Prove if $T\in\mathcal{D}(\mathbb{R})$ and $\mathrm{d}T/\mathrm{d}x=0$ then $T$ is the constant distribution.
I am having some problems both proving this problem, as well as understanding exactly what I am supposed to prove. I guess the starting point is done like this. The statement
$$ \langle T' , \phi \rangle = \langle T , \phi' \rangle = 0 \tag{1} $$
Implises
$$ \langle T , \phi \rangle = C \int_{-\infty}^{\infty} \phi(x) \,\mathrm{d}x \tag{2} $$
Where $\phi$ is the standard testfunction, and $C$ is a constant deppendant on the particular distribution. However I am not sure how to prove that $(2)$ follows from $(1)$.
There is a solution for this problem, here (page 2). However I was completely lost after the first few lines. Could anyone help me start writing a proof for this statement?
A summary of the proof in the link:
Fix a test function $\eta$ with integral $1$ and define $C=\langle T,\eta \rangle$. You have to do this somewhere to identify the constant.
Then, given a test function $\phi$, find a test function $\psi$ such that
$$\psi'(x)=\phi(x)-\eta(x) \int_{-\infty}^\infty \phi(y) dy.$$
One thing the proof in the link does not do is to prove that the resulting $\psi$ is actually a test function. But it is: in particular, its support is contained in $[m,M]$ where $m=\inf \left ( \text{supp} \phi \cup \text{supp} \eta \right )$ and $M=\sup \left ( \text{supp} \phi \cup \text{supp} \eta \right )$. You should check this, as it is an important part of the proof.
Then you use linearity and the definition of the distributional derivative:
$$\langle T',\psi \rangle = 0 = -\langle T,\psi' \rangle = -\langle T,\phi \rangle + \langle T,\eta \rangle \int_{-\infty}^\infty \phi(y) dy = -\langle T,\phi \rangle + C \int_{-\infty}^\infty \phi(y) dy$$
and the result follows by rearranging.