Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Compact Self Adjoint Hilbert Schmidt and Trace Class Operators — Examples

1 · Prerequisites

2 · Summary

The companion computes the theory on concrete operators. The diagonal operator $T e_{n} = d_{n} e_{n}$ on $ℓ^{2}$ is treated first: it is bounded with $∥ T ∥ = \sup_{n} ∣ d_{n} ∣$ exactly when $d \in ℓ^{\infty}$ , compact exactly when $d_{n} \to 0$ , Hilbert–Schmidt against the standard basis exactly when $\sum_{n} ∣ d_{n} ∣^{2} < + \infty$ , and trace class exactly when $\sum_{n} ∣ d_{n} ∣ < + \infty$ , in which case $tr (T) = \sum_{n} d_{n}$ . The Volterra operator $V f (x) = \int_{0}^{x} f$ is the second worked case: its kernel is a square-integrable indicator with $∥ V ∥_{H S}^{2} = \frac{1}{2}$ , so $V$ is compact, the iterated-integration formula produces a factorial operator-norm bound that excludes every nonzero eigenvalue, and Riesz–Schauder then gives $σ (V) = {0}$ : a compact quasinilpotent operator that is not self-adjoint. The rank-one operator $x \mapsto ⟨ x, v ⟩ u$ is followed through its adjoint, its norm $∥ u ∥ ∥ v ∥$ , its single singular value and its trace $⟨ u, v ⟩$ .

The integral-operator example shows how a diagonal trace formula becomes a theorem rather than a definition: for a compact metric space with finite regular Borel measure and a continuous Hermitian positive semidefinite kernel $k$ , the reproducing-kernel space of $k$ is separable, the inclusion $J$ into $L^{2}$ is Hilbert–Schmidt, the operator factors as $T_{k} = J J^{*}$ , and hence $tr (T_{k}) = \int_{X} k (x, x) d μ (x)$ ; the example also records that an arbitrary $L^{2}$ -kernel need not determine diagonal values and that continuity without positivity does not give trace class.

Boundary phenomena are collected as three separating counterexamples and one orientation remark. The diagonal operators $n^{- 1 / 2}$ and $n^{- 1}$ show that compact does not imply Hilbert–Schmidt and Hilbert–Schmidt does not imply trace class; the unilateral shift shows that $S^{*} S = I$ and $S S^{*} = I - P_{0}$ can both fail to be trace class while their difference is rank one with trace $1$ , so cyclicity cannot be extended by subtracting undefined infinite traces. A closing remark records the Schatten scale $S_{p}$ for orientation only, identifying $p = 1, 2, \infty$ with the trace class, Hilbert–Schmidt class and compact operators, and explicitly refusing interpolation, duality and Hölder theory as later material.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Diagonal Schatten class criteria on ell two

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $F \in {R, C}$ , let $ℓ^{2} : = ℓ^{2} (N, F)$ with its standard inner product and norm $∥ a ∥_{2}^{2} = \sum_{n \in N} ∣ a_{n} ∣^{2}$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Real and complex inner-product spaces and their induced length), and let $u_{n}$ be the vector that is $1$ at $n$ and $0$ elsewhere, the standard basis. Given a scalar sequence $d = (d_{n})_{n \in N}$ define, on finite linear combinations, $T (\sum_{n \in F} c_{n} u_{n}) : = \sum_{n \in F} c_{n} d_{n} u_{n} .$ For claims 2–4, saying that $T$ has the indicated operator property includes the existence of its bounded extension. Then:

$T$ extends to a bounded operator on $ℓ^{2}$ if and only if $d \in ℓ^{\infty}$ , that is $\sup_{n} ∣ d_{n} ∣ < + \infty$ , and then $∥ T ∥ = \sup_{n} ∣ d_{n} ∣$ ;
$T$ is compact if and only if $d_{n} \to 0$ ;
$T$ is Hilbert–Schmidt relative to the standard basis (Hilbert–Schmidt operator and Hilbert–Schmidt norm) if and only if $\sum_{n} ∣ d_{n} ∣^{2} < + \infty$ , and then $∥ T ∥_{H S} = (\sum_{n} ∣ d_{n} ∣^{2})^{1 / 2}$ ;
$T$ is trace class (Trace class operator) if and only if $\sum_{n} ∣ d_{n} ∣ < + \infty$ , and then $∥ T ∥_{1} = \sum_{n} ∣ d_{n} ∣$ and $tr (T) = \sum_{n} d_{n}$ (Trace of a trace class operator, Trace is absolutely convergent and basis independent).

Facts & Assumptions

Given: Countable Choice, the inner-product space $ℓ^{2} (N, F)$ with its explicit coordinate vectors $u_{n}$ , and a scalar sequence $d$ .

[A1]

The square-summable-family definition constructs $ℓ^{2}$ as an inner-product space with $⟨ a, b ⟩ = \sum_{n} a_{n} \overline{b_{n}}$ , $∥ a ∥_{2}^{2} = \sum_{n} ∣ a_{n} ∣^{2}$ , and nonnegative sums as suprema of finite subsums. Finite total sums have arbitrarily small tails outside finite sets (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Real and complex inner-product spaces and their induced length). A Hilbert basis is an orthonormal family with dense linear span (Orthonormal families, complete orthonormal systems and Hilbert bases).

[A2]

A linear map is bounded if it has a finite norm bound, and its operator norm is the unit-ball supremum (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Hilbert/Banach completeness means every norm-Cauchy sequence converges (Hilbert space, Banach space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[A3]

Bounded finite-rank operators are compact, and under Countable Choice the operator-norm limit of compact operators into a Banach space is compact (Bounded finite rank operators are compact, Norm limit of compact operators is compact). Compactness gives compact closure of the image of the closed unit ball (Compact linear operator).

[A4]

For a compact operator, $∣ T ∣$ is the unique compact positive square root of $T^{*} T$ ; its positive eigenvalues with multiplicity are the positive-labelled singular values (Absolute value and singular values of a compact operator, Positive square root of a compact positive operator). The adjoint is characterized by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ (The Hilbert-space adjoint of a bounded operator).

[A5]

Relative to a supplied Hilbert basis $E$ , Hilbert–Schmidt membership is finiteness of $\sum_{E} ∥ T e ∥^{2}$ , with norm its square root and with basis independence (Hilbert–Schmidt operator and Hilbert–Schmidt norm, The Hilbert–Schmidt norm is basis independent).

[A6]

Trace class means finite positive singular-value sum, which is its trace norm; for a supplied Hilbert basis the absolutely convergent diagonal sum is the basis-independent trace (Trace class operator, Trace of a trace class operator, Trace is absolutely convergent and basis independent).

[A7]

Real and complex scalar Cauchy sequences converge (The reals are complete, The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts).

Verification

technique · direct

1.1

Completeness and basis. Let $(a^{(j)})$ be norm-Cauchy in $ℓ^{2}$ . It has a common norm bound $B$ : its tail lies within distance $1$ of one term, and the finitely many earlier norms have a maximum. Each coordinate is Cauchy since $∣ a_{n}^{(j)} - a_{n}^{(k)} ∣ \leq ∥ a^{(j)} - a^{(k)} ∥_{2}$ , so [A7] defines its unique limit $a_{n}$ , without any selection of alternative limits. For every finite $F$ , passage to the limit in the finite sum gives $\sum_{n \in F} ∣ a_{n} ∣^{2} \leq B^{2}$ , hence $a \in ℓ^{2}$ . Given $ε > 0$ , choose $N$ so that $∥ a^{(j)} - a^{(k)} ∥_{2} < ε / 2$ for $j, k \geq N$ . Letting $k$ tend to infinity in each finite subsum gives $\sum_{n \in F} ∣ a_{n}^{(j)} - a_{n} ∣^{2} \leq ε^{2} / 4$ for every finite $F$ and $j \geq N$ . Taking the supremum gives $∥ a^{(j)} - a ∥_{2} \leq ε / 2 < ε$ . Thus $ℓ^{2}$ is Hilbert. The $u_{n}$ are orthonormal by their coordinates; finite truncations of any $a$ approximate it in norm by the small-tail assertion of [A1], so their span is dense and they are a Hilbert basis.

