Definition of spectrum $\sigma(T)$

Question

Definition 1: Let $X$ be a normed space and $T:D(T)\subset X\rightarrow X$ be a linear operator. The spectrum of $T$ denoted by $\sigma(T)$ is defined to be $$ \sigma(T):=\{\lambda\in\mathbb{C}:\lambda I-T \quad\text{is not invertible}\} $$ Definition 2: $\rho(T):=\mathbb{C}\backslash\sigma(T)$ is the resolvent set of $T$.

Definition 3: $R(\cdot,T):\rho(T)\rightarrow \mathcal{L}(X)$ with $R(\lambda,T)=(\lambda I- T)^{-1}$ is the resolvent operator.

I always use the fact that the resolvent operator is bounded; however, I don't see why the "not invertible" part of definition 1 is enough to conclude this fact. Why is adding "$\lambda-T$ is bounded" unnecessary? $\quad$

Any hints are appreciated.

Answer 1

It is wrong if you don't assume that $X$ is complete. Let

$$X=c_{00}(\mathbb{R})= \{ (x_i)_{i\geq 1} \in l^{\infty}(\mathbb{R}) \vert \ \exists N\geq 1 \ \forall i\geq N: x_i=0\}$$

with the supremum norm. Consider

$$ T: X \rightarrow X, \quad T((x_i)_{i\geq 1}) = \left( \frac{1}{i} \cdot x_i \right)_{i\geq 1}.$$

This operator is invertible, however, its inverse

$$ T^{-1}: X \rightarrow X, \quad T((x_i)_{i\geq 1}) = \left( i \cdot x_i \right)_{i\geq 1} $$

is not bounded. I.e. $0\in \rho(T)$, but $(T-0 \cdot I)^{-1}$ is not bounded.

Completeness is needed to invoke some kind of closed graph theorem. Usually one works in the setting $X$ Banach space and $T$ has a closed graph (such mappings are called closed), respectively $T$ has a closed extension (such mappings are called closable). In this case we have for $\lambda\in \rho(T)$ that

$$ (\lambda I - T)^{-1} : X \rightarrow D(T)$$

has a closed graph and as both $X$ and $D(X)$ are Banach spaces one concludes by the closed graph theorem that $(\lambda I - T)^{-1}$ is bounded.

Answer 2

From $\sigma(T):=\{\lambda\in\mathbb{C}:\lambda I-T \;\text{is not invertible}\}$ we conclude that $$\rho(T) = \mathbb{C}\backslash\sigma(T)=\{\lambda\in\mathbb{C}:\lambda I-T \;\text{is invertible}\}$$ invertible in the sense of $\mathcal{B}(X)$ (the set of all bounded linear mappings from $X$ to $X$). So if $\lambda \in \rho(T)$ we know by definition that $(\lambda I - T)^{-1}$ exisits and is bounded.

Edit Since there were made some changes in the question and there are so many questions in the comment to my answer i will expand my answer.

First of all I want to regard a more general but simpler Definition.

Definition Let $\mathcal{A}$ be a Semigroup. Then we call $a \in \mathcal{A}$ invertible iff there is a $b \in \mathcal{A}$ such that $$ab = 1 \quad\text{and}\quad ba = 1$$

Strangly I didn't find a Wikipedia definition in the english version for invertible.

So now regard a unital algebra like $\mathcal{B}(X)$ is. Then we can define a spectrum and a resolvent set.

Definition Let $\mathcal{A}$ be a unital algebra, $e$ the unit element and $a\in\mathcal{A}$ then we define \begin{align*} \rho(a) &:= \{\lambda\in\mathbb{C} : \lambda e - a \text{ is invertible}\} \\ \sigma(a) &:= \mathbb{C}\backslash \rho(a)\end{align*}

Now if you set $\mathcal{A} = \mathcal{B}(X)$ you already start with a bounded element!

So if you want to define the spectrum for more general operators you have to change a little bit. You can define the spectrum for linear relations too that would include your case but I think that will confuse you even more so there is something in between.

Now there are more than one way to define the spectrum but you have to decide which assumptions you make

Definition V1 Let $X$ be a Banachspace and $T: D(T) \to X$ a closed linear operator then we define \begin{align}\rho(T) &:= \{\lambda \in \mathbb{C}: \lambda I - T: D(T) \to X \text{ is bijective}\}\\ \sigma(a) &:= \mathbb{C}\backslash \rho(a)\end{align}

or

Definition V2 Let $X$ be a normed vector space and $T: D(T) \to X$ a linear mapping then we define \begin{align}\rho(T) &:= \{\lambda \in \mathbb{C}: (\lambda I - T)^{-1} \in\mathcal{B}(X)\}\\ \sigma(a) &:= \mathbb{C}\backslash \rho(a)\end{align}

Answer 3

not invertible <=> $det(\lambda I - T) = 0$. In points of Resolvent set $det \neq 0$ and we can inverse our matrix. And this inversed called Resolvent operator.