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 $\lim \|x_n\|$ ?
 A: 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 $a\le b$.
A: Because we don't know if the sequence $\{ x_n\}$ converges or not.
Since $\{x_n\}$ weakly converges to $x$, we only know the sequence $\{x_n\}$ bounded in the norm.
We could use an example to illustrate this.
Consider the sequence in $\ell^2$ space:$x^{(i)}=$$\mathop{(0,0,...,1,...)}\limits_{x_1 ^{(i)},x_2 ^{(i)},........,x_i ^{(i)}.........}$, where we label the $j$st component by $x^{(i)}$ $x_j^{(i)}$, and $x_j^{(i)}=\delta^j_i$.
Since in $\ell^2$ space, for every element ${y^{(j)}} \in \ell^2 \quad \forall j $,
$$lim_n y^{j}_n=0.$$
Back to the sequence we chose, we know the sequence $\{x^{(i)}\}$ weakly converges to $0$, but $\{x^{(i)}\}$ converges to $1$ in the norm.
Which is,
$$
lim_i x^{(i)}=(0,0,...) \quad weakly,
$$
$$
lim_i ||x^{(i)}||=1 .
$$
Also we could adjust $x_j^{(i)}$ such that $\{x^{(i)}\}$ doesn't converge in norm, for example by letting $x_j^{(i)}=\delta^j_i$where $i$ is odd, and $x_j^{(i)}=\frac{1}{2}\delta^j_i$ where $i$ is even.
we have
$$
lim_i x^{(i)}=(0,0...) \quad weakly,
$$
$$
limsup_i ||x^{(i)}||=1,
$$
$$
liminf_i ||x^{(i)}||=\frac{1}{2}.
$$
