Square-Integrable Kernels and Hilbert–Schmidt Compactness — Examples

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.

The functions $a={\mathbf{1}}_{[0,1\mathrm{/}2]}$ and $b={\mathbf{1}}_{[0,1]}$ have $\mathrm{\parallel }a{\mathrm{\parallel }}_2^2=\mu ([0,\frac{1}{2}])=\frac{1}{2}$ and $\mathrm{\parallel }b{\mathrm{\parallel }}_2^2=\nu ([0,1])=1$ by [F2], so [F3] gives that $k$ is square integrable with $\mathrm{\parallel }k{\mathrm{\parallel }}_2^2=\frac{1}{2}$, that $(T_{k}f)(x)=⟨f,b⟩{\mathbf{1}}_{[0,1\mathrm{/}2]}(x)$, as an identity of $L^2$ classes. Since $T_{k}b=a\mathrm{\ne }0$ and the range is contained in $\mathbb{C}a$, its range is exactly this one-dimensional subspace, proving rank one.

If $N$ contained a ball $B(p,r)$ with $r>0$, then [F1] and [F6] would produce a measurable rectangle $R\subseteq B(p,r)\subseteq N$ whose completed measure is $\mu (A)\nu (B)>0$, and the disjoint decomposition $N=R\cup (N\setminus R)$, together with the additivity and nonnegativity of [F4], would give $\overline{\mu \times \nu }(N)\ge \overline{\mu \times \nu }(R)>0$, contradicting $\overline{\mu \times \nu }(N)=0$; hence no ball with positive radius is contained in $N$.

Values on the left half. Let $p=(x,y)$ with $0<x<\frac{1}{2}$ and $0<y<1$. For each $j\ge 0$ the ball $B(p,1\mathrm{/}(j+1))$ is not contained in $N$ by [step 1.2], so it contains a point $q_j$ of its complement, and by [F7] the countably many points $q_j$ may be chosen simultaneously; then $q_j\to p$ by construction. For $j$ large enough $q_j$ lies in the rectangle $(0,\frac{1}{2})\times (0,1)$ on which $k=1$, and $q_j\mathrm{\notin }N$ gives $h(q_j)=k(q_j)=1$; sequential continuity [F5] therefore forces $h(p)=1$.

Values on the right half. The same argument with $(0,\frac{1}{2})$ replaced by $(\frac{1}{2},1)$, where $k=0$, and with the same null set $N$, gives $h(p)=0$ for every $p=(x,y)$ with $\frac{1}{2}<x<1$ and $0<y<1$.

Contradiction at the interface. Let $p_0=(\frac{1}{2},\frac{1}{2})$ and let $p_j=(\frac{1}{2}-1\mathrm{/}(j+3),\frac{1}{2})$ and $p_j^{\mathrm{\prime }}=(\frac{1}{2}+1\mathrm{/}(j+3),\frac{1}{2})$; both sequences converge to $p_0$ in $[0,1{]}^2$, [step 2.1] gives $h(p_j)=1$ for every $j$, and [step 2.2] gives $h(p_j^{\mathrm{\prime }})=0$ for every $j$. Sequential continuity [F5] applied to the first sequence gives $h(p_0)=1$ and applied to the second gives $h(p_0)=0$, a contradiction; therefore no continuous $h$ agrees with $k$ almost everywhere.

Steps 1.1 and 3.1 establish all the asserted properties: square integrability with $\mathrm{\parallel }k{\mathrm{\parallel }}_2^2=\frac{1}{2}$, the rank-one form of $T_k$ on the one hand, and the impossibility of a continuous representative on the other.

By [F1] the counting measures are sigma-finite and every subset of ${\mathbb{N}}^2$ is measurable, and by [F2] applied to $\mathrm{\mid }k{\mathrm{\mid }}^2$ one has $\mathrm{\parallel }k{\mathrm{\parallel }}_{L^2(\mathrm{\#}\times \mathrm{\#})}^2={\sum }_m{\sum }_n\mathrm{\mid }k(m,n){\mathrm{\mid }}^2={\sum }_n\mathrm{\mid }{a}_n{\mathrm{\mid }}^2=\mathrm{\parallel }a{\mathrm{\parallel }}_2^2<+\mathrm{\infty }$; hence $k$ is a square-integrable kernel with $\mathrm{\parallel }k{\mathrm{\parallel }}_2=\mathrm{\parallel }a{\mathrm{\parallel }}_2$, and the same computation applies to every diagonal kernel with square-summable coefficients.

