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?
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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 $(1,0)$. 

