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I've been trying to work out the solution to Question 9 in Chapter 5 of Evans, and I'm having some difficulties.

I've been looking at the solution posted here: Question $5.9$ - Evans PDE $2$nd edition

And I can follow it through up to the Holder inequality step, but after that, I'm kind of lost with what I should be doing. The author of that solution says to divide by the gradient term, which I assume means the entire integral, but I'm not 100% sure what this will actually look like, especially when the integrals have exponents on them, and the terms inside each integral have differing exponents.

If anyone could provide a bit of insight into this step, it'd be much appreciated.

Pedro
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Jack
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1 Answers1

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In the end we don't want the gradient term on the RHS of the inequality: $$\int_U|Du|^p\,dx\le C\left(\int_U |u|^\frac{p}{2} |D^2u|^\frac{p}{2}\right)^\frac{2}{p} \left(\int_U |Du|^p \right)^\frac{p-2}{p},$$ dividing by $\left(\int_U |Du|^p \right)^\frac{p-2}{p}$, we obtain $$\left(\int_U |Du|^p \right)^{\frac{2}{p}}\le C\left(\int_U |u|^\frac{p}{2} |D^2u|^\frac{p}{2}\right)^\frac{2}{p}\le C\left(\int_U|u|^p\right)^\frac{1}{p}\left(\int_U|D^2u|^p\right)^\frac{1}{p},$$ taking $p/2$-th power we get the result.

Ellya
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  • In moving from the middle to right inequality on the last line, are you using Holder's Inequality again, or just moving powers around?? – Jack Apr 18 '15 at 14:55
  • Holder's, $\int_U |u|^\frac{p}{2} |D^2u|^\frac{p}{2}=||u|^\frac{p}{2}|D^2u|^\frac{p}{2}|{L^1}\le ||u|^\frac{p}{2}|{L^2}||D^2u|^\frac{p}{2}|{L^2}=|u^p|{L^1}^\frac{1}{2}| |D^2u|^p|_{L^1}^\frac{1}{2}.$ – Ellya Apr 18 '15 at 15:29
  • Just wanted to make sure, thank you!! :) – Jack Apr 18 '15 at 15:29
  • No worries, happy to help :) – Ellya Apr 18 '15 at 15:32