# $Im(K)\subset Y$ closed, infinite-dimensional: $K:X\to Y$ is not a compact operator

Proposition: If $K:X\to Y$ is a bounded linear operator between two Banach spaces $X$ and $Y$ such that $\operatorname{Im}(K)\subset Y$ is an infinite-dimensional closed subspace, then $K$ is NOT compact.

Proof: I have to prove it, but the only think I know is that from the Closed Image theorem: $\operatorname{Im}(K)\subset Y$ is closed if and only if $\exists C>0 \quad \forall x\in X$ s.t. $$\inf_{Ky=0}\Vert x+y \Vert_Y\leq \Vert Kx\Vert_Y.$$

Also I know a Theorem that says: $dim(Z)<\infty$ is equivalent to $B$ compact. Where $B$ is the closed unit ball on Z. But I have no idea how to do the proof formally, and also how to conclude. Can someone help me?

• No no I do mean $Im(K)$ ah yes thank you! I will do some change – MorganeMaPh Jul 23 '16 at 8:02
• Do you mean, then $K$ is not compact. – Peter Jul 23 '16 at 8:04
• Sorry @Petter, I was confused! – MorganeMaPh Jul 23 '16 at 8:07

First, note that $Z:=Im(K)$ is a closed subspace of a Banach space and thus, itself a Banach space. Thus, $K: X\to Z$ is onto. By the open mapping theorem, $K$ is open and hence, $K$ is mapping open sets to open sets.
Now, assume that $K$ is compact and take the image of the open unit ball $C:=K(B_X^\circ)$ which is open in $Z$ and relatively compact in $Y$. But since it is relatively compact in $Y$, it is easy to see that it is relatively compact in $Z$.
As $C$ is open, there is a $r>0$ such that $r B_Z\subset \bar{C}$. Since $r B_Z$ is a closed subset of a compact subset, it is compact. Hence, $B_Z$ is compact which is a contradiction to the fact that $Z$ was an infinite dimensional Banach space.