Is $C[0,1]$ reflexive?

Question

I.e. is the embedding $C[0,1]\hookrightarrow \left( C[0,1] \right)^{**}$ surjective?

I am having a hard time answering that question. It would be enough to find a closed subspace of $C[0,1]$ which is not reflexive since closed subspaces in reflexive spaces are again reflexive. Is there a canonical such subspace? Or alternatively, is there an easy description of the dual/double-dual of $C[0,1]$?

Answer 1

One need not to know the whole dual space of $C[0,1]$ to see the lack of reflexivity.

Simply note that $C[0,1]^*$, opposed to $C[0,1]$, is non-separable because for each $t\in [0,1]$ the map $$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$ is a norm-one linear functional on $C[0,1]$ and $\|\delta_t - \delta_s\| =2$ for distinct $s,t$. To see this, take a norm-one function $f$ such that $f(t)=-1, f(s)=1$. As $\{\delta_t\colon t\in [0,1]\}$ is an uncountable, discrete subset of $C[0,1]^*$, $C[0,1]^*$ cannot be separable.

Answer 2

No it is not.

The dual of $C[0,1]$ is the space $\mathfrak{M}([0,1])$ of complex (or signed) regular Borel measures on $[0,1]$.

The dual of $\mathfrak{M}([0,1])$ contains the set $\mathcal L^\infty([0,1])$ of bounded Borel functions on $[0,1]$.

For example, $\chi_{\{0\}}$, the function which vanishes everywhere, except at $x=0$, where it is equal to $1$, defines a bounded linear functional on $\mathfrak{M}([0,1])$, namely $$ \varphi(\mu)=\int \chi_{\{0\}}\,d\mu=\mu(\{0\}), $$ which is not representable by an element of $C[0,1]$.

Another proof. According to Kakutani's Theorem, A space is reflexive if the closed unit ball of $X$ is compact in the weak topology. Note that, the sequence $\,f_n=\mathrm{e}^{2\pi nx i}\in C[0,1]$ does not possess a weakly converging subsequence.