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I have to prove that a closed unit ball in $C[0,1]$ is not weak-compact. The hint is that I should consider sets: $$V_t=\{f\in C[0,1]:|f(t)|>1/3\}$$ and $$U_t=\{f\in C[0,1]:|f(t)|<2/3\}$$ Now I should show that $$\{V_t:t\in \mathbb{Q}\cap[0,1]\} \cup \{U_t:t\in (\mathbb{R} \setminus \mathbb{Q})\cap[0,1]\}$$ is an open cover of a closed unit ball in weak topology such that we can't choose a finite subcover.

Weak topologies are sth new to me. Can anyone help me? I did only manage to show, that $\phi_t:C[0,1]\ni f \rightarrow f(t)\in K$ are bounded linear functionals.

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As you said, $\phi_t$ is a bounded linear funcional, so as

$$V_t = \phi_t^{-1} \big( \{ |x| > \frac 13 \}\big), $$

then $V_t$ is open in the weak topology. Similarly $U_t$ is also open. It is quite obvious that

$$\{V_t:t\in \mathbb{Q}\cap[0,1]\} \cup \{U_t:t\in (\mathbb{R} \setminus \mathbb{Q})\cap[0,1]\}$$

is an open cover of the unit ball in $C[0,1]$ (Actually, it is an open cover of the whole $C[0,1]$).

Now we show that: for any finite set $t_1, \cdots t_m \in \mathbb Q\cap [0,1]$, $s_1, \cdots s_n \in \mathbb R\setminus \mathbb Q \cap [0,1]$, there is a function $f \in C[0,1]$ such that $||f||\leq 1$ and

$$f\notin \bigcup_{j} V_{t_j} \cup \bigcup_{k} U_{s_k}$$

Well, it just say that I can find a continuous function $f$ such that $|f(t_j)|\leq 1/3$ (so not in $V_{t_j}$) and $|f(s_k)| \geq 2/3$ (so not in $U_{s_k}$) and $||f|| \leq 1$. Such a function can be found by setting

$$f(t_j) =1/3, \ \ \ \ f(s_k) = 2/3$$

and joining line along adjacent $t_j$, $s_k$. This shows that we cannot pick a finite subcover to cover the closed unit ball in $C[0,1]$, thus the unit ball is not weakly compact.

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Although, you want to prove directly, it easily follows from Kakutani's theorem which states that

For a Banach space $X$,

$X$ is reflexive$\iff$ The closed unit ball of $X$ is weakly compact.

But if $X$ is infinite, $C(X)$ is not reflexive. As a result, $C([0,1])$ is not reflexive, So its closed unit ball can not be weakly compact.

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  • $\begingroup$ Should not we require $X$ infinite compact (as a metric space)? $\endgroup$ – Federico May 21 '14 at 21:09
  • $\begingroup$ @Federico I can not remember anything right now! should be? $\endgroup$ – Fermat May 21 '14 at 21:22
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    $\begingroup$ I think yes. I remember a result in my lecture notes that says "If $(K, d)$ is a compact infinite metric space, then $C(K)$ is not reflexive". I'm not sure, however, this is the most general rule. Now it's late for me, but I'll try to check out tomorrow! $\endgroup$ – Federico May 21 '14 at 21:29
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    $\begingroup$ I have not been successful in finding the answer to the question "what if we drop compactness?", so I asked a question: math.stackexchange.com/questions/810425/… Hope could be useful! $\endgroup$ – Federico May 27 '14 at 13:44

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