Reflexivity of $C[a,b]$

Question

I find the statement that the normed, complex or real, linear space $C[a,b]$ is reflexive, i.e. the natural map of the space $C[a,b]$ into the bidual space $C[a,b]^{\ast\ast}$, defined by $\pi:C[a,b]\to C[a,b]^{\ast}$, $x\mapsto \psi_x$, where $\psi_x:C[a,b]^{\ast}\to K$, $f\mapsto f(x)$, is surjective.

I have found nothing about the issue. Can anybody suggest a link with a proof -as elementary as it is possible, since I'm just beginning to study those areas of mathematics, but my book states a lot and proves only a part of what it states- or a direct proof of the statement?

$\infty$ thanks!!!

Answer 1

As pointed out in the comments, $C([a,b])$ is not reflexive. I will consider only $C([0,1])$, since this space is isomorphic to $C([a,b])$. I assume you know that $c_0(\mathbb{N})$ is not reflexive. Since closed subspaces of reflexive spaces are reflexive, it suffices to show that $c_0(\mathbb{N})$ can be embedded isometrically in $C([0,1])$. This can be done explicitly.

Let $x\in c_0(\mathbb{N})$. If $t=\frac 1n$ for some $n$, put $f_x(t)=x_n$. Put $f_x(0)=0$ and extend $f_x$ to a piecewise linear function. This can be done by defining $$f_x(t)=f_x\left(\frac{1}{n+1}\right) +\left(t-\frac{1}{n+1}\right)(n(n+1))\left(f_x\left(\frac1n\right)-f_x\left(\frac{1}{n+1}\right)\right)$$ for $t\in \left[\frac{1}{n+1},\frac1n\right]$. The map $x\mapsto f_x$ is the desired isometry.

Answer 2

Here's another proof of the fact that $C[0,1]$ is not reflexive: $C[0,1]$ is separable but $C[0,1]^{**}$ is not because it contains an uncountable discrete closed set $\{\mathbf{1}_{[0,a]}\colon a\in (0,1]\}$. (Recall that the dual of $C[0,1]$ consists of Radon measures; each bounded Borel function $f$ on $[0,1]$ defines a functional on $C[0,1]^*$ by $\langle f,\mu\rangle = \int_0^1 f(t)\,\mu({\rm d}t)$.)