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Square-Integrable Kernels and Hilbert–Schmidt Compactness

1 · Prerequisites

2 · Summary

The page begins with the Hilbert–Schmidt definition relative to a supplied Hilbert basis: for a bounded operator $T : H \to K$ the square-sum $\sum_{e \in E} ∥ T e ∥^{2}$ is the supremum of its finite subsums, with no enumeration, ordering or countability of the basis assumed and no existence of a basis asserted. The first theorem shows under Countable Choice that this value is basis independent, computing it as the supremum of the matrix coefficients $∣ ⟨ T e, f ⟩ ∣^{2}$ over finite rectangles and identifying it with $\sum_{f \in F} ∥ T^{*} f ∥^{2}$ for a Hilbert basis $F$ of the target; this is plain double Parseval, with the nonnegative finite-supremum interchange proved rather than assumed. Hilbert–Schmidt operators are then shown to be compact: for a finite coordinate set $F$ the operator $T - T P_{F}$ has norm at most the square root of the tail sum, the truncations $T P_{F}$ are compact because the closed unit ball of the finite-dimensional span of $F$ is compact, and a sequence of tail-control sets chosen with Countable Choice puts $T$ in the norm closure of the compact operators of a Banach target.

The second half prepares the analytic input. Finite complex linear combinations of finite-measure rectangle kernels are proved dense in the product $L^{2}$ of two sigma-finite factors, first for the product measure and then for its completion: a finite-measure exhaustion reduces a set of finite product measure to one exhausted rectangle, the generating algebra of finite rectangle unions approximates it in symmetric difference on the trace of that rectangle, and the completion case passes through a base-measurable representative of every completed measurable set.

The kernel theorem then assembles the pair. Under the Axiom of Choice a completed $L^{2}$ class $k$ of the product is represented by a product-measurable kernel of the same norm, the section integral $(T_{k} f) (x) = \int_{Y} k (x, y) f (y) d ν (y)$ is shown to be defined for almost every $x$ , independent of the chosen representatives, measurable, and bounded with $∥ T_{k} ∥ \leq ∥ k ∥_{2}$ . The exact Hilbert–Schmidt norm is computed with the orthonormal family of products $f (x) \overline{e (y)}$ of a pair of Hilbert bases: rectangle kernels lie in its closed span, rectangle density places $k$ there, Bessel's inequality with a finite-Parseval lower bound identifies the norm of the expansion with $∥ k ∥_{2}^{2}$ , and Parseval rewrites that value as $\sum_{e \in E} ∥ T_{k} e ∥^{2} = \sum_{f \in F} ∥ T_{k}^{*} f ∥^{2}$ . Hilbert bases are constructed here by Zorn's lemma, so the statement is unconditional under the Axiom of Choice and never uses the later singular-value or trace-class machinery.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Hilbert–Schmidt operator and Hilbert–Schmidt norm

Definition

Throughout, $H$ and $K$ are real or complex Hilbert spaces with the pairing linear in the first argument and conjugate-linear in the second (Hilbert space), $T \in B (H, K)$ is a bounded linear operator (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators) of operator norm $∥ T ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum), and $E$ is a Hilbert basis of $H$ , that is, a complete orthonormal subset of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases). The basis $E$ is supplied as data; the definitions below are stated relative to it.

The Hilbert–Schmidt square-sum. The family $(∥ T e ∥^{2})_{e \in E}$ consists of nonnegative real numbers and is indexed by the arbitrary set $E$ . Its sum is the supremum of the finite subsums,

$s_{E} (T) : = \sum_{e \in E} ∥ T e ∥^{2} : = \sup {\sum_{e \in F} ∥ T e ∥^{2} : F \subseteq E finite} \in [0, + \infty],$

in the finite-subset-supremum convention of Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ; the empty finite subset contributes the empty sum $0$ , so the supremum is over a nonempty set and exists in $[0, + \infty]$ . No enumeration, ordering or countability of $E$ is used or asserted, and no choice is performed: the supremum ranges over the set of finite subsets of the fixed index set $E$ . For finite $E$ the supremum is the ordinary finite sum over $E$ , because all terms are nonnegative.

Hilbert–Schmidt relative to a basis. The operator $T$ is Hilbert–Schmidt relative to $E$ when $s_{E} (T) < + \infty$ , that is, when the finite subsums are bounded above in $R$ . In that case the Hilbert–Schmidt norm of $T$ relative to $E$ is the nonnegative square root

$∥ T ∥_{H S, E} : = (\sum_{e \in E} ∥ T e ∥^{2})^{1 / 2} \in [0, + \infty) .$

When $s_{E} (T) = + \infty$ , no Hilbert–Schmidt norm relative to $E$ is defined, and $T$ is not Hilbert–Schmidt relative to $E$ .

Degenerate and extreme cases. The zero operator satisfies $s_{E} (0) = 0$ for every basis $E$ , so it is Hilbert–Schmidt relative to every basis with norm $0$ . If $H = {0}$ then the empty family is a Hilbert basis of $H$ , the only finite subset is empty, and $s_{\emptyset} (T) = 0$ for the only linear operator $T : {0} \to K$ ; that operator is Hilbert–Schmidt relative to the empty basis with norm $0$ . If $H$ is finite dimensional and $E$ is finite, $s_{E} (T)$ is an ordinary finite sum of the squared norms $∥ T e ∥^{2}$ , and $T$ is automatically Hilbert–Schmidt relative to $E$ .

The basis is part of the notation. The symbol $∥ T ∥_{H S, E}$ keeps the basis in the subscript on purpose. Until the next result is proved, the phrase " $T$ is Hilbert–Schmidt" is never used without a specified Hilbert basis, and nothing here asserts that a Hilbert basis of $H$ exists: the existence of $E$ is a hypothesis of the definition, and the question of whether the finiteness of $s_{E} (T)$ and the value $∥ T ∥_{H S, E}$ depend on $E$ is taken up in The Hilbert–Schmidt norm is basis independent. In particular this definition makes no basis-existence claim and no comparison with the operator norm $∥ T ∥$ ; every such statement is proved later, where its own hypotheses are displayed.

Notation. For a finite $F \subseteq E$ we write $s_{F} (T) : = \sum_{e \in F} ∥ T e ∥^{2}$ for the finite subsum, so that $s_{E} (T) = \sup {s_{F} (T) : F \subseteq E finite}$ .

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

The Hilbert–Schmidt norm is basis independent

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ and $K$ be real or complex Hilbert spaces (Hilbert space) and let $T \in B (H, K)$ be a bounded linear operator (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum), with Hilbert adjoint $T^{*} \in B (K, H)$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities). Let $E$ be a Hilbert basis of $H$ and $F$ a Hilbert basis of $K$ (Orthonormal families, complete orthonormal systems and Hilbert bases), and let $s_{E} (T) = \sum_{e \in E} ∥ T e ∥^{2}$ and $s_{F} (T^{*}) = \sum_{f \in F} ∥ T^{*} f ∥^{2}$ be the finite-subset-supremum sums of Hilbert–Schmidt operator and Hilbert–Schmidt norm and Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ . Then:

(matrix-coefficient form) the finite-subset supremum $\sup {\sum_{e \in A} \sum_{f \in B} ∣ ⟨ T e, f ⟩ ∣^{2} : A \subseteq E, B \subseteq F finite}$ equals $s_{E} (T)$ , and it also equals $s_{F} (T^{*})$ ;
(basis independence) $s_{E} (T) = s_{E^{'}} (T)$ for every Hilbert basis $E^{'}$ of $H$ , and $s_{E} (T) = s_{F} (T^{*})$ for every Hilbert basis $F$ of $K$ ;
(membership and norms) $T$ is Hilbert–Schmidt relative to $E$ if and only if it is Hilbert–Schmidt relative to every other Hilbert basis of $H$ , and then $∥ T ∥_{H S, E} = ∥ T ∥_{H S, E^{'}} = ∥ T^{*} ∥_{H S, F}$ for all such bases $E, E^{'}$ and every Hilbert basis $F$ of $K$ ; when the common defining sum is $+ \infty$ , none of these Hilbert–Schmidt norms is defined, and $T$ is Hilbert–Schmidt relative to none of the bases.

