I'm having trouble understanding some subtlety of definition of resolvent set for given bounded operator A everywhere defined on some Hilbert space. Book I use (and many other sources) give the following: $\lambda \in \mathbb C $ is in resolvent set if $ R_{ \lambda} = ( \lambda \mathbb I - A ) ^ {-1} $ exists, is bounded and range of $\lambda \mathbb I -A$ is dense. Now my reasoning begins. This range is also domain of resolvent. Since it is bounded operator on dense domain it can be extended to whole Hilbert space by continuity. $ ( \lambda \mathbb I - A ) R_{\lambda}$ is equal to identity on dense subset so its extension is identity on whole Hilbert space. Similarly $R_{\lambda} (\lambda \mathbb I - A ) $ is identity on whole Hilbert space by definition of resolvent. But this means that $A - \lambda \mathbb I$ is bijection because it has left and right inverse. Therefore its range is actually whole Hilbert space. But if that is the case, why everyone demands it to be merely a dense subset?
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Because of unbounded operators. When $A$ is a densely defined unbounded operator, then one wants to have the resolvent set contain the values $\lambda$ where $(\lambda \mathbb{I}- A)^{-1}$ exists, is bounded, and densely defined [i.e. the range of $\lambda\mathbb{I} - A$ is dense], not just where the range is the whole space. – Daniel Fischer Jul 01 '15 at 10:04
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I see. So my reasoning is correct and for bounded operators this definition actually implies bijectivity? – Blazej Jul 01 '15 at 10:09
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1Right. And by the way, not everyone starts with "range is dense", I've seen several texts that define the resolvent set of a continuous linear operator $A$ on a Banach space as the set $\rho(A) = {\lambda \in \mathbb{C} : \lambda \mathbb{I} - A \text{ is bijective}}$. – Daniel Fischer Jul 01 '15 at 10:14
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Oh that is really a big subtlety and usually is mistaken!!!!! See the thread: Operator: Not Closable! – C-star-W-star Jul 01 '15 at 17:48
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@DanielFischer: Can you elaborate on the idea of having only dense resolvent? Thanks! :) (Afaik it is for Riesz-Dunford calculus. But there varying dense domain is bad!)- – C-star-W-star Jul 01 '15 at 19:58
1 Answers
Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting is
$$\rho(A) = \{\lambda\in \mathbb{C} : \lambda\mathbb{I} - A \text{ is bijective}\},$$
and this definition is also given in the literature.
The advantage of the definition you cited is that that definition can be used unchanged for the case of unbounded (densely defined) closable operators on Banach (or more specifically Hilbert) spaces. In the case of unbounded operators, the denseness of the range and boundedness of the inverse do not imply surjectivity, so then the two phrasings of the definition would not be equivalent.
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The definition cannot be used unchanged for the case of unbounded densely-defined operators on Banach spaces!!!! This, sadly, is a big missconception within the math community. For an explicit counterexample see: Operator: Not Closable! (It is due to T.A.E. I had to convince him of this subtlety though.) – C-star-W-star Jul 01 '15 at 19:50
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You're welcome! :) But even for the closable case I think it won't work, or? – C-star-W-star Jul 01 '15 at 20:03
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@Freeze_S Doesn't it? It's so long that I cared about unbounded operators, I don't remember all potential problems. – Daniel Fischer Jul 01 '15 at 20:04
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I'm not sure but I suspect it! In any case these things are very delicate and really require rigorosity. – C-star-W-star Jul 01 '15 at 20:06
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Imagine a closable operator that admits a regular value w.r.t. this definition here. However it cannot have regular values in the usual sense as it is not closed, yet. – C-star-W-star Jul 01 '15 at 20:10
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@Freeze_S I'm having problems imagining that. I'm pretty out of touch with unbounded operators. Have you any idea how such a beast might look? – Daniel Fischer Jul 01 '15 at 20:23
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Sorry to tired to produce something productive right now. But I think one can construct something either by appropriately blending out elements from an ONB or by constructing a suitable spectral measure and then fading out the domain. – C-star-W-star Jul 01 '15 at 21:55
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Indeed, one needs closed and not just closable. If $A$ is closed, then $\lambda-A$ as well so it has closed range as soon as it is bounded below. A counterexample for closable: take for example the restriction of a bounded operator to rational points. Say, the inclusion $\mathbb Q \to \mathbb R$: it is closable, densely defined, has dense range but not closed range. – Bart Michels Oct 24 '18 at 17:33