Enflo's quantitative localized-trace obstruction

Every bounded finite-rank $T$ is an operator-norm limit of finite-rank finite-expansion operators. Indeed, choose a finite basis $f_1,\dots ,f_r$ of $T(B)$ and bounded coefficient functionals $b_j$ with $Tx={\sum }_j{b}_j(x)f_j$. Approximate each $f_j$ by a finite linear combination $f_j^{\mathrm{\prime }}$ of the dense generators so that $T^{\mathrm{\prime }}x={\sum }_j{b}_j(x)f_j^{\mathrm{\prime }}$ has $\mathrm{\parallel }T-T^{\mathrm{\prime }}\mathrm{\parallel }<\epsilon$. Each $T^{\mathrm{\prime }}{e}_k$ has finite expansion. [L1, finite-dimensional coordinates, dense span] (finite-dimensional coordinates; dense span).

If $T$ is finite expansion and $e_k\in M$, property A gives $\mathrm{\mid }{a}_{kk}\mathrm{\mid }\le \frac{\mathrm{\parallel }T{e}_k\mathrm{\parallel }}{\mathrm{\parallel }{e}_k\mathrm{\parallel }}\le \mathrm{\parallel }T{\mathrm{\parallel }}_{(M)}\mathrm{.}$ Averaging over $M$ yields $\mathrm{\mid }\stackrel{}{\mathrm{Tr}}(M,T)\mathrm{\mid }\le \mathrm{\parallel }T{\mathrm{\parallel }}_{(M)}$. [L1, property A] (property A).

If $T$ is also finite rank, its range is spanned by finitely many of the 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 $\stackrel{}{\mathrm{Tr}}(M_k,T)\to 0$. [L1, finite rank, disjointness] (finite rank; disjointness).

Apply step 2.1 to $I-T$ on $M_m$ and telescope step 3.1: $\begin{array}{rl}{\displaystyle \mathrm{\parallel }I-T{\mathrm{\parallel }}_{(M_m)}}& {\displaystyle \ge \mathrm{\mid }1-\stackrel{}{\mathrm{Tr}}(M_m,T)\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle \ge 1-K\mathrm{\parallel }T\mathrm{\parallel }\sum _{k=m}^{\mathrm{\infty }}\frac{1}{\log \mathrm{\mid }{M}_k\mathrm{\mid }}\mathrm{.}}\end{array}$ The growth hypothesis gives $\log \mathrm{\mid }{M}_k\mathrm{\mid }>a^{k-m}\log \mathrm{\mid }{M}_m\mathrm{\mid }$, so the geometric sum is at most $((1-a^{-1})\log \mathrm{\mid }{M}_m\mathrm{\mid }{)}^{-1}$. This proves the estimate for finite-rank finite-expansion $T$. [steps 2.1, 3.1, trace hypothesis, geometric series] (trace hypothesis; geometric series).

For arbitrary bounded finite-rank $T$, take the approximants from step 1.1 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 1.1, 4.1].

If $B$ had $\lambda$-BAP, choose $m$ with $C\lambda \mathrm{/}\log \mathrm{\mid }{M}_m\mathrm{\mid }<1\mathrm{/}2$. The unit ball of finite-dimensional $[M_m]$ is compact, so [L2] would give a finite-rank $T$ with $\mathrm{\parallel }T\mathrm{\parallel }\le \lambda$ and $\mathrm{\parallel }I-T{\mathrm{\parallel }}_{(M_m)}<1\mathrm{/}2$, contradicting step 5.1. [L2, step 5.1, finite-dimensional compactness] (finite-dimensional compactness).