# Unbounded Fredholms operators

Motivated by the situation of bounded Fredholm operators, 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?

• The equality holds for closed unbounded operators, provided one interprets "cokernel" as "quotient by the closure of the image". – André Henriques May 12 '19 at 13:42
• @ André: What is a standard reference for this, or is it easy to see? – Dave Shulman May 12 '19 at 13:44
• 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. – André Henriques May 12 '19 at 13:52
• Please put this as an answer I will accept it! – Dave Shulman May 12 '19 at 14:05

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.