I've been able to show that the extreme points of $C([0,1])$ are the continuous functions that take values on the unit circle. However, I'm not sure how to reason from here as to whether or not it is the dual to any Banach space. Any hints? The standard approach I've been using to show things are not dual to any space has been to use Krein-Milman and Alaoglu to argue by contradiction, but I don't know what the closed convex hull of unitary complex functions is.
Functional Analysis – Is $C([0,1])$ Dual to Any Banach Space?
banach-spacesfunctional-analysis
Best Answer
There are two versions of your question:
Of course, the negative answer to 2. implies the negative answer to 1. However 1. is elementary.
Morally, the answer is no because $C[0,1]$ has too few extreme points.
To be more precise, observe that the extreme points of the unit ball of a $C(K)$-space can take values only from the unit circle. Now use the fact that $[0,1]$ is connected and try to approximate real-valued functions by convex combinations of extreme points... That's impossible as long as the functions are constantly $\pm 1$.
More interestingly, the answer is also no for the isomorphic version of your question too. Even more is true:
This follows from the following standard argumentation.
Take a sequence of disointly supported norm-one functions in $C[0,1]$; their closed linear span is isomorphic to $c_0$. As $C[0,1]$ is separable, by Sobczyk's theorem, this copy of $c_0$ is complemented in $C[0,1]$. However, $c_0$ is not complemented in any dual Banach space because if it were, it would be complemented in its own bidual (see also Example 5.9(i) on p. 22), that is in $\ell_\infty$, which is not the case by the Phillips–Sobczyk theorem.
Compact, Hausdorff spaces $K$ for which $C(K)$ is isometric to a dual Banach space are called hyperstonean. If infinite, they are necessarily non-metrisable. See Chapter 2 of this memoir for more details.