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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.

user3810316
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C-star-W-star
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  • If $T$ is not closable, then $T$ has empty resolvent. So you may as well assume $T$ is closable. If $T$ is closable but not closed, then the second definition of resolvent set is the better one. If $T$ is closed, both definitions are equivalent. By the way, assuming $|(T-\lambda I)x| \ge C|x|$ for some $C > 0$ and all $x \in \mathcal{D}(T)$ implies $\mathcal{N}(T-\lambda I)={0}$. – Disintegrating By Parts Jun 01 '14 at 02:08
  • Why is the latter better if T is closable but not closed? Obviously when I close it the dense range will be closed too and therefore $\mathcal{R}(T)=Y$. But what before doing that? Is it just meant as facilitating step saying before doing so it suffices to check that the range is dense in order to conclude afterwards that it is indeed surjective? What do u mean? – C-star-W-star Jun 01 '14 at 10:16
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    If $(T-\lambda I)$ is closed, has a dense range, and is continuously invertible on its range (equivalently, $|(T-\lambda I)x| \ge C|x|$ for some $C > 0$ and all $x \in X$,) then $T-\lambda I$ is onto. So, if $(T-\lambda I)$ is closable but not closed and is continuously invertible, then $T-\lambda I$ cannot be onto. That's why the second definition is required for a closable, non-closed $T$. – Disintegrating By Parts Jun 01 '14 at 11:46
  • Awesome - that's an answers I was hoping for - Thx a lot!!!!! =D – C-star-W-star Jun 01 '14 at 13:05
  • Waaaaaaiiit a minute if you're saying we rather use the alternative definition, then operators that are not closable may have nonempty spectrum! – C-star-W-star Jun 01 '14 at 14:30
  • Not so. If $\lambda$ is in the resolvent using the second definition, then $T$ is necessarily closable because of the boundedness condition. This is because the inverse is bounded and, thus, extends uniquely by continuity to $\overline{\mathcal{R}(T_{\lambda})}$. Then $T-\lambda$ extends to the inverse of this extension of the inverse, which makes $T$ closable. The first definition gives you empty resolvent for closable, non-closed operators. – Disintegrating By Parts Jun 01 '14 at 15:45
  • But extending doesn't give the inverse of the extension of the inverse in general, e.g. for $T_\lambda:\mathcal{D}\lambda\to\mathcal{R}\lambda$ and $R_\lambda:\mathcal{R}\lambda\to\mathcal{D}\lambda$ with $\mathcal{D}\lambda=\mathcal{H}$ and $\mathcal{R}\lambda\subsetneq\mathcal{H},\overline{\mathcal{R}\lambda}=H$ (sry for the capital 'H' it just doesnt want to compile right) we have $R\lambda=T_\lambda^{-1}$ but $\overline{T_\lambda}^{-1}=T_\lambda^{-1}=R_\lambda\neq\overline{R_\lambda}$. – C-star-W-star Jun 01 '14 at 16:40
  • You're right. I was mistaken in what I was thinking. I guess you have to assume $T$ is closable in order to use the second definition. But I think the first forces $T$ to be closed if the resolvent is non-empty. Does that sound right? – Disintegrating By Parts Jun 01 '14 at 17:11
  • No no ^^ I mean my example is not proven yet to exist I just postulated it... – C-star-W-star Jun 01 '14 at 17:25
  • I think the most reasonable approach is to assume (a) $T$ is closed and to use the first definition of resolvent, or (b) $T$ is closable and to use the second definition of resolvent. Those two formulations are equivalent. I'm not yet sure about your example. – Disintegrating By Parts Jun 03 '14 at 13:03
  • Mhh I'm not really convinced. Weidmann stresses that what we really want is an equation to be uniquely and continuously solvable for all values. So precisely the usual definition and that for any operator not necessarily closed nor closable and then for the special case of closable operators one can use the alternative definition as an intermediate(!) step... I think I'm now happy with it after thinking about for some time ^^ – C-star-W-star Jun 03 '14 at 14:46
  • If $T$ is closable, and if $\lambda$ is in the resolvent in the second sense, then $\lambda$ is in the resolvent of the closure $T^{c}$ in the first sense. So, it makes good sense to assume that $T$ is closed and use only the first definition. By the way, thank you raising this question. – Disintegrating By Parts Jun 03 '14 at 14:57
  • I forgot to explain my previous comment. If $T$ is closable and $\lambda$ is in the resolvent in the second sense, then $T_{\lambda}^{-1}$ extends uniquely by continuity to $S_{\lambda}\in\mathcal{L}(X)$. And $S_{\lambda}$ has trivial null space because $(y,0)$ is in the closure of the graph of $T_{\lambda}^{-1}$ iff $(0,y)$ is in the closure of the graph of $T_{\lambda}$, which is precisely the condition that "$T_{\lambda}$ is closable" prevents. So $S_{\lambda}^{-1}$ exists, is closed and $T_{\lambda}^{c}=S_{\lambda}^{-1}$ in this case. – Disintegrating By Parts Jun 03 '14 at 15:41
  • So can you demask nonclosable operators already in the alternative definition? – C-star-W-star Jun 03 '14 at 16:39
  • I don't see how. As I mentioned before, the problem is that, if $\lambda \in \rho(A)$ in the second sense, then $T_{\lambda}^{-1}$ has a closure, but that closure might have a non-trivial null space (which is equivalent to $T_{\lambda}$ not being closable.) In the case of the first definition, there's no need to assumed closed because if $\rho(T)\ne\emptyset$, then $T$ can be proved to be closed. – Disintegrating By Parts Jun 03 '14 at 17:13
  • So do you know an example of an operator which is injective and has dense range but not being closable? – C-star-W-star Jun 03 '14 at 17:21
  • I was trying to figure out how to take a bounded operator $B$ with a non-trivial null space, restrict to a dense domain that doesn't include anything in the null space, and then let $T=B^{-1}$. That would have to be possible in order to find a counterexample, but I'm not convinced that it is possible. After that I was going to worry about the density of $\mathcal{D}(T)=\mathcal{R}(B)$. Any ideas? – Disintegrating By Parts Jun 03 '14 at 17:43
  • The answer here might help: http://math.stackexchange.com/questions/593857/counterexample-for-the-open-mapping-theorem/658336#658336 I opened a new thread precisely on this problem here: http://math.stackexchange.com/questions/819526/operator-not-closable – C-star-W-star Jun 03 '14 at 17:45

