Spectrum of a linear operator on a Banach space

[AxiomOfChoice]

I'm trying to understand the spectrum and resolvent of a linear operator on a Banach space in as much generality as I possibly can.

It seems that the furthest the concept can be "pulled back" is to a linear operator T: D(T) \to X, where X is a Banach space and D(T)\subseteq X. But here are a few questions:

(1) Doesn't D(T) necessarily have to be a subspace of X for this concept to make any sense? For instance, if it's not a subspace, there can be elements x,y for which Tx and Ty make sense, but for which T(x+y) makes no sense, since x+y might not be an element of D(T).

(2) The Wikipedia article here says that, in order for \lambda \in \rho(T), we need both (1) (T-\lambda)^{-1} exists and (2) (T-\lambda)^{-1} is defined on a dense subset of X. Is this equivalent to saying that \text{Ran} (T-\lambda) must be a dense subset of X and that (T-\lambda): D(T) \to \text{Ran} (T - \lambda) must be a bijection?

 

[micromass]

I'm trying to understand the spectrum and resolvent of a linear operator on a Banach space in as much generality as I possibly can.

It seems that the furthest the concept can be "pulled back" is to a linear operator T: D(T) \to X, where X is a Banach space and D(T)\subseteq X. But here are a few questions:

(1) Doesn't D(T) necessarily have to be a subspace of X for this concept to make any sense? For instance, if it's not a subspace, there can be elements x,y for which Tx and Ty make sense, but for which T(x+y) makes no sense, since x+y might not be an element of D(T).

Yes. And we usually require $D(T)$ to be dense in $T$ as well. It doesn't really matter, if it's not dense, then we can always restrict $X$ to $\overline{D(T)}$ and work with that.

(2) The Wikipedia article here says that, in order for \lambda \in \rho(T), we need both (1) (T-\lambda)^{-1} exists and (2) (T-\lambda)^{-1} is defined on a dense subset of X. Is this equivalent to saying that \text{Ran} (T-\lambda) must be a dense subset of X and that (T-\lambda): D(T) \to \text{Ran} (T - \lambda) must be a bijection?

Yes, this is correct. We also want $(T- \lambda I)^{-1}$ to be bounded though.

If you want to see a very general definition of spectrum, then you should study Banach algebras. But this is a spectrum that coincides with the spectrum of bounded operators.