Suppose $X$ is a Banach space. For any $x \in X$, define the set $\mathcal{F}(X) = \overline{span \{ \delta_x : x \in X \}}$ where $\delta_x(f)=f(x)$ for all $f \in$ $Lip_0(X)$.
The set Lip$_0(X)$ is the set of all real-valued Lipschitz functions which vanish at $0$.
Note that $\delta_x$ is an evaluation functional on Lip$_0(X)$.
The map $\delta: X \rightarrow \mathcal{F}(X)$ is a nonlinear isometry and is nowhere Gateaux differentiable.
Note that we have $\| \delta_x \| = \| x \|$ due to the Hahn-Banach theorem. So the map is an isometry.
Clearly we have $\delta_{x+y} \neq \| \delta_x + \delta_y \|$. So the map is nonlinear.
How to show the map is nowhere Gateaux differentiable? The following is the proof from the paper (Proposition $2.1$):
We only need to remark that if $t \rightarrow \delta(u+tv)$ has any point of differentiability at $0$ then it would follow that for every $f \in Lip_0(X)$ we would have differentiability of $t \rightarrow > f(u+tv)$ at $0$. Consider the map $f(x) = \| x-u \| - \|u\|$.
I don't understand what the proof is doing. Can anyone explain the main idea?
The main idea of the proof is this:
If $\delta$ is Gateaux differentiable at $u\in X$, then there is an homogeneous function $H_u(\cdot):X\to\mathcal{F}(X)$ such that $$ H_u(v)=\lim_{t\to 0}\frac{\delta_{u+tv}-\delta_u}{t} $$ Since $H_u(v)$ is in the dual of $Lip_0(X)$, for all $f\in Lip_0(X)$, $H_u(v)(f)$ is finite. This implies that for all $v\in X$, and all $f\in Lip_0(X)$, the limit $$ \lim_{t\to 0}\frac{\delta_{u+tv}(f)-\delta_u(f)}{t}=\lim_{t\to 0}\frac{f(u+tv)-f(u)}{t} $$ exists and equals $H_u(v)(f)$ which is finite. This shows all $f\in Lip_0(X)$ are Gateaux-differentiable at $u$.
Now, consider $f(x)=\Vert x-u\Vert -\Vert u \Vert$. This is a Lipschitz function (this follows from the triangle inequality) which vanishes at 0. However, this function is not Gateaux differentiable at $u$, because, for all $v\neq 0$, we can show $$\lim_{t\to 0^-}\frac{f(u+tv)-f(u)}{t}=-\Vert v\Vert\neq \Vert v\Vert =\lim_{t\to 0^+}\frac{f(u+tv)-f(u)}{t}$$ This contradicts that all $f\in Lip_0(X)$ are Gateaux differentiable at $u$. This implies $\delta$ is not Gateax differentiable at $u$.