$T \in B(X,Y)$ is an isometry if and only if $T^*$ is an isometry I would like to prove that $T \in \mathscr{B}(X,Y)$ is an isometry of $X$ onto $Y$ if and only if $T^*$ is an isometry of $Y^*$ onto $X^*$. I am not really sure what to do. I started the argument as follows and got stuck.
Suppose that $T \in \mathscr{B}(X,Y)$ is an isometry of $X$ onto $Y$. Then
$$
\|x\| = \|Tx\| = sup \{ |\langle Tx, x^* \rangle| \;x^* \leq 1 \} = sup \{ |\langle x, T^* x^* \rangle | \;x^* \leq 1 \}
$$
I don't know if I can say anything about $x$ here, I would like to finish the equality to say that this is equal to $\|T^*x\|$ but I'm missing something.
Any help is appreciated thanks in advance. 
 A: I am giving you a $\textbf{HINT}$ with some background, so $\textbf{first note that}$ 
$(1)$ $T$ is an isomorphism of $X$ onto $Y$ if and only if $T^*$ is an isomorphism of $Y^*$ onto $X^*$. To prove this one direction is easy, i.e, if $T$ is an isomorphism then $T^*$ is an isomorphism (easy exercise). For the other direction observe that since $T^*$ is an open mapping, there is a constant $c>0$ such that $T^*(B_X) \supseteq \{x^*\in X^*\mid\|x^*\|\leq c\}$. So if $x\in X$, then
\begin{align}
\|Tx\| &= \sup\{|\langle Tx,y^*\rangle|: y^*\in B_{Y^*}\}\\
&= \sup\{|\langle x,T^*y^*\rangle|: y^*\in B_{Y^*}\}\\
&\geq \sup\{|\langle x,x^*\rangle|: x^*\in X^* \text{and } \|x^*\|\leq c\}\\
&=c^{-1}\|x\|.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\text{CORRECTION: It should be $c$ in place of $c^{-1}$})
\end{align}
Thus, $\mathrm{ker}(T) = 0$ and $\mathrm{Range}(T)$ is closed (Why?). On the other hand, $(\mathrm{Range}(T))^{\perp}=\mathrm{ker}(T^*) = \{0\}$ since $T^*$ is invertible. Thus $\mathrm{Range}(T)$ is also dense. This implies that $A$ is surjective and thus invertible.
Now, coming to your question.
$\textbf{HINT}:$ If $T$ is an isometry of $X$ onto $Y$, then by (1) proved above $T^*$ is an isomorphism. Also, $T(B_X) = B_Y$, so \begin{equation}\|T^*(y^*)\| = \sup_\limits{x\in B_X}|\langle x,T^*y^*\rangle| = \|y^*\|.\end{equation} The other direction is similar.
