Is a continuous function with continuous weak derivatives of class $C^1$? Let $f$ be a real valued continuous function of many variables whose weak derivatives of first order are continuous. Is this function equals a.e. function 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. 
By the way, you don't need "equals a.e.": if two continuous functions are equal a.e., they are equal on a dense subset, hence everywhere. 

Remark. The statement remains true if we only assume $f$ to be locally integrable. In this case, item 1 is replaced with "$f* \phi_\epsilon\to f$ a.e.  as $\epsilon \to 0$". Fix a point $x_0$ at which convergence takes place. Since $|f(x_0)|+\sup_{\Omega'} |\nabla f|$ is an equivalent norm on $C^1(\Omega)$, and the rest goes through as before. This time, the a.e. part in "equals  a.e.  to a $C^1$ function" is necessary.
