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.