left( p \right){\text{||}}\hat {U}{\text{|}}{{{\text{|}}}_{p}},~ \\ {{K}_{B}}\left( p \right) = {{\left[ {{{{\left( {\frac{p}{{2\pi }}} \right)}}^{{1/p}}}{{{\left( {\frac{{p'}}{{2\pi }}} \right)}}^{{ - 1/p'}}}} \right]}^{{n/2}}}, \\ \end{gathered} $$

$1 \leqslant p \leqslant 2,\,\,\frac{1}{p} + \frac{1}{{p'}} = 1$, holds with the best Beckner–Babenko constant [1–4].

1.3 Results

Theorem 1.Let$k,\rho ,$and α be the numbers defined above, $U(x) \in {{L}_{2}}({{R}^{n}})$, ${{r}^{{k/2}}}\hat {U}\left( \xi \right) \in {{L}_{2}}({{R}^{n}}),r = \left| \xi \right|$. Then the following multiplicative Sobolev inequality holds:

$${\text{||}}U{\text{|}}{{{\text{|}}}_{{\rho + 2}}} \leqslant {{\bar {K}}_{0}}{\text{||}}{{r}^{{k/2}}}\hat {U}\left( \xi \right){\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}U{\text{|}}{{{\text{|}}}^{{1 - \alpha }}}.$$

((3))

Here, ${{\bar {K}}_{0}} = {{K}_{g}}\left( \alpha \right){{K}_{B}}\left( {\frac{{\rho + 2}}{{\rho + 1}}} \right),$where${{K}_{g}}(\alpha )~$is defined by (1).

We give a scheme for proving (3).

In view of inequality (2), we conclude that

$${\text{||}}\hat {U}{\text{|}}{{{\text{|}}}_{{\frac{{\rho + 2}}{{\rho + 1}}}}} \leqslant {{K}_{g}}(\alpha ){\text{||}}{{\left| \xi \right|}^{{k/2}}}\hat {U}{\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}\hat {U}{\text{|}}{{{\text{|}}}^{{1 - \alpha }}}.$$

((4))

Due to the Plancherel–Parseval theorem, we have

$${\text{||}}\hat {U}{\text{||}} = {\text{||}}U{\text{||}}.$$

((5))

Therefore, under the assumptions of Theorem 1, we deduce that $\hat {U} \in {{L}_{{\frac{{\rho + 2}}{{\rho + 1}}}}}({{R}^{n}})$. Then the Hausdorff–Young inequality implies

$${\text{||}}U{\text{|}}{{{\text{|}}}_{{\rho + 2}}} \leqslant {{K}_{B}}\left( {\frac{{\rho + 2}}{{\rho + 1}}} \right){\text{||}}\hat {U}{\text{|}}{{{\text{|}}}_{{\frac{{\rho + 2}}{{\rho + 1}}}}}.$$

((6))

Inequality (3) follows from (4)–(6).

Assume that k = 2 in Theorem 1. Owing to the relation ${\text{||}}\left| \xi \right|\hat {U}{\text{||}} = {\text{||}}\left. {\nabla U} \right|$, Theorem 1 implies the following corollary.

Corollary 1.Let$\rho \in \left( {0,\infty } \right)$with$n = 1,~2$and$\rho \in \left( {0,\frac{4}{{n - 2}}} \right)~$with$n \geqslant 3,$and let $\propto \, = \,\rho n{\text{/}}[2(\rho + 2)]$. Let $U(x) \in {{H}^{1}}({{R}^{n}})$. Then the following Gagliardo–Nirenberg–Sobolev interpolation inequality holds:

$${\text{||}}U{\text{|}}{{{\text{|}}}_{{\rho + 2}}} \leqslant {{\bar {K}}_{0}}{\text{||}}\nabla U{\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}U{\text{|}}{{{\text{|}}}^{{1 - \alpha }}}.$$

((7))

Here, ${{\bar {K}}_{0}} = {{K}_{g}}\left( \alpha \right){{K}_{B}}\left( {\frac{{\rho + 2}}{{\rho + 1}}} \right)$, where

$$\begin{gathered} {{K}_{g}}\left( \alpha \right) = \frac{1}{\chi }{{\left[ {0.5{{\sigma }_{n}}B\left( {\frac{n}{2},~\,\frac{{n\left( {1 - \alpha } \right)}}{{2\alpha }}} \right)} \right]}^{{\alpha /n}}} \\ = \frac{1}{\chi }{{\pi }^{{\alpha /2}}}{{\left[ {\frac{{{\Gamma }\left[ {\left( {n - n\alpha } \right){\text{/}}2\alpha } \right]}}{{{\Gamma }\left( {n{\text{/}}2\alpha } \right)}}} \right]}^{{\alpha /n}}}. \\ \end{gathered} $$

Assume that k = 4 in Theorem 1. Owing to the relation ${\text{||}}{{\left| \xi \right|}^{2}}\hat {U}{\text{||}} = {\text{||}}\Delta U{\text{||}}$ [5], Theorem 1 implies the following corollary.

