L two kernels give Hilbert–Schmidt operators

Hilbert bases exist. For a real or complex Hilbert space $H$, let $\mathcal{P}$ be the set of orthonormal subsets of $H$ ordered by inclusion; $\mathcal{P}$ is nonempty because $\mathrm{\varnothing }$ is orthonormal, and the union of a chain in $\mathcal{P}$ is orthonormal, because any two of its elements already lie in a common member of the chain, so it is an upper bound. By [F12] there is a maximal element $E$. If the closed linear span $M$ of $E$ were a proper closed subspace, then [F12] applied to some $x\mathrm{\notin }M$ would give $x=m+n$ with $m\in M$ and $n$ orthogonal to $M$ and $n\mathrm{\ne }0$; then $e:=n\mathrm{/}\mathrm{\parallel }n\mathrm{\parallel }$ would satisfy $\mathrm{\parallel }e\mathrm{\parallel }=1$ and be orthogonal to every element of $E\subseteq M$, so $E\cup \{e\}$ would be an orthonormal set strictly containing $E$, contradicting maximality; hence $M=H$ and $E$ is a Hilbert basis of $H$. In particular $L^2(\nu ;\mathbb{C})$ and $L^2(\mu ;\mathbb{C})$, being Hilbert spaces by [F5], admit Hilbert bases.

A base-measurable representative with the same norm. Apply [F8] to the real and imaginary parts of $k$ and replace infinite values of the resulting representatives by $0$; combining them gives an $(\mathcal{A}\otimes \mathcal{B})$-measurable complex function $k_0$ with $k_0=k$ $\bar{\rho }$-almost everywhere. Then $g:=\mathrm{\mid }{k}_0{\mathrm{\mid }}^2$ is $(\mathcal{A}\otimes \mathcal{B})$-measurable and nonnegative. For every nonnegative simple $(\mathcal{A}\otimes \mathcal{B})$-measurable $s\le g$, the integrals against $\rho$ and $\bar{\rho }$ agree, because the two measures agree on the finitely many base-measurable level sets occurring in $s$ by [F2] and [F9]. Conversely, let $t\le g$ be a nonnegative $\bar{\rho }$-measurable simple function and write its positive-level representation as $t={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{E_j}$, where the $c_j>0$ and the completed-measurable sets $E_j$ are pairwise disjoint. By the definition of the completion, write $E_j=A_j\cup N_j$, where $A_j$ is $(\mathcal{A}\otimes \mathcal{B})$-measurable and $N_j$ is contained in a base-measurable $\rho$-null set. Since $A_j\subseteq E_j$, the sets $A_j$ remain pairwise disjoint, and $s:={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{A_j}$ is a base-measurable simple function satisfying $0\le s\le t\le g$. Moreover [F2] and [F9] give $\int sd\rho ={\sum }_j{c}_j\rho (A_j)={\sum }_j{c}_j\bar{\rho }(E_j)=\int td\bar{\rho }$. Thus every completed-simple minorant contributes the value of a base-simple minorant, while every base-simple minorant is also completed-simple; the two suprema in [F9] coincide and $\int gd\rho =\int gd\bar{\rho }$. Finally $\mathrm{\mid }{k}_0\mathrm{\mid }=\mathrm{\mid }k\mathrm{\mid }$ $\bar{\rho }$-almost everywhere, so the norm of the class $k$ gives $\int gd\bar{\rho }=\int \mathrm{\mid }k{\mathrm{\mid }}^{2}d\bar{\rho }<+\mathrm{\infty }$. Hence $k_0\in L^2(\rho ;\mathbb{C})$ and $\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}=\mathrm{\parallel }k{\mathrm{\parallel }}_2$.

Almost every section is square integrable. By [F3] applied to the nonnegative function $(x,y)\mapsto \mathrm{\mid }{k}_0(x,y){\mathrm{\mid }}^2$, whose integral is $\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2<+\mathrm{\infty }$ by [step 1.2], the function $x\mapsto {\int }_Y\mathrm{\mid }{k}_0(x,y){\mathrm{\mid }}^{2}d\nu (y)$ is measurable with finite integral; hence $\mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}<+\mathrm{\infty }$ for $\mu$-almost every $x$, and ${\int }_X\mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}^{2}d\mu (x)=\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2$.

