# Predual to Lipschitz Maps with $p$ Derivatives

Let $$p\in \mathbb{N}$$, and define $$Lip(p)$$ be the collection of functions from $$\mathbb{R}^d$$ to itself, with $$p-1$$ first derivatives bounded and whose $$p^{th}$$ derivative is Lipschitz. This can be made into a Banach space using the norm $$\|f\| \triangleq \max_{| i | \leq p-1} \sup_{x\in\mathbb{R}^d} \left \|(D^if) (x) \right \|_{\mathbb{R}^d} + Lip(f^{(p)}),$$ where $$Lip(h)$$ assigns to a Lipchitz functions $$h$$ 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 Lip(0) is uniquely defined. Is this still the case for positive integers $$p$$?

• Your norm function is not finite for all $f$, so it does not really give you a Banach space.. – Zatrapilla May 7 at 12:39
• I mean the collection on which it is finite. – AIM_BLB May 7 at 12:40
• 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. – Zatrapilla May 7 at 12:44
• In that case, would it have a (non-trivial) pre-dual? – AIM_BLB May 7 at 12:45
• Well, the predual would contain at least all compactly-supported distributions of degree at most p, I think. – Zatrapilla May 7 at 13:28

I think so. There's a criterion for a Banach space $$B$$ to be a dual space: there should exist a set $$E \subseteq B^*$$ which separates the points of $$B$$ and whose weak topology makes the unit ball of $$B$$ compact (see here). For your space we would take the evaluation functionals at points of $$\mathbb{R}^d$$ followed by coordinate functionals (taking $$\mathbb{R}^d$$ to $$\mathbb{R}$$). It's easy to see that these are bounded linear functionals, and trivial that they separate points. Compactness also looks pretty standard, basically an Arzela-Ascoli argument with an induction on $$|i|$$.