Facts & Assumptions

Given: Countable Choice, bounded $T : H \to K$ , a Hilbert basis $E$ of $H$ and a Hilbert basis $F$ of $K$ .

[F1]

Since $F$ is a complete orthonormal family in the Hilbert space $K$ , every $y \in K$ satisfies $∥ y ∥^{2} = \sum_{f \in F} ∣ ⟨ y, f ⟩ ∣^{2}$ , the sum being the finite-subset supremum; similarly for $E$ in $H$ (Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F2]

The Hilbert adjoint satisfies $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ for all $x \in H$ , $y \in K$ , it is the unique such bounded operator, and $T^{*} f \in H$ for every $f \in K$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F3]

For a fixed finite set $A$ , fix one bijection $q : n \to A$ with a von Neumann natural $n$ (The cardinality $∣ A ∣$ of a finite set). Given nonempty sets $C_{e}$ for $e \in A$ , apply Every natural-number-indexed list of nonempty sets has a choice function on its family of values to the function $k \mapsto C_{q (k)}$ on $n$ . Its choice function $c$ on the set of values yields $b_{e} = c (C_{e}) \in C_{e}$ . This transports finite choice to this fixed $A$ ; no enumeration of the entire basis or simultaneous choice of enumerations is asserted.

[F4]

For a nonnegative family $(c_{e})_{e \in E}$ the sum is the supremum of the finite subsums, is monotone in the family, and satisfies $\sum_{e \in E} c_{e} = \sum_{e \in F} c_{e} + \sum_{e \in E ∖ F} c_{e}$ for finite $F \subseteq E$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[F5]

Countable Choice is the hypothesis under which Parseval and the adjoint interface are available (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F6]

The operator $T$ is Hilbert–Schmidt relative to $E$ exactly when $s_{E} (T) < + \infty$ , and then $∥ T ∥_{H S, E} = (s_{E} (T))^{1 / 2}$ ; the same definitions apply to $E^{'}$ and to $T^{*}$ with respect to $F$ (Hilbert–Schmidt operator and Hilbert–Schmidt norm).

Proof

technique · direct

Given: Countable Choice, bounded $T : H \to K$ , Hilbert bases $E$ of $H$ and $F$ of $K$ , and the nonnegative numbers $a_{e, f} : = ∣ ⟨ T e, f ⟩ ∣^{2}$ .

1.1

For every $e \in E$ the vector $T e$ lies in $K$ , so [F1] applied in $K$ to the Hilbert basis $F$ gives $∥ T e ∥^{2} = \sum_{f \in F} a_{e, f}$ , a supremum over finite $B \subseteq F$ .

F1F5

1.2

For every $f \in F$ the vector $T^{*} f$ lies in $H$ by [F2], so [F1] applied in $H$ to the Hilbert basis $E$ gives $∥ T^{*} f ∥^{2} = \sum_{e \in E} ∣ ⟨ T^{*} f, e ⟩ ∣^{2}$ ; since $⟨ T^{*} f, e ⟩ = \overline{⟨ e, T^{*} f ⟩}$ and $⟨ T e, f ⟩ = ⟨ e, T^{*} f ⟩$ by [F2], the moduli agree: $∣ ⟨ T^{*} f, e ⟩ ∣ = ∣ ⟨ T e, f ⟩ ∣ = a_{e, f}^{1 / 2}$ .

F1F2

2.1

The iterated suprema agree with the rectangle supremum. For every finite $A \subseteq E$ the identity $\sup {\sum_{e \in A} \sum_{f \in B} a_{e, f} : B \subseteq F finite} = \sum_{e \in A} \sum_{f \in F} a_{e, f}$ holds. If $A = \emptyset$ , both sides are zero; hence assume $A \neq \emptyset$ . Each row sum is the finite number $∥ T e ∥^{2}$ by step 1.1. The left side is at most the right side because each $B$ gives a subsum, while for the reverse inequality fix a real $η > 0$ and, using [F3], choose for each $e \in A$ a finite $B_{e} \subseteq F$ with $\sum_{f \in B_{e}} a_{e, f} > \sum_{f \in F} a_{e, f} - η$ ; then $B : = ⋃_{e \in A} B_{e}$ is finite and $\sum_{e \in A} \sum_{f \in B} a_{e, f} \geq \sum_{e \in A} \sum_{f \in B_{e}} a_{e, f} > \sum_{e \in A} \sum_{f \in F} a_{e, f} - ∣ A ∣ η$ . Hence the supremum over all finite rectangles $A \times B$ equals $\sup_{A} \sum_{e \in A} \sum_{f \in F} a_{e, f} = \sup_{A} \sum_{e \in A} ∥ T e ∥^{2} = s_{E} (T)$ by [step 1.1] and [F4]; and since every finite $S \subseteq E \times F$ is contained in a rectangle while subsums are monotone, this rectangle supremum is also the supremum over all finite subsets of $E \times F$ .

step 1.1F3F4algebra

2.2

The same computation with the adjoint. By [step 1.2] and the same argument with $E$ and $F$ interchanged, $s_{F} (T^{*}) = \sup_{B} \sum_{f \in B} \sum_{e \in E} ∣ ⟨ T^{*} f, e ⟩ ∣^{2} = \sup_{A, B} \sum_{e \in A} \sum_{f \in B} ∣ ⟨ T^{*} f, e ⟩ ∣^{2} = \sup_{A, B} \sum_{e \in A} \sum_{f \in B} a_{e, f}$ , the last equality by the modulus identity of [step 1.2]; the middle supremum is over finite rectangles, and it is the finite-subset supremum of $E \times F$ because finite subsets of a product lie in rectangles.

step 1.2F3F4

3.1

Conclusion of the matrix-coefficient form. Steps 2.1 and 2.2 identify the rectangle supremum of claim 1 with $s_{E} (T)$ and with $s_{F} (T^{*})$ respectively, so that supremum equals both sums; this proves claim 1.

step 2.1step 2.2

4.1

Basis independence. Let $E^{'}$ be any Hilbert basis of $H$ . Applying [step 3.1] to the pair $(E^{'}, F)$ gives $s_{E^{'}} (T) = s_{F} (T^{*})$ , and applying it to $(E, F)$ gives $s_{E} (T) = s_{F} (T^{*})$ for the same basis $F$ of $K$ ; hence $s_{E} (T) = s_{E^{'}} (T)$ , and also $s_{E} (T) = s_{F} (T^{*})$ for every Hilbert basis $F$ of $K$ , both equalities holding in $[0, + \infty]$ .

