Proposition 19

The mapping $N\mapsto u_N$ is an isometric isomorphism of $L^2(X\times Y)$ onto the space $\mathscr{L}_2(L^2(X); L^2(Y))$ of mappings

of Hilbert–Schmidt deThe linear mapping of L$L^2(X)_2$into$(Y)\otimes L\overset{2}{L}(Y)_2(X)$ into. $\mathscr{L}_2(L^2(X); L^2(Y))$ which associates to $g\otimes f$ the Hilbert–Schmidt mapping $h\mapsto \langle f|h\rangle g$ extends to an isometric isomorphism $\theta_1$ of $L^2(Y)\widehat{\otimes}_2L^2(X)$ onto $\mathscr{L}_2(L^2(X); L^2(Y))$ (EVT, V, p. 52, th. 1).

Let us denote moreover by $\theta_2$ the isometric isomorphism of $L^2(X)\widehat{\otimes}_2L^2(Y)$ onto $L^2(Y)\widehat{\otimes}_2L^2(X)$ which to $f\otimes g$ associates $g\otimes \overline{f}$ for every $f\in L^2(X)$ and every $g\in L^2(Y)$.

The linear mapping $\theta_3=\theta_1\circ \theta_2$ is identified with an isometric isomorphism of $L^2(X\times Y)$ onto $\mathscr{L}_2(L^2(X); L^2(Y))$.

Let $f\in L^2(X)$ and $g\in L^2$(Y), and let N be the element of $L^2(X\times Y)$ identified with $f\otimes g$. From INT, V, p. 95, § 8, n$^o3$, cor. 2 and formula (12) one has

for all $f_1\in L^2(X)$ and $g_1\in L^2(Y)$. The mapping $u=\theta_3(N)$ is the linear mapping $\theta_1(g\otimes f)$; it therefore satisfies $u(h) =\langle f|h\rangle g$ for all $h\in L^2(X)$. Consequently, for all $f_1\in L^2(X)$ and $g_1\in L^2$(Y), one has

whence $\theta_3(N) =u_N$. Since $\theta_3$ and $N\mapsto u_N$ are continuous, it follows that $\theta_3(N) =u_N$ for all $N\in L^2(X\times Y)$, which completes the proof.