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

Question

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. \begin{equation} \inf_{Ky=0}\Vert x+y \Vert_Y\leq \Vert Kx\Vert_Y. \end{equation}

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?

Answer 1

Alternative answer:your operator induces a continuous bijection between quotient space and image subspace. By the open mapping theorem if it were compact then the unit ball in both spaces were compact, thus finite dimensional by Riesz

Answer 2

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.