# 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 $i$th component flipped. Rewrite the previous sum as

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

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