Let $p\in \mathbb{N}$, and define $\mathrm{Lip}_p$ be the collection of functions from $\mathbb{R}^d$ to itself, with $p-1$ first derivatives bounded and whose $p^{\mathrm{th}}$ derivative is Lipschitz. This can be made into a Banach space using the norm $$ \|f\| \triangleq \max_{| i | \leq p-1} \sup_{x\in\mathbb{R}^d} \left \|(D^if) (x) \right \|_{\mathbb{R}^d} + \mathrm{Lip}(f^{(p)}), $$ where $\mathrm{Lip}(h)$ assigns to a Lipschitz function $h$ its minimal Lipschitz constant. (This definition is analogous to that of a Hölder space)

In this article, it was shown that if $p=0$, then the predual of $\mathrm{Lip}_0$ is uniquely defined. Is this still the case for positive integers $p$?

  • $\begingroup$ Your norm function is not finite for all $f$, so it does not really give you a Banach space.. $\endgroup$ – Zatrapilla May 7 '19 at 12:39
  • $\begingroup$ I mean the collection on which it is finite. $\endgroup$ – AIM_BLB May 7 '19 at 12:40
  • $\begingroup$ I think the usual thing is to take seminorms similar to your norm but taking the supremum only over compact sets. The dual of the resulting topological vector space is quite complicated if you do that though. $\endgroup$ – Zatrapilla May 7 '19 at 12:44
  • $\begingroup$ In that case, would it have a (non-trivial) pre-dual? $\endgroup$ – AIM_BLB May 7 '19 at 12:45
  • $\begingroup$ Well, the predual would contain at least all compactly-supported distributions of degree at most p, I think. $\endgroup$ – Zatrapilla May 7 '19 at 13:28

The previous answer I gave here is not correct. I had misread the problem as asking whether this space is a dual space, not whether its predual is unique. Uniqueness is likely to be a lot harder.

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  • $\begingroup$ Wow this is fantastic, thanks you. Also, I very much enjoyed your paper (still reading it at the moment). $\endgroup$ – AIM_BLB May 7 '19 at 15:46
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    $\begingroup$ You are welcome! $\endgroup$ – Nik Weaver May 7 '19 at 16:33

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