Let $A,B\subseteq\mathbb{R}^n$ and let $e_i$ and $f_k$ be the ortonoromal basis of $L_2(A)$ and $L_2(B)$ respectively. How can I get now the orthonormal basis of my new set $L_2(A\times B$)?
Intuitively I would say it should look something like this $\{e_i\cdot f_k\}$, but I can't prove it. Can someone give me maybe a hint on how to start this?
EDIT
Let's say that $g\in L_2(A\times B)$ and lets say $g$ can't be expressed with $e_i(x)\cdot f_k(y)$, then we have that $0\le \lvert\langle g,e_if_k\rangle_{L_1} \vert^2\le\lVert g\rVert_{L_1}\cdot \lVert e_if_k\rVert_{L1}<\infty$, so the value of $\langle g,e_if_k\rangle_{L_1}$ exists and is finite and so we have that $\int_{A\times B} g(x,y)\overline{e_i(x)f_k(y)}d(x,y)=0$ (bcs g cant be expressed as $e_if_k$). From here on I dont know what to do. I got that $\int_A\int_B g(x,y)\overline{f_k(y)}dy\overline{e_i(x)}dx=0$, so we get that $\int_B g(x,y)\overline{f_k(y)}dy=0$, for nearly all $x\in A$, but from there on I dont know what to do. Can someone give me some kind of hint?