For an open set $\Omega$ of class $C^1$, suppose we have $u \in W^{1,p}(\Omega)$ and that $u$ is continuous and all the partial derivatives of $u$ are continuous. I want to show that $u$ is $C^1(\Omega)$ in the classical sense, i.e. that $u$ is continuously differentiable.
I quote here the answer given in the question Is a continuous function with continuous weak derivatives of class $C^1$?
Let $\Omega$ be a domain on which $f$ is defined. Consider $f_\epsilon = f* \phi_\epsilon$, where $\phi_\epsilon $ is a standard mollifier. These are defined on a smaller domain $\Omega'\Subset \Omega$. By the basic properties of convolution,
- $f* \phi_\epsilon\to f$ uniformly as $\epsilon \to 0$.
- $\nabla (f*\phi_\epsilon) = (\nabla f)*\phi_\epsilon$.
- $(\nabla f)*\phi_\epsilon \to \nabla f$ uniformly as $\epsilon \to 0$.
The sequence $f_n=f*\phi_{1/n}$ is Cauchy in the norm of $C^1(\Omega')$. (Indeed, $f_n$ is convergent, hence Cauchy in the uniform norm; the same applies to each partial derivative of $f_n$.) Since $C^1(\Omega')$ is complete, the sequence converges to an element of $C^1(\Omega')$. This element is $f$, due to item 1.
I have some doubts about the proof, the main one being the following:
Doesn't the convolution need the function $f$ to be defined on all $\mathbb{R}^n$? But also,
I have only been able to show the result ($f* \phi_\epsilon\to f$ uniformly as $\epsilon \to 0$) for compact (i.e. closed bounded) sets in $\mathbb{R}^n$. Which is the workaround used here?