# Ngô Quốc Anh

## June 11, 2011

### The Entropy-Logarithmic energy inequality

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 17:39

Recently, I have read a result by J. Demange published in J. Funct. Anal. in 2008 [here]. In that paper,  as a simple consequence, the author proves the following interesting inequality

Theorem. Let $n\geqslant 2$ be an integer and $M$ be a compact connected $n$-dimensional Riemannian manifold with Ricci curvature bounded below by a positive constant $\rho$. The following inequality holds for $d \in \mathbb R$, $d>n$, $d \geqslant 3$, $\alpha=1/2-1/d$, and $f$, a smooth function mapping $M$ onto $\mathbb R_+$

$\displaystyle \left(\int_M f\right)^{2\alpha}-\int_M f^{2\alpha} \leqslant \frac{1}{K(n,d)}\int_M |\nabla f^\alpha|^2$

and

$\displaystyle \left(\int_M f\right)^{2\alpha}-\int_M f^{2\alpha} \leqslant K_1\log\left(1+K_2\int_M |\nabla f^\alpha|^2\right)$

where

$\displaystyle K(n,d)=\frac{\rho (d-2)}{4(1-\frac{1}{n})}, \quad L(n,d)=\frac{d-n}{n+2}\left( 4-9\frac{d-n}{d(n+2)}\right),$

and

$\displaystyle K_2=\frac{L(n,d)d}{4(d-2)K(n,d)(\int_M f)^{2\alpha}},\quad \frac{1}{K_1}=K_2K(n,d).$

In my opinion, the interesting part is that it measures the difference

$\displaystyle \left(\int_M f\right)^{2\alpha}-\int_M f^{2\alpha}.$

In view of the Jensen inequality and since $0<2\alpha<1$, we know that

$\displaystyle {\left( {\frac{1}{{\text{vol}(M)}}\int_M f } \right)^{2\alpha }} \geqslant \frac{1}{{\text{vol}(M)}}\int_M {{f^{2\alpha }}}.$

In other words,

$\displaystyle {\left( {\int_M f } \right)^{2\alpha }} \geqslant {(\text{vol}(M))^{2\alpha - 1}}\int_M {{f^{2\alpha }}} .$

There is a different view. From the Poincare inequality, we all know that

$\displaystyle\int_M {{{\left( {f - \frac{1}{\text{vol}(M)}\int_M f } \right)}^2}} \leqslant \int_M {|\nabla f{|^2}} .$

Expanding the term on the left gives

$\displaystyle\int_M {{f^2}} \leqslant \frac{1}{\text{vol}(M)}{\left( {\int_M f } \right)^2} + \int_M {|\nabla f{|^2}} .$

Therefore, we again have a way to measure the difference

$\displaystyle \left(\int_M f\right)^2-\int_M f^2.$