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Enflo's quantitative localized-trace obstruction

Statement

Let $B$ have a dense linearly independent generator with property A. Suppose there are pairwise disjoint nonempty finite subsets $M_{m}$ of the generator and constants $a > 1$ , $K > 0$ such that

$∣ M_{m + 1} ∣ > ∣ M_{m} ∣^{a}$

and, for every bounded finite-expansion operator $T$ ,

$∣ \overset{}{Tr} (M_{m + 1}, T) - \overset{}{Tr} (M_{m}, T) ∣ \leq \frac{K ∥ T ∥}{\log ∣ M_{m} ∣} .$

Then every bounded finite-rank operator $T$ satisfies

$∥ I - T ∥_{(M_{m})} \geq 1 - \frac{C ∥ T ∥}{\log ∣ M_{m} ∣}, C : = \frac{K}{1 - a^{- 1}} .$

Consequently $B$ has no $λ$ -BAP for any finite $λ$ .

Facts & Assumptions

[L1]

Finite-expansion matrices, normalized localized trace, property A, and $∥ \cdot ∥_{(M)}$ have the fixed-generator meanings of Enflo finite-expansion and localized-trace system.

[L2]

$λ$ -BAP gives norm- $λ$ finite-rank approximations uniformly on each compact set (Approximation property and bounded approximation property).

Proof

technique · direct

Given: The objects and hypotheses in the Statement.

1.1

Every bounded finite-rank $T$ is an operator-norm limit of finite-rank [given, L1] finite-expansion operators. Indeed, choose a finite basis $f_{1}, \dots, f_{r}$ of $T (B)$ and bounded coefficient functionals $b_{j}$ with $T x = \sum_{j} b_{j} (x) f_{j}$ . Approximate each $f_{j}$ by a finite linear combination $f_{j}^{'}$ of the dense generators so that $T^{'} x = \sum_{j} b_{j} (x) f_{j}^{'}$ has $∥ T - T^{'} ∥ < ε$ . Each $T^{'} e_{k}$ has finite expansion.

L1finite-dimensional coordinatesdense span

2.1

If $T$ is finite expansion and $e_{k} \in M$ , property A gives

givenL1step 1.1

$∣ a_{k k} ∣ \leq \frac{∥ T e_{k} ∥}{∥ e_{k} ∥} \leq ∥ T ∥_{(M)} .$

Averaging over $M$ yields $∣ \overset{}{Tr} (M, T) ∣ \leq ∥ T ∥_{(M)}$ . [L1, property A]

3.1

If $T$ is also finite rank, its range is spanned by finitely many of the [given, L1, step 2.1] vectors $T e_{j}$ , hence is contained in the span of a finite subset of the generator. Because the $M_{k}$ are disjoint, its diagonal coefficients on $M_{k}$ vanish for all sufficiently large $k$ . Thus $\overset{}{Tr} (M_{k}, T) \to 0$ .

L1finite rankdisjointness

4.1

Apply step 2.1 to $I - T$ on $M_{m}$ and telescope step 3.1:

givenstep 2.1step 3.1

$\begin{aligned} ∥ I - T ∥_{(M_{m})} & \geq ∣ 1 - \overset{}{Tr} (M_{m}, T) ∣ \\ \geq 1 - K ∥ T ∥ \sum_{k = m}^{\infty} \frac{1}{\log ∣ M_{k} ∣} . \end{aligned}$

The growth hypothesis gives $\log ∣ M_{k} ∣ > a^{k - m} \log ∣ M_{m} ∣$ , so the geometric sum is at most $((1 - a^{- 1}) \log ∣ M_{m} ∣)^{- 1}$ . This proves the estimate for finite-rank finite-expansion $T$ . [steps 2.1, 3.1, trace hypothesis, geometric series]

5.1

For arbitrary bounded finite-rank $T$ , take the approximants from step 1.1 [given, step 1.1, step 4.1] and pass to the limit in the operator norm, the restriction norm, and the right-hand side. This proves the displayed estimate in full generality.

steps 1.14.1

6.1

If $B$ had $λ$ -BAP, choose $m$ with [given, L2, step 5.1] $C λ / \log ∣ M_{m} ∣ < 1 / 2$ . The unit ball of finite-dimensional $[M_{m}]$ is compact, so [L2] would give a finite-rank $T$ with $∥ T ∥ \leq λ$ and $∥ I - T ∥_{(M_{m})} < 1 / 2$ , contradicting step 5.1.

L2step 5.1finite-dimensional compactness∎

Depends on

Used by

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Sources

Per Enflo, A counterexample to the approximation problem in Banach spaces (standard reference, not scraped)