We know that if we have a distribution $T$ in the usual space of distributions $D(\Omega)$, if $T^{\prime}=0$ in the sense of distributions then $T = C$ for some constant function $C$. Now, take the usual cantor function on the interval $[0,1]$ (https://en.wikipedia.org/wiki/Cantor_function). This function has zero derivative a.e., therefore if we take as $T$ the cantor function, $T$ should be constant. This of course is not true. I wondered why. I suspect that is because the distributional derivative of the cantor function does not coincide with the usual derivative a.e.
The fact is that I thought that if a function is a.e. differentiable then it's distributional derivative would coincide with the usual derivative a.e., but this seems to be false now. Can someone confirm it? Thanks!
Best Answer
"The fact is that I thought that if a function is a.e. differentiable then it's distributional derivative would coincide with the usual derivative a.e., but this seems to be false now. Can someone confirm it? Thanks!"
This is indeed not true. The issue is that in this interpretation you are assuming that the distributional derivative is a regular distribution (i.e. given by a function), but this is not the case.
This can be understood better by looking at the Heaviside function, $H(x)=1$ for $x \geq 0$ and $H(X)=0$ otherwise. You have $H'=0$ almost everywhere, but the distributional derivative is not the zero function and is not regular. $H'=\delta_0$.