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by hansen
2912 days ago
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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. |
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