# Question from Evans' PDE book

How do you do the second part of question 8, chapter 5, of Evans' PDE book (first edition)? I have proven the inequality for smooth, compactly supported functions using integration by parts, and I understand why approximating sequences as described in the hint exist, but I can not use the hint to extend the inequality as required.

The question is

"Integrate by parts to prove the interpolation inequality

$$\int_U |Du|^2\,dx\leqslant C\left(\int_Uu^2\, dx\right)^{\frac{1}{2}}\left(\int_U|D^2u|^2\,dx\right)^{\frac{1}{2}}$$

for $u\in C^\infty_c(U)$. By approximation, prove this inequality if $u\in H^2(U)\cap H_0^1(U)$. "

The hint is to approximate $u$ by functions in $C^\infty_c(U)$ which converge to u in $H_0^1(U)$, and to approximate $u$ (also) by functions in $C^\infty(closure(U))$ which converge to $u$ in $H^2(U)$.

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Could you please say what the question in the book is? (There may people who can help but who don't have easy access to the book. Also, in some unfortunate cases, different printings even of the same edition of a book may have differences in exercise numberings.) –  Jonas Meyer Dec 13 '12 at 19:59
@Frank I have added the question. Please let verify that this is, in fact, correct. –  Alex Youcis Dec 13 '12 at 20:04
Yes, that is correct. Many thanks for doing this - embarrassingly, I do not know how to 'type maths.' –  Frank Dec 13 '12 at 20:31

What you need is the following:

Let $v \in C^\infty_c(U)$ and $w\in C^\infty(\bar{U})$, we have

$$\left(\int_U |Dv \cdot Dw| \mathrm{d}x\right)^2 \leq C \int_U |v|^2 \mathrm{d}x \int_U |D^2 w|\mathrm{d}x \tag{*}$$

This follows by directly integrating by parts (the boundary terms vanish as $v$ has compact support).

Now, given $u \in H^1_0(U) \cap H^2(U)$, let $v_i \to u$ in $H^1_0$ and $w_i \to u$ in $H^2(U)$ where $v_i \in C^\infty_c(U)$ and $w_i \in C^\infty(U)$.

By the strong convergence in $H^1_0$ and $H^2$ respectively, we have that for any function $f\in L^2$ we have

$$\lim_{\ell \to \infty}\int_U \partial_{x^j} v_\ell f \mathrm{d}x = \lim_{\ell \to \infty}\int_{U} \partial_{x_j} (v_\ell - u + u) f \mathrm{d}x = \int_{U} \partial_{x^j} u f \mathrm{d}x + \lim_{\ell\to\infty}\int_{U} (\partial_{x_j}v_\ell - \partial_{x_j}u) f \mathrm{d}x$$

The second term on the RHS tends to zero using Cauchy-Schwarz and the assumed convergence of $v_\ell\to u$. Similarly we also have

$$\lim_{\ell \to \infty}\int_U \partial_{x_j} w_\ell f \mathrm{d}x = \int_{U} \partial_{x_j} u f \mathrm{d}x$$

So we have that

$$\int_U |Du|^2 \mathrm{d}x = \lim_{i,j\to \infty} \int_U |Dv_i \cdot D w_j| \mathrm{d}x \leq \lim_{i,j\to\infty} C \|v_i\|_{L^2} \|D^2 w_j\|_{L^2}$$

by (*). Since $v_i \to u$ in $H^1_0$, we also have $v_i \to u$ in $L^2$. Similarly as $w_j \to u$ in $H^2$ we have $D^2w \to D^2 u$ in $L^2$. So the RHS is

$$\lim_{i,j\to\infty} C \|v_i\|_{L^2} \|D^2 w_j\|_{L^2} = C\|u\|_{L^2} \|D^2 u\|_{L^2}$$

and we have the desired result.

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Thanks so much :) –  Frank Dec 17 '12 at 11:48
@WillieWong I have a dumb question: How can we tell that $u$ has compact support? Is it because $u \in C_c^\infty(U)$ (differentiable infinitely many times) and that the bounded set $U$ is compact (well, at least the closure $\bar{U}$ is comapct)? –  les guimauves Jan 12 at 9:24

Assume that $u \in H^2 (U) \cap H_0^1 (U).$ Then $$\exists u_i \in C_c^\infty (U)\;\; s.t. \;\; \| u - u_i \|_{H_0^1} \longrightarrow 0$$ $$\exists u_j \in C^\infty (\overline U) \;\;s.t. \;\; \| u - u_j \|_{H^2} \longrightarrow 0.$$ We have $\int_U |Du_i |^2 \longrightarrow \int_U |Du|^2$ and $\int_U u_i^2 \longrightarrow \int_U u^2$. Thus we have \begin{eqnarray*} \int_U |Du|^2 &=& \lim \int|Du_i |^2 \\ & \leqslant & \lim C \left( \int_U u_i^2 \right)^{1/2} \left( \int_U |D^2 u_i |^2 \right)^{1/2} \\ &=& C \left( \int_U u^2 \right)^{1/2} \left( \lim \int_U |D^2 u_i |^2 \right)^{1/2} \\ & =& C \left( \int_U u^2 \right)^{1/2} \left( \int_U |D^2 u |^2 \right)^{1/2}.\end{eqnarray*} The last equality holds because \begin{eqnarray*} \left| \int_U |D^2 u_i |^2 - \int_U |D^2 u |^2 \right| &\leqslant& \left| \int_U |D^2 u_i | - |D^2 u_j | \right| + \left| \int_U |D^2 u_j | - |D^2 u |\right| \longrightarrow 0\;(i,j \to \infty). \end{eqnarray*}
Your last step is a bit problematic: the last inequality is rather circular. It is better to factor: $$|D^2 u_i|^2 - |D^2 u|^2 = (|D^2 u_i| - |D^2 u|)(|D^2 u_i| + |D^2 u|)$$ –  Willie Wong Dec 14 '12 at 13:04
@Leun But how do you know that the first term on the RHS of the final inequality you write down tends to zero? You do not what will happen to the term $D^2 u_i$. If what you say is true, then you would have that the $u_i$ tend to $u$ in the $H^2$ norm, which is not necessarily true. @ Willie Since we know nothing about the convergence of $D^2 u_i$ I do not see how this factoring helps, as the second multiplier may not be uniformly bounded as $i$ changes. –  Frank Dec 14 '12 at 14:36