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

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

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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.

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A negative answer to your question is contained on page 123 of Godefroy, G.; Kalton, N. J. Lipschitz-free Banach spaces. Dedicated to Professor Aleksander Pełczyński on the occasion of his 70th birthday. Studia Math. 159 (2003), no. 1, 121–141.

If you are interested in the subject, this paper is a strongly recommended reading.

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