A1A2A7

2.1

Boundedness and norm. If $M : = \sup_{n} ∣ d_{n} ∣ < \infty$ , define $T a = (d_{n} a_{n})_{n}$ for every $a \in ℓ^{2}$ . For every finite $F$ , $\sum_{F} ∣ d_{n} a_{n} ∣^{2} \leq M^{2} ∥ a ∥_{2}^{2}$ , so $T a \in ℓ^{2}$ and $∥ T a ∥_{2} \leq M ∥ a ∥_{2}$ . Coordinatewise operations show linearity, and this is the required extension. It is unique because two bounded operators agreeing on the dense finite span have difference zero by the norm bound and approximation in step 1.1. Testing $u_{n}$ gives $∥ T ∥ \geq ∣ d_{n} ∣$ for every $n$ , hence $∥ T ∥ = M$ . Conversely any bounded extension bounds all $∣ d_{n} ∣ = ∥ T u_{n} ∥_{2}$ , so the sequence must be bounded. Equality with a supremum does not assert that the norm is attained at a unit vector.

step 1.1A1A2algebra

3.1

Compactness. If $d_{n} \to 0$ , then $d$ is bounded (a bounded tail and finitely many initial values suffice), so step 2.1 constructs $T$ . Its truncation $T_{N}$ keeping coordinates $0, \dots, N$ has finite rank and $∥ T - T_{N} ∥ = \sup_{n > N} ∣ d_{n} ∣ \to 0$ by that same norm formula. Thus [A3] and step 1.1 make $T$ compact. Conversely, if $T$ is compact but $d_{n} \to̸ 0$ , some $ε > 0$ has infinitely many indices $A$ with $∣ d_{n} ∣ \geq ε$ . For distinct $m, n \in A$ , $∥ T u_{m} - T u_{n} ∥_{2}^{2} = ∣ d_{m} ∣^{2} + ∣ d_{n} ∣^{2} \geq 2 ε^{2}$ . Cover the compact closure of $T$ of the unit ball by all balls of radius $ε / 2$ and take a finite subcover. Each such ball contains at most one of these separated points, a contradiction. This uses no enumeration of A or additional sequential-compactness supplier.

step 1.1step 2.1A1A3algebra

3.2

Summability and Hilbert–Schmidt membership. If $\sum_{n} ∣ d_{n} ∣^{p} < \infty$ for $p = 1$ or $p = 2$ , every term is bounded by that sum and hence $d$ is bounded. Small tails from [A1] show $d_{n} \to 0$ : for any $ε > 0$ take a finite tail-control set for $ε^{p}$ , and every coordinate outside it has $∣ d_{n} ∣ < ε$ . Past its largest index all coordinates are outside it. In particular when the square sum is finite step 2.1 supplies the bounded extension, and $\sum_{n} ∥ T u_{n} ∥_{2}^{2} = \sum_{n} ∣ d_{n} ∣^{2}$ . Conversely Hilbert–Schmidt membership requires that extension and the same finite sum. Formula [A5] gives the stated norm, using the actual standard basis from step 1.1 in both domain and target.

step 1.1step 2.1A1A5

4.1

Absolute value only in the compact case. Assume $T$ is compact. Step 3.1 gives $d_{n} \to 0$ . Step 2.1 constructs bounded diagonal operators with entries $\overline{d_{n}}$ and $∣ d_{n} ∣$ , the latter denoted $D$ . The coordinate pairing in [A1] gives $⟨ T a, b ⟩ = ⟨ a, (\overline{d_{n}} b_{n})_{n} ⟩$ , so the first is $T^{*}$ by [A4]. The operator $D$ is compact by step 3.1, is positive since $⟨ D a, a ⟩ = \sum_{n} ∣ d_{n} ∣ ∣ a_{n} ∣^{2} \geq 0$ , is self-adjoint by the same coordinate pairing, and satisfies $D^{2} = T^{*} T$ . Thus $D = ∣ T ∣$ by [A4]. For $λ > 0$ , the equation $D a = λ a$ says $a_{n} = 0$ wherever $∣ d_{n} ∣ \neq λ$ . There are only finitely many indices with $∣ d_{n} ∣ = λ$ , because $d_{n} \to 0$ . Thus that eigenspace has exactly their coordinate vectors as a finite basis. Repeated moduli contribute their full multiplicity, not multiplicity one.

step 2.1step 3.1A1A4

5.1

Trace class and trace. For compact $T$ , step 4.1 identifies the positive singular-value multiset with the nonzero values $∣ d_{n} ∣$ , counted with their indices. The finite-subset suprema of these nonnegative sums agree: each finite collection of occurrences on either side corresponds to a finite collection on the other side with the same summands. Finite initial segments are cofinal among finite index subsets, so the sums also agree with the ordinary nonnegative series. Consequently [A6] gives trace class exactly when $\sum_{n} ∣ d_{n} ∣ < \infty$ , with $∥ T ∥_{1}$ equal to that sum. If instead that sum is given first, steps 3.2 and 3.1 establish boundedness and compactness before any singular data are used. Finally the standard-basis coefficients are $⟨ T u_{n}, u_{n} ⟩ = d_{n}$ , so their absolutely convergent sum equals $tr (T)$ by [A6].

step 1.1step 3.1step 3.2step 4.1A1A6

6.1

Claims 1–4 follow from steps 2.1, 3.1, 3.2 and 5.1 respectively. The sequence $d = 0$ gives zero norms and trace; repeated entries and finite support are included by the multiplicity argument. The Hilbert basis is explicit and Countable Choice is used only as licensed by the compactness and singular/trace suppliers.

step 2.1step 3.1step 3.2step 5.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Volterra operator is Hilbert Schmidt and quasinilpotent

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $H : = L^{2} ([0, 1], C)$ with the integral pairing linear in the first argument ( $L^{2}$ with the integral pairing is a Hilbert space, The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz) and let $V f (x) : = \int_{0}^{x} f (t) d t (0 \leq x \leq 1)$ be the Volterra operator, that is, the integral operator with kernel $k (x, t) : = 1_{{0 \leq t \leq x}}$ on $[0, 1]^{2}$ . Then:

$V$ is Hilbert–Schmidt with $∥ V ∥_{H S}^{2} = \frac{1}{2}$ , hence compact (Hilbert–Schmidt operator and Hilbert–Schmidt norm, Hilbert–Schmidt operators are compact);
Choosing the integral representatives of the iterates, $V^{n} f (x) = \frac{1}{(n - 1)!} \int_{0}^{x} (x - t)^{n - 1} f (t) d t$ for every $n \geq 1$ , and $∥ V^{n + 1} ∥ \leq \frac{1}{n! \sqrt{2 (n + 1) (2 n + 1)}} \to 0$ ;
$V$ has no nonzero eigenvalue: $\ker (V - λ I) = {0}$ for every $λ \neq 0$ ;
the spectrum of $V$ is $σ (V) = {0}$ (Spectrum and resolvent of a bounded operator); thus $V$ is quasinilpotent, and in particular $V$ is not self-adjoint though it is compact, showing that the compact self-adjoint spectral theorem does not apply.

Facts & Assumptions

Given: AC, the complex Hilbert space $L^{2} ([0, 1])$ , the kernel $k (x, t) = 1_{{0 \leq t \leq x}}$ , and the operator $V$ .

[A1]

The kernel is square-integrable. The function $k$ is measurable on the completed product measure and $\int_{[0, 1]^{2}} ∣ k ∣^{2} d (λ \times λ) = \int_{0}^{1} λ ([0, x]) d x = \int_{0}^{1} x d x = \frac{1}{2}$ , by Tonelli, the measure of intervals and the polynomial integral; the class of $k$ lies in $L^{2}$ of the completed product measure with squared norm $\frac{1}{2}$ (The completed product measure, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous, For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[A2]

Kernel operators. For a kernel class $k$ of finite square norm, the operator $T_{k} f (x) = \int_{0}^{1} k (x, t) f (t) d t$ is bounded with $∥ T_{k} ∥ \leq ∥ k ∥_{2}$ , is Hilbert–Schmidt with $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ , and is therefore compact (L two kernels give Hilbert–Schmidt operators, Hilbert–Schmidt operators are compact, Hilbert–Schmidt operator and Hilbert–Schmidt norm, Compact linear operator, A bounded linear operator between normed spaces).

