A variant of the fundamental lemma of calculus of variation If $F$ is a distribution and its distributional derivative is equal to 0, how can I show that $F$ is (represented by) a constant function i.e. there exists a constant $c$ such that $F(\phi)=c\int\phi$ for all test functions $\phi$.
This question Proof of fundamental lemma of calculus of variation. is also about the fundamental lemma of calculus of variation, but there are no distributions in that question. I'm a little concerned about how to go from locally integrable functions to distributions.
There was also a hint that says first consider the case where $\int\phi=0$. How should I use this hint? Any help is much appreciated.
 A: First, one can prove that
$$\phi \in D(\Bbb R)\mbox{ statisfies } \int_{\Bbb R} \phi(x)dx=0\Leftrightarrow\exists \psi\in D(\Bbb R)\mbox{ such that } \psi'=\phi.$$
Second, fix a test function $\theta$ such that $\int_{\Bbb R} \theta(x)dx=1$. 
Given arbitrary test function $\phi$, we can always say that 
$$\phi(x) =\theta(x)\int_{\Bbb R} \phi(y)dy + \left(\phi(x)-\theta(x)\int_{\Bbb R} \phi(y)dy\right).$$
Clearly, there exists a test function $\phi_1$ such that
$$\phi_1'(x) = \phi(x)-\theta(x)\int_{\Bbb R} \phi(y)dy$$
Finally, we have a distribution $F$ such that $F'=0$.
We write$$
\langle F, \phi\rangle =\left\langle F, \theta(x)\int_{\Bbb R} \phi(y)dy + \left(\phi(x)-\theta(x)\int_{\Bbb R} \phi(y)dy\right) \right\rangle $$
 $$ =\left\langle F, \theta(x)\int_{\Bbb R} \phi(y)dy + \phi_1'   \right\rangle $$
$$ = \int_{\Bbb R} \phi(y)dy\left\langle F, \theta \right\rangle-\left\langle F', \phi_1    \right\rangle = \int_{\Bbb R} \phi(y)dy\left\langle F, \theta \right\rangle $$
Now, the test  function $\theta$ being fixed and not depending on $\phi$, we say that $c=\left\langle F, \theta \right\rangle $ and obtain that
$$ 
\langle F, \phi\rangle = c \int_{\Bbb R} \phi(y)dy,$$
hence, indeed, $F$ is represented by a constant function $c$.