Corollary 2.Let$\rho \in (0,\infty )$with$n \leqslant 4~$and$\rho \in \left( {0,~\frac{8}{{n - 4}}} \right)$with n > 4, and let $\propto \, = n\rho {\text{/}}\left[ {4\left( {\rho + 2} \right)} \right]$. Let$U(x) \in {{L}_{2}}({{R}^{n}})$and $\Delta U \in {{L}_{2}}({{R}^{n}})$. Then the following Sobolev interpolation inequality holds:

$${\text{||}}U{\text{|}}{{{\text{|}}}_{{\rho + 2}}} \leqslant {{\bar {K}}_{0}}{\text{||}}\Delta U{\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}U{\text{|}}{{{\text{|}}}^{{1 - \alpha }}}.$$

((8))

Here,

$${{\bar {K}}_{0}} = {{K}_{g}}\left( \alpha \right){{K}_{B}}\left( {\frac{{\rho + 2}}{{\rho + 1}}} \right),$$

where

$${{K}_{g}}\left( \alpha \right) = \frac{1}{\chi }{{\left[ {\frac{{{{\sigma }_{n}}}}{4}B\left( {\frac{n}{4},~\frac{{n\left( {1 - \alpha } \right)}}{{4\alpha }}} \right)} \right]}^{{2\alpha /n}}}.$$

2 LOGARITHMIC GROSS–SOBOLEV INEQUALITY

Theorem 2.Let k be an arbitrary positive number, $U\left( x \right) \in {{L}_{2}}({{R}^{n}})$, and ${{\left| \xi \right|}^{{k/2}}}\hat {U}\left( \xi \right) \in {{L}_{2}}({{R}^{n}})$. Then the following logarithmic Gross–Sobolev inequality holds:

$$\begin{gathered} \mathop \smallint \limits_{{{R}^{n}}}^{} \frac{{{{{\left| U \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}{\text{ln}}\left( {\frac{{{{{\left| U \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right)dx \\ \leqslant \frac{n}{k}{\text{ln}}\left[ {\frac{{k{{{\left( {\frac{{{{\sigma }_{n}}}}{k}{\Gamma }\left( {\frac{n}{k}} \right)} \right)}}^{{k/n}}}{\text{||}}{{{\left| \xi \right|}}^{{k/2}}}\hat {U}{\text{|}}{{{\text{|}}}^{2}}}}{{n{{\pi }^{k}}{{e}^{{k - 1}}}{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right]. \\ \end{gathered} $$

((9))

Theorem 1 implies the following propositions.

Proposition 1.Let $U(x) \in {{H}^{1}}({{R}^{n}})$. Then the following logarithmic Gross–Sobolev inequality holds:

$$\mathop \smallint \limits_{{{R}^{n}}}^{} \frac{{{{{\left| U \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}{\text{ln}}\left( {\frac{{{{{\left| U \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right)dx \leqslant \frac{n}{2}{\text{ln}}\left( {\frac{{2{\text{||}}\nabla U{\text{|}}{{{\text{|}}}^{2}}}}{{\pi en{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right).$$

((10))

Inequality (10) is sharp: it turns into equality with

$$U\left( x \right) = a~{\text{exp}}( - b{{\left| x \right|}^{2}}),$$

where aandb are arbitrary positive constants.

We put k = 4 in (9).

Proposition 2.Let $U(x) \in {{H}^{2}}({{R}^{n}})$. Then the following logarithmic Gross–Sobolev inequality holds:

$$\begin{gathered} \mathop \smallint \limits_{{{R}^{n}}}^{} \frac{{{{{\left| U \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}{\text{ln}}\left( {\frac{{{{{\left| U \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right)dx \\ \leqslant \frac{n}{4}{\text{ln}}\left[ {\frac{{4{{{\left( {\frac{{{{\sigma }_{n}}}}{4}\Gamma \left( {\frac{n}{4}} \right)} \right)}}^{{4/n}}}{\text{||}}\Delta U{\text{|}}{{{\text{|}}}^{2}}}}{{n{{\pi }^{4}}{{e}^{3}}{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right]. \\ \end{gathered} $$

((11))

Inequality (10) was first proved by Gross [6]. Beckner [7] notes that after Gross found the logarithmic Sobolev inequality, it became folklore. Other inequalities of Gross–Sobolev type were proved in [4, 8–10] and etc.

