Elliptic operators on compact space are Fredholm I have come across this fact in a reading of mine, but I cannot seem to prove it, and I cannot seem to find a proof of it.
Mostly, I am confused why the range of an elliptic operator between appropriate Sobolev spaces has closed range.
Thank you for any help.
 A: Let $X$ and $Y$ be the appropriate Sobolev spaces and $L\colon X \to Y$ the elliptic operator. The two facts we need are that $X$ embeds compactly into $Y$ (this is where we need the domain to be compact), and that $L$ satisfies an estimate of the form $$|u|_X \le C(|Lu|_Y + |u|_Y).$$ 
$\ker L$ is finite-dimensional since, for instance, its closed unit ball is compact. Thus we can write $X$ as a direct sum $X=X_1 \oplus \ker L$. 
Now to show that the range of $L$ is closed. Let $u_n$ be a sequence in $X$ with $Lu_n = f_n \to f$ in $Y$. It's enough to show that $u_n$ has a subsequence converging in $X$. Without loss of generality we can assume $u_n \in X_1$. First let's suppose that the sequence $u_n$ is bounded in $X$. Then since $X \to Y$ is compact, we can extract a subsequence converging in $Y$. Since $$|u_n-u_m|_X \le C(|f_n-f_m|_Y+|u_n-u_m|_Y),$$ this implies that $u_n$ in fact converges in $X_1$. It remains to treat the case where $|u_n|_X \to \infty$. Then we have $$L\frac{u_n}{|u_n|_X} = \frac{f_n}{|u_n|_X} \to 0,$$ and hence by our earlier argument that, after extraction, $u_n/|u_n|_X \to v \in X_1$. But then $v \in X_1$ has $|v|_X = 1$ and $Lv=0$, contradicting the definition of $X_1$.
