# Compact surjective non injective operator

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 ?

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.
• Does $X/kern(A)$ equiped with the norm $|class(x)|=|Ay|$, $y\in class(x)$ ? – user165633 May 20 '14 at 14:10
• norm on $X/kern(A)$ is $\|\hat x\| = \inf_{x\in \hat x}\|x\|_X$ – daw May 20 '14 at 15:08
• I thought $class(x)=\{y\in X, Ax=Ay\}$ – user165633 May 20 '14 at 15:22
• Of course, but the norm of $class(x)$ is the minimal $X$-norm of all elements in the equivalence class. – daw May 22 '14 at 5:47