# 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.

• @John is absolutely right. Distributional derivative does not guarantee the existence of a function that is a weak derivative. I will edit my answer and down vote it. – marwalix Apr 24 '15 at 0:54

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 $f$.

• Thanks! I figured it would work for distributions, but wasn't sure about weak derivatives. Reading through that post, I'm a little confused why $f$ being of bounded variation is relevant? – David Weber Apr 24 '15 at 5:01
• @DavidWeber: That's because they want to show that the function is not even in $BV$, while $BV$ contains all Sobolev space. Thus that function is not a Sobolev function and thus weak derivative does not exist. – user99914 Apr 24 '15 at 5:07

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.