Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 !

share|cite|improve this question
sorry. it should have been epsilon. i've just fixed it. Have you got any idea? Thanks – Franklin McDeover Jul 23 '12 at 14:10
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. – Davide Giraudo Jul 23 '12 at 14:36
awsome! Thanks a lot to you @Davide! – Franklin McDeover Jul 23 '12 at 16:22
@Davide: Perhaps you could post your comment as an answer so the OP can accept it? – Jesse Madnick Jul 23 '12 at 16:25
@JesseMadnick Done now. Thanks! – Davide Giraudo Jul 23 '12 at 16:52

The admitted inequality gives $$\int_{\Omega}\sqrt{\varepsilon^2|\nabla (f^2)|^2+2D^2\varepsilon^2f^4\left(-\log(\varepsilon f^2)\right)}d\mu\geq D\sqrt 2\varepsilon\int_{\Omega}f^2d\mu\sqrt{-\log\left(\int_{\Omega}\varepsilon f^2\right)},$$ hence $$\int_{\Omega}\sqrt{4f^2|\nabla f^2|-2D^2f^4\log(\varepsilon f^2)}d\mu\geq D\sqrt 2\int_{\Omega}f^2d\mu\sqrt{-\log\left(\int_{\Omega}\varepsilon f^2\right)}.$$ The LHS is $$\leq \sqrt{\int_{\Omega}f^2d\mu}\sqrt{\int_{\Omega}(4|\nabla f|^2-2D^2f^2\log(\varepsilon f^2))d\mu}.$$ We get $$\sqrt{\int_{\Omega}(4|\nabla f|^2-2D^2f^2\log(\varepsilon f^2))d\mu}\geq D\sqrt 2\sqrt{\int_{\Omega}f^2d\mu}\sqrt{-\log\left(\int_{\Omega}\varepsilon f^2\right)},$$ and taking the squares, $$\int_{\Omega}(4|\nabla f|^2-2f^2D^2\log(\varepsilon f^2))d\mu\geq 2D^2\int_{\Omega}f^2d\mu\left(-\log\left(\int_{\Omega}\varepsilon f^2\right)d\mu\right).$$ This gives $$\int_{\Omega}|\nabla f|^2d\mu\geq \frac{D^2}2\left(\int_{\Omega}f^2\log(\varepsilon f^2)d\mu-\int_{\Omega}f^2d\mu\log\int_{\Omega}(\varepsilon f^2)d\mu\right).$$ When $\varepsilon \to\color{red}1$, it gives the wanted inequality.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.