Prove that $\left\{\frac{1}{\sqrt{2\pi}^N}e^{ih\cdot x}\right\}_{h \in \mathbb{Z}^N}$ is a maximal orthonormal set in $L^2([-\pi,\pi]^N)$ I know that $\left\{\frac{1}{\sqrt{2\pi}}e^{inx}\right\}_{n \in \mathbb{Z}}$ is a maximal orthonormal set in $L^2(\mathbb{T}) \approx L^2([-\pi,\pi])$. 
How can I use this fact to show that $\left\{\frac{1}{(\sqrt{2\pi})^N}e^{ih\cdot x}\right\}_{h \in \mathbb{Z}^N}$ is a maximal orthonormal set in $L^2([-\pi,\pi]^N)$?
 A: The basis you want to show is a complete orthonormal basis is
$$
             E_{n_1,n_2,\cdots,n_N}= \frac{1}{(2\pi)^{N/2}}e^{in_1 x_1}e^{in_2 x_2}\cdots e^{in_N x_N}.
$$
You can do this by showing that Parseval's identity holds for all $f\in C([-\pi,\pi]^N)$ because this subspace is dense in $L^2([-\pi,\pi]^N)$. Using this subspace makes it easy to deal with Fubini's Theorem:
$$
    \sum_{n_1}\sum_{n_2}\cdots\sum_{n_N}|\hat{f}(n_1,n_2,\cdots,n_N)|^2 \\
    = 2\pi\sum_{n_2}\cdots\sum_{n_N}\int_{-\pi}^{\pi}|\hat{f}(x_1,n_2,\cdots,n_N)|^2 dx_1 \\
   = \cdots =(2\pi)^{N}\int_{-\pi}^{\pi}\cdots\int_{-\pi}^{\pi}|f(x_1,x_2,\cdots,x_n)|^2dx_1 dx_2\cdots dx_N.
$$
The notation is messy because of having discrete transforms mixed with untransformed variables.
For example, start with the case where $N=2$. The functions
$$
      \left\{ \frac{1}{2\pi}e^{in_1x}e^{n_2y}\right\}_{n_1,n_2}
$$
form an orthonormal subset of $L^2([-\pi,\pi]^2)$. Suppose $f$ is continuous on $[-\pi,\pi]^2$. Then Bessel's inequality gives
$$
          \sum_{n_1,n_2}\left|\frac{1}{2\pi}\int\int e^{in_1x}e^{in_2y}f(x,y)dxdy\right|^2 \le \int\int |f(x,y)|^2dxdy.
$$
So the unordered sum on the left is absolutely convergent. This allows you to sum however you want, which gives
$$
   \sum_{n_1,n_2}\left|\frac{1}{2\pi}\int\int e^{-in_1x}e^{-in_2y}f(x,y)dxdy\right|^2 \\
   = \sum_{n_1}\sum_{n_2}\left|\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-in_2y}\left(\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x,y)e^{-in_1 x}dx\right)dy\right|^2
$$
The function $\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x,y)dx$ is a continuous function of $y$ because $f$ is assumed to be jointly continuous in $x,y$ on $[-\pi,\pi]^2$. So it is certainly in $L^2[-\pi,\pi]$, which allows you to apply Parseval's theorem for the variable $y$, which gives the above equal to
$$
    \sum_{n_1}\int_{-\pi}^{\pi}\left|\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-in_1 x}f(x,y)dx\right|^2dy \\
   = \int_{-\pi}^{\pi}\sum_{n_1}\left|\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-in_1 x}f(x,y)dx\right|^2 dy \\
   = \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}|f(x,y)|^2dx dy
$$
