# Finite trigonometric polynomial

I noticed by numerical and some explicit calculations for a few examples that for real-valued finitely supported functions $\phi \in L^2(\mathbb{R})$ we have that

$T(x):= \sum_{n \in \mathbb{Z}} |\hat{\phi}(x+2 \pi n)|^2$ is only a trigonometric polynomial $T(x) = a_0 + a_1 \cos( x) + ...+a_n \cos(n x)$ for some $n \in \mathbb{N}$ and coefficients $a_0,..,a_n.$

Now, the thing is that we apparently only have $T \in L^1([0,2\pi]),$ so it is not even clear to me that there is a Fourier series converging to $T$ at all. Despite, if there is one, then it is clear (by symmetry) that only $\cos$ terms should appear. Nevertheless, the thing that confused me most is that the Fourier expansion is apparently always finite. I guess this must be somehow related to the compact support of $\phi$. But it is not obvious to me where this actually comes from.

Does anybody know why this happens?

A very simple (I tested also harder ones) is

$$\phi(x) = \chi_{[-1,1]}$$

then $$\sum_{n \in \mathbb{Z}} |\sin^2(x+2\pi n)(x+2 \pi n)^{-2}| = \frac{1}{2}+ \frac{1}{2} \cos(x)$$

and now notice that $$\frac{1}{\pi} \int_{-\pi}^{\pi} \sum_{n \in \mathbb{Z}} |\sin^2(x+2\pi n)(x+2 \pi n)^{-2}| \cos(kx)dx$$ is equal to $1$ if $k=0$ and $\frac{1}{2}$ if $k=1$ and zero otherwise. So also the Fourier expansion is finite and agrees with the representation (notice that we have to divide the first Fourier coefficient by $\frac{1}{2}.$)

If we consider $\phi(x) =\chi_{[-n,n]}$ then this shows that exactly the first $k<2n$ integrals are non-zero and the $2n$ th Fourier coefficient will vanish. Thus, it seems as if we could try to state: The larger the support is, the more coefficients are non-zero.

• There is no reason whatsoever for $T$ to be a trigonometric polynomial, $T$ can be any non-negative periodic $L^1$ function. You just took special examples. Also, question of this type are much better suited for math.stackexchange.com. – Christian Remling May 25 '15 at 16:30
• @Christian Remling strange, how do you know that it can be any non-negative periodic $L^1$ function? – Physicist 2.0 May 25 '15 at 16:38
• I overlooked the "hat", this is of course not correct. The first half of that sentence is still true, though. – Christian Remling May 25 '15 at 16:43
• @ChristianRemling and how do you know that the first half is true – Physicist 2.0 May 25 '15 at 17:09

This is well known in basic wavelet theory. Here is an argument. For $f=\sum_n a_n\Phi(x-n)$ and $g=\sum_n b_n \Phi(x-n)$ calculate the scalar product $<f,g>$ using the Plancherel theorem. You get $\int_{-\infty}^\infty \hat f \overline{\hat g}\hat\Phi \overline{\hat \Phi}$. Since $\hat f$ and $\hat g$ are $2\pi$ periodic you write this as $\int_0^{2\pi} \hat f \overline{\hat g}\big(\sum_m|\hat \Phi(\xi+2\pi m)|^2)$. Now you look at $g=\Phi$ and $f=\Phi(x-k)$. Since functions have bounded support for big $k$ we have =0\$. This gives what we want.