The section integral exists almost everywhere and is bounded by the section norm. Let $f\in L^2(\nu ;\mathbb{C})$. For every $x$ with $\mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}<+\mathrm{\infty }$ the section $k_0(x,\cdot )$ is $\mathcal{B}$-measurable by [F3] and $f$ is $\nu$-measurable, so $y\mapsto k_0(x,y)f(y)$ is $\nu$-measurable, and Cauchy–Schwarz in $L^2(\nu )$ by [F5] gives ${\int }_Y\mathrm{\mid }{k}_0(x,y)f(y)\mathrm{\mid }d\nu (y)\le \mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}<+\mathrm{\infty }$ together with $\mathrm{\mid }(T_{k_0}f)(x)\mathrm{\mid }=\mid {\int }_Y{k}_0(x,y)f(y)d\nu (y)\mid \le \mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}$. By [step 2.1] these estimates hold for $\mu$-almost every $x$, which is where $(T_{k_0}f)(x)$ is defined.

Measurability. If $f={\sum }_{j<m}{c}_j{\mathbf{1}}_{B_j}$ is a finite simple function with $\nu (B_j)<+\mathrm{\infty }$, then for every $j$ the indicator ${\mathbf{1}}_{B_j}$ lies in $L^2(\nu ;\mathbb{C})$ because $\nu (B_j)<+\mathrm{\infty }$, so [step 3.1] applied to it shows that the integral $x\mapsto {\int }_Y{k}_0(x,y){\mathbf{1}}_{B_j}(y)d\nu (y)$ is defined and finite for $\mu$-almost every $x$. It is also $\mu$-measurable: the four nonnegative functions $(\mathrm{Re}{g}_j{)}^{+},(\mathrm{Re}{g}_j{)}^{-},(\mathrm{Im}{g}_j{)}^{+},(\mathrm{Im}{g}_j{)}^{-}$, where $g_j(x,y):=k_0(x,y){\mathbf{1}}_{B_j}(y)$ is $(\mathcal{A}\otimes \mathcal{B})$-measurable, have $\mu$-measurable section integrals by Tonelli [F3], and on the conull set where the integral of $g_j$ is finite the real and imaginary parts of that integral are differences of these measurable functions, so zero-extension over the exceptional null set makes the section-integral function measurable; a finite linear combination of these is $\mu$-measurable, so $T_{k_0}f$ is $\mu$-measurable for such $f$. For arbitrary $f\in L^2(\nu ;\mathbb{C})$, [F6] gives finite simple functions $f_j$ with $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_{L^2(\nu )}\to 0$; by [step 3.1], $\mathrm{\mid }{T}_{k_0}(f_j-f)(x)\mathrm{\mid }\le \mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_{L^2(\nu )}\to 0$ for $\mu$-almost every $x$, so $T_{k_0}f$ agrees $\mu$-almost everywhere with the limit superior of the measurable functions $T_{k_0}{f}_j$, which is $\mu$-measurable by [F14].

Independence of representatives and linearity. If $f=f^{\mathrm{\prime }}$ $\nu$-almost everywhere then $k_0(x,\cdot )(f-f^{\mathrm{\prime }})=0$ $\nu$-almost everywhere for every $x$, so $T_{k_0}f=T_{k_0}{f}^{\mathrm{\prime }}$ wherever both are defined, in particular $\mu$-almost everywhere. If $k_1$ is a second $(\mathcal{A}\otimes \mathcal{B})$-measurable representative of $k$ of finite $L^2(\rho )$ norm, then $\mathrm{\mid }{k}_0-k_1\mathrm{\mid }$ has $\rho$-integral $0$, so [F3] gives ${\int }_Y\mathrm{\mid }{k}_0(x,y)-k_1(x,y)\mathrm{\mid }d\nu (y)=0$ for $\mu$-almost every $x$, hence the sections agree $\nu$-almost everywhere for $\mu$-almost every $x$ and $T_{k_0}f=T_{k_1}f$ $\mu$-almost everywhere; linearity in $f$ is linearity of the integral.

