If a continuous function $u:\mathbb R^d\to \mathbb R$ has a weak derivative $\frac{\partial u}{\partial x_j}$ that exists everywhere as a locally integrable function, and it is even continuous, does this imply that the strong derivative $\frac{\partial u}{\partial x_j}$ exists and is equal to the weak? Also, let's make this stronger by assuming that the (strong) partial derivatives of $u$ exist almost everywhere. Can we conclude that they exist, in fact, everywhere?

I have seen this topic function a.e. differentiable and it's weak derivative about approximate derivatives, but I'm asking what we can conclude about the strong derivative, if we are working on weak derivatives.

Edit: I know that we are working on a.e. equivalence classes, but what assumptions can we impose on $u$ (weaker than differentiability) so that the strong derivative exists?

Edit 2: I found that if the weak derivative exists and it is in $L^p$, $\infty>p\geq 1$, then the strong also exists and it is equal to the weak a.e. However, this only gives a.e. existence. (Source : http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/weak-derivatives.pdf , Theorem 19.18)

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    $\begingroup$ Related. If the distributional derivatives are given by continuous functions, the classical partial derivatives exist everywhere and coincide with the distributional derivatives. $\endgroup$ – Daniel Fischer Jun 1 '14 at 22:48
  • $\begingroup$ This perfectly answers the question, thank you! $\endgroup$ – Dimitris Jun 1 '14 at 22:54

The thread pointed out by Daniel Fischer, Various kinds of derivatives, indeed concerns this question. But since it is set in $\mathbb R^1$, and the answers involve indefinite integrals, I give another argument for $\mathbb R^d$.

Consider the sequence of mollified functions $u_k = u*\phi_{1/k}$. These are smooth and satisfy $$\frac{\partial u_k}{\partial x_j} = \frac{\partial u}{\partial x_j} * \phi_{1/k} \tag{1}$$ As $k\to\infty$, the right hand side of (1) converges locally uniformly to $\frac{\partial u}{\partial x_j}$. Therefore, $u_k$ converges in $C^1(\Omega)$ norm on every bounded domain $\Omega$. Let $v$ be its limit. At the same time, $u_k\to u$ in $L^1(\Omega)$. Therefore, $u=v$ a.e., which means $v$ is a $C^1$ representative of $u$.

  • $\begingroup$ That's also a very nice way! $\endgroup$ – Dimitris Jun 2 '14 at 5:41

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