We give a scheme for proving (9). We rewrite (3) in the form

$${\text{||}}U{\text{|}}{{{\text{|}}}_{{\frac{{2n}}{{n - \alpha k}}}}} \leqslant {{K}_{B}}\left( \alpha \right){{K}_{g}}\left( \alpha \right){\text{||}}{{\left| \xi \right|}^{{k/2}}}\hat {U}{\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}U{\text{|}}{{{\text{|}}}^{{1 - \alpha }}},$$

((12))

where

$${{K}_{B}}\left( \alpha \right) = {{\left( {\frac{n}{\pi }} \right)}^{{\frac{{\alpha k}}{2}}}}\frac{{{{{\left( {n - \alpha k} \right)}}^{{\frac{{n - \alpha k}}{4}}}}}}{{{{{\left( {n + \alpha k} \right)}}^{{\frac{{n + \alpha k}}{4}}}}}},$$

$${{K}_{g}}\left( \alpha \right) = \frac{1}{{\sqrt {{{\alpha }^{\alpha }}{{{\left( {1 - \alpha } \right)}}^{{1 - \alpha }}}} }}{{\left[ {\frac{{{{\sigma }_{n}}}}{k}B\left( {\frac{n}{k},~\frac{{n\left( {1 - \alpha } \right)}}{{k\alpha }}} \right)} \right]}^{{\frac{{\alpha k}}{{2n}}}}}.$$

It is straightforward to show that $\mathop {\lim }\limits_{\alpha \to 0 + 0} {{K}_{g}}\left( \alpha \right) = 1$ and ${{K}_{B}}(0) = 1$. Thus, inequality (12) remains valid even when α = 0.

We consider the function

$$f\left( \alpha \right) = {\text{||}}U{\text{|}}{{{\text{|}}}_{{2n/n - \alpha k}}} - {{K}_{0}}{\text{||}}{{\left| \xi \right|}^{{k/2}}}\hat {U}{\text{|}}{{{\text{|}}}^{\alpha }}{\text{||}}U{\text{|}}{{{\text{|}}}^{{1 - \alpha }}},$$

where ${{K}_{0}}\, = \,{{K}_{B}}(\alpha ){{K}_{g}}(\alpha )$. Since $f\left( \alpha \right) \leqslant 0$ for $\alpha \in \left[ {0,} \right.\left. 1 \right)$, we have $f'(0) \leqslant 0$. When calculating $f'(0)$, we should take into account that $K_{B}^{'}(0) = - \ln (\pi e)$ and $K_{g}^{'}(0)$ = $\frac{1}{2}{\text{ln}}\left[ {\frac{{ek{{{\left( {\frac{{{{\sigma }_{n}}}}{k}\Gamma \left( {\frac{n}{k}} \right)} \right)}}^{{k/n}}}}}{n}} \right].$

Inequality (2) is also used to prove the generalized entropy inequality

$$\begin{gathered} - \mathop \smallint \limits_{{{R}^{n}}}^{} \frac{{{{{\left| {U\left( x \right)} \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}{\text{ln}}\left( {\frac{{{{{\left| {U\left( x \right)} \right|}}^{2}}}}{{{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right)dx \\ \leqslant \frac{n}{k}{\text{ln}}\left[ {\frac{{ek{{{\left( {\frac{{{{\sigma }_{n}}}}{k}\Gamma \left( {\frac{n}{k}} \right)} \right)}}^{{k/n}}}{\text{||}}{{r}^{{k/n}}}U{\text{|}}{{{\text{|}}}^{2}}}}{{n{\text{||}}U{\text{|}}{{{\text{|}}}^{2}}}}} \right]. \\ \end{gathered} $$

((13))

Under the condition $U(x)\, \in \,{{L}_{2}}({{R}^{n}})$, we have ${{r}^{{k/2}}}U(x)\, \in$L₂(Rⁿ) for any k > 0. Inequality (13) is sharp: it turns into equality with

$$U\left( x \right) = {{U}_{0}}\left( r \right) = a~{\text{exp}}( - b{{\left| x \right|}^{k}}),$$

where a and b are arbitrary positive constants [11].

Remark 1. Interpolation inequality (7) was also proved in [12] with another constant. This inequality is used to analyze the global solvability of the Cauchy problem for a nonlinear evolution Schrödinger equation [13], as well as in the spectral theory for Schrödinger operators [12].

Remark 2. Inequality (13) with k = 2 was announced in [13] and was proved in [14].