The question is the following:
Show that if the sequence $(x_n)$ in a normed linear space $(X,|| \cdot ||)$ is weakly convergent to $x_0 \in X$, then
$$
||x_0|| \leq \liminf_{n \to \infty} ||x_n||
$$
Here is my attmpt:
Assume that $(x_n)$ in a normed linear space $(X, ||\cdot||)$ is weakly convergent to $x_0 \in X$. Namely, for every bounded linear function $\varphi \in X^*$
$$
\lim_{n \to \infty} \varphi(x_n) = \varphi(x_0)
$$
We now first assume that $(x_n)$ are nonzero. By Hahn-Banach Theorem, choose a bounded linear function $\varphi'$ satisfying
$$
||\varphi'|| = 1 \quad \text{and} \quad \varphi' (x) = ||x||
$$
Therefore
$$
\varphi'(x_n) = ||x_n||\quad \text{and} \quad \varphi'(x_0) = ||x_0||
$$
By the weak convergence, we have
$$
\lim_{n \to \infty} \varphi'(x_n) = \varphi'(x_0) \quad \Longrightarrow \quad \lim_{n \to \infty} ||x_n|| = ||x_0||
$$
Now if $(x_n)$ are zero, then the limit is also zero, and the result is trivial. If $(x_n)$ are nonzero, but the limit is zero, then
$$
\liminf_{n \to \infty}{||x_n||} \geq 0 = ||x_0||
$$
Here we can conclude the result.
which fails because the choice of $\phi'$ is not valid (Thanks @Kabo Murphy). Then how can I do this problem? Any help is appreciated.
Best Answer
Choose $\phi \in X^{*}$ such that $\|\phi \|=1$ and $\phi (x_0)=\|x_0\|$. Then $|\phi(x_n)| \leq \|\phi \| \|x_n\|=\|x_n\|$. Take $lim \inf$ on both sides. Since $\phi(x_n) \to \phi(x_0)$ we get $\|x_0\|=|\phi (x_0)| \leq \lim \inf \|x_n\|$.