For every $m$, the defining integral of the kernel operator is the counting sum $\int k(m,n)f(n)d\mathrm{\#}(n)={\sum }_{n}k(m,n)f(n)$, in which the only possibly nonzero term is $n=m$, so $(T_{k}f)(m)=a_{m}f(m)$; by [F3] this diagonal operator is well defined on $L^2(\mathrm{\#};\mathbb{C})$ and bounded with $\mathrm{\parallel }{T}_k\mathrm{\parallel }\le \mathrm{\parallel }k{\mathrm{\parallel }}_2=\mathrm{\parallel }a{\mathrm{\parallel }}_2$.

The truncated kernels. For fixed $N$ the kernel $k-k_N$ is diagonal with coefficients $a_n{\mathbf{1}}_{n>N}$, a square-summable family, so [step 1.1] applied to it gives $\mathrm{\parallel }{T}_k-T_{k_N}{\mathrm{\parallel }}_{HS}=\mathrm{\parallel }k-k_N{\mathrm{\parallel }}_{L^2(\mathrm{\#}\times \mathrm{\#})}=({\sum }_{n>N}\mathrm{\mid }{a}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ by the exact-norm part of [F3]; moreover $k_N$ is the diagonal kernel with coefficients $a_n{\mathbf{1}}_{n\le N}$, so $T_{k_N}$ is the diagonal operator with those coefficients.

Finite rank of the truncations. By [step 2.2] the range of $T_{k_N}$ is the set of sequences $a_n{\mathbf{1}}_{n\le N}f(n)e_n$, which is exactly the span of $\{e_n:n\in J\}$. Since $J\subseteq \{0,\dots ,N\}$ is finite, write its increasing enumeration as $J=\{j_0<\cdots <j_{r-1}\}$ for some $r\in \mathbb{N}$. The map $q\mapsto e_{j_q}$ with domain the von Neumann natural $r$ is an injective finite list whose image is an orthonormal family, hence is linearly independent, and it spans the range. Thus $(e_{j_q}{)}_{q<r}$ is an ordered basis of the range, and $T_{k_N}$ is compact by [F4], while its Hilbert–Schmidt norm is finite by [step 2.2].

Operator norm of the difference. Let $D:=T_k-T_{k_N}$, so that $(Df)(m)=a_m{\mathbf{1}}_{m>N}f(m)$ by [step 2.2]. For every $f$ in $L^2(\mathrm{\#};\mathbb{C})$ one has $\mathrm{\parallel }Df{\mathrm{\parallel }}^2={\sum }_{m>N}\mathrm{\mid }{a}_m{\mathrm{\mid }}^2\mathrm{\mid }f(m){\mathrm{\mid }}^2\le S_N^2\mathrm{\parallel }f{\mathrm{\parallel }}^2$ where $S_N:={\sup }_{n>N}\mathrm{\mid }{a}_n\mathrm{\mid }$ is the real number of [F7], so $\mathrm{\parallel }D\mathrm{\parallel }\le S_N$ by [F6]; conversely for each $m>N$ the vector $e_m$ has norm one by [F5] and $D{e}_m=a_m{e}_m$, so $\mathrm{\parallel }D\mathrm{\parallel }\ge \mathrm{\mid }{a}_m\mathrm{\mid }$ and hence $\mathrm{\parallel }D\mathrm{\parallel }\ge S_N$. Therefore $\mathrm{\parallel }D\mathrm{\parallel }=S_N={\sup }_{n>N}\mathrm{\mid }{a}_n\mathrm{\mid }$.

Collecting the results, $k$ is a square-integrable diagonal kernel with $\mathrm{\parallel }k{\mathrm{\parallel }}_2=\mathrm{\parallel }a{\mathrm{\parallel }}_2$, $T_k$ is the diagonal operator of [step 2.1], the truncations $T_{k_N}$ are finite rank and compact by [step 3.1], and the two exact truncation errors are [step 2.2] and [step 3.2].

By [F1] the operator $T_k$ is well defined and bounded, and relative to the Hilbert basis $E$ of $L^2(\nu ;\mathbb{C})$ its Hilbert–Schmidt square-sum is $\mathrm{\parallel }k{\mathrm{\parallel }}_2^2<+\mathrm{\infty }$; thus $T_k$ is Hilbert–Schmidt relative to $E$.

The target $L^2(\mu ;\mathbb{C})$ is a Hilbert space, hence a Banach space, and $T_k$ is compact by [F2] applied with [step 1.1], Countable Choice being available by [F4]; since $k$ was an arbitrary square-integrable kernel, every kernel operator of the pair is compact.

In particular, for the kernel of [F3] — square integrable, rank one, and without continuous representative — the hypotheses of [step 2.1] hold, so its integral operator is compact although the kernel is discontinuous; this is the sense in which the present compactness statement strictly extends the continuous-kernel results.

Steps 2.1 and 3.1 prove both assertions: compactness of $T_k$ for every square-integrable kernel over sigma-finite factors, and the specific discontinuous witness.