# Sufficient conditions for an operator to have complemented image

Given a bounded operator $A\colon X\to Y$ ($X$, $Y$ - Banach spaces) with $A^*\colon Y^*\to X^*$ being an isomorphism onto its range.

Under which assumptions on $A:X\to Y$, the range of $A$ is complemented in $Y$?

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Since $A^*$ is an isomorphism on its range then there exist $c>0$ such that for all $y^*\in Y^*$ we have $\Vert A^* y^*\Vert\geq c\Vert y^*\Vert$. Then from theorem 4.15 in W. Rudin Functional analysis we have $\operatorname{Im}(A)=Y$. So the range of $A$ is always complementable.

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What were you assuming about the answers to my two questions? $\:$ (Your answer might be "nothing".) $\;\;$ – Ricky Demer Feb 5 '12 at 23:53
Also, you never used the variable $c$. $\:$ – Ricky Demer Feb 5 '12 at 23:55
1) Which questions? 2) Typo fixed – Norbert Feb 5 '12 at 23:58
"Is $A^*$ supposed to be a $\:$[homeomorphic or isometric]$\:$ isomorphism onto its range? $\hspace{1.4 in}$ If the first of those, what topology do you have on the continuous duals?" $\;\;\;$ – Ricky Demer Feb 6 '12 at 0:13
I have considered strong topology on duals. My answer doesn't depends on the type of isomorphism. If we assume that $A^*$ is isometric, we can additionally say that $A$ is coisometric. As for other topologies I don't know the answer – Norbert Feb 6 '12 at 0:17

Is $A^*$ supposed to be a $\:$[homeomorphic or isometric]$\:$ isomorphism onto its range?
If the first of those, what topology do you have on the continuous duals?
In any case,

$\operatorname{Range}(A)$ is closed in $Y$ $\:$ and $\:$ $Y$ is homeomorphically isomorphic to a Hilbert space
$\implies$
$\operatorname{Range}(A)$ is complemented in $Y$
$\implies$
$\operatorname{Range}(A)$ is closed in $Y$

.

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I am interested mainly in the isomorphic case, that is $A^*$ is just bounded below. Primarily, I am interested in the norm topology mainly but feel free to consider the weak*-topology. – Aleksandras Stulginskis Feb 6 '12 at 0:02
@AleksandrasStulginskis So, have I answered your question? – Norbert Feb 6 '12 at 0:18
Nope, sorry. For Hilbert spaces everything is trivial. – Aleksandras Stulginskis Feb 6 '12 at 0:46
I don't see what that has to do with whether Norbert has answered your question. $\:$ – Ricky Demer Feb 6 '12 at 0:55
Oops, I thought it was you. Pardon. – Aleksandras Stulginskis Feb 6 '12 at 0:56