Multivariable Calculation- Entropy of A Function

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 !

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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.