[A3]

Norm and integral bounds. For $f \in L^{2} [0, 1]$ : Cauchy–Schwarz gives $∣ \int_{0}^{x} g ∣ \leq \sqrt{x} ∥ g ∥_{2}$ and $(\int_{0}^{1} ∣ \int_{0}^{x} g ∣^{2} d x)^{1 / 2} \leq ∥ g ∥_{2}$ ; the operator norm is the unit-ball supremum; and $\int_{0}^{1} x^{m} d x = 1 / (m + 1)$ for integers $m \geq 0$ , computed by the Newton–Leibniz formula applied to the primitive $x^{m + 1} / (m + 1)$ of $x^{m}$ , whose derivative is given by the derivative-of-a-power lemma, with the Riemann integral agreeing with the Lebesgue integral (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous, For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[A4]

Spectrum of a compact operator. Under AC, a nonzero spectral value of a compact operator on a complex Banach space is an eigenvalue of finite algebraic multiplicity, and if the space is infinite dimensional then $0$ belongs to the spectrum (Riesz schauder spectrum of a compact operator, Spectrum and resolvent of a bounded operator, Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, Banach space).

[A5]

Small reciprocal bounds. For every positive real $ε$ , some natural $N \geq 1$ satisfies $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ). In particular $1 / N \to 0$ , and $2^{j} \geq j + 1$ by induction, so a tail bounded by $C 2^{- j}$ tends to zero. The latter implication follows from $C 2^{- j} \leq C / (j + 1)$ and the reciprocal bound.

[A6]

Self-adjointness test. The adjoint is characterized by $⟨ T f, g ⟩ = ⟨ f, T^{*} g ⟩$ , and $T$ is self-adjoint when $T^{*} = T$ (The Hilbert-space adjoint of a bounded operator, Self-adjoint, positive, unitary and normal operators).

[A7]

Complex Fubini. A complex product-measurable function with integrable absolute value on a sigma-finite product has equal double and iterated integrals (Fubini's theorem for L^1 functions on a sigma-finite product). Here both factors are finite Lebesgue measure on $[0, 1]$ .

Verification

technique · direct

Given: AC, the space $L^{2} ([0, 1], C)$ , the Volterra kernel and operator, and the bounds above.

1.1

$V$ is Hilbert–Schmidt with norm $1 / \sqrt{2}$ . By [A1] the class of $k$ has square norm $\frac{1}{2}$ in the completed product measure; the kernel operator of [A2] is $T_{k} f (x) = \int_{0}^{1} k (x, t) f (t) d t = \int_{0}^{x} f (t) d t = V f (x)$ for $f \in L^{2}$ and almost every $x$ , by the definition of $k$ ; Under AC choose a Hilbert basis as supplied by the kernel theorem; hence $V = T_{k}$ is Hilbert–Schmidt with $∥ V ∥_{H S} = ∥ k ∥_{2} = 1 / \sqrt{2}$ and is compact.

A1A2

1.2

Iterated integration and norm decay. By induction on $n$ : for $n = 1$ the formula is the definition of $V$ ; assuming it for $n$ , $V^{n + 1} f (x) = \int_{0}^{x} \frac{1}{(n - 1)!} \int_{0}^{t} (t - s)^{n - 1} f (s) d s d t = \frac{1}{n!} \int_{0}^{x} (x - s)^{n} f (s) d s$ as follows. Cauchy–Schwarz applied to $∣ f ∣$ and $1$ gives $\int_{0}^{1} ∣ f ∣ \leq ∥ f ∥_{2}$ . For each fixed $x$ , the integrand $1_{0 \leq s \leq t \leq x} (t - s)^{n - 1} f (s)$ is product-measurable (put its value zero outside the triangle) and its absolute value is bounded by $∣ f (s) ∣$ . Tonelli thus bounds its double absolute integral by $∥ f ∥_{2} < \infty$ . Complex Fubini [A7] therefore permits reversing the integrals, and the inner integration $\int_{s}^{x} (t - s)^{n - 1} d t = (x - s)^{n} / n$ [A1, A3] gives the displayed identity. Cauchy--Schwarz and Tonelli give $∥ V^{n + 1} f ∥_{2}^{2} \leq \frac{∥ f ∥_{2}^{2}}{(n!)^{2}} \int_{0}^{1} \int_{0}^{x} (x - s)^{2 n} d s d x = \frac{∥ f ∥_{2}^{2}}{(n!)^{2} (2 n + 1) (2 n + 2)},$ so $∥ V^{n + 1} ∥ \leq 1 / (n! \sqrt{(2 n + 1) (2 n + 2)})$ , and the right side tends to $0$ because it is at most $1 / \sqrt{2 n + 2}$ , which tends to zero by [A5]. Changes to $f$ on a null set do not change any integral, so this also identifies the operator classes.

A1A3A5A7algebra

1.3

$V$ is not self-adjoint. Let $f (x) = 1$ and $g (x) = x$ . Then $V f (x) = x$ and $V g (x) = x^{2} / 2$ , so [A3] gives $⟨ V f, g ⟩ = \int_{0}^{1} x^{2} d x = 1 / 3$ but $⟨ f, V g ⟩ = \int_{0}^{1} x^{2} / 2 d x = 1 / 6$ . These values are unequal, whereas [A6] would make them equal if $V = V^{*}$ .

A3A6

2.1

There is no nonzero eigenvalue. Let $V f = λ f$ with $λ \neq 0$ . Iterating, $V^{n} f = λ^{n} f$ for every $n \geq 1$ , so if $f \neq 0$ then step 1.2 gives $1 \leq \frac{∥ V^{n} ∥}{∣ λ ∣^{n}} \leq b_{n} : = \frac{1}{∣ λ ∣^{n} (n - 1)! \sqrt{(2 n - 1) (2 n)}} .$ But $\frac{b_{n + 1}}{b_{n}} = \frac{1}{∣ λ ∣ n} \sqrt{\frac{(2 n - 1) (2 n)}{(2 n + 1) (2 n + 2)}} ⟶ 0,$ Choose $N \geq 1$ with $1 / (∣ λ ∣ N) \leq 1 / 2$ by [A5]. For $n \geq N$ the displayed ratio is at most $1 / 2$ , hence induction gives $b_{N + j} \leq b_{N} 2^{- j} \to 0$ by [A5], contradicting $1 \leq b_{n}$ for every $n$ . Hence $f = 0$ : the kernel of $V - λ I$ is trivial for every $λ \neq 0$ .

step 1.2A5algebra

3.1

The spectrum is ${0}$ . By step 1.1, $V$ is compact on the complex Hilbert, hence Banach, space $L^{2} [0, 1]$ . Thus every nonzero spectral value would be an eigenvalue by [A4], ruled out by step 2.1. To prove $0 \in σ (V)$ directly, for $0 < δ \leq 1$ put $f_{δ} = 1_{[0, δ]} / \sqrt{δ}$ . Then $∥ f_{δ} ∥_{2} = 1$ and $∣ V f_{δ} (x) ∣ = \min (x, δ) / \sqrt{δ} \leq \sqrt{δ}$ , so $∥ V f_{δ} ∥_{2} \leq \sqrt{δ}$ . A bounded inverse with norm $C$ would give $1 \leq C \sqrt{δ}$ for all such $δ$ ; taking $δ = 1 / (j + 1)^{2}$ and using [A5] contradicts this. By the resolvent definition in [A4], $0$ lies in the spectrum. Therefore $σ (V) = {0}$ .

step 1.1step 2.1A1A3A4A5

4.1

Conclusion. Claims 1–4 are [step 1.1], [step 1.2], [step 2.1] and [step 3.1], and [step 1.3] proves the final non-self-adjointness assertion directly.

step 1.1step 1.2step 2.1step 3.1step 1.3∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Adjoint, norm and trace of an operator of rank at most one

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real or complex Hilbert space with the pairing linear in the first argument (Hilbert space, Real and complex inner-product spaces and their induced length), let $u, v \in H$ and let $T : = T_{u, v} \in B (H), T x : = ⟨ x, v ⟩ u .$ Then:

the Hilbert adjoint is $T^{*} x = ⟨ x, u ⟩ v$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities);
$∥ T ∥ = ∥ u ∥ ∥ v ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum);
$T$ is trace class (Trace class operator); if $u \neq 0$ and $v \neq 0$ its singular values are $s_{1} (T) = ∥ u ∥ ∥ v ∥$ and $s_{n} (T) = 0$ for $n \geq 2$ , so it has exactly one nonzero singular value, and $∥ T ∥_{1} = ∥ u ∥ ∥ v ∥$ ; if $u = 0$ or $v = 0$ then $T = 0$ and all singular values vanish;
$tr (T) = ⟨ u, v ⟩$ (Trace is absolutely convergent and basis independent).