step 3.1

5.1

Membership and the norms. By [step 4.1] the sums $s_{E} (T)$ , $s_{E^{'}} (T)$ and $s_{F} (T^{*})$ all equal one extended real number, so they are finite simultaneously; when the common value is finite, taking nonnegative square roots gives $∥ T ∥_{H S, E} = ∥ T ∥_{H S, E^{'}} = ∥ T^{*} ∥_{H S, F}$ by [F6], and when it is $+ \infty$ none of the three norms is defined and $T$ is Hilbert–Schmidt relative to no Hilbert basis of $H$ . This is claim 3.

step 4.1F6∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Hilbert–Schmidt operators are compact

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ and $K$ be real or complex Hilbert spaces (Hilbert space), let $T \in B (H, K)$ be a bounded linear operator (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum), let $E$ be a Hilbert basis of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), and assume that $T$ is Hilbert–Schmidt relative to $E$ (Hilbert–Schmidt operator and Hilbert–Schmidt norm), that is, $s_{E} (T) = \sum_{e \in E} ∥ T e ∥^{2} < + \infty$ in the finite-subset-supremum convention of Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ . For finite $F \subseteq E$ let $P_{F} x : = \sum_{e \in F} ⟨ x, e ⟩ e$ be the coordinate projection (The finite Bessel inequality and best approximation by a finite orthonormal family). Then:

(finite-rank pieces) $P_{F}$ is a bounded linear operator on $H$ with $∥ P_{F} ∥ \leq 1$ , its range lies in the finite-dimensional subspace $span {e : e \in F}$ , and $T P_{F}$ is compact (Compact linear operator);
(norm estimate) $∥ T - T P_{F} ∥ \leq (\sum_{e \in E ∖ F} ∥ T e ∥^{2})^{1 / 2}$ for every finite $F \subseteq E$ , and the right-hand side is arbitrarily small for suitable finite $F$ ;
(compactness) $T$ is a compact operator.

Facts & Assumptions

Given: Countable Choice, bounded $T : H \to K$ , a Hilbert basis $E$ of $H$ with $s_{E} (T) < + \infty$ , and finite sets $F \subseteq G \subseteq E$ .

[F1]

$T$ is compact exactly when $\overline{T ({\overline{B}}_{H})}$ is a compact subset of $K$ , where ${\overline{B}}_{H} = {x \in H : ∥ x ∥ \leq 1}$ (Compact linear operator).

[F2]

For finite $F \subseteq E$ the vector $P_{F} x = \sum_{e \in F} ⟨ x, e ⟩ e$ lies in the span of ${e : e \in F}$ , $∥ P_{F} x ∥^{2} = \sum_{e \in F} ∣ ⟨ x, e ⟩ ∣^{2}$ , the residual $x - P_{F} x$ is orthogonal to every $e \in F$ , and $∥ x - P_{F} x ∥^{2} = ∥ x ∥^{2} - \sum_{e \in F} ∣ ⟨ x, e ⟩ ∣^{2} \leq ∥ x ∥^{2}$ (The finite Bessel inequality and best approximation by a finite orthonormal family).

[F3]

The finite-subset net $(P_{G} x)$ over the finite subsets $G \subseteq E$ , directed by inclusion, converges to $x$ for every $x \in H$ (Fourier expansion in a Hilbert space, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F4]

Since $s_{E} (T) < + \infty$ , for every real $ε > 0$ there is a finite $F \subseteq E$ with $\sum_{e \in E ∖ F} ∥ T e ∥^{2} < ε$ ; for finite $F \subseteq G$ the finite subsum over $G ∖ F$ is at most the sum over $E ∖ F$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[F5]

A finite set $F$ satisfies $F \approx n$ for some $n \in N$ (Finite, countably infinite, countable, uncountable); an orthonormal family is linearly independent, so the image of any enumeration of $F$ is a basis of $span {e : e \in F}$ ; a normed space that admits an ordered basis of finite length has compact closed unit ball (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Orthonormal families, complete orthonormal systems and Hilbert bases, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , The closed unit ball is compact if and only if the normed space is finite-dimensional).

[F6]

A continuous image of a compact set is compact, and a closed subset of a compact metric space is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

[F7]

A bounded linear operator is continuous, so its restriction to any normed subspace is continuous, and every $T P_{F}$ is bounded and linear (For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F8]

A Hilbert space is a Banach space (Hilbert space, Banach space), and under Countable Choice a norm limit of compact operators into a Banach space is compact (Norm limit of compact operators is compact).

[F9]

Countable Choice allows one witness to be selected from each of countably many nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: Countable Choice, bounded $T : H \to K$ , a Hilbert basis $E$ of $H$ with $s_{E} (T) < + \infty$ , and a finite $F \subseteq E$ .

1.1

For every $x \in H$ , [F2] gives $∥ P_{F} x ∥^{2} = \sum_{e \in F} ∣ ⟨ x, e ⟩ ∣^{2} \leq ∥ x ∥^{2}$ and $P_{F} x \in span {e : e \in F}$ ; thus $P_{F}$ is linear by construction, bounded with $∥ P_{F} ∥ \leq 1$ , and its range lies in $span {e : e \in F}$ .

F2algebra

1.2

Since $F$ is finite, [F5] fixes $n \in N$ and a bijection $σ$ from $n$ onto $F$ ; the list $e_{σ (0)}, \dots, e_{σ (n - 1)}$ is injective because the family is orthonormal, and its image spans $Z : = span {e : e \in F}$ , so it is an ordered basis of $Z$ of finite length; therefore ${\overline{B}}_{Z}$ is compact by [F5].

2.1

The set $P_{F} ({\overline{B}}_{H})$ is contained in ${\overline{B}}_{Z}$ by [step 1.1], since $∥ P_{F} x ∥ \leq ∥ x ∥ \leq 1$ and $P_{F} x \in Z$ ; the restriction of $T$ to $Z$ is continuous by [F7], so $T ({\overline{B}}_{Z})$ is compact by [step 1.2] and [F6]; as $T P_{F} ({\overline{B}}_{H}) = T (P_{F} ({\overline{B}}_{H})) \subseteq T ({\overline{B}}_{Z})$ , its closure is a closed subset of the compact set $T ({\overline{B}}_{Z})$ , hence compact by [F6], and $T P_{F}$ is compact by [F1].

step 1.1step 1.2F1F6F7

2.2

For finite $F \subseteq G \subseteq E$ and $x \in H$ we have $P_{F} P_{G} x = P_{F} x$ , because $⟨ P_{G} x, e ⟩ = ⟨ x, e ⟩$ for $e \in G$ ; hence $(I - P_{F}) P_{G} x = P_{G} x - P_{F} x = \sum_{e \in G ∖ F} ⟨ x, e ⟩ e$ and, by linearity of $T$ , $T (I - P_{F}) P_{G} x = \sum_{e \in G ∖ F} ⟨ x, e ⟩ T e$ ; the triangle inequality and the finite Cauchy–Schwarz inequality give $∥ T (I - P_{F}) P_{G} x ∥ \leq (\sum_{e \in G ∖ F} ∣ ⟨ x, e ⟩ ∣^{2})^{1 / 2} (\sum_{e \in G ∖ F} ∥ T e ∥^{2})^{1 / 2} \leq ∥ x ∥ (\sum_{e \in E ∖ F} ∥ T e ∥^{2})^{1 / 2}$ , where the last step uses [F2] for the coefficient factor and [F4] for the tail factor.

