Suppose in a normed space $X$ we have $x_n \rightharpoonup x_0 $ (weakly converge). Prove $\liminf_\limits{n \to \infty} ||x_n|| \ge ||x_0||$.
Here, in a normed space $X$, a sequence $x_n$ weakly converge to $x_0$ is defined by:
$\forall f \in X^*$ (the space of all continuous linear functional), we have $f(x_n)\to f(x_0)$.
My attempt:
By Mazur's lemma, we know $\forall \epsilon > 0, \exists \lambda_i \ge 0 \, (i = 1,2,\dots n)$ and $\sum_i^n \lambda_i =1$ s.t. $||\sum_\limits{ i = 1}^n \lambda_i x_i-x_0|| <\epsilon$. Let $A = \{\lambda_1 x_1 + \cdots+\lambda_k x_k| k \in N^+, \lambda_i \ge 0, i = 1,\dots,k, \sum_\limits{i = 1}^k \lambda_i = 1\}$, then $x_0 \in \bar{A}$. Suppose $y_n = \sum_\limits{i = 1}^{k_n} \lambda_ix_i \to x_0$.
But I don't know how to carry on.
Best Answer
There exists $f \in X^{*}$ such that $\|f\|=1$ and $f(x_0)=\|x_0\|$. So $\|x_0\| =f(x_0)=\lim f(x_n)\leq \lim \inf \|x_n\|$.