Identifying distributional and classical derivative

Question

Let $f \in L^1(\mathbb{R})$ such that its distributional derivative $f'$ is also in $L^1(\mathbb{R})$ (in the sense that there exists a function $h\in L^1(\mathbb{R})$ such that $\langle f', \phi\rangle = -\langle f, \phi'\rangle = \langle h, \phi\rangle $ for all test functions $\phi \in \mathcal{D}(\mathbb{R})$, ie. we can identify the distributional derivative of $f$ with an $L^1$ function).

Now suppose that $f$ and $f'$ are also continuous. How could I prove that $f'$ coincides with the classical derivative of $f$, ie. defining

$$g(x) = \lim_{h\to0} \frac{f(x) - f(x+h)}{h},$$ how can I show that $g(x) = f'(x)$ for all $x\in\mathbb{R}$ ?

Thanks !

Answer 1

Use these two basic facts from distribution theory:

1. If $s\in C^1(\mathbb R)$ then the distributional derivative of $s$ exists and is equal to the ordinary derivative of $s$. This follows from integration by parts.

2. If $r\in\mathcal{D}'(\mathbb R)$ is a distribution and if $r'=0$ (the distributional derivative), then $r$ is a constant. (The main part of the proof is that if $\phi$ is a compactly supported $C^\infty$-function and $\int_{-\infty}^\infty\phi(x)dx=0$, then $\phi=\psi'$ for some compactly supported $\psi$.)

By 1., if $f$ is $C^1$, we are done, but we only know that $f$ is $C^0$. However, the distributional derivative $f'$ is assumed to be $C^0$, so $f'$ admits a primitive function $h\in C^1(\mathbb R)$, $h'=f'$. By 2. this implies that $f$ differs from $h$ by a constant , hence $f$ is $C^1$ (because $h$ is $C^1$), so we are done.