I have a question about weak derivatives.

Let $u,v \in L^{1}_{loc}(U)$ (the space of locally integrable functions) for some open set $\emptyset \neq U \in \mathbb{R}^{n}$. We often say that $v$ is the $\alpha$-th ($\alpha$ is a multi index) weak derivative of $u$ if the following equation holds: \begin{align} \int_{U}u D^{\alpha}\varphi=(-1)^{|\alpha|}\int_{U}v\varphi \end{align} for all $\varphi \in C_{0}^{\infty}(U)$ (the space of infinitely differentiable functions with compact support in $U$).

Let $\emptyset \neq U$ be a bounded open subset of $\mathbb{R}^{n}$ with $C^{1}$-boundary. I want to define weak derivative of $u \in L^{1}(\overline{U})$. How should I define derivatives? I think this notion enables us to define Sobolev spaces on $\overline{U}$.


closed as off-topic by Johannes Hahn, András Bátkai, Stefan Kohl, Alex Degtyarev, Christian Remling Jul 2 '15 at 17:00

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If $\partial U$ has measure zero (which follows from $C^1$ regularity, for example), then $L^1(U)=L^1(\bar U)$. Since $L^1(U)\subset L^1_{\text{loc}}(U)$, $L^1(\bar U)$ is just a special case.

Derivatives in $L^1(\bar U)$ can be defined exactly the same way, taking test functions in $C_0^\infty(U)$. The support of a test function can be arbitrarily close to the boundary, so you have no problem with having a unique weak derivative in all of $\bar U$.

Let $S$ denote either $U$ or $\bar U$. Let $k\in\mathbb N$ and $p\in[1,\infty)$. You can define $W^{k,p}(S)$ as the set of those functions $u\in L^1_{\text{loc}}(S)$ for which there is a weak derivative $\partial^\alpha u\in L^1_{\text{loc}}(S)$ for all $|\alpha|\leq k$ and $$ \sum_{|\alpha|\leq k}\|\partial^\alpha u\|_{L^p(S)} < \infty. $$ Choosing $S=U$ or $S=\bar U$ makes no difference in these definitions, so $W^{k,p}(U)=W^{k,p}(\bar U)$.


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