A square-integrable separable product kernel

Choose finite-valued measurable representatives of $a,b$. The function $(x,y)\mapsto a(x)\overline{b(y)}$ is $(\mathcal{A}\otimes \mathcal{B})$-measurable, and [F1] applied to its squared modulus gives $\mathrm{\parallel }k{\mathrm{\parallel }}_{L^2(\overline{\mu \times \nu })}^2={\int }_X({\int }_Y\mathrm{\mid }a(x){\mathrm{\mid }}^2\mathrm{\mid }b(y){\mathrm{\mid }}^{2}d\bar{\nu }(y))d\mu (x)={\int }_X\mathrm{\mid }a(x){\mathrm{\mid }}^2({\int }_Y\mathrm{\mid }b(y){\mathrm{\mid }}^{2}d\bar{\nu }(y))d\mu (x)$; [F2] rewrites the inner integral as $\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}^2$, so the value is $\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2(\mu )}^2\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}^2$, finite because both factors are $L^2$ classes. The product is measurable because its factors are measurable coordinate pullbacks and scalar multiplication and conjugation are continuous. Replacing representatives by $a\mathrm{\prime },b\mathrm{\prime }$ changes the product by $(a-a\mathrm{\prime })\bar{b}+a\mathrm{\prime }\overline{(b-b\mathrm{\prime })}$; applying the same squared-norm factorization to the two terms gives zero, so the product class is well defined.

For $f\in L^2(\nu ;\mathbb{C})$ the integrand $y\mapsto \overline{b(y)}f(y)$ is $\nu$-integrable with ${\int }_Y\mathrm{\mid }\bar{b}f\mathrm{\mid }d\nu \le \mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}$ by [F3] applied to the real nonnegative functions $\mathrm{\mid }b\mathrm{\mid },\mathrm{\mid }f\mathrm{\mid }$, which are complex $L^2$ functions with the same norms. Hence the product representative has section integral $a(x){\int }_Y\overline{b(y)}f(y)d\nu (y)=a(x)⟨f,b⟩$ wherever its sections represent the completed-product class, and [F4] identifies this function with the $L^2(\mu )$ class $T_{k}f$. Thus $(T_{k}f)(x)=a(x)\beta (f)$ for $\mu$-almost every $x$.

Hence the range of $T_k$ is contained in $\mathbb{C}\cdot a$. If $a\mathrm{\ne }0$ and $b\mathrm{\ne }0$, then $\beta (b\mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}^2)=1$, so the range equals $\mathbb{C}\cdot a$ and $(a)$ is an ordered basis. If $a=0$ or $b=0$, then [step 1.2] makes $T_k$ the zero operator, so its range has the empty ordered basis. Thus the range always has dimension at most one.

Operator norm. By [step 1.2], $\mathrm{\parallel }{T}_{k}f{\mathrm{\parallel }}_{L^2(\mu )}=\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2(\mu )}\mathrm{\mid }⟨f,b⟩\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_{L^2(\mu )}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}$; if $b\mathrm{\ne }0$ then $f_0:=b\mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}$ has norm one and $T_k{f}_0=\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}a$, so $\mathrm{\parallel }{T}_k\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2(\mu )}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}$, while if $b=0$ both sides are zero; the computation also covers $a=0$.

Hilbert–Schmidt norm. Since $k$ lies in $L^2(\overline{\mu \times \nu };\mathbb{C})$ by [step 1.1], [F4] gives $\mathrm{\parallel }{T}_k{\mathrm{\parallel }}_{HS}=\mathrm{\parallel }k{\mathrm{\parallel }}_{L^2(\overline{\mu \times \nu })}$, which is $\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2(\mu )}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2(\nu )}$ by [step 1.1]; this agrees with the operator norm of [step 2.2].

The displayed square-integrability, the rank-at-most-one form of the operator, and the two norm identities are [step 1.1], [step 1.2] with [step 2.1], and [step 2.2] with [step 3.1]; the degenerate cases $a=0$, $b=0$, and $X\times Y$ of measure zero are included in these computations, the empty-list basis of [F5] covering the zero range.