Adjoint of inverse map is the inverse of adjoint map? Let  $E,F$ be Banach normed spaces and $S,T \in L(E,F)$.Denote adjoint of  $T$ as $T^* .$ Prove that  if $   T^{-1}$ exist and $T(E)=F $, then $(T^{-1})^* = (T^*)^{-1}. $
Actually, I could not proceed beyond writing definition of adjoint map.
 A: This should get you started:
For $(T^*)^{-1}$ to exist you, you have to prove there exists a mapping $G \colon E' \to F'$ such that 
$$G \circ T^* = \mathrm{id}_{F'} \text{ and } T^* \circ G = \mathrm{id}_{E'}.$$
In our case the mapping $G = (T^{-1})^*$ looks like the perfect candidate. So let's see if 
$$
(T^{-1})^* \circ T^* = \mathrm{id}_{F'} \tag{1}
$$
Since (1) is an equality of functions you will have to verify (1) in all points of the domain of $(T^{-1})^* \circ T^*$ which is $F'$. So we  know that (1) holds iff
$$
\big((T^{-1})^* \circ T^*\big)(\varphi) = \mathrm{id}_{F'}(\varphi) = \varphi\tag{2}
$$
holds for all $\varphi \in F'$. This is once again an equality of functions. Once again you know that (2) holds iff
$$
\bigg(\big((T^{-1})^* \circ T^*\big)(\varphi)\bigg)(x) = \varphi(x)\tag{3}
$$
holds for all $x \in F$. Now let's simplify (3) using the defining property of the adjoint map:
$$
\bigg(\big((T^{-1})^* \circ T^*\big)(\varphi)\bigg)(x)
= \bigg((T^{-1})^* \big(T^*(\varphi)\big)\bigg)(x)
\overset{\ast}{=} \big(T^*(\varphi)\big)(T^{-1}x)
\overset{\ast}{=}\varphi\big((T \circ T^{-1})x\big)
= \varphi(x).
$$
This proves (3).
Since $x$ was arbitrary, we have proven (2). But since $\varphi$ was arbitrary as well, we have proven (1).
The equality $ T^* \circ (T^{-1})^*$ is proven just like this one.