Facts & Assumptions

Given: Countable Choice, the Hilbert space $H$ , vectors $u, v \in H$ and the operator of rank at most one $T = ⟨ \cdot, v ⟩ u$ .

[A1]

Pairing and adjoint. The pairing is linear in the first argument, conjugate-linear in the second, conjugate symmetric with $⟨ w, w ⟩ = ∥ w ∥^{2} \geq 0$ ; the Hilbert adjoint is characterised by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ and satisfies $T^{* *} = T$ , $(S T)^{*} = T^{*} S^{*}$ (Real and complex inner-product spaces and their induced length, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities, Hilbert space).

[A2]

Cauchy–Schwarz and norm. $∣ ⟨ x, v ⟩ ∣ \leq ∥ x ∥ ∥ v ∥$ , and the operator norm is the unit-ball supremum (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A3]

Compactness and spectral data. Every bounded finite-rank operator is compact (Bounded finite rank operators are compact). A nonzero compact self-adjoint positive operator has a largest eigenvalue equal to its norm with unit eigenvector, its nonzero eigenvalues are positive with finite multiplicities accumulating only at $0$ , its closed span is $(\ker \cdot)^{⊥}$ , and the positive square root is unique; the singular values of a compact operator are the positive eigenvalues of $∣ T ∣$ with multiplicity, in nonincreasing order with zero padding (Norm point of a compact self adjoint operator is an eigenvalue up to sign, Spectral theorem for compact self adjoint operators, Positive square root of a compact positive operator, Absolute value and singular values of a compact operator, Singular value decomposition for compact operators).

[A4]

Trace machinery. A compact operator with $\sum_{n} s_{n} < + \infty$ is trace class with $∥ T ∥_{1} = \sum_{n} s_{n}$ ; for a nuclear representation $T = \sum_{j} ⟨ \cdot, u_{j} ⟩ v_{j}$ the trace is $\sum_{j} ⟨ v_{j}, u_{j} ⟩$ , independently of the representation, and $∣ tr (T) ∣ \leq ∥ T ∥_{1}$ (Trace class operator, Nuclear series characterizes trace norm, Trace is absolutely convergent and basis independent, Trace of a trace class operator).

Verification

technique · direct

Given: Countable Choice, the vectors $u, v$ , the operator $T = ⟨ \cdot, v ⟩ u$ , and the candidate $T^{*} : = ⟨ \cdot, u ⟩ v$ .

1.1

The adjoint. The candidate is linear by first-variable linearity and bounded by $∥ ⟨ x, u ⟩ v ∥ \leq ∥ u ∥ ∥ v ∥ ∥ x ∥$ using [A2]. For all $x, y \in H$ , $⟨ T x, y ⟩ = ⟨ ⟨ x, v ⟩ u, y ⟩ = ⟨ x, v ⟩ ⟨ u, y ⟩$ and $⟨ x, T^{*} y ⟩ = ⟨ x, ⟨ y, u ⟩ v ⟩ = \overline{⟨ y, u ⟩} ⟨ x, v ⟩ = ⟨ u, y ⟩ ⟨ x, v ⟩$ by conjugate symmetry [A1]; the two expressions agree, so by uniqueness of the Hilbert adjoint $T^{*} = ⟨ \cdot, u ⟩ v$ .

A1A2

1.2

The norm. For every $x$ , $∥ T x ∥ = ∣ ⟨ x, v ⟩ ∣ ∥ u ∥ \leq ∥ u ∥ ∥ v ∥ ∥ x ∥$ by [A2], so $∥ T ∥ \leq ∥ u ∥ ∥ v ∥$ ; if $v \neq 0$ then testing $x = v / ∥ v ∥$ gives $∥ T x ∥ = ∥ v ∥ ∥ u ∥$ , whence equality, and if $v = 0$ then $T = 0$ and both sides are $0$ .

A1A2algebra

2.1

The singular value. The range of $T$ is contained in $span {u}$ , when $u, v \neq 0$ , $T (v / ∥ v ∥^{2}) = u$ , so its range has ordered basis $(u)$ ; if either vector is zero its range has the empty basis. Thus the bounded operator $T$ has finite rank and is compact by [A3]. Compute $T^{*} T x = ⟨ x, v ⟩ ∥ u ∥^{2} v$ using [step 1.1] and conjugate linearity in the second argument [A1]; hence $T^{*} T = ∥ u ∥^{2} ∥ v ∥^{2} P$ where $P : = ⟨ \cdot, v / ∥ v ∥ ⟩ v / ∥ v ∥$ is the orthogonal projection onto $span {v}$ when $v \neq 0$ , and put $P = 0$ when $v = 0$ , so the displayed formula holds in that case too. For $v \neq 0$ , writing $e = v / ∥ v ∥$ gives $P^{2} = P$ , $P^{*} = P$ and $⟨ P x, x ⟩ = ∣ ⟨ x, e ⟩ ∣^{2} \geq 0$ directly from [A1]. The operator $S : = ∥ u ∥ ∥ v ∥ P$ is bounded by [A2] and has the one-vector range basis $(v)$ when $u, v \neq 0$ , otherwise the empty range basis. It is therefore compact by [A3], and is self-adjoint and positive with $S^{2} = T^{*} T$ , so $∣ T ∣ = S$ by uniqueness of the positive square root [A3]; its nonzero eigenvalues are the single number $∥ u ∥ ∥ v ∥$ with multiplicity one when $u, v \neq 0$ , and there are none when $u = 0$ or $v = 0$ . By [A3] the singular values of $T$ are exactly this data, and [A4] gives $∥ T ∥_{1} = ∥ u ∥ ∥ v ∥ < + \infty$ , so $T$ is trace class.

step 1.1A1A2A3A4algebra

3.1

The trace. Assume $u, v \neq 0$ (otherwise $T = 0$ and the trace is $0 = ⟨ u, v ⟩$ ). Then, writing $e : = v / ∥ v ∥$ and $s : = ∥ u ∥ ∥ v ∥$ , the identity $T x = s ⟨ x, e ⟩ (u / ∥ u ∥)$ exhibits $T$ as the positive-integer-indexed nuclear representation with $u_{1} : = e$ , $v_{1} : = s u / ∥ u ∥ = ∥ v ∥ u$ and $u_{j} = v_{j} = 0$ for $j \geq 2$ . Its zero-based partial-sum sequence has $R_{0} = 0$ and $R_{m} = T$ for every $m \geq 1$ , so it converges to $T$ exactly as required by [A4]. Therefore $tr (T) = ⟨ v_{1}, u_{1} ⟩ = ⟨ ∥ v ∥ u, v / ∥ v ∥ ⟩ = ⟨ u, v ⟩$ , since scalar multiplication in the first argument and conjugate-linearity in the second give $⟨ ∥ v ∥ u, v / ∥ v ∥ ⟩ = ∥ v ∥ \overline{∥ v ∥^{- 1}} ⟨ u, v ⟩ = ⟨ u, v ⟩$ .

step 2.1A1A4algebra

4.1

Conclusion. Claims 1–4 are [step 1.1], [step 1.2], [step 2.1] and [step 3.1]; in the degenerate cases $u = 0$ or $v = 0$ the operator is $0$ with $∥ T ∥ = ∥ T ∥_{1} = 0$ and $tr (T) = 0 = ⟨ u, v ⟩$ .

step 1.1step 1.2step 2.1step 3.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Integral operator trace under a valid diagonal hypothesis

