Apparently Fredholm operators are usually (at least in Wikipedia and my functional analysis lecture) only defined as operators $T$ between two Banach spaces.

As far as I can see, the definition can be extended to operators between arbitrary normed spaces without any problems, although then the condition that $\operatorname{im} T$ is closed is no longer independent of $\ker T < \infty$ and $\operatorname{coker} T < \infty$.

Is there some reason why this is not usually done?

Are Fredholm operators between non-Banach spaces so much less interesting/useful than those between Banach spaces?

If yes, why?


The way I see it, Fredholm's theory is about existence; if some conditions are satisfied, then the given equation admits at least one solution. All existence results in analysis rely on completeness. These results never build a solution explicitly; rather, they prove that some approximation procedure is convergent, because it is Cauchy. Fredholm's theory is no exception.

  • $\begingroup$ From the point I see... It would be possible to extend the concept of Fredholm Operators to incomplete spaces, even though this would be meaningless as you pointed out. $\endgroup$ – rarc Feb 25 at 11:47
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    $\begingroup$ @rarc: Exactly; that would be like considering "contraction mappings" between incomplete metric spaces. These mappings need not have a fixed point, which is their main feature in the complete case. So, why should we care? The same applies with Fredholm. $\endgroup$ – Giuseppe Negro Feb 25 at 11:52

I don't know if you have already found your answer but... anyway. I think the answer could lie on the closed graph theorem. Think about this: The closed graph theorem states: Let $(E, \|\cdot \|_E)$ and $(F, \|\cdot \|_F)$ be two Banach spaces. if $$T:(E, \|\cdot \|_E) \to (F, \|\cdot \|_F)$$ is a linear mapping and $$\Gamma := \{ (x,Tx) : x \in E \}$$ (its graph) is in $E\times F$ closed, then $T \in \mathcal{L}(E,F)$ (space of linear continuous mappings between E and F). If you consider linear mappings like the derivative between incomplete spaces whose graph is closed, then $T:E \to F$ could be (not necessarily) not continuous (There are many details you have here to consider. This statement is not so trivial since the continuity of a linear functional depends on the topology induced by the norm). Consider the linear the problem (maybe looking for solutions to differential equations) $$Tu = f.$$ Some methods consider solutions approaching them by sequences (really roughly speaking). So, let $\{x_n\}_{n\in \mathbb{N}}, \{y_n\}_{n\in \mathbb{N}}\subset E$ be two sequences with $x_n, y_n \to u$. If $T:E \to F$ is not continuous then the limits of these sequences could be different i.e. $$\lim_{n \to \infty} T(x_n) = f_1 \neq f_2 = \lim_{n \to \infty} T(y_n).$$ So you would not be really able to say much about the "solution" you found approaching it by one specific sequence.

But if you have already found an answer, I would be entirely glad and grateful to read it.


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