The closed unit ball is not compact in infinite dimension spaces. Why? We know that in finite dimension spaces the closed unit ball is compact, that is if H is a finite dimension space, then there exists an $u$ in the closed unit ball in H and $T \in \mathcal{L}(H, \mathbb{R})$ such that $||T|| = |T u|$. 
Why doesn't it happen in infinite dimension spaces? 
For example, let $H = C([a, b])$ with the sup norm. Why doesn't exist an element $u$ in the closed unit ball in $C([a, b])$ such that $T u = ||T||$?
Thank you!
 A: Essentially because sequences can wander off to infinite. Eg take any function with sup norm $1$ supported in the interval $[0,1].$ 
Call it $f.$ 
Define $f_n(x) = f(x-n).$ 
These functions have disjoint support and their distance in the sup norm is $1.$
A: For your second question, I assume you meant to refer to the assertion:

For any linear operator $T \in \mathcal{L} (H, \mathbb R)$, there exists an element $u$ in the unit ball in $H$ (actually the unit sphere), such that $\Vert T \Vert = |Tu|$.

This is true, if and only if the unit ball is compact.
A counter-example in an infinite-dimensional space $H$ is obtained by taking an infinite sequence $(v_n)_{n\in \mathbb{N}}$ of linearly independent elements in $H$ (all of lenght $1$) and defining a linear operator $T \in \mathcal L (H, \mathbb R)$ by 
$$Tv_n = 1- \frac 1 n$$ 
expanding linearly to the span of $(v_n)_{n\in \mathbb{N}}$, i.e.
$$T \left(\sum_{i=1}^k a_k v_{j_i} \right) = \sum_{i=1}^k a_k T (v_{j_i}),\quad k,j_i \in \mathbb N, a_{k} \in \mathbb R $$
and setting $T=0$ otherwise. Then 
$\Vert T \Vert =1$, but $|Tu|<1$ for all $u \in H$.
A: You ask two different questions.  For the second one: yes, given $u \in C([a,b])$, there is a linear functional $T \in \mathcal L(C([a,b]),\mathbb R)$ with $|T(u)| = \|u\|$ and $\|T\|=1$.  But of course, since the unit ball is not compact, you cannot use compactness of the unit ball in the proof...
