Let $X=C^{\alpha}(\Omega,\mathbb{R})$ be the space of Hölder continuous functions. What is its dual?

7$\begingroup$ I assume rather than "what is its dual", you meant to ask "is there a common function space $Y$ that can be identified with the dual of $X$ and what is the corresponding map $Y\to X^*$?" Otherwise the dual of $X$ is just $X^*$... $\endgroup$– Willie WongCommented Apr 10, 2012 at 12:13

1$\begingroup$ Probably some info here ... ams.org/journals/proc/197605702/S00029939197604776737/… $\endgroup$– Gerald EdgarCommented Apr 10, 2012 at 14:32

4$\begingroup$ On a discrete space such as the integers, X is just $\ell^\infty$, so at a bare minimum one is going to get all the axiom of choicerelated pathologies that the space $(\ell^\infty)^*$ has. $\endgroup$– Terry TaoCommented Apr 11, 2012 at 15:42

$\begingroup$ And in fact, for any infinite $\Omega$ a fortiori. $\endgroup$– Pietro MajerCommented Apr 11, 2012 at 17:13
2 Answers
Just a few words about how to get a representation for the dual, since the details on this topic are certainly treated in the literature.
To fix notations, assume e.g. $\Omega\subset\mathbb{R}^n$ be an open neighborhood of $0$ , let $\Delta_\Omega\subset\Omega\times\Omega$ denote the diagonal, and $\tilde\Omega:=(\Omega\times\Omega)\setminus\Delta_\Omega\subset \mathbb{R}^{2n}$.
Let $$\u\_\alpha:= u(0) + \sup _ {(x,y)\in\tilde\Omega}\frac{u(x)u(y)} {xy^\alpha}$$ be the usual $C^\alpha$ norm.
We have therefore an isometric linear embedding $$j: C^ \alpha(\Omega) \to \mathbb{R}\times C^0_b(\tilde\Omega )$$ mapping $u\in C^ \alpha(\Omega)$ to the pair $\left( u(0), \frac{u(x)u(y)} {xy ^ \alpha} \right)$. This presents $C^ \alpha(\Omega)$ as a product of $\mathbb{R}$ and a subspace of the space of bounded continuous functions on the open set $\tilde\Omega$, the dual of which has a wellstudied representation.
Lastly, recall that as a general fact, the dual of a product of two Banach spaces, endowed with the sumnorm, is the product of the duals, with their maxnorm; and that the dual of a subspace $Y$ of a Banach space $X$, is isometrically the quotient of the dual over the annichilator: $Y^* \sim X^*/Y^{\perp}$.
edit. Of course if the uniqueness of representation is not relevant for you, you may skip the quotient. Thus, pairs $(\lambda,\phi)$ of a real numbers $\lambda$ and a linear functional $\phi$ on $C^0_b(\tilde\Omega)$, produce all continuous linear functionals on $C^\alpha(\Omega)$ via
$$\lambda u(0)+\left\langle \phi, \frac{u(x)  u(y)}{xy^\alpha}\right\rangle\ ,$$ for $u\in C^\alpha(\Omega)$.
Your question can be interpreted in several ways, but I guess you are asking "is it a known space, or just some weird new Banach space?"
If $\Omega=R^n$ and $s$ is noninteger, the dual of $C^s$ is known in the above sense (almost). Indeed, one can identify $C^s$ with the Besov space $B^s_{\infty,\infty}$, and duals of Besov spaces are well studied. The precise result is: denote with $\dot C^s$ the closure of the Schwartz space of rapidly decreasing functions in $C^s$, then $(\dot C^s)'=B_{1,1}^{s}$. I bet similar results should be true also on more general open sets $\Omega$ but I'm not sure. A good starting point are Triebel's books (Theory of Function Spaces I, II and III).

$\begingroup$ Good point Piero, even if $\dot C^s$ is quite smaller than $C^s$... $\endgroup$ Commented Apr 11, 2012 at 10:12

$\begingroup$ Sure. Maybe on bounded sets the result is even better, I do not know how much the boundary messes up things $\endgroup$ Commented Apr 11, 2012 at 10:57

$\begingroup$ Let's stay on $\mathbb{R}^n$. For a function in the Schwartz space, $u(x)−u(y)/xy^\alpha=o(1)$ as $x−y\to0$, and $u(x)=o(1)$ as $x\to\infty$. If I'm not wrong, this defines a closed separable subspace $C^\alpha_0$ of $C^\alpha$, thus including $\dot C ^\alpha$ (or maybe coinciding?). Do you agree? And this $C^\alpha_0$ is essentially a space of continuous functions on a compact space... $\endgroup$ Commented Apr 11, 2012 at 11:27

$\begingroup$ It should coincide. As to your last remark, define "essentially" $\endgroup$ Commented Apr 11, 2012 at 12:58

2$\begingroup$ Yes. But what I'm saying is just that the dual of the whole $X:=C^\alpha(\mathbb{R}^n)$ is more complicated. A couple of examples: if $a_k\neq b_k$ and $c_k$ are given sequences in $\mathbb{R}^n$ with $a_kb_k\to0$ and $c_k\to\infty$, and $L$ is a Banach limit, $u\mapsto L (u(a_k) u(b_k))/a_k b_k$ and $u\mapsto L (u(c_k))$ are linear functionals on $X$ (both vanishing on $C^\alpha_0$). $\endgroup$ Commented Apr 11, 2012 at 15:15