Dual of a Banach space is isometric to a subspace of a Lipschitz dual

Question

Suppose $E$ is a Banach space. Denote $E^*$ as a dual space of $E$ and $Lip(E)$ as the Lipschitz dual of $E$. In other words, $Lip(E) = \{ f:E \rightarrow \mathbb{R} | f(0)=0 \}$ and all $f$ is Lipschitz.

In the book Geometric Nonlinear Functional Analysis, volume $1$, page $173$, there is this sentence before Proposition $7.5$:

$E^*$ is naturally isometric to a subspace of $Lip(E)$.

Question: How to show the statement above?

Can give any hint?

Answer 1

Recall that continuous linear functionals are Lipschitz, so every $e^* \in E^*$ is a Lipschitz function, that is we have $E^* \subseteq \def\L{\operatorname{Lip}}\L(E)$. To show the isometry: We have for $e^* \in E^*$ that \begin{align*}\def\norm#1{\left\|#1\right\|} \norm{e^*}_{\L(E)} &= \sup_{e \ne f} \frac{\def\abs#1{\left|#1\right|}\abs{e^*(e) - e^*(f)}}{\norm{e-f}_E}\\ &= \sup_{e\ne f}\frac{\abs{e^*(e-f)}}{\norm{e-f}_E}\\ &= \sup_{e'\ne 0}\frac{\abs{e^*(e')}}{\norm{e'}_E}\\ &=\norm{e^*}_{E^*} \end{align*}