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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.

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.

1 Answers 1

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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 $x$.

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    You are right, ah... I am getting rusty. :-(2012-12-01