$\DeclareMathOperator\AE{AE}\DeclareMathOperator\Lip{Lip}$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](https://arxiv.org/abs/1611.01812) 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](https://en.wikipedia.org/wiki/Lipschitz_continuity) $\Lip(f)$), $\delta\_x$ 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](https://en.wikipedia.org/wiki/Dual_norm) construction; i.e.:
$$
\|F-G\|:=\inf\_{f \in \Lip\_0(X),\, \Lip(f)\leq 1} F(f)-G(f).
$$



Differentiability of the map $x\mapsto \delta_x$ in the Arens-Eells/Lipschitz-free space
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.


