# If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$

How does one show that if $x_n \to x_0$ weakly, then $\liminf ||x_n|| \geq ||x_0||$?

I'm just doing additional problems from a book to try and prepare for a final. Unfortunately, I have really gotten nowhere with this one, so I'm afraid I just have to ask for help from the get go. Any suggestions?

• $\langle x_n,x_0\rangle\to \langle x_0,x_0\rangle$ May 8, 2013 at 21:14
• Choose a subsequence $(x_{n_k})$ with $\Vert x_{n_k}\Vert \rightarrow \liminf\limits_{n\rightarrow\infty} \Vert x_n\Vert$. Using Hahn-Banach, find $f^*$ with $\Vert f^*\Vert\le1$ and $|f^*(x)|=\Vert x\Vert$. Now consider the quantities $|f^*(x_{n_k})|$. May 8, 2013 at 21:35
• @HagenvonEitzen I have been confused by that notation. I am use to <.,.> denoting an inner product, but this need not be an inner product space. Could you explain more? May 8, 2013 at 23:27

Suppose $x_n \to x$ weakly. Fix $x^* \in X^*$. Then $$|x^*(x)| = \lim_{n \to \infty} | x^*(x_n)| \leq \liminf_{n \to \infty} ||x^*||\ ||x_n|| = ||x^*||\liminf_{n \to \infty} ||x_n|| .$$ Now take the supremum of both sides over all $x^* \in X^*$ with $||x^*||\leq 1$, giving $$||x|| \leq \liminf_{n \to \infty}||x_n||.$$
(Recall that $||x|| = \sup \{|x^*(x)| : x^* \in X^*, ||x|| \leq 1\}$; this follows quickly from the Hahn-Banach theorem. This is not definition, it is a (very useful) theorem which one proves after one has Hahn-Banach at one's disposal).
Suggestion: To see if you really understand this proof, try this similar problem: If $x_n^* \to x^*$ weakly$^*$, then $||x^*|| \leq \liminf_{n \to \infty} ||x_n^*||$.
• can you explain the step from $lim |x*(x_n)| \le lim inf ||x*||||x_n||$? I understand that $|x*(x_n)| \le ||x*||||x_n||$ I guess I am confused about the lim inf, it seems like the lim should exist which would mean lim inf = lim. May 8, 2013 at 23:25
• @Fractal20 As for your first comment, we know $||x^*(x_n)|| \leq ||x^*||\ ||x_n||$ for all $n$. However, we do not know that $||x_n||$ converges. Thus, the best we can say is that $\lim |x^*(x_n)| = \liminf |x^*(x_n)| \leq \liminf ||x^*||\ ||x_n||$. May 9, 2013 at 2:46
• @Fractal20 For your second comment, yes $||x^*|| = \sup \{ x^*(x), x \in X, ||x||\leq 1\}$ by definition. However, it follows quickly from the Hahn-Banach theorem that for any $x \in X$ there is an $x^* \in X^*$ such that $x^*(x) = ||x||$. Thus it follows (convince yourself of this) that for any $x \in X$ we have $||x|| = \sup\{x^*(x) : x^* \in X^*, ||x^*|| \leq 1\}$. May 9, 2013 at 2:50
• @Fractal20 The statement $x_n \to x$ weakly by definition means $x^*(x_n) \to x^*(x)$ for every $x^* \in X^*$. The absolute value function $|\cdot|$ is continuous on $\mathbb{R}$ so we also have $|x^*(x_n)| \to |x^*(x)|$. Thus the first limit is defined. May 10, 2013 at 19:34
• Then from here, I am using that if $\lim a_n$ exists, then $\lim a_n = \liminf a_n$, and if $a_n \leq b_n$ for all $n$ , then $\liminf a_n \leq \liminf b_n$. These are basic facts from real analysis which you should be very familiar with before attempting the more subtle theory involved in functional analysis. May 10, 2013 at 19:48