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

Let $f:\mathbb{R}\to\mathbb{R}$ be a $T$-periodic function, that is $f(t+T)=f(t)$ for all $t\in \mathbb{R}$. Assume that $$\int_0^{+\infty}|f(s)|ds<+\infty.$$ Now if we assume in addition that $f$ is continuous, my intuition tells me that we must have necessarily $f=0$, is this correct ?

share|cite|improve this question
Yes, that is correct (for $T \neq 0$). – Daniel Fischer Jun 27 '14 at 14:42
up vote 3 down vote accepted


$$\int_0^{\infty} |f(s)| ds = \sum_{k = 0}^{\infty} \int_{kT}^{(k + 1)T} |f(s)| ds$$

share|cite|improve this answer

This is correct. The way you can see this is by considering the maximum of $|f|$, call it $L$. For any $x$ such that $|f(x)|=L$, we have that $|f(y)| > \frac{L}{2}$ for all $|x-y| < \delta$ (for a sufficient choice of $\delta$). Can you see how to argue it from here?

share|cite|improve this answer

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.