# Show that $g_n(x)g_m(y)$ forms an orthonormal basis for $L^2(\Omega \times \Omega)$ when $g_n$ is an orthonormal basis of $L^2(\Omega)$

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 $$\langle f,g_ng_m\rangle = 0$$ for all $$n,m$$ iff $$\|f\| = 0$$. I was able to show that $$\langle f,g_ng_m\rangle = 0$$ using the linear combination thingy above, but now I'm stuck to arguing that $$\|f\| = 0$$. Since $$\langle f,g_ng_m\rangle = 0$$ we can write $$\|f\| = \|f-\sum_{n=1}^\infty\sum_{m = 1}^\infty\langle f,g_ng_m\rangle g_ng_m\|$$

• math.stackexchange.com/questions/105451/… Commented Feb 4 at 11:48
– jd27
Commented Feb 4 at 12:41
• 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 functions $\phi_n, \psi_n \in L^2(\Omega)$ such that $\|f-\sum_{n = 1}^N\phi_n \psi_n\|_{L^2(\Omega \times \Omega)} < \varepsilon$ ? Commented Feb 4 at 12:48
• sorry @jd27 im not familiar with the tensor thingy, but your idea seems to be similiar compared to mine. Commented Feb 4 at 13:46
• @voroshilov in the context of my answer in the linked post, simply take $\mathcal{H}_1 = L^2(\Omega)$ and $\mathcal{H}_2 = L^2(\Omega)$, and $\mathcal{H}_1 \hat{\otimes} \mathcal{H}_2 = L^2( \Omega \times \Omega)$. In that context if $f,g \in L^2 (\Omega)$, then $f \otimes g$ simply denotes the function $(x,y) \mapsto f(x) g(x)$. Also take $M_1 = M_2$ the orthonormal basis of $L^2(\Omega)$.
– jd27
Commented Feb 4 at 13:55

Since $$\overline{\text{span}(\phi\psi : \phi,\psi \in L^2(\Omega))} = L^2(\Omega \times \Omega)$$ it is enough to show that $$\text{span}(\phi\psi : \phi,\psi \in L^2(\Omega)) \subset \overline{\text{span}( \phi\psi: \phi \in M_1, \psi \in M_2 )}$$ where $$M_1 \subset L^2(\Omega)$$ , $$M_2 \subset L^2(\Omega)$$ , $$\overline{\text{span}(M_1)} = L^2(\Omega)$$ and $$\overline{\text{span}(M_2)} = L^2(\Omega)$$.
• $\overline{L^2(\Omega) \times L^2(\Omega)} = L^2(\Omega \times \Omega)$ is not correct. The correct statement is $\overline{\operatorname{span} \{ f \otimes g : f,g \in L^2 (\Omega)\}} = L^2(\Omega \times \Omega)$. So it is enough to show that $\operatorname{span} \{ f \otimes g : f,g \in L^2 (\Omega) \subset \overline{ \operatorname{span} \{ \varphi \otimes \psi : \varphi \in M_1 , \psi \in M_2 \} }$