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Holub proved that Fredholm operators are stable under compact perturbations. I am interested in a slight refinement of this theorem.

Suppose we have two operators $T_1$ and $T_2$ acting on a primary Banach space $E$. Assume that the range $(T_1+T_2)(E)$ is isomorphic to $E$ and complemented in $E$. Is it true that there exists $i\in \{1,2\}$ such that the range $T_i(E)$ is isomorphic to $E$ and complemented in $E$?

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    Sure, but I am not interested in trivial cases o$f$ course.2012-03-25

2 Answers 2

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The answer is no. This isn't true even for prime spaces, which is stronger than primary. Take, for example $\ell_p$. Take your favorite operator $T$ which is not closed range. Define $T_1$ on $\ell_p=\ell_p\oplus \ell_p$ to be $T_1=(I-T)\oplus T$, and $T_2=T \oplus (I-T)$.

It is easy to see that neither operator is closed range, so neither image can be isomorphic to $\ell_p$. But $T_1+T_2=I$, which has range isomorphic to $\ell_p$ and complemented in $\ell_p$.

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Let $F=\ell^1\oplus\ell^2$, $\pi_1,\pi_2$ the coordinate projections. Then $\pi_1+\pi_2$ is the identity, so its range is trivially isomorphic with $F$, but neither projection has range isomorphic to $F$ (since $\ell^2$ is a Hilbert space, $\ell^1$ isn't, and complemented subspaces of $\ell^1$ are isomorphic to $\ell^1$).

Now let $E=F\oplus F$, $T_i=\pi_i\oplus 0$. Note that $E\simeq F$, because both $\ell^1$ and $\ell^2$ are isomorphic to their squares. So $T_1+T_2$ has range isomorphic to $E$ and complemented. But neither range of $T_1,T_2$ is isomorphic to $E$, as $E$ cannot be isomorphic to neither $\ell^1$ nor $\ell^2$.

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    Banach space $E$ is primary, whenever we have $E=X\oplus Y$, then at least $X$ or $Y$ is isomorphic to $E$.2012-03-25