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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∎

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Sources

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.6, Hilbert–Schmidt but not trace-class diagonal operators (standard reference, not scraped)