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $(X, d)$ be a compact metric space, let $μ$ be a finite regular Borel measure on $X$ (Finite, sigma-finite, and semifinite measures, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ) and let $k : X \times X \to C$ be continuous, Hermitian, $k (y, x) = \overline{k (x, y)}$ , and positive semidefinite, that is $\sum_{i, j = 1}^{n} c_{i} \overline{c_{j}} k (x_{i}, x_{j}) \geq 0$ for all finite families $x_{1}, \dots, x_{n} \in X$ and scalars $c_{1}, \dots, c_{n} \in C$ . Let $T_{k}$ be the integral operator on the complex Hilbert space $L^{2} (X, μ; C)$ ( $L^{2}$ with the integral pairing is a Hilbert space, The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz) $T_{k} f (x) : = \int_{X} k (x, y) f (y) d μ (y) .$ Then:

$T_{k}$ is a bounded Hilbert–Schmidt operator with $∥ T_{k} ∥_{H S} = ∥ k ∥_{L^{2} (μ \times μ)}$ and $∥ T_{k} ∥ \leq ∥ k ∥_{L^{2} (μ \times μ)}$ , hence compact (L two kernels give Hilbert–Schmidt operators, Hilbert–Schmidt operator and Hilbert–Schmidt norm, Compact linear operator);
$T_{k}$ is self-adjoint and positive: $T_{k}^{*} = T_{k}$ and $⟨ T_{k} f, f ⟩ \geq 0$ for all $f$ (Self-adjoint, positive, unitary and normal operators);
$T_{k}$ is trace class and $tr (T_{k}) = \int_{X} k (x, x) d μ (x)$ (Trace class operator, Trace is absolutely convergent and basis independent);
two boundaries are part of the statement. First, a class in $L^{2} (X \times X, μ \times μ)$ does not in general determine diagonal values: when $μ$ is nonzero and nonatomic, representatives may be changed on the product-null diagonal, changing their diagonal integrals. Thus the displayed identity is a theorem under the continuity and positivity hypotheses and is not a definition of the trace. Second, continuity of $k$ alone does not imply trace class: it only gives Hilbert–Schmidt, and a continuous Hermitian kernel that is not positive semidefinite may fail to be trace class.

Facts & Assumptions

Given: AC, the compact metric space $X$ , the finite regular Borel measure $μ$ , the continuous Hermitian positive semidefinite kernel $k$ , and the symbols $k_{x} : = k (\cdot, x)$ .

[A1]

Kernel arithmetic. $k$ is bounded and measurable on $X \times X$ with $∥ k ∥_{2}^{2} = \int_{X \times X} ∣ k ∣^{2} d (μ \times μ) \leq μ (X)^{2} \sup ∣ k ∣^{2} < + \infty$ , the completed product measure is finite and Tonelli applies to nonnegative measurable functions (The completed product measure, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Finite, sigma-finite, and semifinite measures, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ).

[A2]

Kernel operators. For $k$ of finite square norm the operator $T_{k}$ is bounded with $∥ T_{k} ∥ \leq ∥ k ∥_{2}$ , is Hilbert–Schmidt with $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ , and is compact by Hilbert–Schmidt operators are compact applied to the Hilbert basis supplied under AC by the kernel theorem; the complex space $L^{2} (X, μ; C)$ with $⟨ f, g ⟩ = \int f \overline{g} d μ$ is a Hilbert space (L two kernels give Hilbert–Schmidt operators, $L^{2}$ with the integral pairing is a Hilbert space, The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz, Hilbert–Schmidt operator and Hilbert–Schmidt norm, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Hilbert space, Banach space, Compact linear operator).

[A3]

The reproducing-kernel space. On the complex span $E_{0}$ of the functions $k_{x}$ put $⟨ \sum_{i} a_{i} k_{x_{i}}, \sum_{j} b_{j} k_{y_{j}} ⟩_{0} : = \sum_{i, j} a_{i} \overline{b_{j}} k (y_{j}, x_{i})$ . Positive semidefiniteness and Hermitian symmetry make this a positive semidefinite Hermitian form, so $∣ ⟨ h, g ⟩_{0} ∣^{2} \leq ⟨ h, h ⟩_{0} ⟨ g, g ⟩_{0}$ and the null set $N : = {h : ⟨ h, h ⟩_{0} = 0}$ is a subspace on which the form vanishes identically and whose elements are exactly the functions vanishing on $X$ , because $∣ h (x) ∣ = ∣ ⟨ h, k_{x} ⟩_{0} ∣ \leq ∥ h ∥_{0} ∥ k_{x} ∥_{0} = ∥ h ∥_{0} \sqrt{k (x, x)}$ for $h \in E_{0}$ ; the quotient $E_{0} / N$ with the induced inner product has a completion $H_{k}$ , a Hilbert space (The norm completion of an inner-product space is a Hilbert space, Real and complex inner-product spaces and their induced length, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Orthogonality and the orthogonal complement). In $H_{k}$ the reproducing identity $h (x) = ⟨ h, k_{x} ⟩$ and the bound $∣ h (x) ∣ \leq ∥ h ∥ \sqrt{k (x, x)}$ hold, the inclusion $J : H_{k} \to L^{2} (X, μ)$ , $J h : = h$ , is a well-defined bounded linear map with $∥ J ∥ \leq \sqrt{μ (X) \sup_{x} k (x, x)}$ , and $T_{k} = J J^{*}$ (Hilbert-adjoint identities, The Hilbert-space adjoint of a bounded operator).

[A4]

A finite or countable orthonormal basis of $H_{k}$ . By compactness $X$ is totally bounded, so for each integer $n \geq 1$ there is a finite $(1 / n)$ -net of $X$ (A compact metric space is complete and totally bounded, and neither implication uses any choice principle); AC chooses one net for each $n \geq 1$ , their union $D$ is at most countable and dense, and the $Q (i)$ -span of ${k_{d} : d \in D}$ is an at most countable dense subset of $H_{k}$ , because $∥ k_{x} - k_{y} ∥^{2} = k (x, x) - k (x, y) - k (y, x) + k (y, y) \to 0$ as $y \to x$ by continuity and Hermitian symmetry. The separable-basis theorem therefore provides a Hilbert basis $(e_{i})_{i \in I}$ of $H_{k}$ , where $I$ is empty, finite, or countably infinite (A Hilbert space with a dense sequence has a finite or countable orthonormal basis, Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, Finite, countably infinite, countable, uncountable, Every subset of an at most countable set is at most countable, Countable unions of at most countable sets, assuming ${A C}_{ω}$ , Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Orthonormal families, complete orthonormal systems and Hilbert bases). Using its canonical order, write the basis as $(e_{1}, \dots, e_{r})$ when it is finite and as $(e_{j})_{j \geq 1}$ when it is infinite, and define a positive-integer-indexed family $(h_{j})_{j \geq 1}$ by $h_{j} = e_{j}$ on the existing indices and $h_{j} = 0$ after $r$ in the finite case (all terms are zero when $I = \emptyset$ ).

[A5]

Nuclear series, Parseval and Tonelli. A positive-integer-indexed nuclear family whose shifted coefficient-norm series is summable has zero-based partial sums converging in operator norm and defines a trace-class operator, whose trace is the corresponding shifted sum $\sum_{j \geq 1} ⟨ v_{j}, u_{j} ⟩$ with $∥ T ∥_{1} \leq \sum_{j \geq 1} ∥ u_{j} ∥ ∥ v_{j} ∥$ ; and for the Hilbert basis $(e_{i})_{i \in I}$ of $H_{k}$ , Parseval gives $\sum_{i \in I} ∣ e_{i} (x) ∣^{2} = \sum_{i \in I} ∣ ⟨ k_{x}, e_{i} ⟩ ∣^{2} = k (x, x)$ for every $x$ , while Tonelli for this at most countable nonnegative family gives $\int_{X} \sum_{i \in I} ∣ e_{i} (x) ∣^{2} d μ (x) = \sum_{i \in I} ∥ J e_{i} ∥^{2}$ (Nuclear series characterizes trace norm, Trace is absolutely convergent and basis independent, Trace of a trace class operator, Parseval equivalences for an orthonormal family, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability).

Verification

technique · direct

Given: AC, the data above, the space $H_{k}$ with its basis $(e_{i})_{i \in I}$ , its zero-padded positive enumeration $(h_{j})_{j \geq 1}$ , and the inclusion $J$ .

