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Prove $$\|Du\|_{L^{2p}(U)} \le C\|u\|_{{L^\infty}(U)}^{1/2} \|D^2 u\|_{L^p(U)}^{1/2}$$ for $1 \le p < \infty$ and all $u \in C_c^\infty(U)$.

This is PDE Evans, 2nd edition: Chapter 5, Exercise 10(b).

Here is what I did so far: \begin{align} \int_U |Du|^{2p} \, dx &= \sum_{n=1}^\infty \int_U u_{x_i} |Du|^{2p-1} \, dx \\ &= -\sum_{i=1}^n \int_U u(x_i)(2p-1)|Du|^{2p-2} |D^2 u| \, dx \\ &\le C\int_U u|Du|^{2p-2} |D^2 u| \, dx \\ &\le C\left(\int_U u|D^2 u|^{p} \right)^{\frac 1{p}} \left(\int_U \left||Du|^{2p-2}\right|^{\frac{2p}{2p-2}} \, dx\right)^{\frac{2p-2}{2p}} \\ \end{align} where the last step is trying to use Hölder's inequality. The $2p-1$'s (or $2p-2$'s) in the exponent of the second integrand are supposed to cancel, but they aren't. Where can I go from here?

Assuming my previous work is correct so far (probably not), we divide both sides of our inequality by $\left(\int_U u|D^2 u|^{2p} \right)^{\frac 1{2p}}$ to obtain \begin{align} \left(\int_U |Du|^{2p} \, dx \right)^{\frac{2p-1}{2p}} &\le C \left(\int_U \left||Du|^{2p-2}\right|^{\frac{2p}{2p-2}} \, dx\right)^{\frac{2p-2}{2p}} \\ &= C \left(\int_U |Du|^{2p} \, dx\right)^{\frac{2p-2}{2p}} \end{align}

By the way, since this is part (b) of Exercise 10 in the 2nd edition book, I suspected this was similar to part (a) of the exercise. So I suspected this method I utilized here should be similar to the method presented in part (a).

Anyways, how can I show that $$\left(\int_U |Du|^{2p} \, dx \right)^{\frac 2{2p}} \le C\left(\text{ess sup}_U |u| \right) \left(\int_U |D^2 u| \, dx \right)^{\frac 1p},$$ from which the result follows by taking the square root of both sides?

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Cookie
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  • You are on the right trick. Just double check your calculation. – spatially Jan 13 '15 at 18:04
  • Did I mess up at the line in which I applied Hölder's inequality? – Cookie Jan 13 '15 at 18:51
  • I am on my phone so I can not look at it carefully... I will write you an answer tonight if you still need it. – spatially Jan 13 '15 at 19:01
  • Maybe try to start with $|\nabla u|^{2p}=|\nabla u|^{2p-2}\nabla u\nabla u$ – spatially Jan 13 '15 at 23:13
  • I had tried that too. In either case, I had trouble getting Hölder's inequality to work for the RHS of the inequality $$\int_U |Du|^{2p} , dx \le C \int_U |D^2u| |Du|^{2p-2} , dx$$ (if I even arrived at this inequality correctly...) – Cookie Jan 13 '15 at 23:20
  • Ok. I'll work on it later – spatially Jan 14 '15 at 00:39
  • The beginning of your proof isn't correct; see my answer here –  Jul 11 '15 at 17:15
  • @1999 I thought the gradient is denoted as $Du = \sum_{i=1}^n u_{x_i}$. – Cookie Jul 11 '15 at 17:25
  • We all were mistaken once. Please see the definition of a gradient. –  Jul 11 '15 at 17:28
  • @1999 Ohhh... $Du = \langle u_{x_1,},\ldots,u_{x_n} \rangle$ is a vector. But I hope this is correct: $|Du|^2 = \sum_{i=1}^n u_{x_i}^2$. It's just like saying $|x|^2=\sum_{i=1}^n x_i^2$ (for $x =(x_1,\ldots,x_n)$), but we certainly cannot say $|x|=\sum_{i=1}^n x_i$. – Cookie Jul 11 '15 at 17:31
  • Yes this is correct. This is what I use in the linked answer. –  Jul 11 '15 at 17:39

1 Answers1

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Notice that rewriting what you have on the second to last line of teh first chain of inequalities (you applied Holder wrong in the last line): $$ \| Du\|_{2p}^{2p} \leq C\int |u||D^2 u||Du|^{2p-2} \leq C\| u\|_{\infty} \| D^2 u\|_{p}\| Du\|_{2p}^{2p-2}, $$ since $2p/(2p-2)=p/(p-1)$ is the conjugate exponent of $p$. In other words, $$ \| Du\|_{2p}^2 \leq C\| u\|_{\infty} \| D^2 u\|_p. $$

Jose27
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  • Ohh wow, so it was just a simple algebra 1 mistake in finding the conjugate. :| – Cookie Jan 14 '15 at 19:32
  • By the way, when we invoke Hölder's inequality on $$C \left( \int_U |u \cdot D^2u |^p \right)^{1/p},$$ am I going to be using the conjugates $1$ and $\infty$ for $\frac 1{\infty}+\frac 1p=1$? Hence, I am going to receive an $L^\infty$ norm along the way in my long chain of inequalities? – Cookie Jan 14 '15 at 19:40
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    Yes, you're using Holder with $1$ and $\infty$, that's where the $| u|_\infty$ term comes from. – Jose27 Jan 14 '15 at 22:15