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This question is similar to other questions on MSE, but none of them has an answer which satisfies me. Given an infinite dimensional complex Banach space $X$, $T \in B(H)$ is Fredholm iff the cokernel and the kernel are finite dimensional (if the dimension of the cokernel is finite dimensional, the range is closed, so I have not written this last condition because it is redundant). Now, for Hilbert spaces it can be shown, using decompositions in direct orthogonal sums, that $\dim \operatorname{coker} T < \infty$ is equivalent to $\dim \ker T^* < \infty$., where $T^*$ is the adjoint of $T$. The adjoint can be defined even for general Banach spaces. Moreover, we can define a notion of orthogonal complement in Banach spaces using the dual space $X^*$. When $X$ is reflexive, we can obtain properties of orthogonality which are similar to Hilbert spaces (when $X$ is not reflexive, some analoguous facts which hold for Hilbert spaces are not anymore valid). So, I would expect that $\dim \operatorname{coker} T < \infty \Leftrightarrow \dim \ker T^* < \infty$ could be true for $X$ reflexive. But what happens for general Banach spaces? A proof as in the case of Hilbert spaces seem to be not anymore possible, but maybe something else could work. So my question is: does this equivalence still hold in general? If the answer is yes, could you please provide some reference with a proof of this fact?

EDIT: @s.harp Even though your proof seems correct to me, consider the Toeplitz operator with symbol $(z-1)$ on $H^2$. The kernel of this operator and the kernel of its adjoint are both trivial, so they have dimension $0$. This would imply that the dimension of the cokernel is finite, which implies that the range of the operator is closed. However, this Toeplitz operator has a dense - but not closed - range. Maybe this fact depends on the logical axioms used (as, for instance, in the case of Whitehead problem)?

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The relation $\dim(\mathrm{coker}(T))<\infty\iff\dim(\ker(T^*))<\infty$ remains true in arbitrary Banach spaces:

If $T$ has finite dimensional co-kernel then $\overline{\mathrm{im}(T)}$ admits a finite dimensional complement, choose one such complement and call it $V$. Note that $T^*(f)=0$ iff $f(Tv)=0$ for all $v\in X$, ie iff $f\lvert_{\overline{\mathrm{im}(T)}}=0$. As such $f$ is uniquely determined by its values on $V$, ie the map $\ker(T^*)\to V^*$ given by $f\mapsto f\lvert_V$ is injective. But $V^*$ is finite dimensional, so $\ker(T^*)$ also is finite dimensional.

On the other hand if the co-kernel is not finite dimensional then $\overline{\mathrm{im}(T)}$ admits infinite dimensional (not necessarily closed though) complements. Now you can check for any $V$ finite dimensional and linearly independent to $\overline{\mathrm{im}(T)}$ that for any $f\in V^*$ the map $V\oplus \mathrm{im}(T)\to\Bbb C, (v,x)\mapsto f(v)$ is continuous. In particular it admits Hahn Banach extensions with domain all of $X$. But any such Hahn Banach extension is $0$ on $\overline{\mathrm{im}(T)}$ hence lies in $\ker(T^*)$. Since you can do this for all $f\in V^*$ where $V$ has arbitrary finite dimension you must find that $\ker(T^*)$ is infinite dimensional.

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  • $\begingroup$ I have edited the question with a possible counterexample. Do you know how to explain this? $\endgroup$ – Manuel Norman Jul 16 '20 at 14:26
  • $\begingroup$ @ManuelNorman the cokernel (in the category of Banach spaces!) is not given by $X/\mathrm{im}(T)$, but by $X/\overline{\mathrm{im}(T)}$, since if you do not take the closure the quotient space will not be Hausdorff. $\endgroup$ – s.harp Jul 16 '20 at 17:59
  • $\begingroup$ Oh right, ok, thanks! $\endgroup$ – Manuel Norman Jul 16 '20 at 18:04

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