# Question about lemma concerning uniformly convex Banach spaces?

I was going through Jurgen Jost's book on the Calculus of Variations. In his discussion of uniformly convex Banach spaces he states the following result.

Given a sequence $\{f_n\} \subset V$ where $V$ is a unifomly convex Banach space, if

(a). $\lim\limits_{n \to \infty} \sup ||f_n|| \leq 1$

(b). $\lim\limits_{n,m \to \infty}\Bigg|\Bigg|(f_n+f_m)/2\Bigg|\Bigg|=1$

, then $\exists f \in V, ||f||=1$ s.t. $||f_n -f|| \to 0$.

The proof of the result is pretty straightforward. I was just wondering if (b). alone guarantees $||f_n|| \to 1$. Also struggling to make sense of the notation used for the double limit. Does it mean $\forall \epsilon >0, \exists A$ s.t. $n,m \geq A \implies |a_{nm}-1|<\epsilon$ where $a_{nm}$ is $\Bigg|\Bigg|(f_n+f_m)/2\Bigg|\Bigg|$.

It is then not hard to show from this (by taking $m= n$) that $\|f_n\| \to 1$. Hence, the first assumption is superfluous.
• Thanks for the help. Although superfluous for the $||f_n||\to 1$ just thought I'd mention (a). is needed for the full result. Cheers! – almosteverywhere May 20 '17 at 21:01
• @almosteverywhere: What I meant is that if $\|f_n \| \to 1$, then $\lim\sup \|f_n\| = \lim \| f_n \| = 1 \leq 1$, so (a) is also superfluous for the full result. – PhoemueX May 21 '17 at 5:39