1
$\begingroup$

I'm having trouble understanding the following calculation we've learned in class. The objective was to prove a log-Sobolev inequality.

We did it this way: We know that $$ \int_\Omega \sqrt{ | \nabla(\epsilon f^2 ) | ^2 + (D \cdot I(\epsilon f^2) )^2 } d\mu \geq D I\left(\int_\Omega \epsilon f^2 d \mu\right) $$ for some constant $D>0$ ($ \Omega$ is a region, $\mu$ is some measure on it, and $I(\epsilon ) = \sqrt{2} \epsilon \sqrt{\log(1/\epsilon)} $ when $\epsilon \to 0 $ ). The lecturer then said that by taking $ \epsilon \to 0 $ , we get: $$ \frac{D^2 } {2} \left( \int_\Omega f^2 \log f^2 d\mu -\int_\Omega f^2 d\mu \cdot \log \int_\Omega f^2 d\mu \right)\leq \int _\Omega | \nabla f| ^2 d\mu $$ (The LHS is excatly the entropy of $f^2$)

Can someone help me understand the calculation?

Thanks !

  • 0
    sorry. it should have been epsilon. i've just fixed it. Have you got any idea? Thanks2012-07-23
  • 0
    If $\Omega$ is of measure $1$. First, write what $I(\cdot)$ corresponds, with the definition. The $\varepsilon$ behind the square root will be simplified. Then write the gradient of the square as $2f\cdot\nabla f$, use Cauchy Schwarz inequality.2012-07-23
  • 0
    awsome! Thanks a lot to you @Davide!2012-07-23
  • 0
    @Davide: Perhaps you could post your comment as an answer so the OP can accept it?2012-07-23
  • 0
    @JesseMadnick Done now. Thanks!2012-07-23

1 Answers 1