The definition I have found in a couple of places is the following: We have [itex]\lambda \in \rho(T)[/itex] for a bounded linear operator [itex]T[/itex] on a Banach space [itex]X[/itex] iff [itex](T-\lambda)[/itex] is bijective with a bounded inverse. (This seems to be equivalent to just saying that [itex](T-\lambda)[/itex] is both injective and surjective, since this implies [itex](T-\lambda)^{-1}[/itex] is bounded by the inverse mapping theorem.) But in class, my professor said that [itex]\lambda \in \rho(T)[/itex] iff [itex](T-\lambda)^{-1}[/itex] exists and [itex]\mathcal R(T-\lambda) = X[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

I don't see why my professor's definition doesn't contain superfluous information. Obviously, if [itex]\mathcal R(T-\lambda) = X[/itex], then [itex](T-\lambda)[/itex] is surjective. But doesn't saying "[itex](T-\lambda)^{-1}[/itex] exists"containthe surjectivity statement? I mean, why not just say: "We need (1) [itex](T-\lambda)[/itex] injective and (2) [itex](T-\lambda)[/itex] surjective" and be done with it? It just seems he's said more than he really needs to.

Of course, if [itex]\mathcal R(T-\lambda) \neq X[/itex], I guess that [itex](T-\lambda)^{-1}[/itex] only makes sense as a map [itex]\mathcal R(T-\lambda) \to X[/itex]...so you can't really just say "[itex]\lambda \in \rho(T)[/itex] iff [itex](T-\lambda)^{-1}[/itex] exists as a bounded operator" without there being some ambiguity...because can't [itex](T-\lambda)^{-1}: \mathcal R(T-\lambda) \to X[/itex] be bounded without having [itex]\mathcal R(T-\lambda) = X[/itex]?

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# Confusing definition of resolvent set

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