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

Question

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.

Answer 1

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.