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 $H^2(U)$.

  • $\begingroup$ 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.) $\endgroup$ – Jonas Meyer Dec 13 '12 at 19:59
  • $\begingroup$ @Frank I have added the question. Please let verify that this is, in fact, correct. $\endgroup$ – Alex Youcis Dec 13 '12 at 20:04
  • $\begingroup$ Yes, that is correct. Many thanks for doing this - embarrassingly, I do not know how to 'type maths.' $\endgroup$ – 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

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

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 $w_i \in C^\infty(\bar{U})$.

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.

| cite | improve this answer | |
  • $\begingroup$ @Cookie The meaning of $u \in C^{\infty}_{c}(U)$ is that $u$ is infinitely differentiable in $U$ and has support compactly contained in $U$. $\endgroup$ – mlg4080 Feb 7 '16 at 18:00
  • $\begingroup$ When $v\neq w$, how do you apply integration by parts in (*) when there is an absolute value under the integral sign? $\endgroup$ – Jack Sep 12 '16 at 14:49
  • $\begingroup$ Also, I think you need Hölder there? $\endgroup$ – Jack Sep 12 '16 at 14:52
  • $\begingroup$ @Jack: you are right, I don't need the absolute value sign for the argument. $\endgroup$ – Willie Wong Sep 12 '16 at 15:25
  • $\begingroup$ I guess $C$ could be safely omitted. Isn't it just $1$ there? $\endgroup$ – Jack Sep 13 '16 at 14:44

How about this?

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*}

| cite | improve this answer | |
  • $\begingroup$ Your last step is a bit problematic: the last inequality is rather circular. It is better to factor: $$ |D^2 u_i|^2 - |D^2 u|^2 = (|D^2 u_i| - |D^2 u|)(|D^2 u_i| + |D^2 u|) $$ $\endgroup$ – Willie Wong Dec 14 '12 at 13:04
  • $\begingroup$ @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. $\endgroup$ – Frank Dec 14 '12 at 14:36
  • $\begingroup$ @WillieWong @ Frank So I think we need some help from Wille Wong. $\endgroup$ – Leun Kim Dec 14 '12 at 15:13
  • $\begingroup$ @Frank: ah I see your objection. I didn't see that at first because of the notation used. Let me write an answer instead. $\endgroup$ – Willie Wong Dec 17 '12 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.