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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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up vote 7 down vote accepted

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 $(1,0)$.

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Yes, it is, thanks for your nice example. –  Wang Ming May 5 '12 at 12:14
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Just to add to Nik answer, reflexivity is invariant to equivalent renormings (topologic property), while uniform convexity is not, it depends on the geometry of the norm (geometric property). The hypotheses that $||f_n||\to||f||$ is also a geometric property, and the conclusion that $f_n\to f$ is again topologic. This should give you the intuition, before any concrete counter-example, that the property shouldn't hold under the assumption of reflexivity, or any other topologic property for that matter. –  Adi Tcaciuc May 9 '12 at 17:49
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