I have been asked this question at my course on Functional Analysis, to tell something about an operator that is both compact and Fredholm. The answer needs to be related to the spaces between which our operator acts. What I know is that a Fredholm operator is already a bounded linear map between Banach Spaces (let's call them $X$ and $Y$).
I also know that the spectrum of a compact operator is the same as the point spectrum, and that it is at most countable. I was hoping to somehow show using these facts that $Y$ must be finite dimensional. Maybe we could show that the image is finite dimensional, and hence show that $Y$ is also. But for now this is just speculation.
Thanks in advance!
Best Answer
A Fredholm operator $T$ is an operator that is invertible modulo the compacts. That is, there exists $S:Y\to X$ such that $I_X-ST$ and $I_Y-TS$ are compact. Which can be phrased as saying that $K_1=I_X-ST$ and $K_2=I_Y-TS$ are compact.
If $T$ is also compact, then $ST$ and $TS$ are compact, and then $$ I_X=ST+K_1\qquad\text{ and } \qquad I_Y=TS+K_2 $$ are compact. This forces $X$ and $Y$ to be finite-dimensional.
Alternatively one can argue that if $T$ is Fredholm then it has finite-dimensional kernel, finite-dimensional co-kernel, and closed range. If $T$ is also compact, any compact operator with closed range has finite-dimensional range. As $\dim Y/\operatorname{ran}T<\infty$, we get that $\dim Y<\infty$. Since $\dim\ker T<\infty$, it is complemented in $X$. On the complement, $T$ is injective, which makes this complement finite-dimensional (as $\dim\operatorname{ran}T<\infty$). Then $\dim X<\infty$.