- Let \(\{ \varphi _ { k } \} _ { k = 1 } ^ { \infty }\) be an orthonormal basis for \(L ^ { 2 } ( \mathbb { R } ^ { d } )\) and define \(\varphi _ { k , j } ( x , y ) = \varphi _ { k } ( x ) \varphi _ { j } ( y )\) Prove that \(\{ \varphi _ { k , j } \} _ { k , j = 1 } ^ { \infty }\) is an orthonormal basis of \(L ^ { 2 } ( \mathbb R ^ { d } \times \mathbb R ^ { d } )\)
Solution
1. Prove orthonormality.
For indices \(k,j,k',j'\), Fubini’s theorem gives \[ \begin{aligned} \langle \varphi_{k,j},\varphi_{k',j'}\rangle &= \int_{\mathbb R^d}\int_{\mathbb R^d} \varphi_k(x)\varphi_j(y) \overline{\varphi_{k'}(x)\varphi_{j'}(y)}\,dy\,dx\\ &=\langle\varphi_k,\varphi_{k'}\rangle \langle\varphi_j,\varphi_{j'}\rangle\\ &=\delta_{kk'}\delta_{jj'}. \end{aligned} \] Thus the product family is orthonormal.
2. Let a function be orthogonal to every product basis vector.
Suppose \[ F\in L^2(\mathbb R^d\times\mathbb R^d) \] and \[ \langle F,\varphi_{k,j}\rangle=0 \] for every \(k,j\).
For fixed \(j\), define for almost every \(x\) \[ H_j(x):= \int_{\mathbb R^d}F(x,y)\overline{\varphi_j(y)}\,dy. \] By Cauchy–Schwarz in the \(y\) variable, \[ |H_j(x)|^2 \le \left(\int|F(x,y)|^2\,dy\right) \left(\int|\varphi_j(y)|^2\,dy\right). \] Since \(\|\varphi_j\|_2=1\), Fubini yields \[ \|H_j\|_2^2 \le \|F\|_2^2. \] Thus \(H_j\in L^2(\mathbb R^d)\).
3. Use completeness of the one-variable basis twice.
For every \(k\), \[ \begin{aligned} \langle H_j,\varphi_k\rangle &=\int_{\mathbb R^d} \left(\int_{\mathbb R^d}F(x,y)\overline{\varphi_j(y)}\,dy\right) \overline{\varphi_k(x)}\,dx\\ &=\langle F,\varphi_{k,j}\rangle=0. \end{aligned} \] Since \((\varphi_k)\) is an orthonormal basis, this implies \[ H_j=0 \] almost everywhere, for every \(j\).
By Fubini, for almost every \(x\), the section \[ F_x(y):=F(x,y) \] belongs to \(L^2(\mathbb R^d)\). Removing the countable union of the exceptional null sets for the identities \(H_j=0\), we obtain that for almost every \(x\), \[ \langle F_x,\varphi_j\rangle=0 \] for every \(j\). Completeness of \((\varphi_j)\) then gives \[ F_x=0 \] in \(L^2(\mathbb R^d)\) for almost every \(x\).
Hence \(F=0\) in \[ L^2(\mathbb R^d\times\mathbb R^d). \] Therefore the orthonormal family \((\varphi_{k,j})\) is complete, and hence is an orthonormal basis.