The orthonormal family of product kernels and the pairing identity. Fix a Hilbert basis $E$ of $L^2(\nu ;\mathbb{C})$ and a Hilbert basis $F$ of $L^2(\mu ;\mathbb{C})$, both of which exist by [step 1.1], and for $f\in F$, $e\in E$ let ${\psi }_{f,e}$ be the class in $L^2(\rho ;\mathbb{C})$ of $(x,y)\mapsto f(x)\overline{e(y)}$; this function is $(\mathcal{A}\otimes \mathcal{B})$-measurable with $\mathrm{\parallel }{\psi }_{f,e}{\mathrm{\parallel }}_{L^2(\rho )}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\mu )}^2\mathrm{\parallel }e{\mathrm{\parallel }}_{L^2(\nu )}^2=1$ by Tonelli [F3], and for $f,f^{\mathrm{\prime }}\in F$, $e,e^{\mathrm{\prime }}\in E$ the pairing $⟨{\psi }_{f,e},{\psi }_{f^{\mathrm{\prime }},e^{\mathrm{\prime }}}⟩={\int }_{X}f\overline{f^{\mathrm{\prime }}}d\mu \cdot \stackrel{‾}{{\int }_{Y}e\overline{e^{\mathrm{\prime }}}d\nu }=⟨f,f^{\mathrm{\prime }}{⟩}_{L^2(\mu )}\overline{⟨e,e^{\mathrm{\prime }}{⟩}_{L^2(\nu )}}$ vanishes unless $f=f^{\mathrm{\prime }}$ and $e=e^{\mathrm{\prime }}$, again by [F3]; so $({\psi }_{f,e})$ is an orthonormal family in $L^2(\rho ;\mathbb{C})$. Moreover $⟨T_{k}e,f{⟩}_{L^2(\mu )}={\int }_X({\int }_Y{k}_0(x,y)e(y)d\nu (y))\overline{f(x)}d\mu (x)={\int }_{X\times Y}{k}_0(x,y)e(y)\overline{f(x)}d\rho (x,y)=⟨k,{\psi }_{f,e}{⟩}_{L^2(\rho )}$, where the middle equality is Fubini [F4] applied to the $L^1(\rho )$ function $(x,y)\mapsto k_0(x,y)e(y)\overline{f(x)}$, whose absolute value has $\rho$-integral at most $\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}\mathrm{\parallel }e{\mathrm{\parallel }}_{L^2(\nu )}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\mu )}$ by Cauchy–Schwarz and Tonelli [F3], [F5].

Boundedness. For $f\in L^2(\nu ;\mathbb{C})$, the class of $T_{k_0}f$ is in $L^2(\mu ;\mathbb{C})$ and $\mathrm{\parallel }{T}_{k_0}f{\mathrm{\parallel }}_{L^2(\mu )}^2={\int }_X\mathrm{\mid }(T_{k_0}f)(x){\mathrm{\mid }}^{2}d\mu (x)\le {\int }_X\mathrm{\parallel }{k}_0(x,\cdot ){\mathrm{\parallel }}_{L^2(\nu )}^{2}d\mu (x)\cdot \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}^2=\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}^2$ by [step 2.1] and [step 3.1], using measurability from [step 4.1] to integrate the squared estimate; hence $T_k:=T_{k_0}$ is linear and bounded with $\mathrm{\parallel }{T}_k\mathrm{\parallel }\le \mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}=\mathrm{\parallel }k{\mathrm{\parallel }}_2$ by [step 1.2].

Rectangle kernels lie in the closed span of the family. Let $A\in \mathcal{A}$, $B\in \mathcal{B}$ have finite measure, so ${\mathbf{1}}_A\in L^2(\mu ;\mathbb{C})$ and ${\mathbf{1}}_B\in L^2(\nu ;\mathbb{C})$; by [F10] the finite-subset nets ${\sum }_{f\in G}⟨{\mathbf{1}}_A,f⟩f$ over finite $G\subseteq F$ and ${\sum }_{e\in H}⟨{\mathbf{1}}_B,e⟩e$ over finite $H\subseteq E$ converge to ${\mathbf{1}}_A$ and ${\mathbf{1}}_B$. For such finite $G,H$ the function $u_G(x)\overline{v_H(y)}$ with $u_G:={\sum }_{f\in G}⟨{\mathbf{1}}_A,f⟩f$ and $v_H:={\sum }_{e\in H}⟨{\mathbf{1}}_B,e⟩e$ is a finite linear combination of the functions $f(x)\overline{e(y)}$, hence its class lies in the span of the family $({\psi }_{f,e})$; and $\mathrm{\parallel }u\bar{v}-u_G\overline{v_H}{\mathrm{\parallel }}_{L^2(\rho )}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(\mu )}\mathrm{\parallel }v-v_H{\mathrm{\parallel }}_{L^2(\nu )}+\mathrm{\parallel }u-u_G{\mathrm{\parallel }}_{L^2(\mu )}\mathrm{\parallel }{v}_H{\mathrm{\parallel }}_{L^2(\nu )}\to 0$ by Tonelli [F3] and convergence of the two nets, so the rectangle kernel ${\mathbf{1}}_A(x){\mathbf{1}}_B(y)={\mathbf{1}}_A(x)\overline{{\mathbf{1}}_B(y)}$ lies in the closed span of $({\psi }_{f,e})$.

The kernel lies in that closed span. By [step 5.2] and [F7] every class of $L^2(\rho ;\mathbb{C})$ — in particular $k_0$, whose $L^2(\rho )$ class exists by [step 1.2] — lies in the closed span of the orthonormal family $({\psi }_{f,e})$.

