# property of uniformly convex Banach Space

###### Theorem 1.

Let $X$ be a uniformly convex Banach space. Let $(x_{n})$ be a sequence in $X$ such that $\lim x_{n}=x$ in the weak-topology $(w(X,X^{*}))$ and $\limsup\|x_{n}\|\leq\|x\|$.Then $x_{n}$ converges to $x$.

###### Proof.

For $x=0$ the claim is obvious, so suppose that $x\neq 0$.The sequence $({x_{n}})_{n\geq 1}$ converges to $x$ for $w-topology$ $\Rightarrow\|x\|\leq\liminf\|x_{n}\|$.So let $\lambda_{n}=\max\{\|x\|,\|x_{n}\|\}$ and we have that $\lim\lambda_{n}=\|x\|$. Define $y_{n}=\frac{x_{n}}{\lambda_{n}}$ and $y=\frac{x}{\|x\|}$.Then $y_{n}$ converges to $y$ in $w-topology$. We conclude that $\|y\|\leq{\liminf{\|\frac{y_{n}+y}{2}\|}}$. Also, $\|y\|=1,\|y_{n}\|\leq 1$ so we have that $\lim\|\frac{y_{n}+y}{2}\|=1$. As the Banach space is uniformly convex one can easily see that $\lim\|y_{n}-y\|=0$. Therefore $x_{n}$ converges to $x$. The proof is complete. ∎

Title property of uniformly convex Banach Space PropertyOfUniformlyConvexBanachSpace 2013-03-22 15:14:18 2013-03-22 15:14:18 georgiosl (7242) georgiosl (7242) 20 georgiosl (7242) Theorem msc 46H05