# A question on when Fourier Transform on the circle get vanished.

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 $d$?

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• 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 $x$.
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