Let $\Omega \subset \mathbb{R}^n$. Show that $g_n(x)g_m(y)$ forms an orthonormal basis for $L^2(\Omega \times \Omega)$ for all $n,m \geq 1$ when $g_n:\Omega \rightarrow \mathbb{C}$ is an orthonormal basis of $L^2(\Omega)$ for all $n \geq 1$.
Fact that $g_mg_n$ is orthonormal is easy to see, but the completeness of the orthonormal basis is harder...
I think i could argue that $\|f\| = 0$ and then apply the completeness lemma for an orthonormal basis of a Hilbert space. To argue that $\|f\| = 0$ i think i need to apply the fact that linear combinations functions $g(x)f(y)$ are dense in $L^2(\Omega \times \Omega)$ and then apply that $g_n$ is an orthonormal basis of $L^2(\Omega)$. Any suggestions?
By the linear combination thingy i think my lecturer means that for any $f \in L^2(\Omega \times \Omega)$ and for all $\varepsilon > 0$ there is a linear combination of function $\phi_n, \psi_n \in L^2(\Omega)$ such that $\|f-\sum_{n=1}^N\phi_n \psi_n\| < \varepsilon$.
My idea is to show that $f\in L^2(\Omega)$ there holds $<f,g_ng_m> = 0$ for all $n,m$ iff $\|f\| = 0$. I was able to show that $<f,g_ng_m> = 0$ using the linear combination thingy above, but now im stuck to arguing that $\|f\| = 0$. Since $<f,g_ng_m> = 0$ we can write $\|f\| = \|f-\sum_{n=1}^\infty\sum_{m = 1}^\infty<f,g_ng_m>g_ng_m\|$
Since $\overline{span(\phi\psi : \phi,\psi \in L^2(\Omega))} = L^2(\Omega \times \Omega)$ it is enough to show that $span(\phi\psi : \phi,\psi \in L^2(\Omega)) \subset \overline{span( \phi\psi: \phi \in M_1, \psi \in M_2 )}$ where $M_1 \subset L^2(\Omega)$ , $M_2 \subset L^2(\Omega)$ , $\overline{span(M_1)} = L^2(\Omega)$ and $\overline{span(M_2)} = L^2(\Omega)$.