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 $\lim \|x_n\|$ ?

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}.
$$
