Dual of the space of Hölder continuous functions? Let $X=C^{\alpha}(\Omega,\mathbb{R})$ be the space of Hölder continuous functions. What is its dual?
 A: 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 $$\|u\|_\alpha:= |u(0)| + \sup _ {(x,y)\in\tilde\Omega}\frac{|u(x)-u(y)|} {|x-y|^\alpha}$$
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 $\left( u(0), \frac{u(x)-u(y)} {|x-y| ^ \alpha} \right)$. 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 well-studied representation.
Lastly, recall that as  a general fact, the dual of a product of two Banach spaces, endowed with the sum-norm, is the product of the duals, with their max-norm; 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
$$\lambda u(0)+\left\langle \phi, \frac{u(x) - u(y)}{|x-y|^\alpha}\right\rangle\  ,$$ for  $u\in  C^\alpha(\Omega)$.
A: 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).
