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cheeez
Feb26-11, 10:25 PM
$\frac{d}{dp}(\Vert f\Vert_{p}^{2})\mid_{p=2}=\frac{d}{dp}((\int_{\mat hbb{\mathbb{R}}^{n}}\mid f\mid^{p}dx)^{\frac{2}{p}})\ldots=\frac{1}{2}\int_ {\mathbb{\mathbb{R}}^{n}}\mid f(x)\mid^{2}ln\left(\frac{\mid f(x)\mid^{2}}{\Vert f\Vert_{2}^{2}}\right)$

this is the derivative evaluated at p=2.

does anyone see how this works? I started with the e^ln(whole integral) and then chain rule and so on...then plugged in p=2 at the end but couldn't get it.





wow i cannot get the d/dp to show up properly something is wrong with the latex....

Char. Limit
Feb27-11, 03:00 AM
\frac{d}{dp}(\Vert f\Vert_{p}^{2})\mid_{p=2}=\frac{d}{dp}((\int_{\mat hbb{\mathbb{R}}^{n}}\mid f\mid^{p}dx)^{\frac{2}{p}})\ldots=\frac{1}{2}\int_ {\mathbb{\mathbb{R}}^{n}}\mid f(x)\mid^{2}ln\left(\frac{\mid f(x)\mid^{2}}{\Vert f\Vert_{2}^{2}}\right)$

this is the derivative evaluated at p=2.

does anyone see how this works? I started with the e^ln(whole integral) and then chain rule and so on...then plugged in p=2 at the end but couldn't get it.





wow i cannot get the d/dp to show up properly something is wrong with the latex....

Did that fix it?