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Let $k\colon\mathbb{R}\to\mathbb{C}$ be a 1-periodic function with $k|[0,1]\in L^2([0,1])$. Define the convolution operator $T$ as $f\mapsto\int\limits_{[0,1]}k(s-t)f(t)\, dt$. Develop $k$ in a Fourierseries and with this find the eigenvalues and eigenfunctions.

Could anyone please help me? I never did something like this before.

I guess the development of k in a Fourierseries is

$k(t)=\frac{a_0}{2}+\sum\limits_{k=1}^{\infty}(a_k\cos(k2\pi t)+b_k\sin(k2\pi t))$.

And what is the next step?

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very unfortunate choice of summation index – joriki Dec 15 '12 at 13:09
What do you know about the relationship between convolution, multiplication and Fourier transforms? – joriki Dec 15 '12 at 13:10
you are right its not very smart to use k as function and as summation index. i change the summation index to i. – math12 Dec 15 '12 at 14:26
there is the convolution formula for fourier transformation but do i really need fourier transformation here? – math12 Dec 15 '12 at 14:27
up vote 0 down vote accepted

I thought about it and came to the conclusion that it might be more appropriate to develop the function $k$ in a complex fourier series, i.e.

$k(t)=\sum\limits_{n=-\infty}^{\infty}c_n e^{2\pi int}$

with coefficients

$c_n=\int\limits_0^1 k(t)e^{-2\pi int}\, dt$.


$\int\limits_0^1 k(s-t)f(t)\, dt=\int\limits_0^1 \sum\limits_{n=-\infty}^{\infty}c_n e^{2\pi in (s-t)}f(t)\, dt$

I think now i can change the integral and the sum (Lebesgue) getting

$\sum\limits_{n=-\infty}^{\infty}c_n e^{2\pi ins}\int\limits_0^1 e^{2\pi int}f(t)\, dt$.

Is $f$ a 1-periodic function, too? Then the integrals above are the fourier coefficients of $f$, aren't they?

Can anybody help me to find the eigenvalues and eigenfunctions?

Thank you very much!

greetings math12

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