property of uniformly convex Banach Space

Theorem 1.

Let $X$ be a uniformly convex Banach space. Let $\mathrm{(}{x}_n\mathrm{)}$ be a sequence in $X$ such that $\lim \mathit{}{x}_n\mathrm{=}x$ in the weak-topology $\mathrm{(}w\mathit{}\mathrm{(}X\mathrm{,}{X}^{\mathrm{*}}\mathrm{)}\mathrm{)}$ and $\mathrm{lim\; sup}\mathit{}\mathrm{\parallel }{x}_n\mathrm{\parallel }\mathrm{\le }\mathrm{\parallel }x\mathrm{\parallel }$.Then $x_n$ converges to $x$.

Proof.

For $x=0$ the claim is obvious, so suppose that $x\ne 0$.The sequence ${(x_n)}_{n\ge 1}$ converges to $x$ for $w-topology$ $\Rightarrow \parallel x\parallel \le lim\; inf\parallel x_n\parallel$.So let ${\lambda }_n=\max \{\parallel x\parallel ,\parallel x_n\parallel \}$ and we have that $lim{\lambda }_n=\parallel x\parallel$.Define $y_n=\frac{x_n}{{\lambda }_n}$ and $y=\frac{x}{\parallel x\parallel }$.Then $y_n$ converges to $y$ in $w-topology$.We conclude that $\parallel y\parallel \le lim\; inf\parallel \frac{y_n+y}{2}\parallel$. Also, $\parallel y\parallel =1,\parallel y_n\parallel \le 1$ so we have that $lim\parallel \frac{y_n+y}{2}\parallel =1$. As the Banach space is uniformly convex one can easily see that$lim\parallel y_n-y\parallel =0$. Therefore $x_n$ converges to $x$. The proof is complete.∎