Let $u \in D'(\mathbb{R})$ be a distribution and suppose that its distributional derivative $u'$ can be identified with an $L^1_{\mathrm{loc}}$ function. Can the distribution $u$ itself then also be identified with an $L^1_{\mathrm{loc}}$ function?
Stated differently, if $u$ is a distribution which is not a function, can its distributional derivative be a function?
Best Answer
Indeed, $u$ can be identified with an absolutely continuous function.
Let $v(x) = \int_0^x u'(y)\,dy$, where the integral makes sense when we treat $u'$ as a locally integrable function. Then by the fundamental theorem of calculus for Lebesgue integrals, $v$ is absolutely continuous and $v' = u'$ almost everywhere. Since $v'$ is also the distributional derivative of $v$, we have that the distribution $u-v$ satisfies $(u-v)'=0$. This implies that $u-v$ is some constant $c$, so we have $u=v+c$ as distributions.