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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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    What were you assuming about the answers to my two questions? $\:$ (Your answer might be "nothing".) $\;\;$2012-02-05
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    Also, you never used the variable $c$. $\:$2012-02-05
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    1) Which questions? 2) Typo fixed2012-02-05
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    "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?" $\;\;\;$2012-02-06
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    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 answer2012-02-06
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    Is your "strong topology" the norm topology or the SOT (http://en.wikipedia.org/wiki/Strong_topology)?2012-02-06
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    Norm topology, of course2012-02-06
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    I think that something's wrong: Let $A\colon \ell_1\to \ell_1$ be an isomorphism onto uncomplemented subspace of $\ell_1$ isomorphic to $\ell_1$. On the other hand, any subspace of $\ell_1^*=\ell_\infty$ isomorphic to $\ell_\infty$ is complemented, whence $A^*$ satisfies the assumptions, whereas $A$ fails the claim.2012-02-06
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    If $A$ is an isomorphism onto its range, then $A^*$ is not an isomorphis onto its range.2012-02-06
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    Why not? Take the identity operator.2012-02-06
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    Ok, I meant operators with non dense images2012-02-06
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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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    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.2012-02-06
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    @AleksandrasStulginskis So, have I answered your question?2012-02-06
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    Nope, sorry. For Hilbert spaces everything is trivial.2012-02-06
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    I don't see what that has to do with whether Norbert has answered your question. $\:$2012-02-06
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    Oops, I thought it was you. Pardon.2012-02-06