Let T and T' be regular distributions. T=T$_{f}$ and $T'=T_{g}$ . Assume both f and g are continous. Show that f is continuously differntiable and f'=g.
I define
$\langle T_{f},\Phi \rangle = $$\int_\mathbb{R} f(x)\phi(x) dx$
$\langle T_{g},\Phi \rangle = $$\int_\mathbb{R} g(x)\phi(x) dx$
Now i use the derivation rules for Distributions and get from $T'_{f}(\Phi)$
$\langle T'_{f},\Phi \rangle =-$$\int_\mathbb{R} f(x)\phi'(x) dx=-[f(x)\phi(x)]^{a}_{-a} +$$\int_\mathbb{R} f'(x)\phi(x) dx$
And $-[f(x)\phi(x)]^{a}_{-a} = 0 $ for $supp(\phi) \subset [-a,a]$
Hence
$ $$\int_\mathbb{R} f'(x)\phi(x) dx =$$\int_\mathbb{R} g(x)\phi(x) dx$
Then
$ $$\int_\mathbb{R} (f'(x)-g(x))\phi(x) dx =0$ with Fundamental lemma of calculus of variations
implies that $f'(x)-g(x)=0$
so $f'(x)=g(x)$.
This should be correct. My Question is how i get that f is continuously differentiable so i can use the Lemma? Does this follow immediately from the derivation rules for distributions??
First of all, you can't do integration by parts when $f$ is not differentiable! now we have
$$\langle T_{g},\phi' \rangle=\langle T',\phi \rangle =-\langle T,\phi' \rangle =-\langle T_{f},\phi' \rangle$$ so
$$\int_\mathbb{R} g(x)\phi(x) dx=-\int_\mathbb{R} f(x)\phi'(x) dx$$ then by substitution $$\int_\mathbb{R} g(x)\phi(x) dx=-\int_\mathbb{R} \lim_{h \rightarrow 0}f(x) \dfrac{\phi(x+h)-\phi(x)}{h} dx=\int_\mathbb{R} \lim_{h \rightarrow 0}\phi(x) \dfrac{f(x-h)-f(x)}{-h} dx $$
from this we have
$$\lim_{h \rightarrow 0} \dfrac{f(x-h)-f(x)}{-h}=g(x) \quad a.e$$ finely by the continuity of $g$ we deduce that $f$ is continuously differentiable and $f'=g$