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 $C^1$ ?
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1Weak derivatives as "distribution derivatives"? Then $f$ is $C^1$. – Daniel Fischer Sep 18 '13 at 18:06
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1Maybe any idea how to obtain it. – user111 Sep 18 '13 at 18:52
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Integrate the weak derivative and you get a continuous function which you can view as another distribution with the same distributional derivatives. Two distributions with the same distributional first derivative differ by a constant. – Chris Janjigian Dec 02 '13 at 03:19
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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.
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