1.1

The operator and its factorization. By [A1] the class of $k$ has finite square norm, so [A2] makes $T_{k}$ a bounded Hilbert–Schmidt compact operator with the stated norms. By [A3] the inclusion $J$ is bounded with $T_{k} = J J^{*}$ : for $f \in L^{2}$ and every $x$ , $(J J^{*} f) (x) = ⟨ J^{*} f, k_{x} ⟩ = ⟨ f, J k_{x} ⟩_{L^{2}} = \int_{X} f (y) \overline{k (y, x)} d μ (y) = \int_{X} k (x, y) f (y) d μ (y) = (T_{k} f) (x)$ , using the reproducing identity and Hermitian symmetry.

A1A2A3

1.2

$J$ is Hilbert–Schmidt and $T_{k}$ is positive. By [A4] the family $(e_{i})_{i \in I}$ is a Hilbert basis of the domain $H_{k}$ of $J$ , so $\sum_{i \in I} ∥ J e_{i} ∥^{2} = \int_{X} \sum_{i \in I} ∣ e_{i} (x) ∣^{2} d μ (x) = \int_{X} k (x, x) d μ (x)$ by [A5], and the right-hand side is finite because $k (x, x)$ is continuous on the compact space $X$ ; hence $J$ is Hilbert–Schmidt relative to this supplied basis, including the finite and zero-dimensional cases. Moreover $T_{k}^{*} = (J J^{*})^{*} = J J^{*} = T_{k}$ and $⟨ T_{k} f, f ⟩ = ⟨ J^{*} f, J^{*} f ⟩ \geq 0$ for all $f$ by [A3], so $T_{k}$ is self-adjoint and positive.

A1A3A4A5

1.3

Trace class and the trace formula. Expanding $J^{*}$ in the Hilbert basis and then using the zero-padded enumeration of [A4] gives $J^{*} f = \sum_{i \in I} ⟨ J^{*} f, e_{i} ⟩ e_{i} = \sum_{j \geq 1} ⟨ f, J h_{j} ⟩ h_{j},$ where the first expression is a finite-subset net and the second is its ordinary positive-indexed enumeration (eventually zero in finite dimension). Applying $J$ gives the positive-indexed nuclear representation $T_{k} = J J^{*} = \sum_{j \geq 1} ⟨ \cdot, J h_{j} ⟩ J h_{j} .$ Its zero-based partial sums converge in operator norm by [A5], and its shifted coefficient-norm series satisfies $\sum_{j \geq 1} ∥ J h_{j} ∥^{2} = \sum_{i \in I} ∥ J e_{i} ∥^{2} = \int_{X} k (x, x) d μ (x) < + \infty$ by [step 1.2]. Hence [A5] makes $T_{k}$ trace class with $∥ T_{k} ∥_{1} \leq \int_{X} k (x, x) d μ (x)$ and $tr (T_{k}) = \sum_{j \geq 1} ⟨ J h_{j}, J h_{j} ⟩ = \sum_{i \in I} ∥ J e_{i} ∥^{2} = \int_{X} k (x, x) d μ (x)$ .

step 1.2A4A5

2.1

Conclusion and both boundaries. Claims 1–3 are [step 1.1], [step 1.2] and [step 1.3]. For the first boundary, the diagonal $Δ$ is closed and product-measurable (a compact metric space has a countable base). If $μ$ is nonatomic, Tonelli in [A1] gives $(μ \times μ) (Δ) = \int_{X} μ ({x}) d μ (x) = 0$ . For nonzero $μ$ , the representatives $k$ and $k + 1_{Δ}$ therefore give the same $L^{2}$ class but their diagonal integrals differ by $μ (X) > 0$ . This is a failure in general, not in every measure space: on a singleton with unit mass the kernel class does determine its diagonal value. For the second boundary, here is a continuous Hermitian kernel whose operator is not trace class. For each $n \geq 1$ put $N_{n} = 2^{4 n}$ and choose the explicit finite cluster $X_{n} = {2^{- n} (1 + \frac{j}{2 N_{n}}) : 0 \leq j < N_{n}}, X = {0} \cup ⋃_{n \geq 1} X_{n} .$ The clusters are disjoint, all their points are isolated, and their only accumulation point is $0$ , so $X$ is compact. Give each point of $X_{n}$ mass $2^{- n} / N_{n}$ and give $0$ mass zero. This defines a finite Borel measure of total mass $1$ , regular because finite subsets approximate the mass of any set from inside, and complements of finite subsets of its complement approximate it from outside. Index $X_{n}$ by binary vectors $u \in {0, 1}^{4 n}$ in lexicographic order, and set $H_{n} (u, v) = (- 1)^{u \cdot v}, k (x_{u}, x_{v}) = 2^{- n} H_{n} (u, v) (x_{u}, x_{v} \in X_{n}),$ with $k = 0$ on different clusters and whenever either coordinate is $0$ . The dot product in the exponent is taken modulo $2$ . This real symmetric kernel is continuous: away from $0$ points are isolated; near $(0, 0)$ nonzero block values have modulus $2^{- n} \to 0$ ; near $(0, x)$ or $(x, 0)$ with $x \neq 0$ the kernel is eventually zero. A vector $u$ of odd parity gives $k (x_{u}, x_{u}) = - 2^{- n}$ , so the kernel is not positive semidefinite. Pairing binary vectors differing in a coordinate where $u \neq w$ shows $\sum_{v} (- 1)^{(u + w) \cdot v} = 0$ ; for $u = w$ the sum is $N_{n}$ . Thus $H_{n}^{2} = N_{n} I$ . The normalized singleton indicators form a complete orthonormal basis of this atomic $L^{2}$ space (truncating a square-summable atomic integral proves completeness). On its $X_{n}$ block the operator matrix is $2^{- n} (2^{- n} / N_{n}) H_{n} = 2^{- 6 n} H_{n}$ . Consequently $T_{k}^{*} T_{k}$ is $2^{- 8 n} I$ on that block and $∣ T_{k} ∣$ is $2^{- 4 n} I$ . The block singular values are $2^{- 4 n}$ , repeated $N_{n} = 2^{4 n}$ times. Their squared sum is $\sum_{n \geq 1} 2^{- 4 n} < \infty$ ; their sum is $\sum_{n \geq 1} 1 = \infty$ , so $T_{k}$ is not trace class by Trace class operator. Compactness follows from [A2], or directly because the block norms tend to zero and finite block truncations have finite rank. This proves the second boundary while retaining the positive-kernel conclusion: positivity supplies trace-class membership here; continuity alone does not.

step 1.1step 1.2step 1.3A1A2algebra∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Compact does not imply Hilbert Schmidt

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $ℓ^{2} : = ℓ^{2} (N, F)$ with standard basis $(u_{n})_{n \in N}$ and define the diagonal operator $T u_{0} : = 0, T u_{n} : = n^{- 1 / 2} u_{n} (n \geq 1),$ extended linearly and by continuity. Then $T$ is compact (Compact linear operator) but not Hilbert–Schmidt relative to any Hilbert basis (Hilbert–Schmidt operator and Hilbert–Schmidt norm); that is, compactness does not imply the Hilbert–Schmidt property.

Facts & Assumptions

Given: Countable Choice, the space $ℓ^{2}$ with its standard basis $(u_{n})$ , and the diagonal operator with $d_{0} = 0$ , $d_{n} = n^{- 1 / 2}$ for $n \geq 1$ .

[A1]

Diagonal criteria. For a diagonal operator with bounded sequence $d$ , boundedness, compactness, the Hilbert–Schmidt criterion $\sum_{n} ∣ d_{n} ∣^{2} < + \infty$ relative to the standard basis, and the trace-class criterion $\sum_{n} ∣ d_{n} ∣ < + \infty$ hold as in the diagonal example; the Hilbert–Schmidt property and norm are basis-independent, and the standard basis is orthonormal with $∥ u_{n} ∥_{2} = 1$ (Diagonal Schatten class criteria on ell two, Hilbert–Schmidt operator and Hilbert–Schmidt norm, The Hilbert–Schmidt norm is basis independent, Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[A2]

Divergence and convergence of $p$ -series. For rational $p > 1$ the series $\sum_{k \geq 1} k^{- p}$ converges, while at $p = 1$ the harmonic series $\sum_{k \geq 1} 1 / k$ diverges; in particular $\sum_{n \geq 1} n^{- 1 / 2}$ is not summable because its terms dominate the harmonic terms for $n \geq 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[A3]

Countable Choice is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Counterexample

technique · direct

Given: Countable Choice, the sequence $d_{0} = 0$ , $d_{n} = n^{- 1 / 2}$ , and the diagonal operator $T$ .

