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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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Compact does not imply Hilbert Schmidt

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Let 2:=2(N,F) with standard basis (un)nN and define the diagonal operator Tu0:=0,Tun:=n1/2un(n1), 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 2 with its standard basis (un), and the diagonal operator with d0=0, dn=n1/2 for n1.

[A1]

Diagonal criteria. For a diagonal operator with bounded sequence d, boundedness, compactness, the Hilbert–Schmidt criterion ndn2<+ relative to the standard basis, and the trace-class criterion ndn<+ hold as in the diagonal example; the Hilbert–Schmidt property and norm are basis-independent, and the standard basis is orthonormal with un2=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 2(I)).

[A2]

Divergence and convergence of p-series. For rational p>1 the series k1kp converges, while at p=1 the harmonic series k11/k diverges; in particular n1n1/2 is not summable because its terms dominate the harmonic terms for n1 (For rational p>0, 1/kp converges iff p>1).

[A3]

Countable Choice is the standing hypothesis (The Axiom of Countable Choice (ACω)).

Counterexample

technique · direct

Given: Countable Choice, the sequence d0=0, dn=n1/2, and the diagonal operator T.

1.1

T is compact. The sequence (dn)nN tends to 0 (given rational ε>0, choose a natural m>1/ε2; then n1/2<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, nTun22=n1n1=+ 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

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