Deduce if $\dfrac{dT}{dx} \in \mathcal{E}(\mathbb{R})$ for $T \in \mathcal{D}'(\mathbb{R})$ then $T \in \mathcal{E}(\mathbb{R})$.
My attempt :
We know for any $T \in \mathcal{D}'(\mathbb{R})$ , if $T'(\phi)= -T(\phi')\ \ $ for $\ \phi \in \mathcal{D}(\mathbb{R})$ then, $T' \in \mathcal{D}'(\mathbb{R})$.
Here it is given that $T' \in \mathcal{E}(\mathbb{R})$, i.e from definition, $T' = T'_f \ $ for some $f$ such that $f$ is continuous function with compact support.
And, $T'_f (\phi) = \displaystyle\int_{\mathbb{R}} f(x)\phi(x)dx = - \int_{\mathbb{R}} \phi'(x) \Big( \int f(x)dx \Big)dx$
Now, choose $\psi(x) = \displaystyle\int_{-\infty}^{x} f(t)dt$ is the antiderivative function. $\psi$ is a $C^1$ function.
But what I need is $\psi$ to be a function with compact support right?
Then I can conclude, $T'_f (\phi) = -T_\psi (\phi') $ which implies $T=T_\psi \in \mathcal{E}(\mathbb{R})$ .
So please help me out.
$T'\in\mathcal{E}(\mathbb{R})$ means that $T'(\phi)=\int f(x)\,\phi(x)\,dx$ for some $f\in \mathcal{E}(\mathbb{R})=C^\infty(\mathbb{R}).$
Let $F(x)=\int_0^x f(t)\,dt$ and define $S\in\mathcal{D}'(\mathbb{R})$ by $S(\phi)=\int F(x)\,\phi(x)\,dx.$ Show that $S'=T'.$
Then there is a theorem saying that if $S'=T'$ then $S-T$ is constant, which of course is in $\mathcal{E}(\mathbb{R}).$