contuous weak derivative $\Rightarrow$ classic derivative ?

• Apr 21st 2011, 07:24 AM
qwertyuio
continuous weak derivative $\Rightarrow$ classic derivative ?
Hello! I'm new in this formus, so I'm sorry if I make some mistake..

Let $u$ be a weakly differentiable function on $\Omega\subset\R^n$ open bounded.
If $grad u$ is continuous, can I say $u\in\C^1(\Omega)$, i.e. $u$ strongly differentiable? (better: u a.e. equal to a $C^1$ function)

Reading Evans' book the answer seems to be yes, but it's not so obvious to me.
I made a proof in the case $n=1$, but I'm not able to generalize it. I sum it up:

Since $C^\infty(\bar\Omega)$ is dense in $W^{1,1}(\Omega)$, you can show that for a.e. $a\in\Omega$
$u(x)=u(a)+\int_a^x u'(t) dt$ for a.e. $x\in\B(a,r)\subset\Omega$.
But if $u'$ is continuous, the integral function is $C^1$. Thus $u$ is a.e. equal to a $C^1$ function on $B(a,r)$.
Conclude by arbitrariety of $a$.

The problem is that in dimension $n>1$ I don't manage anymore to write $u$ as a $C^1$ function. Can you help me?
Can you generalize the proof for $n>1$, or do you have another idea to prove this result?
• Apr 21st 2011, 02:12 PM
Jose27
Use the usual regularization trick: In a sufficiently small ball a subsequence of these regularizations converges pointwise a.e. to u, and the derivatives also converge but uniformly in this ball. Then it's standard to prove that the convergence is uniform in both cases and so the limit is a C^1 function.
• Apr 22nd 2011, 02:37 AM
qwertyuio
It works, thank you a lot!