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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.

Esat Koç
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  • Try to prove that $(T^{-1})^* ; T^* = \mathrm{id}{F'}$ and that $T^, (T^{-1})^ = \mathrm{id}{E'}$. Note that these are equalities of functions, i.e. you have to evaluate them in elements of their domain. – el_tenedor Mar 28 '17 at 19:49

1 Answers1

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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.

el_tenedor
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