# Limit inferior taken on the norm of a sequence

Let $E$ a normed vector space and let $(x_n)$ be a sequence in $E$. Suppose that $x_n$ converges weakly (i.e. wrt the weak topology) to $x$. Why is it that from the inequality $$|f(x_n)| \leq \|f\| \|x_n\|,$$ passing to the limit we obtain $$|f(x)| \leq \|f\| \lim\inf\|x_n\|$$ ?

Particularly, why can't we simply write $\lim \|x_n\|$ ?

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You do not know that $\lim\limits_{n\rightarrow\infty} \Vert x_n\Vert$ exists. For instance, the sequence $(e_1, 2e_2, e_3, 2e_4,\ldots)$ converges weakly to $0$ in $\ell_2$.
But, as the $\Vert x_n\Vert$ are reals, $\liminf\limits_{n\rightarrow\infty}\Vert x_n\Vert$ exists, and you can find a subsequence $\Vert x_{n_k}\Vert$ converging to its value. Then since $(x_{n_k})$ converges weakly to $x$ $$\tag{1} |f(x)|= \lim\limits_{k\rightarrow\infty}|f(x_{n_k})| \le \lim\limits_{k\rightarrow\infty}(\,\Vert x\Vert\Vert x_{n_k}\Vert\,) =\Vert x\Vert \liminf\limits_{n\rightarrow\infty}\Vert x_{n }\Vert.$$
Here we are just using the result for real numbers: Suppose $a_n\le b_n$ for each $n$. Then if $a_n\rightarrow a$ and if $b_n\rightarrow b$, it follows that $a\le b$.
Thanks, the fact that the limit could not exist is a clear motivation to the question. May I ask you why the sequence you have shown does converge to $0$? (Probably I have not understood the meaning of the $e_i$s) –  Oo3 Apr 28 '12 at 10:33
@Oo3 The sequence has terms $x_1=(1,0,\ldots)$, $x_2=(0,2,0,\ldots)$, $x_3=(0,0,1,0,\ldots)$.. Given $f=(\alpha_1,\alpha_2,\dots)$ in the dual, $f(x_i)$ is either $\alpha_i$ or $2\alpha_i$; so, since $f\in\ell_2$, $f(x_i)\rightarrow0$. –  David Mitra Apr 28 '12 at 13:57