step 1.1F2F4algebra

3.1

As $G$ runs over the finite subsets of $E$ containing $F$ , the net $(I - P_{F}) P_{G} x = (P_{G} x - P_{F} x)$ converges to $(I - P_{F}) x$ , by [F3] and the boundedness of $P_{F}$ from [step 1.1]; the continuous operator $T$ of [F7] therefore carries this net to a net converging to $T (I - P_{F}) x = (T - T P_{F}) x$ , while [step 2.2] bounds every term of that net by $∥ x ∥ (\sum_{e \in E ∖ F} ∥ T e ∥^{2})^{1 / 2}$ ; the norm being continuous, the limit obeys the same bound, and taking the supremum over $∥ x ∥ \leq 1$ gives $∥ T - T P_{F} ∥ \leq (\sum_{e \in E ∖ F} ∥ T e ∥^{2})^{1 / 2}$ .

step 1.1step 2.2F3F7algebra

4.1

Given a real $ε > 0$ , [F4] provides a finite $F \subseteq E$ with $\sum_{e \in E ∖ F} ∥ T e ∥^{2} < ε^{2}$ ; then $∥ T - T P_{F} ∥ < ε$ by [step 3.1], and $T P_{F}$ is compact by [step 2.1], so for every positive tolerance there is a compact operator $T P_{F}$ within that tolerance of $T$ .

step 2.1step 3.1F4

4.2

By [F9] applied to the countably many nonempty sets of finite $F \subseteq E$ satisfying $\sum_{e \in E ∖ F} ∥ T e ∥^{2} < (n + 1)^{- 2}$ for $n \in N$ — each nonempty by [F4] — there is a sequence $(F_{n})$ of finite subsets of $E$ with these tails; then $∥ T - T P_{F_{n}} ∥ \leq (n + 1)^{- 1} \to 0$ by [step 3.1], and each $T P_{F_{n}}$ is compact by [step 2.1].

step 2.1step 3.1F4F9choose

5.1

The target $K$ is a Banach space by [F8], so the norm limit $T$ of the compact operators $T P_{F_{n}}$ is compact by [F8]; this proves claim 3, while claims 1 and 2 are [step 1.1] with [step 2.1] and [step 3.1] with [step 4.1].

step 2.1step 3.1step 4.2F8∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Product rectangle kernels are dense in product L two

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(X, A, μ)$ and $(Y, B, ν)$ be sigma-finite measure spaces (Finite, sigma-finite, and semifinite measures), let $μ \times ν$ be the product measure on the product sigma-algebra $A \otimes B$ (The product sigma-algebra and its finite iterates, For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique), and let $\overline{μ \times ν}$ be its completion (The completed product measure). Write $1_{A} (x) 1_{B} (y)$ for the rectangle kernel of a measurable rectangle $A \times B$ (Measurable rectangles in a product of measurable spaces) with $μ (A) < + \infty$ and $ν (B) < + \infty$ . Then the set of finite complex linear combinations of such rectangle kernels is dense both

in $L^{2} (μ \times ν; C)$ , and
in $L^{2} (\overline{μ \times ν}; C)$ (The space $L^{p} (μ)$ as the quotient by null functions).

Facts & Assumptions

Given: Countable Choice and two sigma-finite measure spaces $(X, A, μ)$ and $(Y, B, ν)$ .

[F1]

Sigma-finiteness provides a sequence $(X_{k})$ in $A$ with $μ (X_{k}) < + \infty$ and $X = ⋃_{k} X_{k}$ , and likewise a sequence $(Y_{l})$ for $ν$ ; finite unions of sets of finite measure again have finite measure (Finite, sigma-finite, and semifinite measures, Finite and countable subadditivity of measures).

[F2]

The product measure is the unique measure on $A \otimes B$ with $(μ \times ν) (A \times B) = μ (A) ν (B)$ ; it is sigma-finite, and its completion $\overline{μ \times ν}$ extends it, agreeing with it on every $A \otimes B$ -measurable set (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, Assuming countable choice, every measure space has a unique complete extension to its completion).

[F3]

For an increasing sequence of measurable sets the measure of the union is the supremum of the measures, and measures are finitely and countably subadditive (Continuity from below for measures, Finite and countable subadditivity of measures).

[F4]

Finite disjoint unions of measurable rectangles form an algebra of subsets of $X \times Y$ generating $A \otimes B$ ; in particular a finite union of measurable rectangles is a finite disjoint union of measurable rectangles (Finite disjoint unions of measurable rectangles form an algebra generating the product sigma-algebra, Algebras of subsets).

[F5]

If a finite measure space carries an algebra generating its sigma-algebra, then every measurable set is approximable in symmetric difference by an element of that algebra (Approximation in symmetric difference by a generating algebra).

[F6]

Complex finite simple functions with finite-measure nonzero sets are dense in $L^{p}$ for every exponent $1 \leq p < \infty$ , on every measure space (Complex finite-simple and smooth compact-support density for finite p).

[F7]

Under Countable Choice, a function measurable for a completion is almost everywhere equal to a function measurable for the original sigma-algebra, and the completion of a measure agrees with it on the original measurable sets (A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra, Assuming countable choice, every measure space has a unique complete extension to its completion).

[F8]

If a measurable set $E$ has $ρ (E) < + \infty$ , then $1_{E}$ has an $L^{2}$ class and $∥ 1_{E} ∥_{2}^{2} = ρ (E)$ . More generally, if measurable $E, C$ satisfy $ρ (E △ C) < + \infty$ , then $1_{E} - 1_{C}$ has an $L^{2}$ class with squared norm $ρ (E △ C)$ (The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · direct

Given: Countable Choice, sigma-finite $(X, A, μ)$ and $(Y, B, ν)$ , and the increasing finite-measure exhaustions $X_{n} : = ⋃_{k \leq n} X_{k}$ , $Y_{n} : = ⋃_{l \leq n} Y_{l}$ , $Z_{n} : = X_{n} \times Y_{n}$ of [F1], with $ρ : = μ \times ν$ .

1.1

Each $Z_{n}$ is a measurable rectangle of finite product measure, $Z_{n} \subseteq Z_{n + 1}$ , and $⋃_{n} Z_{n} = X \times Y$ ; moreover $ρ (E \cap Z_{n}) ↑ ρ (E)$ for every $E \in A \otimes B$ by [F3], so for $ρ (E) < + \infty$ and real $δ > 0$ there is $n$ with $ρ (E ∖ Z_{n}) < δ / 2$ .

F1F2F3

1.2

A local algebra on each exhausted rectangle. Fix $n$ and let $G_{n} : = {F \subseteq Z_{n} : F = E \cap Z_{n} for some E \in A \otimes B}$ be the trace sigma-algebra, and let $C_{n}$ be the family of finite unions of rectangles $A \times B$ with $A \in A$ , $A \subseteq X_{n}$ , $B \in B$ , $B \subseteq Y_{n}$ . Then $C_{n}$ is an algebra of subsets of $Z_{n}$ : it contains $\emptyset$ , it is closed under finite unions by definition, and for $A \times B \subseteq Z_{n}$ the complement in $Z_{n}$ is $((X_{n} ∖ A) \times Y_{n}) \cup (A \times (Y_{n} ∖ B))$ , a union of two rectangles inside $Z_{n}$ , while complements of finite unions follow by De Morgan and the closure of products of intersections; every element of $C_{n}$ is a finite disjoint union of rectangles by [F4]. Furthermore $σ (C_{n}) = G_{n}$ : the inclusion $\subseteq$ is clear since each generator of $C_{n}$ lies in $G_{n}$ , and conversely ${E \in A \otimes B : E \cap Z_{n} \in σ (C_{n})}$ is a sigma-algebra containing every measurable rectangle, because $(A \times B) \cap Z_{n} = (A \cap X_{n}) \times (B \cap Y_{n}) \in C_{n}$ , hence it contains $A \otimes B$ and therefore $G_{n}$ . Finally the trace measure $ρ_{n} (F) : = ρ (F)$ on $G_{n}$ is a finite measure because $ρ_{n} (Z_{n}) = μ (X_{n}) ν (Y_{n}) < + \infty$ by [F2].

