# Does the space of Lipschitz functions have the Radon-Nikodym property?

Context.
Space of Lipschitz functions. Denote by $$Lip_0(D)$$ the space of all Lipschitz functions on a metric space $$D$$ vanishing at some base point $$e \in D$$. The norm in $$Lip_0$$ is defined as follows $$\|f\|_{Lip_0} := Lip(f),$$ where $$Lip(f)$$ denotes the Lipschitz constant of $$f$$.

Radon-Nikodym property (RNP). There are many equivalent definitions of the RNP, I will give two of them.
Definition 1. Let $$\Sigma$$ be the $$\sigma$$-algebra of subsets of a set $$\Omega$$. A Banach space $$X$$ is said to have the RNP if for any measure $$\mu \colon \Sigma \to X$$ of bounded variation with values in $$X$$, and any finite positive scalar measure $$\lambda \colon \Sigma \to \mathbb R$$ such that $$\mu$$ is absolutely continuous w.r.t. $$\lambda$$, there exists a $$\lambda$$-Bochner integrable function $$f$$ such that $$\mu(E) = \int_E f \,d\lambda$$ for all $$E \in \Sigma$$.
Theorem 1. A Banach space $$X$$ has the RNP if and only if every Lipschitz function $$\mathbb R \to X$$ is differentiable almost everywhere.

Question. Does the $$Lip_0$$ space have the Radon-Nikodym property?

I have tried the following sources, but wasn't able to find an answer: Weaver, Lipschitz Algebras; Ryan, Introduction to Tensor Products of Banach Spaces; Diestel&Uhl, Vector Measures.

Any help will be much appreciated.

Let $$X$$ be a metric space consisting of a countable set of points, the distance between any two of which is $$2$$, together with one additional point $$e$$ whose distance to any of the other points is $$1$$. Then $${\rm Lip}_0(X)$$ is isometrically isomorphic to $$l^\infty$$, which fails the RNP.
Another example: let $$X = [0,1]$$ with $$e = 0$$. Then $${\rm Lip}_0(X)$$ is isometrically isomorphic to $$L^\infty[0,1]$$, which fails the RNP.