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

$$\DeclareMathOperator\AE{AE}\DeclareMathOperator\Lip{Lip}$$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.: $$\|F-G\|:=\inf_{f \in \Lip_0(X),\, \Lip(f)\leq 1} F(f)-G(f).$$

• Note that $\delta_{x+h}-\delta_x\not\to0$ as $h\to0$ Commented Jul 13, 2021 at 9:47
• @PieroD'Ancona So silly of me! I meant to put the Arens-Eells space (I realized my typo after drinking some coffee :/ ) Commented Jul 13, 2021 at 10:09

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.