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Nuclear bounded cyclicity for trace-class operators

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space) and for $u, v \in H$ let $T_{u, v} x = ⟨ x, v ⟩ u$ (Rank-one operator calculus).

(a) For every bounded $B \in B (H)$ (A bounded linear operator between normed spaces),

$B T_{u, v} = T_{B u, v}, T_{u, v} B = T_{u, B^{*} v},$

with the Hilbert adjoint $B^{*}$ (The Hilbert-space adjoint of a bounded operator).

(b) Let $T \in B (H)$ be trace class and let $T = \sum_{j} T_{u_{j}, v_{j}}$ be a nuclear representation, i.e. the partial sums converge to $T$ in operator norm and $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ (Nuclear series characterizes trace norm, Trace class operator). Then for every bounded $B$

$tr (B T) = \sum_{j} ⟨ B u_{j}, v_{j} ⟩, tr (T B) = \sum_{j} ⟨ u_{j}, B^{*} v_{j} ⟩,$

both series converging absolutely with $∣ tr (B T) ∣ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ , and $tr (B T) = tr (T B)$ (Trace is absolutely convergent and basis independent). In particular $tr (B T) = tr (T B)$ for every trace-class $T$ and bounded $B$ , and $tr (T_{u, v}) = ⟨ u, v ⟩$ . No cyclic permutation of unbounded operators is asserted.

Facts & Assumptions

Given: A complex Hilbert space $H$ ; vectors $u, v, u_{j}, v_{j} \in H$ ; bounded operators $B$ ; a trace-class $T$ with nuclear representation $T = \sum_{j} T_{u_{j}, v_{j}}$ .

[F1]

$T_{u, v} \in B (H)$ is trace class, $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ , $tr (T_{u, v}) = ⟨ u, v ⟩$ , and $T_{u, v}^{*} = T_{v, u}$ (Rank-one operator calculus).

[F2]

$T$ is trace class exactly when it has a nuclear representation $T = \sum_{j} T_{u_{j}, v_{j}}$ with $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ converging in operator norm; and then $∥ T ∥_{1} = \inf \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ over all such representations (Nuclear series characterizes trace norm).

[F3]

The trace-class operators form a two-sided Banach operator ideal: $B T \in S_{1} (H)$ for bounded $B$ and trace-class $T$ , with $∥ B T ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1}$ for the operator norm (Trace class is a two sided Banach operator ideal, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

The trace is linear on trace-class operators and $∣ tr S ∣ \leq ∥ S ∥_{1}$ (Trace is absolutely convergent and basis independent).

[F5]

Bounded adjoints obey $∥ B^{*} ∥ = ∥ B ∥$ (Hilbert-adjoint identities). The Hilbert adjoint satisfies $⟨ B x, y ⟩ = ⟨ x, B^{*} y ⟩$ and $T_{u, v}^{*} = T_{v, u}$ (The Hilbert-space adjoint of a bounded operator, Rank-one operator calculus).

Proof

1.1givenF5algebra

[Rank-one multiplication] For $x \in H$ , $(B T_{u, v}) x = B (⟨ x, v ⟩ u) = ⟨ x, v ⟩ B u = T_{B u, v} x$ , so $B T_{u, v} = T_{B u, v}$ ; and $(T_{u, v} B) x = ⟨ B x, v ⟩ u = ⟨ x, B^{*} v ⟩ u = T_{u, B^{*} v} x$ by [F5], so $T_{u, v} B = T_{u, B^{*} v}$ . Both are bounded because $B$ is and the displayed maps have the stated form.

2.1step 1.1F1F2F3algebra

[Trace-norm convergence of the multiplier series] Write $T_{n} : = \sum_{j \leq n} T_{u_{j}, v_{j}}$ , so $∥ T - T_{n} ∥ \to 0$ . Then $B T_{n} = \sum_{j \leq n} T_{B u_{j}, v_{j}}$ by step 1.1, and $\sum_{j} ∥ T_{B u_{j}, v_{j}} ∥_{1} = \sum_{j} ∥ B u_{j} ∥ ∥ v_{j} ∥ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ by [F1]; since the trace-class operators are complete in trace norm [F3], the series $\sum_{j} T_{B u_{j}, v_{j}}$ converges in trace norm, and its limit agrees with $B T$ because trace-norm convergence implies operator-norm convergence and $B T_{n} \to B T$ in operator norm. Hence $B T = \sum_{j} T_{B u_{j}, v_{j}}$ in trace norm, with $∥ B T ∥_{1} \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ .

3.1step 2.1F1F4algebra

[Trace of the left product] By [F4], $tr (B T) = \lim_{n} tr (B T_{n}) = \lim_{n} \sum_{j \leq n} tr (T_{B u_{j}, v_{j}}) = \lim_{n} \sum_{j \leq n} ⟨ B u_{j}, v_{j} ⟩$ , because the trace is linear on the finite sums $B T_{n}$ and continuous for the trace norm, and $B T_{n} \to B T$ in trace norm by step 2.1. Thus $tr (B T) = \sum_{j} ⟨ B u_{j}, v_{j} ⟩$ , and the series converges absolutely since $∣ ⟨ B u_{j}, v_{j} ⟩ ∣ = ∣ tr (T_{B u_{j}, v_{j}}) ∣ \leq ∥ B u_{j} ∥ ∥ v_{j} ∥$ by [F1]; the bound $∣ tr (B T) ∣ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ follows from [F4] and step 2.1.

4.1step 1.1step 2.1step 3.1F1F2F3F4F5algebra∎

[Right multiplication and cyclicity] By step 1.1, $T_{n} B = \sum_{j \leq n} T_{u_{j}, B^{*} v_{j}}$ , and $\sum_{j} ∥ u_{j} ∥ ∥ B^{*} v_{j} ∥ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ by [F5]. Exactly as in step 2.1, trace-norm completeness gives convergence of this series to $T B$ , since its operator-norm limit is $T B$ . Trace continuity and the rank-one trace formula therefore give $tr (T B) = \sum_{j} ⟨ u_{j}, B^{*} v_{j} ⟩$ with absolute convergence. Each summand equals $⟨ B u_{j}, v_{j} ⟩$ by the adjoint identity, so step 3.1 gives $tr (T B) = tr (B T)$ for the original arbitrary nuclear representation. Every trace-class operator has such a representation by [F2]; the stated rank-one trace identity is [F1]. Only bounded multipliers were used.

Nuclear bounded cyclicity for trace-class operators

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources