Derivation under the integral sign

Question

From the Theory of Distributions, we know that if $u \in L^1_{loc}(\mathbb{R})$ and $v(x)=\int_0^x u(t)dt$ then $v$ is continuous on $\mathbb{R}$ and $v'=u$ in the sens of distributions.

Now, if I suppose that $u$ is continuous on some interval $[a,b]$ with $x\leq b$, we know from the classical "fundamental theorem of calculus" that $v'(x)=u(x)$ in the strong sens.

How can we prove rigorously the second result above (so with the case $u$ is continuous) with arguments from distribution theory (I guess involving the derivative of indicator function) ?

Answer 1

I believe that all continuous functions are the uniform limit of distributions. It would follow from that and the uniform convergence theorem for integrals.

Answer 2

Let $u\in C[0,1]$ and $v\in C^1[0,1]$ such that $v(x) = \int_0^x u(s)ds$. We want to show that the derivative of $v$ in the sense of $D'$ is indeed $u$.

Take a test function $\phi \in D(0,1)$, and apply $v'$ to it ($v$ is $L^1_{loc}$, so the integral representation of $v$ exists): $$\langle v',\phi\rangle = -\langle v,\phi'\rangle = \int_0^1\left( \int_0^x u(s)ds\right) \phi'(x)dx.$$ Since $x\to \int_0^x u(s)ds$ is a $C^1$ function, as well as $x\to\phi'(x)$, we can say that (after integration by parts) $$\langle v',\phi\rangle =- \int_0^1 u(s)ds \phi(1) + \int_0^0 u(s)ds \phi(0) + \int_0^1 u(x) \phi(x) dx = \int_0^1 u(x) \phi(x) dx.$$ As the above identities hold for any test function $\phi\in D(0,1)$, we can now conclude that $v'=u$ in the sense of distributions.