Fredholm operators, rank, and closure

Question

Let $D$ be a densely defined symmetric unbounded operator on a Hilbert space $H$. Assume that the kernel and cokernel of $D$ are finite dimensional, and hence that we can assign a well-defined index to $D$ by $$ Index == Dim(kernel) - Dim(cokernel). $$ Denoting the closure of $D$ by $D^c$, is it true that $D^c$ also has finite dimensional kernel and cokernel? Moreover, is it true that $$ Index(D) = Index(D^c)? $$ (Note that the index of $D^c$ is also defined to be the difference in dimensions between its kernel and cokernel.

Answer 1

I guess with $D$ having a finite-dimensional cokernel you mean that $H = ran\,D\dotplus N$ with some finite-dimensional subspace $N$ ($\dotplus$ denoting the direct sum). Actually, I will not answer your question here but try to shed a little more light on the whole thing.

Assuming your hypothesis on $D$, I will show that the index of $D$ must be non-positive and that the closure $S = D^c$ (which is also symmetric) must be Fredholm with also non-positive index.

First, as in the comments above, $ker\,S\subset ker\,S^* = (ran\,S)^\perp\subset(ran\,D)^\perp$. From this we infer $\dim ker\,S < \infty$, but also that $$ \dim\,(H/ran\,D)\,\ge\,\dim(H/\overline{ran\,D}) = \dim\,(ran\,D)^\perp\,\ge\,\dim ker\,S\,\ge\,\dim ker\,D. $$ Hence, $ind\,D\le 0$. Also, $ran\,S$ as a superset of $ran\,D$ must be finite-codimensional. By a theorem of Kato, $ran\,S$ is closed. So, $S$ is indeed Fredholm. Moreover, \begin{align*} ind\,S &= \dim ker\,(S) - \dim\,(ran\,(S))^\perp\\ &= \dim ker\,(S) - \dim ker\,(S^*), \end{align*} which is non-positive as $ker\,S\subset ker\,S^*$.