Basis of $L^{2}(\mathbb{R}^{2})$ formed by the products of basis functions for $L^2(\mathbb R^1)$

Question

Let $\{\phi_{n}\}_{n\in\mathbb{N}}$ be a orthogonal basis of $L^{2}(\mathbb{R}^{1})$.

Is it true that

$\{\phi_{n}(x)\phi_{m}(y)\}_{n,m\in\mathbb{N}}$ is a basis of $L^{2}(\mathbb{R}^{2})$?

How could we verify this?

I've been looking through Rudin's PMA, but couldn't find any related theorems.

If it's okay, suggest a recommendation book or stuff.

Answer 1

Orthogonality is easy; the issue is the density of linear span. For any $f,g\in L^2(\mathbb{R}^1)$ we can approximate $f$ by a finite sum $\sum a_m \phi_m$ and $g$ by $\sum b_n \phi_n$; then $\sum_{m,n} a_mb_n \phi_m(x)\phi_n(y)$ approximates $f(x)g(y)$. So the problem boils down to showing that the linear span of functions of the form $f(x)g(y)$, with $f,g\in L^2(\mathbb{R}^1)$, is dense in $L^2(\mathbb{R}^2)$. This linear span is usually denoted $L^2(\mathbb{R}^1)\otimes L^2(\mathbb{R}^1)$, the tensor product of two vector spaces.

Observe that the characteristic function of any rectangle $[a,b]\times[c,d]\subset\mathbb{R}^2$ is in $L^2(\mathbb{R}^1)\otimes L^2(\mathbb{R}^1)$, being the product $\chi_{[a,b]}(x)\chi_{[c,d]}(y)$. Every open set $\Omega\subset\mathbb{R}^2$ is a union of disjoint rectangles (except for their boundaries), for example one can take the union of all maximal dyadic squares contained in $\Omega$. This implies that $\chi_{\Omega}$ is in the closure of $L^2(\mathbb{R}^1)\otimes L^2(\mathbb{R}^1)$ whenever $\Omega$ is an open set of finite measure. An arbitrary measurable set $E\subset\mathbb{R}^2$ can be approximated by open sets from the outside, hence $\chi_E$ is also in the closure of $L^2(\mathbb{R}^1)\otimes L^2(\mathbb{R}^1)$. From characteristic functions we get simple functions $\sum c_n \chi_{E_n}$, which are known to be dense in $L^2(\mathbb{R}^2)$.

Answer 2

For $f,g\in L^2(\mathbb{R})$, functions $f(x)g(y)$ are in $L^2(\mathbb{R}^2)$. If $\{ \phi_n \}$ is a complete orthonormal basis of $L^2(\mathbb{R})$, then $\{ \phi_n(x)\phi_m(y)\}$ is a complete orthonormal basis of $L^2(\mathbb{R}^2)$ by an application of Parseval's identity to $h\in L^2(\mathbb{R}^2)$. Clearly $h(x,y)\phi_n(x)\phi_m(y) \in L^1(\mathbb{R}^2)$, which allows the use of Fubini's Theorem to prove Parseval's identity: $$ \sum_{n,m}\left|\iint h(x,y)\phi_n(x)\phi_m(y)dxdy\right|^2 \\ =\sum_{n,m} \left|\int\left(\int h(x,y)\phi_n(x)dx\right)\phi_m(y)dy\right|^2\\ =\sum_{n}\int\left|\int h(x,y)\phi_n(x)dx\right|^2dy \\ =\int\int |h(x,y)|^2dxdy. $$ You just have to be careful about a.e. issues.