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Seems simple enough but I can't quite get it. $X$ is a complex Banach space, and $T\in B(X)$, $T'\in B(X')$ is its adjoint. Suppose $T'$ is invertible. How can we show that $T$ is invertible?

I have actually already shown $T$ to be 1-1, and that $T^{-1}$, were it to exist, would be bounded. So it actually only remains to show that $T$ is onto.

EDIT: I suspect the following simple result, which I have shown to be true, might help:

$\|Tx\| ≥{\|(T')^{-1}\|}^{-1} \|x\|$

This is how I know that $ker(T)=\{0\}$, hence 1-1. I think that were we to assume $T$ were not onto, we could somehow show $T'$ is not 1-1 (by showing it has non zero kernel), contradiction since $T'$ invertible. Can anyone see exactly how this might work?

Svetoslav
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Rfult
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  • For a given $x\in X$, we will get an element $\xi$ of $X''$ such that $T''\xi=x''$. Do we know that $X$ is reflexible? Or, possible there is some more trick in there.. – Berci Mar 30 '13 at 16:04
  • No we don't have reflexivity. I have already shown T invertible implies T' invertible, that was pretty easy. If we had reflexivity, my question above would follow immediately but sadly we don't. Please see my 'EDIT' above for additional information which I think might help – Rfult Mar 30 '13 at 16:21

1 Answers1

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Let $T:X\longrightarrow Y$ be a bounded linear operator between two normed vector spaces, and let $T^*:Y^*\longrightarrow X^*$ denote its adjoint. By the open mapping theorem, $T^*$ is invertible if and only if it is bijective, as $X^*$ and $Y^*$ are automatically Banach spaces. This is not true for $T$ in general, but it is true if we assume $X,Y$ to be Banach spaces.

Of course, if $T$ is invertible, $TS=ST=I$ implies $S^*T^*=T^*S^*=I$ hence $T^*$ is invertible with inverse $(T^{-1})^*$. You want the converse, which amounts to: does $T^*$ bijective imply $T$ bijective? Note that if $X,Y$ are supposed to be reflexive, then the result follows trivially via the identification $T^{**}\simeq T$ and the observation above.

Recall that the orthogonal of a set $S$ in $X$ is defined to be $S^\perp:=\{\phi\in X^*\;;\;\phi(x)=0\;\forall x\in S\}$. This is a weak* closed subspace of $X^*$. Now $$ \mbox{Im}\:T^*\subseteq (\mbox{Ker}\;T)^\perp. $$ Indeed, if $x^*=T^*y^*$, then $(x^*,x)=(T^*y^*,x)=(y^*,Tx)=0$ for every $x\in\mbox{Ker}\;T$.

As a consequence, if $T^*$ is onto, then $(\mbox{Ker}\;T)^\perp=X^*$. Now if we had $\mbox{Ker}\;T\neq\{0\}$, Hahn-Banach would provide us with a bounded linear functional $x^*\in X^*$ which does not vanish on $\mbox{Ker}\;T$, contradicting $(\mbox{Ker}\;T)^\perp=X^*$. Hence

$$T^*\;\mbox{surjective}\quad\Rightarrow\quad T\;\mbox{injective}.$$

Now observe that in general $$ \mbox{Ker}\;T^*=(\mbox{Im}\;T)^\perp. $$ Indeed, $T^*y^*=0$ if and only if $(T^*y^*,x)=0$ for all $x\in X$: that is $(y^*,Tx)=0$ for all $x\in X$, i.e. $(y^*,y)=0$ for all $y\in \mbox{Im}\;T$. So if $T^*$ is injective, we get $(\mbox{Im}\;T)^\perp=\{0\}$: i.e. every bounded linear functional $y^*\in Y^*$ which vanishes on $\mbox{Im}\;T$ must vanish on $Y^*$. By Hahn-Banach, this yields $\overline{\mbox{Im}\;T}=Y$. The good way to see this is to consider the annihilator: if $S\subseteq Y^*$, denote $S^o:=\{y\in Y\;;(y^*,y)=0\;\forall y^*\in S\;$. Then for any subspace $F$ in $Y$, $(F^\perp)^o=\overline{F}$. Hence $$ (\mbox{Ker}\;T^*)^o=\overline{\mbox{Im}\;T}. $$ By Hahn-Banach, $S^o=Y$ if and only if $S=\{0\}$. So we get an equivalence:

$$T^*\;\mbox{injective}\quad\iff\quad \overline{\mbox{Im}\;T}=Y.$$

Finally, and that's the tricky part of the whole thing, actually: when $X,Y$ are Banach spaces

$$\mbox{Im}\;T\;\mbox{is closed}\quad\iff\quad \mbox{Im}\;T^*\;\mbox{is closed}.$$

Proof: see here, page 77.

Note: it can probably be simplified here given that we assume $T^*$ to be invertible, but I think it is good to remember this general fact.

Conclusion: these three pieces altogether show that when $X,Y$ are Banach spaces, $T^*$ is invertible if and only if $T$ is invertible.

Julien
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  • Thanks. It turns out that using the fact I stated in the EDIT in the question, you can show T(X) to be closed quite easily by considering a cauchy sequence in T(X) – Rfult Mar 31 '13 at 00:23
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    You said that "By the open mapping theorem, T ∗ is invertible if and only if it is bijective, as X ∗ and Y ∗ are automatically Banach spaces.". But, I am wondering why you say "By the open mapping theorem". Isn't ANY map invertible if and only if it is bijective, by definition of invertibility? – mononono Mar 03 '15 at 16:43