TITLE: Differentiability of the map $x\mapsto \delta_x$ in the Arens-Eells/Lipschitz-free space
QUESTION [5 upvotes]: $\DeclareMathOperator\AE{AE}\DeclareMathOperator\Lip{Lip}$Let $\AE(X)$ denote the Arens-Eells space on a Banach space $X$.  Consider the map:
$$
\begin{aligned}
\delta: X & \rightarrow \AE(X)
\\
x&\mapsto \delta_x
\end{aligned}
$$
Is the map $\delta$ ever Gâteaux (or Fréchet) differentiable?

Recall that $\AE(X)$ is the/a pre-dual of the Banach space $\Lip_0(X)$ whose elements are Lipschitz functions sending $0\in X$ to $0\in \mathbb{R}$ (with norm sending any $f\in \Lip_0(X)$ to its (unique) Lipschitz constant $\Lip(f)$), $\delta_x$ denotes the evaluation map defined on Lipschitz functions $f\in \Lip_0(X)$ by
$$
\delta_x(f):= f(x),
$$
and $\AE(X)$ is normed using the dual-norm construction; i.e.:
$$
\|F-G\|:=\inf_{f \in \Lip_0(X),\, \Lip(f)\leq 1} F(f)-G(f).
$$

REPLY [7 votes]: This fails for $X = \mathbb{R}$, and hence for every nonzero Banach space, since they all contain copies of $\mathbb{R}$. If the map $t \mapsto \delta_t$ were differentiable in either sense then for every bounded linear functional $F$ on $AE(\mathbb{R})$ the map $t\mapsto F(\delta_t)$ would be differentiable. Recalling that the dual of $AE(\mathbb{R})$ is ${\rm Lip}_0(\mathbb{R})$, differentiability at $t$ would imply that every Lipschitz function on $\mathbb{R}$ is differentiable at $t$, which is obviously false.