Weak Convergence and its Relationship to a Sequence of Norms

Question

"Prove that $x_n\xrightarrow{w} x$ implies lim inf$||x_n|| \geq ||x||$."

I'm trying to understand weak convergence better through this exercise.

Here, $\xrightarrow{w}$ means weakly convergent, i.e. $\forall f \in X'$ (dual space of bounded functionals), $f(x_n) \to f(x)$.

There is a proof that shows $||x_n||$ is bounded. In the proof, $g_n \in X''$ (double dual of $X$) is defined as follows: $g_n(f):=f(x_n)$. The lemma $||g_n||=||x_n||$ is used to complete the proof which comes from a corollary of the Hahn-Banach Theorem. I was trying to use these two facts in the proof of the above by substituting $||x_n||$ with $||g_n||$ but couldn't make much progress.

Any help is appreciated.

Answer 1

Let $x$ be the weak limit of a weakly convergent sequence $(x_n)$ in a normed space $X$. The inequality is trivially true when $x=0$ so it can be assumed $x\neq 0$. By a corollary of Hahn-Banach theorem there is a $f\in X'$ such that $f(x)=\|x\|$ and $\|f\|=1$. $f(x_n)\to f(x)=\|x\|$ in scalars ($\mathbb R$ or $\mathbb C$) by weak convergence of the sequence $(x_n)$ to $x$. It follows that $|f(x_n)|\to |f(x)|=\|x\|$. this implies for each $\epsilon>0$ there is a $n_0\in\mathbb N$ such that $\|x\|-\epsilon\leq |f(x_n)|$ whenever $n\geq n_0$. Since $f$ is a continuous linear functional and $\|f\|=1$, $|f(x_n)|\leq\|x_n\|$ for all $n$ so $\|x\|-\epsilon \leq \|x_n\|$ whenever $n\geq n_0$. This implies $\|x\|\leq \liminf \|x_n\|$.