F1F2F4

2.1

Approximation of sets of finite product measure. Let $E \in A \otimes B$ with $ρ (E) < + \infty$ and let $δ > 0$ . Choose $n$ with $ρ (E ∖ Z_{n}) < δ / 2$ by [step 1.1]; then $E \cap Z_{n} \in G_{n}$ , so [F5] applied to the finite measure space $(Z_{n}, G_{n}, ρ_{n})$ and its generating algebra $C_{n}$ of [step 1.2] gives $C \in C_{n}$ with $ρ_{n} ((E \cap Z_{n}) △ C) < δ / 2$ . Since $E △ C \subseteq (E ∖ Z_{n}) \cup ((E \cap Z_{n}) △ C)$ , [F3] gives $ρ (E △ C) < δ$ , and $C$ is a finite union of rectangles with $μ (A) < + \infty$ , $ν (B) < + \infty$ as a subset of $Z_{n}$ .

step 1.1step 1.2F3F5

3.1

Indicator approximation. For $E$ and $C$ as in [step 2.1], $1_{C}$ is a finite sum of rectangle kernels by [F4] and [step 2.1], and by [F8] the difference of the classes of $1_{E}$ and $1_{C}$ has squared $L^{2} (ρ)$ -norm $ρ (E △ C) < δ$ .

step 2.1F4F8

4.1

Density in the product space. Let $h$ be a class in $L^{2} (ρ; C)$ and let $ε > 0$ . By [F6] with $p = 2$ there is a complex finite simple function $s$ with $∥ h - s ∥_{2} < ε / 2$ . If $s = 0$ , take the zero rectangle combination. Otherwise write $s = \sum_{j < m} c_{j} 1_{E_{j}}$ using only its nonzero values, so every $c_{j} \neq 0$ and every $E_{j}$ has finite measure; put $B : = \sum_{j < m} ∣ c_{j} ∣ > 0$ and $δ : = (ε / (2 B))^{2}$ . For each $j$ , [step 3.1] gives a set $C_{j}$ that is a finite union of finite-measure rectangles and satisfies $∥ 1_{E_{j}} - 1_{C_{j}} ∥_{2} < ε / (2 B)$ . Then $R : = \sum_{j} c_{j} 1_{C_{j}}$ is a finite complex linear combination of rectangle kernels and the triangle inequality gives $∥ h - R ∥_{2} \leq ∥ h - s ∥_{2} + B \max_{j} ∥ 1_{E_{j}} - 1_{C_{j}} ∥_{2} < ε$ .

step 3.1F6

5.1

Density in the completed space. Let $h$ be a class in $L^{2} (\overline{μ \times ν}; C)$ and let $ε > 0$ . By [F6] applied in the completed measure space there is a complex finite simple function $s$ with $∥ h - s ∥_{\overline{μ \times ν}} < ε / 2$ . If $s = 0$ , take the zero rectangle combination. Otherwise, using the same nonzero-value representation and coefficient bookkeeping as in [step 4.1], write $s = \sum_{j < m} c_{j} 1_{E_{j}}$ with every $c_{j} \neq 0$ and every $E_{j}$ of finite completed measure, and put $B : = \sum_{j < m} ∣ c_{j} ∣ > 0$ and $δ^{'} : = (ε / (2 B))^{2}$ . For each $j$ , [F7] applied to the indicator of $E_{j}$ provides $A_{j} \in A \otimes B$ with $\overline{μ \times ν} (E_{j} △ A_{j}) = 0$ , hence $ρ (A_{j}) = \overline{μ \times ν} (A_{j}) = \overline{μ \times ν} (E_{j}) < + \infty$ by [F2]. By [step 2.1] there is a set $C_{j}$ that is a finite union of finite-measure rectangles with $ρ (A_{j} △ C_{j}) < δ^{'}$ , so the classes satisfy $∥ 1_{E_{j}} - 1_{C_{j}} ∥_{\overline{μ \times ν}}^{2} = \overline{μ \times ν} (E_{j} △ C_{j}) \leq \overline{μ \times ν} (E_{j} △ A_{j}) + ρ (A_{j} △ C_{j}) < δ^{'}$ by [F3] and [F8]. The triangle inequality in the completed space therefore gives $∥ h - \sum_{j} c_{j} 1_{C_{j}} ∥_{\overline{μ \times ν}} < ε$ , and the approximant is a finite complex linear combination of rectangle kernels.

step 2.1step 4.1F2F3F6F7F8

6.1

Steps 4.1 and 5.1 give the two density assertions of the statement, for an arbitrary class and arbitrary positive tolerance in each of the two spaces; all approximations are finite complex linear combinations of rectangle kernels with $μ (A) < + \infty$ and $ν (B) < + \infty$ .

step 4.1step 5.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

L two kernels give Hilbert–Schmidt operators

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $(X, A, μ)$ and $(Y, B, ν)$ be sigma-finite measure spaces (Finite, sigma-finite, and semifinite measures), let $μ \times ν$ be their product measure (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique), let $\overline{μ \times ν}$ be its completion (The completed product measure), and let $k$ be a class in $L^{2} (\overline{μ \times ν}; C)$ . Write $L^{2} (ν; C)$ and $L^{2} (μ; C)$ for the complex $L^{2}$ spaces of the original measures, with $⟨ f, g ⟩ = \int f \overline{g}$ linear in the first variable (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz, $L^{2}$ with the integral pairing is a Hilbert space). Then:

(representative of finite norm) there is a $(A \otimes B)$ -measurable representative $k_{0}$ of $k$ with $\int ∣ k_{0} ∣^{2} d (μ \times ν) < + \infty$ , and its norm equals $∥ k ∥_{2}$ ;
(the kernel operator) for every $f \in L^{2} (ν; C)$ the section integral $(T_{k} f) (x) : = \int_{Y} k_{0} (x, y) f (y) d ν (y)$ converges for $μ$ -almost every $x$ , agrees almost everywhere with a $μ$ -measurable function, and its class in $L^{2} (μ; C)$ depends only on the classes of $f$ and $k$ ; the resulting map $T_{k} : L^{2} (ν; C) \to L^{2} (μ; C)$ is linear and bounded with $∥ T_{k} ∥ \leq ∥ k ∥_{2}$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum);
(exact Hilbert–Schmidt norm) every Hilbert space admits a Hilbert basis under the Axiom of Choice (Orthonormal families, complete orthonormal systems and Hilbert bases), and for every Hilbert basis $E$ of $L^{2} (ν; C)$ and every Hilbert basis $F$ of $L^{2} (μ; C)$ , $\sum_{e \in E} ∥ T_{k} e ∥^{2} = \sum_{f \in F} ∥ T_{k}^{*} f ∥^{2} = ∥ k ∥_{2}^{2},$ the sums being finite-subset suprema (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ); consequently $T_{k}$ is Hilbert–Schmidt relative to every such basis and $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ (Hilbert–Schmidt operator and Hilbert–Schmidt norm, The Hilbert–Schmidt norm is basis independent).

