Timeline for Dual of the space of Hölder continuous functions?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2012 at 7:29 | vote | accept | warsaga | ||
| May 1, 2012 at 9:34 | |||||
| Apr 12, 2012 at 7:29 | vote | accept | warsaga | ||
| Apr 12, 2012 at 7:29 | |||||
| Apr 11, 2012 at 15:15 | comment | added | Pietro Majer | 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 $C^\alpha_0$). | |
| Apr 11, 2012 at 14:25 | comment | added | Piero D'Ancona | 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 | |
| Apr 11, 2012 at 13:36 | comment | added | Pietro Majer | 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. | |
| Apr 11, 2012 at 12:58 | comment | added | Piero D'Ancona | It should coincide. As to your last remark, define "essentially" | |
| Apr 11, 2012 at 11:27 | comment | added | Pietro Majer | 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... | |
| Apr 11, 2012 at 10:57 | comment | added | Piero D'Ancona | Sure. Maybe on bounded sets the result is even better, I do not know how much the boundary messes up things | |
| Apr 11, 2012 at 10:12 | comment | added | Pietro Majer | Good point Piero, even if $\dot C^s$ is quite smaller than $C^s$... | |
| Apr 10, 2012 at 17:45 | history | answered | Piero D'Ancona | CC BY-SA 3.0 |