Lemma 10

The mapping $\theta$ defines by passing to quotients a linear mapping $\widetilde{\theta}$ from $L^2(X)\otimes L^2(Y)$ into $L^2(X\times Y)$, and there exists a unique isometric isomorphism from $L^2(X)\widehat{\otimes}_2L^2(Y)$ onto $L^2(X\times Y)$ which coincides with this one on $L^2(X)\otimes L^2(Y)$.

The first assertion is elementary. Let us prove the second.

Let $(f_i)_{i\in I}$ and $(g_j)_{j\in J}$ be orthonormal bases of $L^2(X)$ and $L^2$(Y), respectively. The family $(f_i\otimes g_j)_{(i,j)\in I\times J}$ is an orthonormal basis of $L^2(X)\widehat{\otimes}_2L^2(Y)$ (EVT, V, p. 29, cor. 1) and the family $(\widetilde{\theta}(\overline{f}_i\otimes g_j))_{(i,j)\in I\times J}$ is orthonormal in $L^2(X\times Y)$ (INT, V, p. 95, § 8, n$^o3$, cor. 2). The mapping $\widetilde{\theta}$ therefore extends by continuity to an isometric linear mapping from $L^2(X)\widehat{\otimes}_2L^2(Y)$ into $L^2(X\times Y)$. It remains to prove that this extension is surjective.

For this, it is enough to prove that the image of $\widetilde{\theta}$ is dense in $L^2(X\times Y)$. Let N be an element of $L^2(X\times Y)$ orthogonal to the image of $\widetilde{\theta}$. For all $i\in I$ and $j\in J$, one has $\langle g_j|u_N(f_i)\rangle =\langle \widetilde{\theta}(\overline{f}_i\otimes g_j)|N\rangle = 0$ (formula (12)), whence $u_N= 0$, and therefore N = 0.

The preceding lemma establishes the assertion stated in EVT, V, p. 29, example 2.

Henceforth we shall identify $L^2(X)\widehat{\otimes}_2L^2(Y)$ and $L^2(X\times Y)$ by means of the isometric isomorphism of Lemma 10.