Dual operator of an isometry

Question

If $X,Y$ are Banach spaces and $\phi:X\to Y$ is an isometry, show that $\phi^*$ is surjective.

I can use the equality $^\perp(ran \phi^*) = \ker\phi=\{0\}$, and also use the fact that $ran \phi^*$ is closed to show that $\phi^*$ is surjective. But I want to show it by using Hahn- Banach theorem, while I do not have any idea about it. Please give me a hint. Thanks for your help.

Answer 1

Since $\phi$ is a linear isometry, $\phi(X)$ is a subspace imbedded in $Y$. Let $f\in X^{\ast}$. We want to show $\exists g\in Y^{\ast}$ such that $\phi^{\ast}(g)=f$.

Given $f \in X^{\ast}$, define $h_{f}\in \phi(X)^{\ast}$ by $h_{f}(\phi(x))=f(x), x\in X$. $h_{f}$ is continuous since $\phi$ is an isometry. By Hahn-Banach extension theorem, $\exists g\in Y^{\ast}$ such that $g\restriction{\phi(X)}=h_{f}$, i.e., $g(\phi(x))=h_{f}(\phi(x))=f(x) \forall x\in X$.

Hence $\phi^{\ast}(g)=f$.