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I was looking at the following definition of the weak derivative:

Let $\Omega$ be a domain (ie an open connected subset of $\mathbb{R}^n$). Suppose $u,v \in L_{1,loc}(\Omega)$ and

\begin{equation} \int_{\Omega}u(x)\partial^{\alpha}\eta(x)dx = (-1)^{|\alpha|}\int_{\Omega}v(x)\eta(x)dx, \forall \eta \in C^{\infty}_c(\Omega) \end{equation} where $\alpha$ is a multiindex. Then $v$ is called the weak partial derivative of $u$ in $\Omega$, and is denoted by $\partial^{\alpha}u$.


Now suppose the classical derivative $\partial^{\alpha}u$ exists and is continuous on some domain $\Omega$. Is the classical derivative the weak derivative on $\Omega$?

I suppose this reduces to the integration by parts formula (ie the Guass-Green Theorem). But I'm having trouble justifying its use since for any particular $\eta \in C^{\infty}_c(\Omega)$ I seem to require a subdomain $\Omega' \subset \Omega$ such that $\text{supp} \eta \subset \Omega'$ and $\partial\Omega'$ is $C^{|\alpha|}$. Is there a standard construction, or do I need to have stronger restrictions on the domain $\Omega?$

Edit: The $\partial^{\alpha}u$ exists and is continuous on all of $\Omega$, not all of $\mathbb{R}^n$ as initially stated.

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If $\eta$ has compact support, then let $b$ be a smooth function that is one on $S=\operatorname{supp} \eta$, and is zero on the complement of some compact set that contains $S$ in its interior. Then we have $u(x) = u(x) b(x)$ for all $x \in S$, that is, $ \operatorname{supp} (u \cdot b) \subset S$ ($\cdot$ means pointwise multiplication). (Note that we have $\operatorname{supp} D^\alpha \eta \subset S$ as well.)

Then, for appropriate $\alpha$, we have $\int_\Omega u D^\alpha \eta = \int_\Omega (u b)D^\alpha \eta = (-1)^{|\alpha|} \int_\Omega D^\alpha (u \cdot b) \eta= \int_\Omega (D^\alpha u)\eta$.

Hence the classical derivative is the weak derivative.

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  • $\begingroup$ I think I understand. There's no need to integrate over only the domain when we've extended our functions by zero to all of R^n. To apply integration by parts we need only choose a large enough ball. $\endgroup$ – Dogomien Dec 20 '13 at 18:21
  • $\begingroup$ Well, sort of. We show that for any particular $\eta$, the integral is equivalent to one where both functions have compact support, and hence the equality follows. $\endgroup$ – copper.hat Dec 20 '13 at 18:23
  • $\begingroup$ Sorry I'm a bit confused. How does one apply the "integration by parts" step? $\endgroup$ – Dogomien Dec 20 '13 at 18:48
  • $\begingroup$ I don't have a straightforward answer for that. Maybe math.stackexchange.com/a/171925/27978 might be of help? $\endgroup$ – copper.hat Dec 20 '13 at 19:48

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