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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.

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Aden Dong
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1 Answers1

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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 $c$.

TZakrevskiy
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