The operator is interpreted through the representative $k_{0}$ of claim 1, and claim 2 asserts that no other choice of representative changes the resulting classes; this is the sense in which a kernel of the completed product defines an operator on the $L^{2}$ spaces of the original factors.

Facts & Assumptions

Given: The Axiom of Choice, sigma-finite $(X, A, μ)$ and $(Y, B, ν)$ , their product $ρ : = μ \times ν$ and completed product $\overline{ρ}$ , and a class $k \in L^{2} (\overline{ρ}; C)$ .

[F2]

The product measure is the unique measure on $A \otimes B$ with $(μ \times ν) (A \times B) = μ (A) ν (B)$ on measurable rectangles and is sigma-finite; the completed product is its completion, so it is a complete measure extending $ρ$ and agrees with $ρ$ on every $(A \otimes B)$ -measurable set (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, The completed product measure, Assuming countable choice, every measure space has a unique complete extension to its completion).

[F3]

Tonelli applies to a nonnegative $(A \otimes B)$ -measurable function $g$ : the section-integral function is measurable and $\int g d ρ = \int_{X} (\int_{Y} g_{x} d ν) d μ$ ; sections of $(A \otimes B)$ -measurable sets are measurable, $ρ$ is countably additive and monotone, and a nonnegative measurable function has zero integral exactly when it vanishes almost everywhere (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Every section of a product-measurable set is measurable, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[F4]

Fubini applies to every $g \in L^{1} (ρ; C)$ : for $μ$ -almost every $x$ the section $g_{x}$ is $ν$ -integrable, the section integrals form a $μ$ -integrable function after zero extension, and the iterated integral equals $\int g d ρ$ (Fubini's theorem for L^1 functions on a sigma-finite product).

[F5]

On complex $L^{2}$ the pairing $⟨ h_{1}, h_{2} ⟩ = \int h_{1} \overline{h_{2}}$ is representative-independent, linear in the first variable, conjugate-symmetric and positive definite, satisfies Cauchy–Schwarz, and complex $L^{2}$ of a measure space is a Hilbert space under Countable Choice (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz, $L^{2}$ with the integral pairing is a Hilbert space).

[F6]

Finite complex simple functions with finite-measure nonzero sets are dense in complex $L^{2}$ (Complex finite-simple and smooth compact-support density for finite p).

[F7]

Finite complex linear combinations of finite-measure rectangle kernels are dense in $L^{2} (ρ; C)$ and in $L^{2} (\overline{ρ}; C)$ (Product rectangle kernels are dense in product L two).

[F8]

Under Countable Choice every $\overline{ρ}$ -measurable real function agrees $\overline{ρ}$ -almost everywhere with an $(A \otimes B)$ -measurable function (A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra).

[F9]

The nonnegative integral is the supremum of the integrals of the nonnegative simple functions it dominates, and the integral of a nonnegative simple function is the corresponding finite sum of set values (The nonnegative Lebesgue integral, The integral of a nonnegative simple function).

[F10]

For a Hilbert basis $G$ of a Hilbert space and $h$ in it, $∥ h ∥^{2} = \sum_{g \in G} ∣ ⟨ h, g ⟩ ∣^{2}$ , and the finite-subset net of partial sums converges to $h$ (Parseval equivalences for an orthonormal family, Fourier expansion in a Hilbert space).

[F11]

For an orthonormal family $(ψ_{i})$ in an inner-product space and $h$ in it, $\sum_{i} ∣ ⟨ h, ψ_{i} ⟩ ∣^{2} \leq ∥ h ∥^{2}$ ; if $h$ lies in the span of a finite orthonormal family $A$ , then $\sum_{ψ \in A} ∣ ⟨ h, ψ ⟩ ∣^{2} = ∥ h ∥^{2}$ (the finite Parseval identity) (The Bessel inequality for an arbitrary orthonormal family, The finite Bessel inequality and best approximation by a finite orthonormal family).

[F12]

Assuming Choice, every nonempty poset in which every chain has an upper bound has a maximal element; and in a Hilbert space a proper closed subspace has a nonzero orthogonal vector, every vector decomposing as $h = m + n$ with $m$ in the subspace and $n$ orthogonal to it (Zorn's lemma, Orthogonal decomposition by a closed subspace).

[F13]

Choice implies Countable Choice and Dependent Choice; Countable Choice is the hypothesis consumed by the completion-representative interface, the Hilbert structure of $L^{2}$ , and Parseval, and Riesz representation supplies the Hilbert adjoint $T_{k}^{*}$ with $⟨ T_{k} e, f ⟩ = ⟨ e, T_{k}^{*} f ⟩$ (AC supplies the countable and dependent choices used in Banach integration, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F14]

A pointwise almost-everywhere limit of a sequence of measurable functions is measurable when represented by its limit superior (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable).

[F15]

The operator $T_{k}$ is Hilbert–Schmidt relative to a Hilbert basis $G$ of $L^{2} (ν; C)$ exactly when $\sum_{e \in G} ∥ T_{k} e ∥^{2} < + \infty$ , its Hilbert–Schmidt norm is then the square root of that sum, and the finiteness and the value are independent of $G$ (Hilbert–Schmidt operator and Hilbert–Schmidt norm, The Hilbert–Schmidt norm is basis independent).

Proof

technique · direct

Given: Choice, sigma-finite $(X, A, μ)$ , $(Y, B, ν)$ , the product measure $ρ = μ \times ν$ with completion $\overline{ρ}$ , a class $k \in L^{2} (\overline{ρ}; C)$ , and the complex $L^{2}$ spaces $L^{2} (ν; C)$ , $L^{2} (μ; C)$ with their first-variable-linear pairings.

1.1

Hilbert bases exist. For a real or complex Hilbert space $H$ , let $P$ be the set of orthonormal subsets of $H$ ordered by inclusion; $P$ is nonempty because $\emptyset$ is orthonormal, and the union of a chain in $P$ 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 $x \notin M$ would give $x = m + n$ with $m \in M$ and $n$ orthogonal to $M$ and $n \neq 0$ ; then $e : = n / ∥ n ∥$ would satisfy $∥ e ∥ = 1$ 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 $L^{2} (ν; C)$ and $L^{2} (μ; C)$ , being Hilbert spaces by [F5], admit Hilbert bases.

F5F12F13

1.2

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 $(A \otimes B)$ -measurable complex function $k_{0}$ with $k_{0} = k$ $\overline{ρ}$ -almost everywhere. Then $g : = ∣ k_{0} ∣^{2}$ is $(A \otimes B)$ -measurable and nonnegative. For every nonnegative simple $(A \otimes B)$ -measurable $s \leq g$ , the integrals against $ρ$ and $\overline{ρ}$ agree, because the two measures agree on the finitely many base-measurable level sets occurring in $s$ by [F2] and [F9]. Conversely, let $t \leq g$ be a nonnegative $\overline{ρ}$ -measurable simple function and write its positive-level representation as $t = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ , 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 $(A \otimes B)$ -measurable and $N_{j}$ is contained in a base-measurable $ρ$ -null set. Since $A_{j} \subseteq E_{j}$ , the sets $A_{j}$ remain pairwise disjoint, and $s : = \sum_{j = 1}^{m} c_{j} 1_{A_{j}}$ is a base-measurable simple function satisfying $0 \leq s \leq t \leq g$ . Moreover [F2] and [F9] give $\int s d ρ = \sum_{j} c_{j} ρ (A_{j}) = \sum_{j} c_{j} \overline{ρ} (E_{j}) = \int t d \overline{ρ}$ . 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 $\int g d ρ = \int g d \overline{ρ}$ . Finally $∣ k_{0} ∣ = ∣ k ∣$ $\overline{ρ}$ -almost everywhere, so the norm of the class $k$ gives $\int g d \overline{ρ} = \int ∣ k ∣^{2} d \overline{ρ} < + \infty$ . Hence $k_{0} \in L^{2} (ρ; C)$ and $∥ k_{0} ∥_{L^{2} (ρ)} = ∥ k ∥_{2}$ .

