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.


1 Answer 1


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 $x$.

  • $\begingroup$ 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. $\endgroup$
    – bosco98
    Feb 28, 2021 at 17:07

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