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I did not find a reference for the following question, so I will pose it here. I think the answer should be elementary.

Let $F:X\rightarrow Y$ be a semi-Fredholm operator between Banach spaces, i.e. $\ker F$ is finite-dimensional and the image of $F$ is closed. Usually one wants to argue that $F$ is even Fredholm, thus exhibiting $\mathrm{coker} F:=Y/\mathrm{im} F$ as finite-dimensional. The usual argument involves the annihilator of the image of $F,$ which is easily seen to be isomorphic to $(\mathrm{coker}F)'$. More precisely, the annihilator of the image is connected to weak solutions of the (formally) adjoint operator. Once one has shown that the annihilator of the image is finite-dimensional (usually via ellipticity), it follows that the annihilator of the image is isomorphic to the cokernel of $F$.

But I have also read sources where the cokernel is directly related to the annihilator of the image. Is it true that $(\mathrm{im} F)°\cong \mathrm{coker}F$ for the situation described above? It is certainly true if $X$ and $Y$ are Hilbert spaces.

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    $\begingroup$ Your question has nothing to do with semi-Fredholm operators. If $Y$ is a closed subspace of Banach space $X$, then the annihilator of $Y$ in $X^{*}$ is canonically isometrically isomorphic to the dual of $X/Y$. Look in any basic text in functional analysis--this will either be in the text or in the problems. $\endgroup$
    – Bill Johnson
    Commented Oct 18, 2011 at 20:06
  • $\begingroup$ @Bill: I know, but that was not the question. The question was whether one can always identify the annihilator of $Y$ with $X/Y$. I assume this is false, since one can simply take some Banach space $Z$ which is not isomorphic to its dual and consider $Y=0$ and $X=Z.$ Am I correct? $\endgroup$
    – Orbicular
    Commented Oct 18, 2011 at 20:42
  • $\begingroup$ Yes, that is right. $\endgroup$
    – Bill Johnson
    Commented Oct 19, 2011 at 0:08

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