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by hansen 2912 days ago
If you want Cᵏ to be a normed space you do

|f| = ∑ᵢ₌₀ᵏ sup|f⁽ⁱ⁾|.

On open domains you can also use the topology that forces uniform convergence on all compact subsets. But this will only give you a metric space, no Banach space (but you’ll include unbouded functions). This is needed for studying Brownian motions with an unbounded time domain.

Regarding the other question: If you take Cᵏ⁻¹ as a the co-domain differentiation will be continuous. IIRC to get unbounded linear maps defined on the whole Banach space you need the axiom of choice, you won’t be able to write one down.

The problem with differential operators is that they are usually only defined on a dense subset of the domain, and there they are not bounded. E.g. in quantum mechanics the space of states is L² but all the interesting observables are differential operators. You can weasel out of this situation by defining them on a subset of “physical states” (e.g. smooth wave functions of rapid decay). But they aren’t continuous anymore (the spectrum is unbounded). But on Hilbert spaces everything mostly works out fine. Physicist usually ignore those technical problems and still don’t make mistakes.