Continuous function without a weak derivative Let $f:\Omega\to\mathbb{R}$ be a continuous function. Is it necessarily true that $f$ has a derivative in the weak sense? That is, is there some $v:\Omega\to\mathbb{R}$ such that for every test function $\phi\in C^\infty_C(\Omega)$
$$\int_\Omega \phi'(x)f(x)~\mathrm{d} x = -\int_\Omega \phi(x) v(x)~\mathrm{d} x$$
I know that there are functions such as the Weierstrass function which are continuous, but nowhere differentiable. I'm wondering if a similar idea holds in the realm of weak derivatives.
 A: It's not true. Well, one can always define a distributional derivative: for any $f \in L^1_{loc} (\Omega)$, define $Df$ as a functional $Df : C_c^\infty (\Omega) \to \mathbb R$ given by 
$$Df[\phi] := - \int_\Omega f \phi' .$$
So what you are asking is whether or not a distributional derivative (a functional) is represented by an function $v\in L^1(\Omega)$
$$Df [\phi]= \int_\Omega v \phi, \ \ \ \ \forall \phi$$
(Note if this is true, we call $v$ the weak derivative and $f$ is said to be in the Sobolev space $W^{1, 1}(\Omega)$). 
There are continuous function $f$ on $(-1, 1)$ so that $Df$ is not a function. A lot of example is suggested in 
https://mathoverflow.net/questions/38751/a-h%C3%B6lder-continuous-function-which-does-not-belong-to-any-sobolev-space
even for Holder continuous function $f$. 
A: Any continuous function is locally integrable (meaning on a compact set) and therefore is infinitely derivable in the distribution sense. However as @John describes it in his answer that distribution is not necessarily a function.
