# Integral of test function equals 0 implies antiderivative has compact support

I was reading through the Accepted Answer to this question and am having trouble with one part of the reasoning given. To quote part of the answer:

Detail: We used the following fact above: Given a test function ϕ on R, there exists a test function ψ with ϕ=ψ′ if and only if ∫ϕ=0. In case this is not clear: First, if ϕ=ψ′ then ∫ϕ=∫ψ′=0 because ψ has compact support. Suppose on the other hand that ∫ϕ=0, and define ψ(x)=∫x−∞ϕ. Then ψ′=ϕ and hence ψ is infinitely differentiable, while the fact that ∫ϕ=0 shows that ψ has compact support.

I understand that since $$\psi' = \phi$$, $$\psi$$ is infinitely differentiable since $$\phi$$ is infinitely differentiable. However, why does the fact that $$\int\phi = 0$$ imply that $$\psi$$ has compact support?

I think that by using the fundamental theorem of calculus, since $$\psi' = \phi$$, we can show that $$\lim_{x \to +/-\infty} \psi(x) = 0$$, but some digging around suggests that this is not enough to conclude that $$\psi$$ would have compact support.

There is an assumption in the question which is very much needed. This is that $$\phi$$ is a test function so in particular it has compact support.
We have defined $$\psi(x) = \int_{-\infty}^x \phi(x) dx$$ and since $$\phi$$ have compact support we have for all sufficiently large $$x$$ that the function $$\psi(x)$$ is constant since $$\phi(x) = 0$$ there. That $$\int_{-\infty}^\infty \phi(x)dx = 0$$ tells us that this constant is zero. Thus $$\psi(x) = 0$$ for all sufficiently large $$x$$.
• I see now, thank you! It all comes down to the fact that $\psi(x)$ is zero for sufficiently large $x$, no need to bring limits into it. I have accepted your answer. Feb 28, 2021 at 17:07