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3. The product of square integrable functions is the space of square integrable functions on the product

The product of square integrable functions is the space of square integrable functions on the product

Theorem : Given two separable Hilbert spaces $\mathcal{H}_1 = L^2(X, \mu)$ and $\mathcal{H}_2 = L^2(Y, \lambda)$, there exists a unique isometric isomorphism

\begin{equation} L^2(X, \mu) \otimes L^2(Y, \lambda) \cong L^2(X \times Y, \mu \times \lambda) \end{equation}

which maps $f \otimes g$ to $f(x) g(y)$.

Proof : Being separable Hilbert spaces, $\mathcal{H}_1$ and $\mathcal{H}_2$ can be given a basis, $\{ \psi^1_i(x) \}$ and $\{ \psi^2_i(y) \}$. We need to show first that $\{ \psi^1_i(x) \psi^2_j(y) \}$ is a basis of $L^2(X \times Y, \mu \times \lambda)$. This is true if this set obeys Parseval's identity : for any $v \in L^2(X \times Y, \mu \times \lambda)$,

\begin{equation} \sum_{i,j} |\langle v, \psi^1_i \psi^2_j \rangle|^2 = \langle v, v \rangle \end{equation}

Consider our function $v(x,y)$. Then :

\begin{eqnarray} \langle v, \psi^1_i \psi^2_j \rangle &=& \int v^*(x,y) \psi^1_i(x) \psi^2_j(y) d\mu(x) d\lambda(y) \end{eqnarray}