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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 $\|f\| = \|f-\sum_{n=1}^\infty\sum_{m = 1}^\infty\langle f,g_ng_m\rangle g_ng_m\|$

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  • $\begingroup$ math.stackexchange.com/questions/105451/… $\endgroup$ Commented Feb 4 at 11:48
  • $\begingroup$ See also my answer here. $\endgroup$
    – jd27
    Commented Feb 4 at 12:41
  • $\begingroup$ 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 $\|f-\sum_{n = 1}^N\phi_n \psi_n\|_{L^2(\Omega \times \Omega)} < \varepsilon$ ? $\endgroup$
    – voroshilov
    Commented Feb 4 at 12:48
  • $\begingroup$ sorry @jd27 im not familiar with the tensor thingy, but your idea seems to be similiar compared to mine. $\endgroup$
    – voroshilov
    Commented Feb 4 at 13:46
  • $\begingroup$ @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 $L^2(\Omega)$. $\endgroup$
    – jd27
    Commented Feb 4 at 13:55

1 Answer 1

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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 $\overline{\text{span}(M_2)} = L^2(\Omega)$.

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  • $\begingroup$ $\overline{L^2(\Omega) \times L^2(\Omega)} = L^2(\Omega \times \Omega)$ 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 $\operatorname{span} \{ f \otimes g : f,g \in L^2 (\Omega) \subset \overline{ \operatorname{span} \{ \varphi \otimes \psi : \varphi \in M_1 , \psi \in M_2 \} }$ $\endgroup$
    – jd27
    Commented Feb 4 at 19:50
  • $\begingroup$ thanks for pointing out. ill fix my answer. $\endgroup$
    – voroshilov
    Commented Feb 4 at 20:08

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