# Definition of spectrum $\sigma(T)$

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.

• What is $X$? Is it a Hilbert space? – Severin Schraven May 27 '17 at 16:34
• @SeverinSchraven $X$ is a normed space – janko May 27 '17 at 16:41

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.

• No it's just not the right definition for spectrum in this setting. I have edited my answer and pointed out the different approaches. – Nathanael Skrepek May 27 '17 at 17:15
• @NathanaelSkrepek This answer was by no means against your answer! In the question is says "Why is adding $\lambda I - T$ invertible unnecessary" and this is wrong in such generality as shown in my answer. – Severin Schraven May 27 '17 at 17:19
• @SeverinSchraven Great! Thank you very much. – janko May 27 '17 at 17:40
• @janko You're welcome. It's a very interesting question. – Severin Schraven May 27 '17 at 17:41

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}

• lol why the hell gave me somebody a downvote – Nathanael Skrepek May 27 '17 at 15:32
• Thank you for the answer. I understand that for all $\lambda \in \rho(T)$, $\lambda-T$ is invertible, but why does this imply boundedness? This looks like a demand on $\lambda-T$. – janko May 27 '17 at 15:38
• It is not. There is the so-called inverse mapping theorem telling you that the inverse of a bijective, linear, bounded mapping is always bounded. – Severin Schraven May 27 '17 at 15:50
• @SeverinSchraven Yes, but the definition does not say $\lambda - T$ is bounded? – janko May 27 '17 at 15:55
• @janko $\lambda I$ is bounded and $T$ is bounded so $\lambda I - T$ is bounded too. – Nathanael Skrepek May 27 '17 at 16:20

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.

• This works on finite-dimensional spaces, but judging by the "functional analysis" tag on the question the OP is interested in the infinite-dimensional case. – eepperly16 May 27 '17 at 15:31
• hmm, ok i missed it. But definitions are fine. And operator is bounded like $||\lambda I - T|| \leq C(\lambda)$. If you $\lambda \to s$ where $s \in σ(T)$. Your constant will raise. – duncan May 27 '17 at 15:38