Orthonormal basis of $L^2(X \times Y)$

Question

Let $(X, \mathscr{S}, \mu)$ and $(Y, \mathscr{T}, \lambda)$ be two $\sigma$-finite measure spaces. Let $\{ \phi_i\}_{\in I}$, $\{\psi_j\}_{j \in J}$ be two orthonormal bases for $L^2(X,\mu)$ and $L^2(Y, \lambda)$ respectively. For $i \in I$, $j \in J$, let $$\theta_{ij}(x,y):=\phi_i(x)\psi_j(y), \,\,\,\, (x\in X, y \in Y).$$

Is it true that $\{\theta_{ij}\}_{i \in I, j \in J}$ is an orthonormal basis for $L^2(X \times Y, \mu \times \lambda)$ ?

Answer is yes whenever either $L^2(X)$ or $L^2(Y)$ is separable. See for instance the following.

Orthonormal basis for product $L^2$ space

The question is precisely for the case when both $L^2(X)$ and $L^2(Y)$ are not separable i.e. both $I$ and $J$ are uncountable.

Answer 1

This follows from a general result on the tensor product of Hilbert spaces.

Proposition: Let $\mathcal{H}_{1}, \mathcal{H}_{2}$ be Hilbert spaces and $M_{i} \subset \mathcal{H}_{i}$ with $\overline{ \operatorname{span} M_{i}} = \mathcal{H}_{i}$ for $i \in \{1,2\}$. Then $\overline{ \operatorname{span} \{ \varphi \otimes \psi : \varphi \in M_{1} , \psi \in M_{2} \} } = \mathcal{H}_{1} \hat{ \otimes} \mathcal{H}_{2} $.

Here $\mathcal{H}_{1} \hat{ \otimes} \mathcal{H}_{2} $ denotes the Hilbert space tensor product.

Proof: Let $\sum_{i=1}^{n} \varphi_{i} \otimes \psi_{i} \in \mathcal{H}_{1} \otimes \mathcal{H}_{2}$ and $\varepsilon >0$. For $i \in \{1 , \dots , n\}$ let $\varphi_{i}^{\prime} \in \operatorname{span} M_{1} $ and $\psi_{i}^{\prime} \in \operatorname{span} M_{2} $ with $\| \varphi_{i} - \varphi_{i}^{\prime} \| \| \psi_{i} \| < \varepsilon$ and $\| \psi_{i} - \psi_{i}^{\prime} \| \| \varphi^\prime_{i} \| < \varepsilon$ ( first choose $\varphi^\prime_i$, then $\psi^\prime_i$). Then: \begin{align*} &\bigg \| \sum_{i=1}^{n} \varphi_{i} \otimes \psi_{i} -\sum_{i=1}^{n} \varphi^{\prime}_{i} \otimes \psi^{\prime}_{i} \bigg \| \leq \sum_{i=1}^{n} \| \varphi_{i} \otimes \psi_{i} - \varphi^{\prime}_{i} \otimes \psi^{\prime}_{i} \|\\ &=\sum_{i=1}^{n} \| (\varphi_{i} - \varphi^{\prime}_{i}) \otimes \psi_{i} + \varphi^{\prime}_{i} \otimes( \psi_{i} - \psi^{\prime}_{i}) \| \\ &\leq \sum_{i=1}^{n} \big( \| (\varphi_{i} - \varphi^{\prime}_{i}) \otimes \psi_{i} \| + \| \varphi^{\prime}_{i} \otimes( \psi_{i} - \psi^{\prime}_{i}) \| \big) < 2n \varepsilon. \end{align*} The proposition follows since $\sum_{i=1}^{n} \varphi^{\prime}_{i} \otimes \psi^{\prime}_{i} \in \operatorname{span} \{ \varphi \otimes \psi: \varphi \in M_{1}, \psi \in M_{2} \}$ and because $\mathcal{H}_{1} \otimes \mathcal{H}_{2} $ is dense in $\mathcal{H}_{1} \hat{\otimes} \mathcal{H}_{2} $.

It follows that $ ( \psi_i \otimes \varphi_j)_{(i,j) \in I \times J}$ is an orthonormal basis of $\mathcal{H}_1 \hat{\otimes}\mathcal{H}_2$ for an orthonormal basis $(\psi_i)_{i \in I}$ of $\mathcal{H}_1$ and $(\varphi_j)_{j \in J}$ of $\mathcal{H}_2$ with arbitrary index sets $I,J$.

Since $L^2(\mu) \hat{\otimes} L^2(\lambda) = L^2( \mu \times \lambda)$ the original statement follows. For a proof see my post here.