Almost norming functionals, Part 2

Let E be a real Banach space with the dual E^*. Fix \delta\in (0,1) and call a linear functional e^*\in E^* almost norming for e if |e|=|e^*|=1 and e^*(e)\ge \delta. In Part 1 I showed that in any Banach space there exists a continuous selection of almost norming functionals. Here I will prove that there is no uniformly continuous selection in \ell_1.

Claim. Let S be the unit sphere in \ell_1^n, the n-dimensional \ell_1-space.  Suppose that f\colon S\to \ell_{\infty}^n is a map such that f(e) is almost norming e in the above sense. Then the modulus of continuity \omega_f satisfies \omega_f(2/n)\ge 2\delta.

(If an uniformly continuous selection was available in \ell_1, it would yield selections in \ell_1^n with a modulus of continuity independent of n.)

Proof. Write f=(f_1,\dots,f_n). For any \epsilon\in \{-1,1\}^n we have n^{-1}\epsilon \in S, hence

\sum\limits_{i=1}^n \epsilon_i f_i(n^{-1}\epsilon)\ge n\delta for all \epsilon\in \{-1,1\}^n. Sum over all \epsilon and change the order of summation:

\sum\limits_{i=1}^n \sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon)\ge n2^n\delta

There exists i\in\{1,2,\dots,n\} such that

\sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon) \ge 2^n \delta

Fix this i from now on. Define \tilde \epsilon to be the same \pm vector as \epsilon, but with the ith component flipped. Rewrite the previous sum as

\sum\limits_{\epsilon} -\epsilon_i f_i(n^{-1}\tilde \epsilon)\ge 2^n\delta

and add them together:

\sum\limits_{\epsilon}\epsilon_i [f_i(n^{-1}\epsilon)-f_i(n^{-1}\tilde \epsilon)]\ge 2^{n+1}\delta

Since \|n^{-1}\epsilon-n^{-1}\tilde \epsilon\|=2/n, it follows that 2^n \omega_f(2/n) \ge 2^{n+1}\delta, as claimed.

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