# 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". May 12 '19 at 13:42
• @ André: What is a standard reference for this, or is it easy to see? 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. May 12 '19 at 13:52
• Please put this as an answer I will accept it! 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.
Claim: $$\mathrm{ker}(D^*)=\mathrm{Im}(D)^{\perp}$$ for any $$D$$ densely defined. If $$x\in \mathrm{ker}(D^*)$$ and $$a\in \mathrm{Dom}(D)$$, then $$(x,Da)=(D^*x,a)=0$$, so LHS in RHS. Conversely, if $$y\in \mathrm{Im}(D)^{\perp}$$, then $$(y,D\cdot)=0$$ on $$\mathrm{Dom}(D)$$, so $$y\in \mathrm{Dom}(D^*)$$ and $$D^*y=0$$, hence $$y\in \mathrm{ker}(D^*)$$, ie RHS in LHS. Now recall $$\mathrm{Im}(D)^{\perp}=H_2/\overline{\mathrm{Im}(D)}$$.