Let $X$ be a Banach space. Let $\{x_n\}$ be a sequence in $X$ converging weakly to $x \in X$. Then by using Hahn-Banach theorem to get $f \in X^*$ with $||f|| \le 1$ and $|f(x)|=||x||$, we see that $||x|| \le \lim \inf ||x_n||$. So if moreover $||x_n|| \le ||x||, \forall n \in \mathbb N$, then we see that $||x_n|| \to ||x||$ . Moreover, if $X$ were actually a Hilbert space, then this would imply $x_n \to x$ in norm . So my question is the following :
Let ${x_n}$ be a sequence in a Banach space $X$ such that $\{x_n\}$ converges weakly to $x \in X$ and $||x_n|| \le ||x||, \forall n \in \mathbb N$ . Then is it true that $x_n \to x$ in norm ?
Counterexample in $C([0,1])$: Let $f_n$ be the piecewise-linear function interpolating the values $f_n(0)=1$, $f_n(1/n)=0$, $f_n(2/n)=1$, $f_n(1)=1$. Dominated convergence shows that $f_n\to1$ weakly.