# How to think about dual space of a certain space of Lipschitz functions

Consider the following Banach space (for concreteness):

$$X=Lip(\bar{\mathbb{B}}^n)=\{f\in C^0(\bar{\mathbb{B}}^n): \Vert f \Vert_L<\infty \}$$ where $$\bar{\mathbb{B}}^n=\{\mathbf{x}\in \mathbb{R}^n: |\mathbf{x}|\leq 1\}$$ is the closed ball and $$\Vert f \Vert_L=\sup_{\bar{\mathbb{B}}^n}|f| +\sup_{\mathbf{x}\neq \mathbf{y}\in \bar{\mathbb{B}}^n} \frac{|f(\mathbf{x})-f(\mathbf{y})|}{|\mathbf{x}-\mathbf{y}|}$$ is one of the usual versions of a lipschitz norm.

I'm curious what is the best way to think about the (topological) dual space $$X^*$$ of $$X$$ as this space is a bit mysterious to me.

For instance, it's clear that any finite (signed) measure on $$\bar{\mathbb{B}}^n$$ can be thought of as an element of $$X^*$$, but one should also have elements that look like differences of infinite measures whose supports are sufficiently close and whose relative mass" is finite. Are there other natural elements?

Any references (in particular those that are more concrete and less abstract) would be appreciated.

Edit: I guess (please correct me if I am wrong) a fairly pathological element in the dual would be something like $$\mu=\sum_{i=1}^\infty \left( 2^i \delta_{2^{-2i}}-2^{i} \delta_{-2^{-2i}}\right)$$ on $$\mathbb{B}=[-1,1]$$.

Are there any natural subsets on which one can restrict and have a less crazy situation? For instance, fix a Radon measures $$\mu$$ on $$\mathbb{B}^n$$ (the open ball) with $$\mu(\mathbb{B}^n)=\infty$$. Is the subspace

$$Z=\{T\in X^*: T=\nu-\mu: \nu \mbox{ a Radon measure}\}$$ any nicer?

• The space you describe contains an isomorphic copy of $l^\infty$. Its dual is wild. – Nik Weaver Jan 3 '19 at 23:00
• @NikWeaver Any interesting ways to tame it? For instance, I'm actually interested in something like the following: I have a fixed Radon measure $\mu$ (of infinite mass) on $\mathbb{B}^n$ (the open ball) and am interested in the set of measures $\nu$ on $\mathbb{B}^n$ so that $\nu-\mu$ lives in $X^*$. – RBega2 Jan 3 '19 at 23:12
• I think you want to look at the predual, which is variously known as the "Arens-Eels" or "Lipschitz-free" space. For specific information about the measures which live there see Functional Analysis by Kantorovich and Rubenstein. I don't have the reference handy but it can be found in my book on Lipschitz algebras. – Nik Weaver Jan 3 '19 at 23:58
• As a linear space, $Z$ is isomorphic to a subspace of Radon measures, so yes, it is nicer. I suppose you would like to say something nice about the norm on $Z$. – user95282 Jan 8 '19 at 11:23
• I think the topological dual of your space is the space of measures on $\overline{\mathbb B}^n$, equipped with the Kantorovich-Rubenstein norm. See my answer below. – dohmatob Nov 6 '19 at 18:34

This would be a really long comment, so I've decided to post it as an answer. Hope it helps!

Disclaimer. I'm still learning FA, and my answer is based on my blurred understanding of the subject. I hope experts here will (in)validate this.

On a metric space $$X=(X,d)$$, let $$\mathcal M(X)$$ be the set of all measures and $$\mathcal M_0(X)$$ be the subset of measures $$\mu$$ for which $$\mu(X)=0$$. Define $$\|\cdot\|_{KR} \rightarrow \mathbb R$$ by

$$\|\mu\|_{KR} := \inf_{\nu \in \mathcal M_0(X)}\|\nu\|_0 + \|\mu-\nu\|_{TV},$$

where

• $$\|\mu-\nu\|_{TV}$$ is the total-variation between $$\mu$$ and $$\nu$$
• $$\|\nu\|_0 := \underset{\lambda \in \Phi_\nu}{\inf}\int_{X \times X}d(x,x')d\gamma(x,x')$$ and $$\Phi_\nu$$ is the subset of nonnegative measures $$\gamma$$ on $$X \times X$$ such that $$\gamma(X \times B) - \gamma(B \times X) = \nu(B)$$ every Borell subset $$B$$ of $$X$$.

Let $$\|f\|_L := \max(\|f\|_\infty, L(f))$$ be the bounded-Lipschitz norm on $$Lip(X)$$. Then

Theorem. $$(\mathcal M(X),\|\cdot\|_{KR})$$ is a normed vector space and the dual pairing $$\langle f, \mu\rangle := \int_{X} fd\mu,\;for\;(f,\mu) \in Lip(X) \times \mathcal M(X)$$ establishes an isometric isomorphism between $$(Lip(X), \|.\|_L)$$ and the topological dual of $$(\mathcal M(X),\|\cdot\|_{KR})$$.

This result goes back to Kantorovich and Rubenstein (hence "KR"), and is well-documented in Hanin '92 (see Theorem 0 there).

• I don't think Hanin is the appropriate citation. He quotes this result but it was originally proven by Kantorovich and Rubenstein (hence "KR"). – Nik Weaver Nov 6 '19 at 18:11
• Sorry, I failed to mention that KR = Kantorovich-Robenstein. – dohmatob Nov 6 '19 at 18:29
• Doesn't $\mu(X) = 0 \implies \mu = 0$? – ogogmad Sep 13 '20 at 7:36
• I guess you mean signed measures – ogogmad Sep 13 '20 at 9:12
• Yep, by "measures" I mean "signed measures". – dohmatob Sep 13 '20 at 11:02