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Let $E$ be a reflexive Banach space and consider a sequence $\{x_n\}_{n\in\mathbb{N}}\subset E$ weakly converging to some $x\in E$. Moreover, assume that the sequence satisfies $$ \Vert x_n\Vert_E=1 \quad \hbox{for all }\,n\in\mathbb{N}. $$ Does that implies $x_n\to x$ strongly in $E$? I know that weakly convergence plus $$ \limsup\Vert x_n\Vert_E\leq \Vert x\Vert_E, $$ implies strong convergence, so I would like to use something like that. On the other hand, I don't know anything about the norm of $x$, and I think that this is not true in the case of the weak-* topology. I mean, for any such a Banach space, there always exists sequences of constant norm weakly-* converging to zero.

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No. For instance, the canonical Hilbert basis $\{e_n\}_{n\in\Bbb N}$ of $\ell^2$, as a sequence, converges weakly to $0$.

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For a counterexample which underlying space is not a Hilbert space, consider any $1<p\ne 2<\infty$ that $L^{p}(\mathbb{R})$, this post exhibits some weakly convergent sequence which fails to be strongly convergent, one can normalize it to have norm $1$.

Note that $L^{p}(\mathbb{R})$ is reflexive for all $1<p<\infty$, this is due to Riesz Representation Theorem.

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