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 $Y$?
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 $Y$?
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.
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 $Y$
.