2 Answers2

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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.

Disintegrating By Parts
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  • Perfect awesome: =D The restricted left shift is densely defined and invertible, its closure is precisely the uniform extension that is not invertible so the inverse of the restriction is not closable but due to the preceding its resolvent set is not empty. Btw you could as well just extend e_1 to a Hamel basis and then do the same... So thank you very much!!! =D ...one last plea: Can u put that answer also to: http://math.stackexchange.com/questions/819526/operator-not-closable – C-star-W-star Jun 03 '14 at 18:25
  • You're right about the Hamel basis. Thanks for pressing on this; I learned something fun and interesting while trying to work this out. I went ahead and copied this answer to your linked question. – Disintegrating By Parts Jun 03 '14 at 18:34
  • Good so :) and I'm finally lightened ^^. Thanks for the good work together. – C-star-W-star Jun 03 '14 at 18:38
  • By the way, I just noticed that $T$ is densely-defined because the range of $S$ includes all standard basis elements. Yes, we work good together. – Disintegrating By Parts Jun 03 '14 at 18:41
  • Not only densely defined but even everywhere defined if u use the Hamel basis approach... – C-star-W-star Jun 03 '14 at 18:52
  • Boundedness of $S$ might be a problem to prove for general bases. – Disintegrating By Parts Jun 03 '14 at 19:10
  • Not really since the left shift operator is bounded so also its restriction to any set. Seen from the topological side this is just $T$ continuous then $T\circ\iota$ continuous since the embedding of subspace $\iota$ is continuous. – C-star-W-star Jun 03 '14 at 19:25
  • I'm not sure about the boundedness of the left shift for a general basis other than an orthogonal one. – Disintegrating By Parts Jun 03 '14 at 19:28
  • You don't have to ask boundedness for bases but subsets (in our case span of the basis) and then this is just the preceding comment – C-star-W-star Jun 03 '14 at 19:47
  • You have to be able to show $|Sx|\le C|x|$ for $x$ equal to an arbitrary finite linear combination of elements $e_{1}+\frac{1}{2}e_{2}$, $e_{2}+\frac{1}{3}e_{3}$, etc. That's a tall order, because any (large) finite number of these could be involved, and the bound $|Sx|\le C|x|$ must be uniform (i.e., $C$ does not depend on that arbitrary finite linear combination.) It's fine for orthogonal, but I don't know what to do with a Schauder basis or, more generally, a Hamel basis. Do you see how to argue this? – Disintegrating By Parts Jun 03 '14 at 20:17
  • Yes I perfectly agree with you - I still have that problem when trying to show that the annihilation and creation operators are bounded (at least on the n-fold tensor product). However in the case of the left shift operator with its domain given by the span of a Hamel basis reduced by the first basis vector this becomes the question: Is it bounded for any $(0,a_2,a_3,\ldots)$? – C-star-W-star Jun 05 '14 at 17:32
  • I'm sry I mean for any $(0,a_2,a_3,\ldots)$ with $a_k\neq 0$ for infinitely many - but better approach this problem as I mentioned before by simply looking at all $(a_1,a_2,a_3,\ldots)$ and then deduce that this holds especially for any $(0,a_2,a_3,\ldots)$ with $a_k\neq 0$ for infinitely many – C-star-W-star Jun 05 '14 at 22:09
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The business is rather wolly but basically everything boils down to closed operators...

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'.

C-star-W-star
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