1.1

$T$ is compact. The sequence $(d_{n})_{n \in N}$ tends to $0$ (given rational $ε > 0$ , choose a natural $m > 1 / ε^{2}$ ; then $n^{- 1 / 2} < ε$ for $n > m$ ), so by the diagonal compactness criterion [A1] the operator $T$ is compact.

A1A2algebra

1.2

$T$ is not Hilbert–Schmidt. Relative to the standard basis, $\sum_{n} ∥ T u_{n} ∥_{2}^{2} = \sum_{n \geq 1} n^{- 1} = + \infty$ by the divergence of the harmonic series [A2], so $T$ is not Hilbert–Schmidt relative to the standard basis by the diagonal criterion [A1]; since the Hilbert–Schmidt property and its norm are independent of the chosen Hilbert basis [A1], $T$ is not Hilbert–Schmidt relative to any Hilbert basis.

A1A2

2.1

Conclusion. The operator $T$ is compact by [step 1.1] and fails to be Hilbert–Schmidt by [step 1.2]; hence compactness does not imply the Hilbert–Schmidt property.

step 1.1step 1.2A3∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Hilbert Schmidt does not imply trace class

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $ℓ^{2} : = ℓ^{2} (N, F)$ with standard basis $(u_{n})_{n \in N}$ and define the diagonal operator $T u_{0} : = 0, T u_{n} : = n^{- 1} u_{n} (n \geq 1),$ extended linearly and by continuity. Then $T$ is Hilbert–Schmidt relative to the standard basis (Hilbert–Schmidt operator and Hilbert–Schmidt norm) but not trace class (Trace class operator); that is, the Hilbert–Schmidt property does not imply the trace-class property.

Facts & Assumptions

Given: Countable Choice, the space $ℓ^{2}$ with its standard basis $(u_{n})$ , and the diagonal operator with $d_{0} = 0$ , $d_{n} = n^{- 1}$ for $n \geq 1$ .

[A1]

Diagonal criteria. For a diagonal operator with bounded sequence $d$ : boundedness with $∥ T ∥ = \sup_{n} ∣ d_{n} ∣$ , compactness in the case $d_{n} \to 0$ and only there, the Hilbert–Schmidt criterion $\sum_{n} ∣ d_{n} ∣^{2} < + \infty$ with $∥ T ∥_{H S}^{2} = \sum_{n} ∣ d_{n} ∣^{2}$ relative to the standard basis, and the trace-class criterion $\sum_{n} ∣ d_{n} ∣ < + \infty$ with $∥ T ∥_{1} = \sum_{n} ∣ d_{n} ∣$ (Diagonal Schatten class criteria on ell two, Hilbert–Schmidt operator and Hilbert–Schmidt norm, Trace class operator, Absolute value and singular values of a compact operator).

[A2]

$p$ -series. For rational $p > 1$ the series $\sum_{k \geq 1} k^{- p}$ converges, and at $p = 1$ the harmonic series diverges; in particular $\sum_{k \geq 1} k^{- 2} < + \infty$ and $\sum_{k \geq 1} k^{- 1} = + \infty$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[A3]

Countable Choice is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Counterexample

technique · direct

Given: Countable Choice, the sequence $d_{0} = 0$ , $d_{n} = n^{- 1}$ , and the diagonal operator $T$ .

1.1

$T$ is Hilbert–Schmidt. The sequence $(d_{n})$ is bounded by $1$ and $\sum_{n \geq 1} ∣ d_{n} ∣^{2} = \sum_{n \geq 1} n^{- 2} < + \infty$ by [A2]; hence $T$ is Hilbert–Schmidt relative to the standard basis with $∥ T ∥_{H S}^{2} = \sum_{n \geq 1} n^{- 2}$ by the diagonal criterion [A1].

A1A2

1.2

$T$ is not trace class. The positive singular values of the diagonal operator are $n^{- 1}$ , $n \geq 1$ , in nonincreasing order with multiplicity [A1]. Their series $\sum_{n \geq 1} n^{- 1}$ diverges by [A2], so the finiteness condition in the trace-class definition fails and $T$ is not trace class. No value of $∥ T ∥_{1}$ is assigned, because that norm is defined only for trace-class operators.

A1A2

2.1

Conclusion. $T$ is Hilbert–Schmidt by [step 1.1] and not trace class by [step 1.2]; hence the Hilbert–Schmidt property does not imply the trace-class property.

step 1.1step 1.2A3∎

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

The unilateral shift obstructs a cyclic linear trace extension

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $ℓ^{2} : = ℓ^{2} (N, F)$ with standard basis $(u_{n})_{n \in N}$ given by $u_{n} (m) = δ_{m n}$ , and let $S \in B (ℓ^{2})$ be the unilateral forward shift $S u_{n} : = u_{n + 1} (n \in N),$ extended linearly and by continuity, with $P_{0} : = ⟨ \cdot, u_{0} ⟩ u_{0}$ . Then $S^{*} S = I, S S^{*} = I - P_{0}, S^{*} S - S S^{*} = P_{0},$ neither $S^{*} S = I$ nor $S S^{*} = I - P_{0}$ is trace class (Trace class operator), while $P_{0}$ is rank one with $tr (P_{0}) = 1$ (Adjoint, norm and trace of an operator of rank at most one). Consequently there is no linear functional $τ$ on a linear subspace of $B (ℓ^{2})$ that contains the trace-class operators, $S^{*} S$ , and $S S^{*}$ , agrees with the usual trace on trace-class operators, and satisfies $τ (S^{*} S) = τ (S S^{*})$ . Thus the cyclicity identity $tr (S T) = tr (T S)$ of Cyclicity of the trace has no linear cyclic extension whose domain contains this pair of nonsummable products.

Facts & Assumptions

Given: Countable Choice, the space $ℓ^{2}$ with its standard basis $(u_{n})$ , the forward shift $S$ , and the projection $P_{0} = ⟨ \cdot, u_{0} ⟩ u_{0}$ .

[A1]

The standard basis and shifts. The vectors $u_{n} \in ℓ^{2}$ satisfy $⟨ u_{m}, u_{n} ⟩ = δ_{m n}$ and $∥ u_{n} ∥_{2} = 1$ , and if $a \in ℓ^{2}$ has $⟨ a, u_{n} ⟩ = 0$ for every $n$ then $a = 0$ , so the zero-complement characterisation makes $(u_{n})$ a complete orthonormal family, a Hilbert basis of $ℓ^{2}$ ; hence every $x \in ℓ^{2}$ equals $\sum_{n} ⟨ x, u_{n} ⟩ u_{n}$ and two vectors with equal coefficients coincide (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Parseval equivalences for an orthonormal family, Fourier expansion in a Hilbert space, Orthonormal families, complete orthonormal systems and Hilbert bases, Real and complex inner-product spaces and their induced length). The forward shift has $∥ S u_{n} ∥ = 1$ , is an isometry, and its adjoint satisfies $S^{*} u_{0} = 0$ , $S^{*} u_{n} = u_{n - 1}$ for $n \geq 1$ , the adjoint being characterised by $⟨ S x, y ⟩ = ⟨ x, S^{*} y ⟩$ (Hilbert-adjoint identities, The Hilbert-space adjoint of a bounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A2]

Trace-class diagonal test. If $T$ is trace class and $E$ is a Hilbert basis, then $\sum_{e \in E} ∣ ⟨ T e, e ⟩ ∣ \leq ∥ T ∥_{1} < + \infty$ ; hence an operator $T$ for which some Hilbert basis has infinitely many $e$ with $⟨ T e, e ⟩ = 1$ is not trace class (Trace of a trace class operator, Trace is absolutely convergent and basis independent, Trace class operator, Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[A3]

Rank-one operators. For $u, v \in H$ the operator $⟨ \cdot, v ⟩ u$ has adjoint $⟨ \cdot, u ⟩ v$ , norm $∥ u ∥ ∥ v ∥$ , and trace $⟨ u, v ⟩$ ; in particular $P_{0} = ⟨ \cdot, u_{0} ⟩ u_{0}$ has trace $⟨ u_{0}, u_{0} ⟩ = 1$ (Adjoint, norm and trace of an operator of rank at most one, Orthonormal families, complete orthonormal systems and Hilbert bases).

