# Dual of the space of Hölder continuous functions?

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

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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 $X^*$... –  Willie Wong Apr 10 '12 at 12:13
Probably some info here ... ams.org/journals/proc/1976-057-02/S0002-9939-1976-0477673-7/… –  Gerald Edgar Apr 10 '12 at 14:32
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 choice-related pathologies that the space $(\ell^\infty)^*$ has. –  Terry Tao Apr 11 '12 at 15:42
And in fact, for any infinite $\Omega$ a fortiori. –  Pietro Majer Apr 11 '12 at 17:13

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, linear functionals $\phi$ on $C^0_b(\tilde\Omega)$, produce all continuous linear functionals on $C^\alpha(\Omega)$ via
$$\left\langle \phi, \frac{u(x)}{|x-y|^\alpha}\right\rangle\\ ,$$ for $u\in C^\alpha(\Omega)$.

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I'd be grateful to the user who down-voted, or to whoever has an idea about it, if (s)he left a few lines of explanation. –  Pietro Majer Apr 12 '12 at 14:26
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).
Good point Piero, even if $\dot C^s$ is quite smaller than $C^s$... –  Pietro Majer Apr 11 '12 at 10:12
Let's stay on $\mathbb{R}^n$. For a function in the Schwartz space, $|u(x)−u(y)|/|x-y|^\alpha=o(1)$ 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... –  Pietro Majer Apr 11 '12 at 11:27
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 $|a_k-b_k|\to0$ 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 $C^\alpha_0$). –  Pietro Majer Apr 11 '12 at 15:15