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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) ?

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    So, basically you want to show that for $C^1$ functions the derivatives in the sense of distributions and in the classical sense are equivalent? – 2017-01-27
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    @TZakrevskiy I would like the proof of: If $u$ is continuous (so $C^0$), then $v'(x)=u(x)$ (in the strong sens), with the proof using distribution theory arguments and eventually proving the (strong) equality in the special case where u is more than $L^1_{loc}$, that is continuous. – 2017-01-27

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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.

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    @Henry does this satisfy you? – 2017-01-31
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I believe that all continuous functions are the uniform limit of distributions. It would follow from that and the uniform convergence theorem for integrals.

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    I believe you and I am asking for the proof. – 2017-01-26
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    @Henry Are you unsure how to turn this explanation into a formal proof, or are you simply demanding that I do it. – 2017-01-26