Let $X$ be a compact metric space with infinitely many points, then show $C(X)$ is not reflexive. I think I see why this is the case for $X=[a,b]$, but I don't see how one can extend it. Is there a way to construct a continuous function which doens't attain its norm, because this is my intuition of how to proceed.
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A strongest statement is given in the answer of Tomek Kania here – nowhere dense Feb 07 '19 at 23:59
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Do you mean you want a continuous linear functional on $C(X)$ that doesn't attain its norm? – Nate Eldredge Feb 08 '19 at 03:10
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Let $x_1,...,x_n$ be distinct points we can suppose it converges towards $x$ (up to a subsequence). Let $Y=\{x_1,...,x_n\}\cup\{x\}$ it is closed, consider the function $f_n$ defined on $Y$ whose restriction to $\{x_1,..,x_n\}$ is $1$ and whose restriction to $\{x_{n+1},...\}\cup\{x\}$ is $0$, we can extend it to $X$ by Tietze-Uryshon in $g_n$, which has norm $1$
Since $C(X)$ is reflexive, the unit ball is weakly compact, so we can suppose up to a subsequence that $g_{n_k}$ converges weakly towards $g$, this implies that $ev_{x_{n_k}}(g_{n_k}))$ converges towards $ev_{x_{n_k}}(g)$, we deduce that $g(x_{n_k})=1$, it results that $g(x)=1$, but $lim_{n_k}ev_x(g_{n_k})=0$. Contradiction.
Tsemo Aristide
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To follow your original approach, take an infinite set $\{x_1, x_2, \dots\}$ and let $x$ be a limit point of this set (which must exist by compactness). Consider the linear functional $$\ell(f) = \sum_{n=1}^\infty 2^{-n} f(x_n) - f(x).$$ It is clear that $\|\ell\| \le 2$. Moreover, by Tietze, for any $k$ you can find a continuous function $f$ with $\|f\| \le 1$ having $f(x_1) = \dots = f(x_k) = 1$ and $f(x) = -1$, so that $|\ell(f)| \ge 2 - 2^{-k}$. Hence $\|\ell\| = 2$. On the other hand, if some $f$ with $\|f\| = 1$ is to attain this norm, evidently it must have $f(x_n) = 1$ for every $n$ and $f(x) = -1$ (or vice versa), which contradicts continuity.
Nate Eldredge
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