Question: If $H$ is a Hilbert space and $T:H\rightarrow H$ is a bounded linear operator, $\mathrm{ran}\,T$ is closed, and
$$\dim \ker T=d<\infty$$
if $T-T^*$ is compact, prove that $T$ is a Fredholm operator with index $0$.
My attempt Since $T=(T-T^*)+T^*$, by Fredholm theory, if $$\mathrm{codim}\,\mathrm{ran}\,T<\infty$$
then $T$ is Fredholm, so $$\mathrm{ind}\,T=\mathrm{ind}\,T^*$$
however, by definition we have $\mathrm{ind}\,T=-\mathrm{ind}\,T^*$, so $\mathrm{ind}\,T=0$. So the remaining problem is to prove
$$\mathrm{codim}\,\mathrm{ran}\,T<\infty$$
I do not know how to begin, any help will be appreciated.
Best Answer
Since $\dim \ker T<\infty$ and $\mathrm{ran}\,T$ is closed, it results that $T$ is semi-Fredholm. Hence $T^*$ is semi-Fredholm with $\mathrm{codim}\,\mathrm{ran}\,T^*<\infty$ (and $\mathrm{ran}\,T^*$ is closed). Furthermore we have
$$(1) \quad \mathrm{ind}\,T=-\mathrm{ind}\,T^*.$$
Now let $K:= T -T^*$. Since $K$ is compact, $T-K$ is semi- Fredholm with
$$(2) \quad \mathrm{ind}\,T=\mathrm{ind}\,(T-K).$$
But we have $T^*=T-K$, hence
$$(3) \quad \mathrm{ind}\,T=\mathrm{ind}\,T^*.$$
From $(1)$ and $(3)$ we now see, that $ \mathrm{ind}\,T=\mathrm{ind}\,T^*$ is finite $=0$ and therefore $\mathrm{codim}\,\mathrm{ran}\,T<\infty.$