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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?

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

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  • $\begingroup$ Just to be sure, unlike for the Fredholm case, there exist closed unbounded operators whose image is not closed? $\endgroup$ – Dave Shulman May 12 at 14:15
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    $\begingroup$ A closed self-adjoint operator rarely has closed image. The image is closed iff zero is an isolated point in the spectrum (a property which holds true for Fredholm operators). $\endgroup$ – André Henriques May 12 at 14:19
  • $\begingroup$ Ok, I see - thanks a lot! $\endgroup$ – Dave Shulman May 12 at 14:26

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