# Weak derivative

Let $u \in C(\Omega)$ be a function with weak derivative $Du \in C(\Omega)^n$. How does one prove that $Du$ coincides with the classical derivative?

Is the mean value theorem for integration helpful?

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What is $\Omega$? (I guess it's an open subset of $\mathbb R^n$, but I want to be sure) You have to use the fact that if $f$ is locally integrable and $\int_{\Omega} f\varphi dx=0$ for each $\varphi\in\mathcal D(\Omega)$, then $f=0$. –  Davide Giraudo Dec 3 '11 at 12:33
this is done in any book about distributions –  Glougloubarbaki Dec 3 '11 at 12:44
Yes, $\Omega$ is an open subset of $\mathbb R^n$. I only found proofs for the claim, that a function is weakly differentiable if it is differentiable in the classic sense and both derivates coincides. But I did not found a proof for my claim above. The definition I use for weak derivative: A function $u \in L^1_{\mathrm{loc}}(\Omega)$ has a $\alpha$'th weak derivative in $\Omega$ if there is a function $u_\alpha \in L^1_{\mathrm{loc}}(\Omega)$ with: $\int_\Omega u D^\alpha \phi\ dx=(-1)^{\left | \alpha \right |} \int_\Omega u_\alpha \phi\ dx \ \ \forall \phi \in C^\infty_0(\Omega)$. –  David75 Dec 3 '11 at 14:44
Well, the classical derivative also satisfies the definition above for a weak derivative, so you just need to use the comment by Davide Giraudo. The mean value theorem is not helpful here. –  Jeff Dec 3 '11 at 19:00
Isn't this just integration by parts? –  julien Mar 7 at 21:30