Exact norm of the family expansion. Bessel's inequality [F11] applied to $k_0$ and the orthonormal family $({\psi }_{f,e})$ gives ${\sum }_{f,e}\mathrm{\mid }⟨k_0,{\psi }_{f,e}⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2$. For the reverse inequality fix a real $\epsilon >0$ and, by [step 6.1], a vector $h$ in the span of finitely many ${\psi }_{f,e}$ with $\mathrm{\parallel }h-k_0{\mathrm{\parallel }}_{L^2(\rho )}<\epsilon$; if $\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}=0$ then $k_0$ is the zero class and both sides vanish, and otherwise $h\mathrm{\ne }0$ for small $\epsilon$ and the finite Parseval identity and Cauchy–Schwarz on the finitely many coefficients give ${\sum }_{f,e}\mathrm{\mid }⟨k_0,{\psi }_{f,e}⟩{\mathrm{\mid }}^2\ge \mathrm{\mid }⟨k_0,h⟩{\mathrm{\mid }}^2\mathrm{/}\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(\rho )}^2\ge \mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2(\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}-\epsilon {)}^2\mathrm{/}(\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}+\epsilon {)}^2$, where the last bound uses $\mathrm{\mid }⟨k_0,h⟩\mathrm{\mid }\ge \mathrm{\parallel }{k}_0{\mathrm{\parallel }}^2-\mathrm{\parallel }{k}_0\mathrm{\parallel }\epsilon$ and $\mathrm{\parallel }h\mathrm{\parallel }\le \mathrm{\parallel }{k}_0\mathrm{\parallel }+\epsilon$ and tends to $\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2$ as $\epsilon \to 0$; hence ${\sum }_{f,e}\mathrm{\mid }⟨k_0,{\psi }_{f,e}⟩{\mathrm{\mid }}^2=\mathrm{\parallel }{k}_0{\mathrm{\parallel }}_{L^2(\rho )}^2=\mathrm{\parallel }k{\mathrm{\parallel }}_2^2$ by [step 1.2].

Transfer to the basis sums. For each $e\in E$ the vector $T_{k}e$ lies in $L^2(\mu ;\mathbb{C})$ by [step 5.1], so [F10] applied with the Hilbert basis $F$ gives $\mathrm{\parallel }{T}_{k}e{\mathrm{\parallel }}_{L^2(\mu )}^2={\sum }_{f\in F}\mathrm{\mid }⟨T_{k}e,f⟩{\mathrm{\mid }}^2$, and [step 4.3] identifies each summand with $\mathrm{\mid }⟨k_0,{\psi }_{f,e}⟩{\mathrm{\mid }}^2$; since every finite subset of $E\times F$ is contained in a rectangle and all terms are nonnegative, taking suprema over finite subsets gives ${\sum }_{e\in E}\mathrm{\parallel }{T}_{k}e{\mathrm{\parallel }}^2={\sum }_{f,e}\mathrm{\mid }⟨k_0,{\psi }_{f,e}⟩{\mathrm{\mid }}^2=\mathrm{\parallel }k{\mathrm{\parallel }}_2^2$ by [step 7.1]. The same computation with the roles of $E$ and $F$ interchanged, using the adjoint identity $⟨T_{k}e,f⟩=⟨e,T_k^{\ast }f⟩$ of [F13] and Parseval with respect to $E$ applied to the vectors $T_k^{\ast }f$, gives ${\sum }_{f\in F}\mathrm{\parallel }{T}_k^{\ast }f{\mathrm{\parallel }}^2=\mathrm{\parallel }k{\mathrm{\parallel }}_2^2$ as well.

Conclusion. The bases $E$ and $F$ in [step 8.1] were arbitrary Hilbert bases of $L^2(\nu ;\mathbb{C})$ and $L^2(\mu ;\mathbb{C})$, and by [step 1.1] such bases exist; by [step 8.1] every one of them realizes the value $\mathrm{\parallel }k{\mathrm{\parallel }}_2^2$, so by [F15] the operator $T_k$ is Hilbert–Schmidt with $\mathrm{\parallel }{T}_k{\mathrm{\parallel }}_{HS}=\mathrm{\parallel }k{\mathrm{\parallel }}_2$, the operator itself being independent of the representative $k_0$ by [step 4.2] and bounded with $\mathrm{\parallel }{T}_k\mathrm{\parallel }\le \mathrm{\parallel }k{\mathrm{\parallel }}_2$ by [step 5.1]. This proves all three claims.