if $\frac{dT}{dx}=T_{f}$ for some $f$, show that $T=T_{g}$ for some $g$.

Question

Prove that if $T\in D^{'}(\mathbb{R})$, where $D^{'}(\mathbb{R})$ denotes the space of distributions on $D(\mathbb{R})$, and $\frac{dT}{dx}=T_{f}$ for some $f\in C^{\infty}(\mathbb{R})$, then show that $T=T_{g}$ for some $g\in C^{\infty}(\mathbb{R})$.

I don't have any idea about how to solve this. By definition $-T(\phi^{'})=\int f\phi dx$ for all $\phi \in D(\mathbb{R})$. I have to produce a $g\in C^{\infty}(\mathbb{R})$ such that $T(\phi)=\int g\phi dx$. But I am stuck.

Thanks in advance for any help.

Answer 1

$T'(\phi)=\int f\phi$. Take $g$ a primitive of $f$, $T'_g(\phi)=-\int g\phi'=\int fg$ by using integration by part. This implies that the derivative of $T_g-T$ is zero. Now you can use this to conclude:

Solving $ T' = 0 $ for distributions in $\mathbb{R}^n$