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Let $X$ and $Y$ be normed spaces, both real or both complex, let $B(X,Y)$ denote the space of all the bounded linear operators $T \colon X \to Y$, and let $T^\times$ denote the adjoint operator of $T$.

Then how to prove the following?

If $T \in B(X,Y)$ is bijective and $T^{-1} \in B(Y,X)$, then $T^\times$ is also bijective and $$(T^{-1})^\times = (T^\times)^{-1}.$$

Let $X^\prime$ denote the space of all the bounded linear functionals with domain $X$, and let $Y^\prime$ denote the space of all the bounded linear functionals with domain $Y$.

Then $T^\times \in B(Y^\prime, X^\prime)$ such that $$ (T^\times g ) (x) = g( Tx) \ \mbox{ for all } \ x \in X \ \mbox{ and for all } \ g \in Y^\prime.$$

Also, we have $$\Vert T^\times \Vert = \Vert T \Vert. $$

And, $$I^\times = I,$$ where $I$ is the identity operator on a normed space.

Moreover, if $X, Y, Z$ are normed spaces, all real or all complex, and if $T \in B(X,Y)$ and $S \in B(Y,Z)$, then $ST \in B(X,Z)$ and $$(ST)^\times = T^\times S^\times.$$

Now as $$T T^{-1} = I = T^{-1} T,$$ so $$(T^{-1})^\times T^\times = I = T^\times (T^{-1})^\times.$$

Thus, the operator $T^\times \colon Y^\prime \to X^\prime$ is bijective and we have $$(T^\times)^{-1} = (T^{-1})^\times. $$

Is this reasoning correct?

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Yes, your reasoning is correct.

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