F2F8F9F13

2.1

Almost every section is square integrable. By [F3] applied to the nonnegative function $(x, y) \mapsto ∣ k_{0} (x, y) ∣^{2}$ , whose integral is $∥ k_{0} ∥_{L^{2} (ρ)}^{2} < + \infty$ by [step 1.2], the function $x \mapsto \int_{Y} ∣ k_{0} (x, y) ∣^{2} d ν (y)$ is measurable with finite integral; hence $∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)} < + \infty$ for $μ$ -almost every $x$ , and $\int_{X} ∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)}^{2} d μ (x) = ∥ k_{0} ∥_{L^{2} (ρ)}^{2}$ .

step 1.2F3

3.1

The section integral exists almost everywhere and is bounded by the section norm. Let $f \in L^{2} (ν; C)$ . For every $x$ with $∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)} < + \infty$ the section $k_{0} (x, \cdot)$ is $B$ -measurable by [F3] and $f$ is $ν$ -measurable, so $y \mapsto k_{0} (x, y) f (y)$ is $ν$ -measurable, and Cauchy–Schwarz in $L^{2} (ν)$ by [F5] gives $\int_{Y} ∣ k_{0} (x, y) f (y) ∣ d ν (y) \leq ∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)} ∥ f ∥_{L^{2} (ν)} < + \infty$ together with $∣ (T_{k_{0}} f) (x) ∣ = ∣ \int_{Y} k_{0} (x, y) f (y) d ν (y) ∣ \leq ∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)} ∥ f ∥_{L^{2} (ν)}$ . By [step 2.1] these estimates hold for $μ$ -almost every $x$ , which is where $(T_{k_{0}} f) (x)$ is defined.

step 2.1F3F5

4.1

Measurability. If $f = \sum_{j < m} c_{j} 1_{B_{j}}$ is a finite simple function with $ν (B_{j}) < + \infty$ , then for every $j$ the indicator $1_{B_{j}}$ lies in $L^{2} (ν; C)$ because $ν (B_{j}) < + \infty$ , so [step 3.1] applied to it shows that the integral $x \mapsto \int_{Y} k_{0} (x, y) 1_{B_{j}} (y) d ν (y)$ is defined and finite for $μ$ -almost every $x$ . It is also $μ$ -measurable: the four nonnegative functions $(Re g_{j})^{+}, (Re g_{j})^{-}, (Im g_{j})^{+}, (Im g_{j})^{-}$ , where $g_{j} (x, y) : = k_{0} (x, y) 1_{B_{j}} (y)$ is $(A \otimes B)$ -measurable, have $μ$ -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 $μ$ -measurable, so $T_{k_{0}} f$ is $μ$ -measurable for such $f$ . For arbitrary $f \in L^{2} (ν; C)$ , [F6] gives finite simple functions $f_{j}$ with $∥ f_{j} - f ∥_{L^{2} (ν)} \to 0$ ; by [step 3.1], $∣ T_{k_{0}} (f_{j} - f) (x) ∣ \leq ∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)} ∥ f_{j} - f ∥_{L^{2} (ν)} \to 0$ for $μ$ -almost every $x$ , so $T_{k_{0}} f$ agrees $μ$ -almost everywhere with the limit superior of the measurable functions $T_{k_{0}} f_{j}$ , which is $μ$ -measurable by [F14].

step 3.1F3F5F6F14

4.2

Independence of representatives and linearity. If $f = f^{'}$ $ν$ -almost everywhere then $k_{0} (x, \cdot) (f - f^{'}) = 0$ $ν$ -almost everywhere for every $x$ , so $T_{k_{0}} f = T_{k_{0}} f^{'}$ wherever both are defined, in particular $μ$ -almost everywhere. If $k_{1}$ is a second $(A \otimes B)$ -measurable representative of $k$ of finite $L^{2} (ρ)$ norm, then $∣ k_{0} - k_{1} ∣$ has $ρ$ -integral $0$ , so [F3] gives $\int_{Y} ∣ k_{0} (x, y) - k_{1} (x, y) ∣ d ν (y) = 0$ for $μ$ -almost every $x$ , hence the sections agree $ν$ -almost everywhere for $μ$ -almost every $x$ and $T_{k_{0}} f = T_{k_{1}} f$ $μ$ -almost everywhere; linearity in $f$ is linearity of the integral.

step 3.1F3F5

4.3

The orthonormal family of product kernels and the pairing identity. Fix a Hilbert basis $E$ of $L^{2} (ν; C)$ and a Hilbert basis $F$ of $L^{2} (μ; C)$ , both of which exist by [step 1.1], and for $f \in F$ , $e \in E$ let $ψ_{f, e}$ be the class in $L^{2} (ρ; C)$ of $(x, y) \mapsto f (x) \overline{e (y)}$ ; this function is $(A \otimes B)$ -measurable with $∥ ψ_{f, e} ∥_{L^{2} (ρ)}^{2} = ∥ f ∥_{L^{2} (μ)}^{2} ∥ e ∥_{L^{2} (ν)}^{2} = 1$ by Tonelli [F3], and for $f, f^{'} \in F$ , $e, e^{'} \in E$ the pairing $⟨ ψ_{f, e}, ψ_{f^{'}, e^{'}} ⟩ = \int_{X} f \overline{f^{'}} d μ \cdot \overline{\int_{Y} e \overline{e^{'}} d ν} = ⟨ f, f^{'} ⟩_{L^{2} (μ)} \overline{⟨ e, e^{'} ⟩_{L^{2} (ν)}}$ vanishes unless $f = f^{'}$ and $e = e^{'}$ , again by [F3]; so $(ψ_{f, e})$ is an orthonormal family in $L^{2} (ρ; C)$ . Moreover $⟨ T_{k} e, f ⟩_{L^{2} (μ)} = \int_{X} (\int_{Y} k_{0} (x, y) e (y) d ν (y)) \overline{f (x)} d μ (x) = \int_{X \times Y} k_{0} (x, y) e (y) \overline{f (x)} d ρ (x, y) = ⟨ k, ψ_{f, e} ⟩_{L^{2} (ρ)}$ , where the middle equality is Fubini [F4] applied to the $L^{1} (ρ)$ function $(x, y) \mapsto k_{0} (x, y) e (y) \overline{f (x)}$ , whose absolute value has $ρ$ -integral at most $∥ k_{0} ∥_{L^{2} (ρ)} ∥ e ∥_{L^{2} (ν)} ∥ f ∥_{L^{2} (μ)}$ by Cauchy–Schwarz and Tonelli [F3], [F5].