[A4]

Cyclicity theorem. If $T$ is trace class and $S$ is bounded then $S T$ and $T S$ are trace class and have equal traces (Cyclicity of the trace).

Counterexample

technique · direct

Given: Countable Choice, the shift $S$ , the projection $P_{0}$ , and the standard basis.

1.1

The products. Since $S$ is an isometry, $⟨ S^{*} S x, y ⟩ = ⟨ S x, S y ⟩ = ⟨ x, y ⟩$ for all $x, y$ by [A1], so $S^{*} S = I$ . Similarly $⟨ S S^{*} x, u_{n} ⟩ = ⟨ S^{*} x, S^{*} u_{n} ⟩$ for $n \geq 1$ equals $⟨ S^{*} x, u_{n - 1} ⟩ = ⟨ x, u_{n} ⟩$ , while $⟨ S S^{*} x, u_{0} ⟩ = ⟨ S^{*} x, S^{*} u_{0} ⟩ = 0$ ; hence $S S^{*}$ fixes each $u_{n}$ with $n \geq 1$ and annihilates $u_{0}$ , that is $S S^{*} x = x - ⟨ x, u_{0} ⟩ u_{0} = (I - P_{0}) x$ for all $x$ , and $S^{*} S - S S^{*} = P_{0}$ .

1.2

Neither product is trace class. For $I$ with the Hilbert basis $(u_{n})$ , every diagonal coefficient is $⟨ I u_{n}, u_{n} ⟩ = 1$ , so $\sum_{n} ∣ ⟨ I u_{n}, u_{n} ⟩ ∣ = + \infty$ and $I$ is not trace class by [A2]. For $I - P_{0}$ , the coefficients at $u_{n}$ with $n \geq 1$ are $⟨ (I - P_{0}) u_{n}, u_{n} ⟩ = 1$ , again infinitely many equal to $1$ , so $I - P_{0}$ is not trace class by [A2].

A1A2

1.3

The difference has trace one. $P_{0} = ⟨ \cdot, u_{0} ⟩ u_{0}$ is a rank-one operator whose trace is $⟨ u_{0}, u_{0} ⟩ = 1$ by [A3]; note that $u_{0} \neq 0$ because it is a unit vector.

2.1

Conclusion. Suppose that a linear functional $τ$ on a linear subspace containing the trace-class operators, $S^{*} S$ , and $S S^{*}$ agreed with the usual trace on trace-class operators and satisfied $τ (S^{*} S) = τ (S S^{*})$ . By linearity, [step 1.1], and [step 1.3], $0 = τ (S^{*} S) - τ (S S^{*}) = τ (P_{0}) = tr (P_{0}) = 1,$ a contradiction. Hence no such cyclic linear extension exists. The products themselves are not trace class by [step 1.2], so [A4] neither asserts nor assigns their individual traces.

step 1.1step 1.2step 1.3A4assume-contradischarge-contradiction∎

RemarkRemark: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Schatten p classes

Remark

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ and $K$ be real or complex Hilbert spaces and let $T \in B (H, K)$ be compact (Compact linear operator) with zero-padded singular-value sequence $(s_{n} (T))_{n \geq 1}$ (Absolute value and singular values of a compact operator). For a real $1 \leq p < + \infty$ the Schatten $p$ -class $S_{p} (H, K)$ is defined to be the set of compact operators with $∥ T ∥_{p} : = (\sum_{n \geq 1} s_{n} (T)^{p})^{1 / p} < + \infty .$ This is an orientation remark only. It records the scale of ideals without developing any of its theory.

The two endpoints that this page does develop are recognised as follows. $S_{1} (H, K)$ is exactly the trace-class ideal and $∥ \cdot ∥_{1}$ is the trace norm, by definition of the latter (Trace class operator), including the zero padding and the finite-rank case. $S_{2} (H, K)$ is exactly the class of operators that are Hilbert–Schmidt relative to a supplied Hilbert basis $E$ of $H$ : for such a basis, let $(e_{j})_{j \in J} \subseteq H$ and $(f_{j})_{j \in J} \subseteq K$ be the right and left singular families supplied by the SVD. Its expansion gives $T e = \sum_{j \in J} s_{j} ⟨ e, e_{j} ⟩ f_{j},$ so orthonormality of the $f_{j}$ , Parseval for the supplied basis $E$ , and the interchange of the nonnegative finite-subset suprema give $\sum_{e \in E} ∥ T e ∥^{2} = \sum_{n \geq 1} s_{n} (T)^{2},$ since $\sum_{e \in E} ∥ T e ∥^{2} = \sum_{j \in J} s_{j}^{2} \sum_{e \in E} ∣ ⟨ e, e_{j} ⟩ ∣^{2} = \sum_{j \in J} s_{j}^{2},$ and the last sum is the zero-padded singular-value sum (Singular value decomposition for compact operators, Hilbert–Schmidt operator and Hilbert–Schmidt norm, Parseval equivalences for an orthonormal family, Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Orthonormal families, complete orthonormal systems and Hilbert bases), so that, when the common sum is finite, $∥ T ∥_{2} = ∥ T ∥_{H S, E}$ . For clarity, the interchange uses only finite rectangles: every finite subset of $E \times J$ lies in the product of its two finite projections, and finitely many finite sets of indices have finite union. Thus both iterated nonnegative suprema equal the supremum over finite rectangles, even when they are infinite. The norm identity for each $T e$ follows first for finite orthogonal sums and then by norm convergence of its SVD expansion. Parseval is applied in $H$ to $e_{j}$ , with $∣ ⟨ e, e_{j} ⟩ ∣ = ∣ ⟨ e_{j}, e ⟩ ∣$ . Conversely, every bounded operator Hilbert–Schmidt relative to $E$ is compact by Hilbert–Schmidt operators are compact, so the same calculation applies to it and puts it in $S_{2} (H, K)$ . No basis of $K$ is required, and the norm retains the notation $∥ T ∥_{H S, E}$ of its definition. The formula proves the same value for every supplied basis of $H$ in this compact-operator setting; it does not assert existence of such a basis. Empty singular families and an empty domain basis contribute zero. For $p = + \infty$ one writes $S_{\infty} (H, K)$ for the compact operators with the operator norm $∥ T ∥_{S_{\infty}} : = ∥ T ∥ = s_{1} (T)$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Deliberate boundaries. Nothing here asserts completeness of $∥ \cdot ∥_{p}$ , Hölder or Young inequalities, duality, interpolation, or the identification of $S_{p}$ with a space of sequences; no monotonicity of the norms beyond $∥ T ∥ \leq ∥ T ∥_{1}$ from the trace-class page is claimed. Later items must not use this remark as a supplier: it is recorded for orientation, exactly as the functional-analysis plan's FA-16 boundary requires, and the general theory of Schatten classes belongs to a later, separate development.

Compact Self Adjoint Hilbert Schmidt and Trace Class Operators — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Diagonal Schatten class criteria on ell two

Example

Facts & Assumptions

Verification

Volterra operator is Hilbert Schmidt and quasinilpotent

Example

Facts & Assumptions

Verification

Adjoint, norm and trace of an operator of rank at most one

Example

Facts & Assumptions

Verification

Integral operator trace under a valid diagonal hypothesis

Example

Facts & Assumptions

Verification

Compact does not imply Hilbert Schmidt

Statement refuted

Facts & Assumptions

Counterexample

Hilbert Schmidt does not imply trace class

Statement refuted

Facts & Assumptions

Counterexample

The unilateral shift obstructs a cyclic linear trace extension

Statement refuted

Facts & Assumptions

Counterexample

Schatten p classes

Remark

Sources