Motivated by the situation of bounded [Fredholm operators](https://en.wikipedia.org/wiki/Fredholm_operator), I have the following question about "unbounded Fredholm operators".


Let $\mathcal{H}\_1$ and $\mathcal{H}\_2$ be two Hilbert spaces, and 
$$
D: \mathrm{dom}(D) \subset \mathcal{H}\_1 \to \mathcal{H}\_2
$$
be a densely defined a unbounded operator. If we assume that the image is closed, and that $\mathrm{ker}(D)$ and $\mathrm{cokernel}(D)$ are both finite dimensional, is true that 
$$
\mathrm{cokernel}(D) = \mathrm{ker}(D^\*),
$$
where $D$ is the adjoint of $D$? Can you relax the assumptions on closure of the image, and/or dimensionality and still produce the smae result?



Unbounded Fredholms operators
The equality holds for closed unbounded operators, provided one interprets "cokernel" as "quotient by the closure of the image".


You can check it directly for closed self-adjoint operators by using the spectral theorem (i.e. check it for multiplication operators on $L^2(X)$ for some measure space $X$). Then use polar decomposition to reduce the case of an arbitrary closed operator to that of a self-adjoint operator.


PS: this has nothing to do with the operator being Fredholm or not.


