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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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    Yes, that is correct. Many thanks $f$or doing this - embarrassingly, I do not know how to 'type maths.'2012-12-13

2 Answers 2

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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.

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    There's a $C$ in the original question. I keep the $C$ to be consistent.2016-09-13
-1

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*}

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    @Frank: ah I see your objection. I didn't see that at first because of the notation used. Let me write an answer instead.2012-12-17