Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can we prove that if $f$ is compactly supported and $g$ is periodic with period $P$ then $f*g$ exists and is also P-periodic ? thanks.

share|cite|improve this question
up vote 0 down vote accepted

\begin{align} (f*g)(x+P) & = \int_{-\infty}^\infty f(t)g(x-t)\,dt = \int_{-A}^A f(t)g(x+P-t)\,dt \\[6pt] & {}\qquad \text{(where $f$ is supported on $[-A,A]$)} \\[10pt] & = \int_{-A}^A f(t) g(x-t)\,dt = \int_{-\infty}^\infty f(t)g(x-t)\,dt = (f*g)(x). \end{align}

share|cite|improve this answer
okay ! the integral does not exist anymore if we just assume that the function is continuous, we need the function to be compactly supported (or at least it works...). It seems like the hypothesis "compactly supported" is not really used to prove the periodicity of the function but rather to make the integral finite. Thanks. – Stacy Feb 27 '14 at 23:50
@Stacy : It makes the integral of the absolute value finite, and that implies that the value of the integral without the absolute value actually exists. – Michael Hardy Feb 27 '14 at 23:53
yes the $L^1$ norm, thanks ! – Stacy Feb 27 '14 at 23:57


Write down the definition of the convolution product: $$ (f*g)(x):=\int_{-\infty}^{\infty}f(t)\,g(x-t)\,dt. $$ How much of the range $(-\infty,\infty)$ do you actually need, given that $f$ is compactly supported?

From here, write down an expression for $(f*g)(x+P)$, and see if you can use the fact that $g(w)=g(w+P)$ for all $w$ to show that $(f*g)(x+P)=(f*g)(x)$.

share|cite|improve this answer
I found $$ (fg)(x):=\int_{-A}^{A}f(t)\,g(x-t)\,dt. $$ and $$ (fg)(x+P):=\int_{-A}^{A}f(t)\,g(x+P-t)\,dt. $$ since g in $P$-periodic then $g(x+P-t)=g(x-t)$ thus $f*g$ is also $P$-periodic but which is strange is that I do not need the hypothesis that $f$ is compactly supported... – Stacy Feb 27 '14 at 22:31
@Stacy The fact that $f$ is compactly supported is the reason that you were able to reduce from integration over $t\in(-\infty,\infty)$ to integration over $t\in[-A,A]$. – Nick Peterson Feb 27 '14 at 22:45
ok but, is there some trouble if we integrate over all $\mathbb R$ ? we still have $g(x+P-t)=g(x-t)$, this still works even if the function is assumed to be just continuous ? – Stacy Feb 27 '14 at 22:57
@Stacy: What if $f(x)=x^2$ and $g(x)=\sin x$? Then $\int_{-\infty}^\infty |f(t)g(x-t)|\,dt=\infty$. You can just drop the hypothesis of compact support. (You could weaken it, though, and still get the result.) – Michael Hardy Feb 27 '14 at 23:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.