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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 $Y$?

2 Answers 2

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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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    @Aleksandras, in your "counter-example" $A^*$ is not injective, so it does not satisfy the hypotheses of your question. As Norbert has in effect said: if $A:X\to Y$ and $A^*:Y^*\to X^*$ is injective with norm-closed range, then $A$ is surjective -- this is proved in Rudin, see the reference given.2012-02-08
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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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    Oops, I thought it was you. Pardon.2012-02-06