If $T\in B(X,Y)$ and $T$ is bijective is $T^*$ also bijective?

Question

1) This first link seems to provide a proof that seems to work for all normed vector spaces.

If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism?

2) This second link provides a counter example.

I can't seem to find any differences in the assumptions. Could someone explain whats the difference and if the statement is rue in general for all normed vector spaces? I know in the first link it mentions linear homeomorphisms and hilbert spaces but the proof provided seems to hold for any normed spaces(not just banach)and the the continuity of the inverse is not used in the proof.

Kavi Rama Murthy · Accepted Answer · 2019-02-26T12:24:14.480

1

In the first post the spaces are complete and in the second they are not. In the first post completeness is necessary. It is assumed that the inverse $S$ of $T$ is a bounded operator. (If it is just a linear map we cannot even define its adjoint). So Open Mapping Theorem is used here.

edited Feb 26 '19 at 12:24

answered Feb 26 '19 at 12:20

Kavi Rama Murthy

311,013

Yes but the proof in the first post hold for all normed spaces not just for banach spaces. – Jhon Doe Feb 26 '19 at 12:21
@JhonDoe I have added some details to my answer. – Kavi Rama Murthy Feb 26 '19 at 12:24
The first link shows: if $T$ has an inverse, then... However, merely "bijective" is not enough to imply an inverse when the spaces are not complete. – GEdgar Feb 26 '19 at 12:30
I see. So both spaces have to be banach for the inverse to be bounded. – Jhon Doe Feb 26 '19 at 12:35
@GEdgar i think the inverse exists just that it need not be bounded – Jhon Doe Feb 26 '19 at 12:37

If $T\in B(X,Y)$ and $T$ is bijective is $T^*$ also bijective?

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