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We know if $u:\mathbb{R}^{n}\to \mathbb{R}$ is a smooth function with compact support, then we can show, via integration by part, that $$ ||\Delta u||=||{\nabla}^2u|| $$ where $||\cdot||$ is the $L^2(\mathbb{R}^n)$ norm.

Now we let $U$ be a pre-compact subset of $\mathbb{R}^n$, and let $u$ be defined on $\bar{U}$, where the boundary $\partial U$ is sufficiently smooth. I want to show the following inequality holds $$ ||{\nabla}^2u||_{L^2(U)}\leq C||\Delta u||_{L^2(U)} $$ where the constant $C$ does not depend on $u$.

My proof goes as follows: Choose bounded set $V$ such that $U \subset \subset V$, there exists an extension $Eu=\tilde{u}$ of $u$ satisfying

1.$Eu=u$ a.e. in $U$

2.$Eu$ is supported in $V$

3.$||u||_{H^{2}(U)}\leq ||Eu||_{H^{2}(\mathbb{R}^n)} \leq C ||u||_{H^{2}(U)}$, the constant $C$ does not depend on $u$.

In particular, we can construct the extension by higher-order reflection so that derivatives of $u^+=u$ and the reflected part $u^-$ of order no more than 2 agree on the boundary (e.g., Evans PDE, chapter 5). The explicit formula of $\tilde{u}$ is as follows, and WLOG, we assume $\partial U$ is flat, i.e., $\partial U=\{x\in \mathbb{R}^n|x_{n}=0\}$. In addition, we write $x=(x',x_{n})$ for convenience. Assume there is an open ball, centered at $x^{0}\in \partial U$, such that $B^+=B\cap \{x_n\geq 0\}\subset \bar{U}$ and $B^-=B\cap \{x_n\leq 0\}\subset \mathbb{R}^n-U$ $$ \tilde{u}= u(x)\quad \text{if}\quad x\in B^+ $$ and $$ \tilde{u}=-au(x'-x_n)+bu(x',-\frac{x_n}{2})+cu(x',-\frac{x_n}{3})\quad \text{if}\quad x\in B^- $$

where $u^+=\tilde{u}|_{B^+}$,$u^-=\tilde{u}|_{B^-}$,and $a,b,c$ are chosen so that $\nabla^{\alpha}u^-|_{x_n=0}=\nabla^{\alpha}u^+|_{x_n=0}$, for all multiindex $\alpha$ with order no greater than $2$. Now, we have $$ ||{\nabla}^2u||_{L^2(U)}\leq ||{\nabla}^2 \tilde{u}||_{L^2(\mathbb{R}^n)}=||\Delta \tilde{u}||_{L^2(\mathbb{R}^n)}\leq C ||\Delta u||_{L^2(U)} $$ I'm not quite sure about the last inequality, I think it follows from the construction of the extension $\tilde{u}$, because $u^-$ is a linear combination of $u$ and the second order derivative agrees on the boundary. Can anyone help me to check my proof? thanks

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  • $\begingroup$ don't you need some boundary conditions on $u$ to get any estimates? $\endgroup$
    – Craig
    Commented Mar 18, 2015 at 4:44
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    $\begingroup$ You have to estimate the commutator of the extension operator and partial differentiation. I suggest looking at the extension operator defined in Stein's book, Singular Integrals and Differentiability Properties of Functions. $\endgroup$
    – Deane Yang
    Commented Mar 18, 2015 at 11:47
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    $\begingroup$ Mmm, I don't see how that can be true. Take $u$ the solution to $$ \Delta u=0 \text{ in }U $$ and $$ u=g(x,y) \text{ on }\partial U $$. Here $g$ is a smooth function. Then the right hand side vanishes, but the function $u$ is not necessarily constant. $\endgroup$
    – guacho
    Commented Mar 18, 2015 at 21:14
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    $\begingroup$ Just a side comment (all norms are over $U$). Take $ \Omega $ smooth bounded in $R^N$ and consider $ u_m(x)=|x-x_m|^{2-N}$ where $ x_m \rightarrow x_0 \in \partial \Omega$ (but $ x_m$ not in the closure of $\Omega$). This examples shows you can't get an estimate of the form $$ \| D^2 u \|_{L^2} \le C \| \Delta u \|_{L^2} + C_1 \| u \|_{L^2}.$$ Depending on dimension, it even shows you can get $ \| \nabla u\|_{L^2} \le C_1 \| \Delta u \|_{L^2} + C_2 \|u\|_{L^2}$.... $\endgroup$
    – Craig
    Commented Mar 20, 2015 at 23:40
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    $\begingroup$ yes, or even $ \| D^2 u \|_{L^2} \le C \|\Delta u\|_{L^2}$ if you have $u=0 $ on $ \partial \Omega$ provided $\Omega$ is bounded and has sufficiently smooth boundary (this is just the usual $H^2$ regularity theory, say from Evans). $\endgroup$
    – Craig
    Commented Mar 23, 2015 at 0:32

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