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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 $<f,g_ng_m> = 0$ 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 $\|f\| = \|f-\sum_{n=1}^\infty\sum_{m = 1}^\infty<f,g_ng_m>g_ng_m\|$

voroshilov
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  • https://math.stackexchange.com/questions/105451/orthonormal-basis-for-product-l2-space – Evangelopoulos Foivos Feb 04 '24 at 11:48
  • See also my answer here. – jd27 Feb 04 '24 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 $|f-\sum_{n = 1}^N\phi_n \psi_n|_{L^2(\Omega \times \Omega)} < \varepsilon$ ? – voroshilov Feb 04 '24 at 12:48
  • sorry @jd27 im not familiar with the tensor thingy, but your idea seems to be similiar compared to mine. – voroshilov Feb 04 '24 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 Feb 04 '24 at 13:55
  • ah i see you just use the definitions of spans and the density fact. – voroshilov Feb 04 '24 at 14:53

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

voroshilov
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  • $\overline{L^2(\Omega) \times L^2(\Omega)} = L^2(\Omega \times \Omega)$ 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 $\operatorname{span} { f \otimes g : f,g \in L^2 (\Omega) \subset \overline{ \operatorname{span} { \varphi \otimes \psi : \varphi \in M_1 , \psi \in M_2 } }$ – jd27 Feb 04 '24 at 19:50
  • thanks for pointing out. ill fix my answer. – voroshilov Feb 04 '24 at 20:08