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The question comes from the following link on page 25: https://www.ucl.ac.uk/~ucahad0/3103_handout_3.pdf

They prove $(T^{-1})^*=(T^*)^{-1}$, but I don't see how it proves $T^*$ is a topological isomorphism.

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    $\begingroup$ You need $T^*$ to have an inverse, and lo and behold, it's $(T^{-1})^*$. $\endgroup$ – Randall Apr 22 at 15:16
  • $\begingroup$ We have that the adjoint is bounded and linear, hence continuous. The thing I'm not too certain about is where to get it to also be bijective. $\endgroup$ – HCS Apr 22 at 15:22
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    $\begingroup$ the claimed identity proves that it is bijective. $\endgroup$ – Randall Apr 22 at 15:23
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    $\begingroup$ I can't believe I forgot about this! Thank you! $\endgroup$ – HCS Apr 22 at 15:28
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Thanks to Randall, I was able to figure it out.

$T^*$ having an inverse implies it is bijective. We also know the adjoint is bounded and linear. We can use this to show $T^*$ is a topological isomorphism.

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