Are Fourier series a basis for $L^2((-\pi,\pi)^m)$?

It is a classical result that Fourier series are a basis for $L^2((-\pi,\pi))$. The proof can be found in the Rudin and uses isometry properties of the Fourier transform.

Now, I am struggling to see if this proof can generalize for higher dimensions : are Fourier series also a basis for $L^2((-\pi,\pi)^m)$ for $m\geq 1$ ?

In other words, for $x \in \mathbb{R}^m$ and $n \in \mathbb{Z}^m$, I can define $e_n(x) = e^{in.x}$ (Fourier series basis vector n), and the question is can $(e_n)_{n \in \mathbb{Z}^m}$ span $L^2((-\pi,\pi)^m)$.

I suppose that the proof will use isometry properties of Fourier series (Parseval theorem), but it seems that the problem lies in the approximation of functions.

Solution 1:

Yes. The $e^{in.x}(2\pi)^{-m/2}$ are an orthogonal family so they generate a closed subspace $H$ (which means a sub-Hilbert-space).

Take $f\in L^2((-\pi,\pi)^m)$, let $h$ be its Fourier series, if $h=0$ then let $g=f$, otherwise let $g=f- h\langle f,\frac{h}{\|h\|^2}\rangle$.

$g$ is orthogonal to $H$ and so are its translates, and hence so is $g_k=g\ast e^{-k^2|x|^2}$ (convolution).

If $g\ne 0$ then for $k$ large enough $g_k$ is non-zero.

The point is that $g_k$ is also continuous, if $g_k(a)\ne 0$ then $\int_{(-\pi,\pi)^m} g_k(x)\prod_{r=1}^m \cos^{2l}(x_r-a_r)dx$ will be non-zero for $l$ large enough, so that some Fourier coefficients of $g_k$ will be non-zero, contradicting that $g_k$ is orthogonal to $H$.

Whence in fact $g_k=0$ and $g=0$ and $f$ is equal to its Fourier series and $H$ is the whole of $L^2((-\pi,\pi)^m)$.