# Equivalent definitions of Fredholm operators on infinite dimensional Banach spaces

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)?

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.
• @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. Jul 16, 2020 at 17:59