How to show the dual map of an isometric isomorphism is an isometric isomorphism?

Question

Let $X$ and $Y$ be normed spaces and $T : X \to Y$ an isometric isomorphism. Then how to show the dual map $T^* : Y^* \to X^*$ is an isometric isomorphism?

The map $T^*$ acts as $T^*(f) = f \circ T$ for all $f \in Y^*$.

Answer 1

Tsemo is correct, in that the fact that $T$ maps the unit sphere of $X$ bijectively onto the unit sphere of $Y$ implies that $T$ is an isometry. Furtermore, note that \begin{align*} \operatorname{Ran}(T^*)_\perp&=\{x\in X:\forall f\in Y^*,\langle T^*f,x\rangle=0\}\\ &=\{x\in X:\forall f\in Y^*, \langle f,Tx\rangle=0\}\\ &=\ker(T)\\ &=\{0\} \end{align*} Thus $\operatorname{Ran}(T^*)$ is weak$^*$-dense in $X^*$. But as $T^*$ is an isometry, it's range is norm-closed, hence weak$^*$-closed, and thus equals $X^*$. Therefore $T^*$ is surjective.

Answer 2

Since $T$ is an isomorphism and an isometry, $T^{-1}$ is also an isometry. We deduce that $T(\{x\in X:\|x\|=1\})=\{y\in Y:\|y\|=1\}$ This implies that $\|T^*(f)\|=sup_{\|x\|=1}||f(T(x))|=sup_{\|y\|=1}|f(y)|=\|f\|$.