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

Fan Zheng
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warsaga
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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
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    Probably some info here ... http://www.ams.org/journals/proc/1976-057-02/S0002-9939-1976-0477673-7/home.html – Gerald Edgar Apr 10 '12 at 14:32
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    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

2 Answers2

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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 $u\in C^\alpha(\Omega)$.

Pietro Majer
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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).

Piero D'Ancona
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  • Good point Piero, even if $\dot C^s$ is quite smaller than $C^s$... – Pietro Majer Apr 11 '12 at 10:12
  • Sure. Maybe on bounded sets the result is even better, I do not know how much the boundary messes up things – Piero D'Ancona Apr 11 '12 at 10:57
  • 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
  • It should coincide. As to your last remark, define "essentially" – Piero D'Ancona Apr 11 '12 at 12:58
  • Any $u\in C^\alpha 0$ writes uniquely as $u(x)=|x-y|^\alpha g(x,y)$ with $g$ in a closed subspace $Y$ of $C^0(K)$ with $K:=\mathbb{R}^{2n}\cup(\infty)$ (in particular, $g$ vanishing at infinity and on the diagonal, etc). This gives a representation for $(C^\alpha _0)^*$ via Riesz-Markov in terms of Radon measures $m$ on $\mathbb{R}^{2n}$, in the sense that any linear form is $\int u(x)/|x-y|^\alpha dm(x,y)$. I guess this writes $\int u(x)d\mu(x)$ with special measures $\mu$ obtained by disintegration from $|x-y|^{-\alpha}\cdot m$: elements of $B^{-\alpha}{\infty,\infty}$ I guess. – Pietro Majer Apr 11 '12 at 13:36
  • I think you are correct. The nice fact about the Besov space approach is that you have a completely independent characterization of the elements of the dual, in terms of dyadic decomposition, and this seems difficult to get via Riesz-Markov – Piero D'Ancona Apr 11 '12 at 14:25
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    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