If $f$ is differentiable everywhere, is $f'$ the weak derivative?

Question

Let $f \in C^0([0,T])$ be such that $f'$ exists in the classical sense everywhere, but $f'$ may not be continuous. Is it true that $f'$ is the weak derivative of $f$ too, if it exists?

I know this is true if $f \in C^1$, but I don't have that..

Answer 1

In order for $f'$ to be the weak derivative of $f$ (i.e., weak derivative in the sense of distributions) the integral $$ \int_0^1 \varphi(x)\, f'(x)\,dx, \quad \varphi\in C^\infty_0(0,1), $$ should make sense, and although $f'$ is not in general locally $L^1$, the above is alternatively defined to make sense as $$ -\int_0^1 \varphi'(x)\, f(x)\,dx, $$ which has the same value with the first integral if $f'$ is continuous.

However, $f'$ is not necessarily a strong derivative of $f$ (strong is weaker than classical but stronger than weak), as strong derivatives belong to $L^p$ spaces.