Deriviative estimates in finite $L^p$ space

Question

I'm stuck as to how to solve the following exercise:

If $U$ is open and bounded with smooth boundary, $1<p<\infty$, $\epsilon>0$, and $u\in C^{\infty}(\bar U)$, show $\exists C$ s.t. $||Du||_p \leq \epsilon ||D^2u||_p + C||u||_p$.

Here $||\cdot||_p = ||\cdot||_{L^p(U)}$. Normally I'd try integration by parts but as $u$ doesn't necessarily have compact support I'm at a loss.

I don't like to completely spoil problems for myself: I'd appreciate a sketch of the proof or a to-do list more than a detailed solution; just something so that I know where to head.

Answer 1

Here is a sketch of a proof, you'll need to fill in a lot of details. I'll be more precise on the first step which is the crucial one.

• $\color{blue}{\text{Case}\ U = (0,b):}$ Fix $s \in (0,b/3)$, $t \in (2/3b,b)$. By the MVT there exists $\xi \in (s,t)$ for which $$u'(\xi) = \frac{u(t) - u(s)}{t -s}.$$ Then by the FTC, for $x \in (0,b)$ we obtain $$u'(x) = u'(\xi) + \int_{\xi}^xu''(y)\,dy = \frac{u(t) - u(s)}{t -s} + \int_{\xi}^xu''(y)\,dy.$$ Using that $t - s \ge b/3$ and Holder's inequality, we can estimate $|u'(x)|$ as follows: \begin{align} |u'(x)| \le &\ \frac{3}{b}(|u(t)| + |u(s)|) + \int_0^b|u''(y)|\, dy\\ \le &\ \frac{3}{b}(|u(t)| + |u(s)|) + C(b,p)\Big(\int_0^b|u''(y)|^p\, dy\Big)^{\frac 1p}. \end{align} Now raise everything to the $p$ and use convexity of $t \mapsto |t|^p$ with weights $1 - \epsilon$ and $\epsilon$ to get $$|u'(x)|^p \le C(b,p)(|u(t)|^p + |u(s)|^p) + \epsilon C(b,p)\int_0^b|u''(y)|^p\,dy.$$ Finally, average on $s$ and $t$ first and then on $x$ to get the desired estimate in this case.
• $\color{blue}{\text{Case}\ U = \prod_{i = 1}^n(a_i,b_i):}$ Use the previous result together with Fubini's theorem.
• $\color{blue}{\text{Case}\ U\ \text{is open and bounded:}}$ For any $U' \subset \subset U$ find a cover made of rectangles with sides parallel to the axes and contained in $U$. Use the previous result and then let $U' \uparrow U$.