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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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  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jul 1 '15 at 10:04
  • $\begingroup$ I see. So my reasoning is correct and for bounded operators this definition actually implies bijectivity? $\endgroup$ – Blazej Jul 1 '15 at 10:09
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    $\begingroup$ Right. 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 $\rho(A) = \{\lambda \in \mathbb{C} : \lambda \mathbb{I} - A \text{ is bijective}\}$. $\endgroup$ – Daniel Fischer Jul 1 '15 at 10:14
  • $\begingroup$ Oh that is really a big subtlety and usually is mistaken!!!!! See the thread: Operator: Not Closable! $\endgroup$ – C-Star-W-Star Jul 1 '15 at 17:48
  • $\begingroup$ @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!)- $\endgroup$ – C-Star-W-Star Jul 1 '15 at 19:58
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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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  • $\begingroup$ 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.) $\endgroup$ – C-Star-W-Star Jul 1 '15 at 19:50
  • $\begingroup$ Oy, thanks. Forgot the closability :( $\endgroup$ – Daniel Fischer Jul 1 '15 at 20:01
  • $\begingroup$ You're welcome! :) But even for the closable case I think it won't work, or? $\endgroup$ – C-Star-W-Star Jul 1 '15 at 20:03
  • $\begingroup$ @Freeze_S Doesn't it? It's so long that I cared about unbounded operators, I don't remember all potential problems. $\endgroup$ – Daniel Fischer Jul 1 '15 at 20:04
  • $\begingroup$ I'm not sure but I suspect it! In any case these things are very delicate and really require rigorosity. $\endgroup$ – C-Star-W-Star Jul 1 '15 at 20:06

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