step 1.1step 3.1F3F4F5

5.1

Boundedness. For $f \in L^{2} (ν; C)$ , the class of $T_{k_{0}} f$ is in $L^{2} (μ; C)$ and $∥ T_{k_{0}} f ∥_{L^{2} (μ)}^{2} = \int_{X} ∣ (T_{k_{0}} f) (x) ∣^{2} d μ (x) \leq \int_{X} ∥ k_{0} (x, \cdot) ∥_{L^{2} (ν)}^{2} d μ (x) \cdot ∥ f ∥_{L^{2} (ν)}^{2} = ∥ k_{0} ∥_{L^{2} (ρ)}^{2} ∥ f ∥_{L^{2} (ν)}^{2}$ 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 $∥ T_{k} ∥ \leq ∥ k_{0} ∥_{L^{2} (ρ)} = ∥ k ∥_{2}$ by [step 1.2].

step 1.2step 2.1step 3.1step 4.1

5.2

Rectangle kernels lie in the closed span of the family. Let $A \in A$ , $B \in B$ have finite measure, so $1_{A} \in L^{2} (μ; C)$ and $1_{B} \in L^{2} (ν; C)$ ; by [F10] the finite-subset nets $\sum_{f \in G} ⟨ 1_{A}, f ⟩ f$ over finite $G \subseteq F$ and $\sum_{e \in H} ⟨ 1_{B}, e ⟩ e$ over finite $H \subseteq E$ converge to $1_{A}$ and $1_{B}$ . For such finite $G, H$ the function $u_{G} (x) \overline{v_{H} (y)}$ with $u_{G} : = \sum_{f \in G} ⟨ 1_{A}, f ⟩ f$ and $v_{H} : = \sum_{e \in H} ⟨ 1_{B}, e ⟩ e$ is a finite linear combination of the functions $f (x) \overline{e (y)}$ , hence its class lies in the span of the family $(ψ_{f, e})$ ; and $∥ u \overline{v} - u_{G} \overline{v_{H}} ∥_{L^{2} (ρ)} \leq ∥ u ∥_{L^{2} (μ)} ∥ v - v_{H} ∥_{L^{2} (ν)} + ∥ u - u_{G} ∥_{L^{2} (μ)} ∥ v_{H} ∥_{L^{2} (ν)} \to 0$ by Tonelli [F3] and convergence of the two nets, so the rectangle kernel $1_{A} (x) 1_{B} (y) = 1_{A} (x) \overline{1_{B} (y)}$ lies in the closed span of $(ψ_{f, e})$ .

step 4.3F3F10

6.1

The kernel lies in that closed span. By [step 5.2] and [F7] every class of $L^{2} (ρ; C)$ — in particular $k_{0}$ , whose $L^{2} (ρ)$ class exists by [step 1.2] — lies in the closed span of the orthonormal family $(ψ_{f, e})$ .

step 1.2step 5.2F7

7.1

Exact norm of the family expansion. Bessel's inequality [F11] applied to $k_{0}$ and the orthonormal family $(ψ_{f, e})$ gives $\sum_{f, e} ∣ ⟨ k_{0}, ψ_{f, e} ⟩ ∣^{2} \leq ∥ k_{0} ∥_{L^{2} (ρ)}^{2}$ . For the reverse inequality fix a real $ε > 0$ and, by [step 6.1], a vector $h$ in the span of finitely many $ψ_{f, e}$ with $∥ h - k_{0} ∥_{L^{2} (ρ)} < ε$ ; if $∥ k_{0} ∥_{L^{2} (ρ)} = 0$ then $k_{0}$ is the zero class and both sides vanish, and otherwise $h \neq 0$ for small $ε$ and the finite Parseval identity and Cauchy–Schwarz on the finitely many coefficients give $\sum_{f, e} ∣ ⟨ k_{0}, ψ_{f, e} ⟩ ∣^{2} \geq ∣ ⟨ k_{0}, h ⟩ ∣^{2} / ∥ h ∥_{L^{2} (ρ)}^{2} \geq ∥ k_{0} ∥_{L^{2} (ρ)}^{2} (∥ k_{0} ∥_{L^{2} (ρ)} - ε)^{2} / (∥ k_{0} ∥_{L^{2} (ρ)} + ε)^{2}$ , where the last bound uses $∣ ⟨ k_{0}, h ⟩ ∣ \geq ∥ k_{0} ∥^{2} - ∥ k_{0} ∥ ε$ and $∥ h ∥ \leq ∥ k_{0} ∥ + ε$ and tends to $∥ k_{0} ∥_{L^{2} (ρ)}^{2}$ as $ε \to 0$ ; hence $\sum_{f, e} ∣ ⟨ k_{0}, ψ_{f, e} ⟩ ∣^{2} = ∥ k_{0} ∥_{L^{2} (ρ)}^{2} = ∥ k ∥_{2}^{2}$ by [step 1.2].

step 1.2step 6.1F11

8.1

Transfer to the basis sums. For each $e \in E$ the vector $T_{k} e$ lies in $L^{2} (μ; C)$ by [step 5.1], so [F10] applied with the Hilbert basis $F$ gives $∥ T_{k} e ∥_{L^{2} (μ)}^{2} = \sum_{f \in F} ∣ ⟨ T_{k} e, f ⟩ ∣^{2}$ , and [step 4.3] identifies each summand with $∣ ⟨ k_{0}, ψ_{f, e} ⟩ ∣^{2}$ ; since every finite subset of $E \times F$ is contained in a rectangle and all terms are nonnegative, taking suprema over finite subsets gives $\sum_{e \in E} ∥ T_{k} e ∥^{2} = \sum_{f, e} ∣ ⟨ k_{0}, ψ_{f, e} ⟩ ∣^{2} = ∥ k ∥_{2}^{2}$ by [step 7.1]. The same computation with the roles of $E$ and $F$ interchanged, using the adjoint identity $⟨ T_{k} e, f ⟩ = ⟨ e, T_{k}^{*} f ⟩$ of [F13] and Parseval with respect to $E$ applied to the vectors $T_{k}^{*} f$ , gives $\sum_{f \in F} ∥ T_{k}^{*} f ∥^{2} = ∥ k ∥_{2}^{2}$ as well.

step 5.1step 4.3step 7.1F10F13

9.1

Conclusion. The bases $E$ and $F$ in [step 8.1] were arbitrary Hilbert bases of $L^{2} (ν; C)$ and $L^{2} (μ; C)$ , and by [step 1.1] such bases exist; by [step 8.1] every one of them realizes the value $∥ k ∥_{2}^{2}$ , so by [F15] the operator $T_{k}$ is Hilbert–Schmidt with $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ , the operator itself being independent of the representative $k_{0}$ by [step 4.2] and bounded with $∥ T_{k} ∥ \leq ∥ k ∥_{2}$ by [step 5.1]. This proves all three claims.

step 1.1step 4.2step 5.1step 8.1F15∎

5 · Examples, counterexamples and false statements

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Square-Integrable Kernels and Hilbert–Schmidt Compactness

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Hilbert–Schmidt operator and Hilbert–Schmidt norm

Definition

The Hilbert–Schmidt norm is basis independent

Statement

Facts & Assumptions

Proof

Hilbert–Schmidt operators are compact

Statement

Facts & Assumptions

Proof

Product rectangle kernels are dense in product L two

Statement

Facts & Assumptions

Proof

L two kernels give Hilbert–Schmidt operators

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources