Resolvent: Definition

Question

Given a Banach space $E$.

Consider linear operators: $$T:E\supset\mathcal{D}(T)\to E:\quad T(\kappa x+\lambda y)=\kappa T(x)+\lambda T(y)$$

(No other assumptions on the operator!)

Denote for shorthand: $$R_\lambda:\mathcal{R}_\lambda\to\mathcal{D}_\lambda:\quad R_\lambda:=(\lambda-T)^{-1}$$

The standard definition for the resolvent set: $$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}_\lambda=(0),\mathcal{R}_\lambda=E,\|R_\lambda \|<\infty\}$$ An alternative definition for the resolvent set: $$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}_\lambda=(0),\overline{\mathcal{R}_\lambda}=E,\|R_\lambda\|<\infty\}$$ (See Werner or Weidmann resp. Kubrusly or Kreyszig.)

Do these definitions really agree?

Clearly for closed operators they do.

Answer 1

The business is rather wolly but basically everything boils down to closed operators...

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Before, let's set our framework:

Framework:

As always we consider Banach spaces: $X$, $Y$

First let's define the resolvent set properly:

Definition:

The resolvent set is defined as those values where the linear problem can be solved always, uniquely and continuously: $$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}(T-\lambda)=\{0\},\mathcal{R}(T-\lambda)=Y,C\|(T-\lambda)x\|\geq\|x\|\}$$

Next let's state some general theorems on closed operators:

Theorem 1:

A continuous operator with closed domain is closed: $$\mathcal{D}(T)\text{ closed}:\quad T\text{ continuous}\Rightarrow T\text{ closed}$$

Theorem 2:

A closed operator with complete codomain is continuous: $$\mathcal{D}(T)\text{ complete}:\quad T\text{ closed}\Rightarrow T\text{ continuous}$$

Theorem 3:

The domain of a continuous closed operator with complete codomain is closed: $$T\text{ closed, continuous}\Rightarrow\mathcal{D}(T)\text{ closed}\qquad(\mathcal{C}(T)\text{ complete})$$

Now let's investigate our situations:

Consequence 1:

The resolvent set of unclosed operators is empty: $$T\text{ not closed}\Rightarrow\rho(T)=\varnothing$$

Observation:

A cobound automatically guarantees the existence of the resolvent: $$C\|(T-\lambda)x\|\geq\|x\|\Rightarrow\mathcal{N}(T-\lambda)=\{0\}$$

Consequence 3:

The domain of a continuous resolvent is automatically closed: $$T\text{ closed}:\quad C\|(T-\lambda)x\|\geq\|x\|\Rightarrow\mathcal{R}(T-\lambda)\text{ closed}$$

Consequence 2:

An everywhere defined resolvent is automatically continuous: $$T\text{ closed}:\quad\mathcal{N}(T-\lambda)=\{0\},\mathcal{R}(T-\lambda)\text{ closed}\Rightarrow C\|(T-\lambda)x\|\geq\|x\|$$

Finally let's conclude:

Conclusion:

An alternative definition therefore could address the resolvent: $$R(\lambda)\text{ exists, densely defined, closely defined, continuous}$$ So spectrum then could be splits up into: $$\lambda\in\sigma_p(T):\iff R(\lambda)\text{ not exists}$$ $$\lambda\in\sigma_r(T):\iff R(\lambda)\text{ exists, not densely defined}$$ $$\lambda\in\sigma_c(T):\iff R(\lambda)\text{ exists, densely defined, not closely defined}$$ and the resolvent set becomes: $$\lambda\in\rho(T):\iff R(\lambda)\text{ exists, everywhere defined, continuous}$$ Note that the remaining combination for closed operators does not occur: $$T\text{ closed}: R(\lambda)\text{ exists, everywhere defined}\Rightarrow R(\lambda)\text{ continuous}$$

Outlook:

A more detailed distinction is listed in Kubrusly's 'The Elements of Operator Theory'.

Answer 2

Here's an idea. Let $X=l^{2}$ be the space of sequences $\{ x_{n}\}_{n=1}^{\infty}$ of square-summable sequences of complex numbers. Let $e_{n}$ the standard basis element which is a sequence with all 0's except for a $1$ in the n-th place. Let $M$ be the subspace spanned by finite linear combinations of $\{ e_{1}+\frac{1}{n}e_{n}\}_{n=2}^{\infty}$. The closure $M^{c}$ includes $e_{1}$ and, because of that, also includes every $e_{n}$ for $n \ge 2$. So $M^{c}=X$. Define $S$ to be the restriction of the left shift to $M$. $S$ is bounded on $M$, and $\mathcal{N}(S)=\{0\}$ because $e_{1} \notin M$. Let $T$ be the inverse of $S$. Then $T : \mathcal{D}(T) \rightarrow X$ has domain $\mathcal{D}(T)=S(M)$, and the range of $T$ is $M$ which is dense in $X$. And $S=T^{-1}$ is bounded by $1$, which gives $\|Tx\| \ge \|x\|$ for all $x \in \mathcal{D}(T)$. So $0 \in \rho(T)$ using the second definition of resolvent, which is very wrong from my point of view. However, $T$ is not closable because $(e_{1},0)$ is in the closure of the graph of $S$ and, so, $(0,e_{1})$ is in the closure of the graph of $T$. I think this works, but please check my details.