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

I am given $f\in L^1 (\mathbb{T})$, and $f(x+\frac{2\pi}{k})=f(x)$ for some natural number $k$.

I want to show that $f$'s Fourier transform gets vanished for $n=rk+d$ where $1\leq d<k$.

So here's what I did so far (I write without the normalization factor cause it doesn't make a difference here):

$$\hat{f}(n)= \int_{-\pi}^{\pi} f(x) e^{-inx}dx = \int_{-\pi}^{\pi} f(x+\frac{2\pi}{k}) e^{-inx}dx = \int_{\frac{2\pi}{k}-\pi}^{\pi+\frac{2\pi}{k}} f(x) e^{-inx} e^{-i\frac{2\pi}{k} d}dx$$

Now here's what I thought if $d$ were odd and $k$ an even integer then we can repeat the above $\frac{k}{2}$ to get:

$$\int_{0}^{2\pi} f(x)e^{-inx} e^{-i\pi d} dx = -\int_{0}^{2\pi} f(x) e^{-inx}dx$$ Which is zero, but then how would you show show for other cases of $k$ and $d$?

Thanks in advance.

share|cite|improve this question
up vote 1 down vote accepted

You are on the right track and close to an answer.

  • The bounds $-\pi+\tfrac{2 \pi}{k}$ and $\pi +\tfrac{2 \pi}{k}$ in your last integral for $\hat{f}(n)$ can be replaced by $-\pi$ and $\pi$ since the integrand is periodic with period $2\pi$.

  • The term $e^{-i \frac{2 \pi}{k}d}$ can be placed outside the integral as a factor since it does not depend on $x$.

share|cite|improve this answer
Your answer doesn't really help, I mean I need to show that $$e^{-i\frac{2\pi}{k} d}$$ is minus 1; otherwise I cannot get that $\hat{f}(n)=-\hat{f}(n)=0$. – MathematicalPhysicist Dec 1 '12 at 7:56
@MathematicalPhysicist If you followed both steps you should end up with something of the form $\hat{f}(n) = \textrm{factor} \cdot \hat{f}(n)$. Note that the factor does not have to be $-1$: not equal to $1$ is sufficient. – WimC Dec 1 '12 at 8:08
You are right, ah... I am getting rusty. :-( – MathematicalPhysicist Dec 1 '12 at 8:37

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.