# Expansion theorem or Poisson Summation Formula? - Basis of eigenfunctions gives rise to a Fourier series

Does anyone could explain to me why in the Semiclassical's answer on the question Wave kernel for the circle $\mathbb{S}^1$ - Poisson Summation Formula, the basis gives a series of the form $\displaystyle \frac{1}{\sqrt{2\pi}}a_0+\frac{1}{\sqrt{\pi}}\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right)$? Is it because of the Expansion theorem?

• I don't understand the question. Are you asking why these are the eigenfunctions of the Laplacian on the unit circle? Or why they form a complete basis for sufficiently smooth periodic functions? – user7530 May 24 '16 at 3:04
• @user7530 My apologie for the quality of the question. I mean I know that $\left\{\dfrac1{\sqrt{2\pi}},\dfrac1{\sqrt{\pi}}\cos x,\dfrac1{\sqrt{\pi}}\sin x,\dfrac1{\sqrt{\pi}}\cos 2x,\dfrac1{\sqrt{\pi}}\sin 2x,\cdots\right\}$ form a basis, but dont know why we could construct the Fourier series $\displaystyle \frac{1}{\sqrt{2\pi}}a_0+\frac{1}{\sqrt{\pi}}\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right)$ with that basis, and that Fourier series is the same as $W(t,x,y)$ (wave kernel). – user332990 May 24 '16 at 3:12
• I'm still missing the actual question here: if $\{b_0, b_1, \ldots\}$ is a basis for a vector space $V$, then every element of $V$ can be expressed as $\sum a_i b_i$ for some coefficients $a_i$... that is all that is going on here. – user7530 May 24 '16 at 5:39

I don't fully understand the question: if you agree that $\left\{\frac{1}{\sqrt{2\pi}}, \frac{1}{\sqrt{\pi}}\cos kx, \frac{1}{\sqrt{\pi}}\sin kx\right\}$ are a basis for the function space of $2\pi$-periodic functions, it follows immediately that any function in this space can be expressed as a linear combination of these basis functions (and the $a_i$ and $b_i$ are these coefficients).... that's just what it means for the functions to be a basis.
1. Computing that the eigenfunctions $\mu$ of the Laplacian $f'' = \lambda f$ with periodic boundary conditions are precisely the sines and cosines listed above;
2. The eigenvalue corresponding to $\frac{1}{\sqrt{\pi}}\cos kx$ and $\frac{1}{\sqrt{\pi}}\sin kx$ is $-k^2$;
• Thanks for you answer! However, why could we say that $W(t,x,y) = \displaystyle \frac{1}{\sqrt{2\pi}}a_0+\frac{1}{\sqrt{\pi}}\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right)$ ? – user332990 May 24 '16 at 11:36
• Huh? That's not the wave kernel. You had the right formula in your own post, $W(t,x,y)= \sum_{k \geq 1} e^{it \sqrt{- \lambda_k}} \mu_k(x) \mu_k(y).$ Now plug in $\lambda$ and $\mu$ and use $\cos(x-y) = \cos x \cos y + \sin x \sin y$. Fourier series, Poisson summation, etc are all red herrings here. – user7530 May 24 '16 at 14:59
• Do you have an idea why if $$\sum_{k \geq 0} e^{it \sqrt{-\lambda_k}}=\int_{-\infty}^{\infty} (\sum_{k \geq 0} e^{it \sqrt{-\lambda_k}} \mu_k^2(x))dx= \sum_{k \geq 0} e^{it \sqrt{-\lambda_k}} \int_{-\infty}^{\infty}\mu_k^2(x)dx,$$ then necessarily, $\int_{-\infty}^{\infty}\mu_k^2(x)dx=1$? This is come from the same question – user332990 May 24 '16 at 15:03
• The definition of $\mu_k$ is the $k$th eigenvalue of $\Delta$, normalized so that $\langle \mu_k, \mu_k\rangle = 1$. – user7530 May 24 '16 at 15:10