# Strong convergence for constant norm sequences

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.

## 2 Answers

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

For a counterexample which underlying space is not a Hilbert space, consider any $$1 that $$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, this is due to Riesz Representation Theorem.