Proof of several facts on bounded operators $A : X\to Y, B : Y \to X$ such that $AB$ is Fredholm

by L.F. Cavenaghi   Last Updated May 22, 2020 23:20 PM - source

Let $$A : X\to Y$$ and $$B : Y \to X$$ be bounded operators between Banach spaces $$X,Y$$. Assume that $$AB : Y\to Y$$ is Fredholm. I would like to prove that $$A(X)$$ and $$B(Y)$$ are closed. Furthermore, possibly conclude that $$A$$ is also Fredholm.

I made some progress on it. For instant, since $$\dim \ker AB = \dim B(Y)\cap \ker A$$, then this intersection has finite dimension. It would be very nice to conclude that $$\ker A$$ is finite dimensional. However, I don't see how it is possible from here. But note that if $$B$$ is Fredholm, then necessarily $$\dim B(Y) = \infty$$, and hence, $$\dim \ker A < \infty$$.

On the other hand, if I show that $$A(X)$$ is closed, then since $$\infty > \dim Y/AB(Y) \geq \dim Y/A(X)$$ it would follow that $$A$$ is Fredholm.

However, I was not able to show that $$A(X)$$ and $$B(Y)$$ are closed, neither that is the case that $$\dim B(Y) = \infty$$ (in general).

Any hints?

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