Adjoint Operator between Banach Spaces is isomorphic iff Operator is If $T: X \to Y$ is an [isometric] isomorphism between normed spaces $X$ and $Y$, then also $T^{\prime}:Y^{\prime} \to X^{\prime}$ is an [isometric] isomorphism. 
If $X$ and $Y$ are Banach Spaces, even the converse holds.
$X^{\prime}$ stands for the dual space of $X$.
I already proved the statement for the non-banach case. For the banach space case (converse direction), im pretty clueless.
Applying the proven statement again to $T^{\prime}$ yields that $ T^{\prime \prime}: X^{\prime \prime} \to Y^{\prime \prime}$ is isomorphic, but thats not quite want i need.
I hope you can help me out.
Thank you.
 A: As you know, there is a natural isometric embedding $\Phi_X$ from $X$ into $X''$, defined by $(\Phi_X x)(f) = f(x)$ for $f \in X'$, and likewise $\Phi_Y$ from $Y$ into $Y''$.  You can verify that
$$T'' \Phi_X = \Phi_Y T.$$
Now suppose $T'$ is an isomorphism; then by the forward direction of the statement, $T''$ is too.  Then you can check that $T$ is injective and bounded, and that if $T''$ is an isometry then so is $T$.  You can also show that $T$ has dense image, using Hahn-Banach: suppose that $f \in Y'$ satisfies $f(Tx) = 0$ for all $x \in X$, and conclude that $f=0$.
It remains to check that $T$ is surjective and that $T^{-1}$ is bounded.  Here you need the assumption that $X$ is Banach.  In this case you can show that the range of $T'' \Phi_X$ is closed in $Y''$.  In particular it is complete, and so you can show that the range of $T$ is also complete.  Since the range was already dense, this shows that $Y$ is Banach and that $T$ is surjective.  Now the open mapping theorem lets you conclude that $T^{-1}$ is bounded.
Note it isn't necessary to assume that $Y$ is Banach, but it follows from the other assumptions.
