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Let $X$ be an infinite dimensional Banach space. I know that every compact operator $A$ is not bijective or $0\in\sigma(A)$.

Fox example the compact operator $A$ defined on $X=C([0,1],\mathbb{R})$ (equipped with the supremum norm) for each $x\in X$ by $$(Ax)(t)=\int_0 ^t x(s)ds, \ \ \ \forall t\in \mathbb{R}.$$ This operator is injective and non surjective. Can we find an example of a compact operator which is not injective but surjective ?

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Let $X,Y$ be a infinite-dimensional Banach space.

Then compact operators from $X$ to $Y$ cannot be surjective: This would imply that their range is closed. Then the canonical injection $$ \hat A: X/kern(A) \to R(A) $$ would be continuously invertible and compact - a contradiction.

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  • $\begingroup$ I just saw that if a compact operator is surjective then it is open by the open map theorem, so the image of an open ball is open and precompact at the same time which is not possible in infinite dimensions, because there's no open (nonempty) set inside a compact. $\endgroup$ – user165633 May 20 '14 at 13:54
  • $\begingroup$ Does $X/kern(A)$ equiped with the norm $|class(x)|=|Ay|$, $y\in class(x)$ ? $\endgroup$ – user165633 May 20 '14 at 14:10
  • $\begingroup$ norm on $X/kern(A)$ is $\|\hat x\| = \inf_{x\in \hat x}\|x\|_X$ $\endgroup$ – daw May 20 '14 at 15:08
  • $\begingroup$ I thought $class(x)=\{y\in X, Ax=Ay\}$ $\endgroup$ – user165633 May 20 '14 at 15:22
  • $\begingroup$ Of course, but the norm of $class(x)$ is the minimal $X$-norm of all elements in the equivalence class. $\endgroup$ – daw May 22 '14 at 5:47

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