3
$\begingroup$

Given Holder space $C^{\alpha}(\mathbb{R}^n)$, does there exist a Banach space $X$ such that the dual of $X$ is $C^{\alpha}(\mathbb{R}^n)$?

What I can imagine is that such $X$ must contain the fractional sobolev space $W^{-s,p}$ with $s>0$ and $sp' \ge n+\alpha$, where $p'$ is the conjugate number of $p$. This is because $(W^{-s,p})^*=W^{s,p'}$, which contains the $C^{\alpha}$ space by sobolev embedding theorem. However, I've no idea how to characterize the predual of $C^{\alpha}$ spaces.

Can anyone give me some references? Any comment and ideas are really appreciated.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Yes, indeed every Lipschitz space is a dual space, a fact which has been rediscovered (in varying levels of generality) several times. The earliest proof is due to Arens and Eells.

Holder spaces are special cases of Lipschitz spaces because a function is $\alpha$-Holder continuous for a metric $\rho$ if and only if it is Lipschitz for the metric $\rho^\alpha$.

${\rm Lip}_0(X)$ is the space of Lipschitz functions on $X$ vanishing at a base point. This is the most general class of Lipschitz spaces; other Lipschitz spaces are special cases of these. The predual of ${\rm Lip}_0(X)$ is simply characterized as the universal Banach space containing an isometric copy of $X$ such that the base point of $X$ corresponds to $0$.

There's a lot of information about the predual in my book Lipschitz Algebras (and there will be even more in a soon forthcoming second edition). A recent result is that for a large class of metric spaces the predual of ${\rm Lip}_0(X)$ is unique.

$\endgroup$
2
  • $\begingroup$ What about the Holder spaces $C^{k,\alpha}(\mathbb{R}^d;\mathbb{R}^D)$, for some $k>0$? $\endgroup$
    – ABIM
    Commented May 7, 2019 at 11:49
  • $\begingroup$ Answered at your posted question. $\endgroup$
    – Nik Weaver
    Commented May 7, 2019 at 15:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .