Let $(X, \|\cdot\|)$ be an Banach space. Assume that a sequence $f\_n \rightarrow f$ weakly in $X$, and $\|f\_n\| \rightarrow \|f\|$ as $n \rightarrow \infty$. It's known that if $X$ is a uniformly convex Banach space, then we have $f\_n \rightarrow f$ strongly in $X$. Is this true for an reflexive Banach space?



Strong convergence in reflecxive Banach space
I take it "strongly" means "in norm"? No, this can fail for reflexive Banach spaces.


For a counterexample, let $X$ be the $l^\infty$ direct sum of ${\bf R}$ and $l^2$. (I'm doing this for real Banach spaces, but the same counterexample works in the complex case too.) Thus a typical element of $X$ looks like $(a,f)$ where $a \in {\bf R}$ and $f \in l^2$, and its norm is ${\rm max}(|a|, \|f\|\_2)$. Since $X$ is a direct sum of two reflexive spaces, it is reflexive. Now consider the sequence $(1, e\_n)$ where $(e\_n)$ is the standard basis of $l^2$. It's easy enough to check that this sequence converges weakly, but not in norm, to $(1,0)$.


