Discussion on Lecture 04

Anna Muranova, 2022/11/27 11:44

Dear all,

I have two more questions.

How does polarization work in Proposition 3.15?

and

How one shows the very last part in Proposition 3.17? Does $g \in E (B) H \Rightarrow g \in H$ ?

Thanks, Anna

Christian Seifert, 2022/11/29 17:12

Dear Anna,

Polarization for some form $q$ means that you can describe $q (f, g)$ for $f, g \in D (q)$ by quadratic terms, see e.g. the bottom of page 61. Now, look at Proposition 3.13 (b) to see that you can represent the suitable quadratic terms via integrals w.r.t. some spectral measures. Then you just need to use linearity to obtain the measure.

$E (B)$ is an orthogonal projection (onto $E (B) H$ ), so $g \in E (B) H$ yields $g = E (B) H$ . Combined with the first part (with $f := g$ ) we obtain the assertion.

Best, Christian

Zulaihat Hassan, 2023/02/03 19:05

Dear Christian,

can you please provide a detail on how polarization was used in 3.15?

Thank you so much.

Peer Kunstmann, 2022/11/19 20:46

Thank you very much for the lecture.

I must confess that I am somehow irritated by Remark 3.8 as it stands. I think the argument that shows $φ_{z} (A) = (A - z)^{- 1}$ can only work for $z \in ρ (A)$ , for which $φ_{z}$ is bounded on $σ (A)$ .

I guess that this is what you wanted to say.

Best regards, Peer

Christian Seifert, 2022/11/20 14:04

Dear Peer,

Many thanks, and of course, this is what we intended to say.

Best, Christian

EL-Houcine OUALI, 2022/11/16 16:50

Thanks for this lecture.

Anna Muranova, 2022/11/14 12:03

Dear lecturers,

thank you for the 4th lecture!

Since I am not familiar with operator theory, the proof of Lemma 3.3 was too short for me (particularlz, the jumps from balls to inverse operators and vice versa). Could smbd, please, explain it in more details?

Best, Anna

Patrizio Bifulco, 2022/11/15 16:43, 2022/11/17 11:52

Dear Anna,

I think you can argue as follows:

First, if $λ \notin e s s r a n u$ , then there exists $ε > 0$ such that $μ (u^{- 1} (B_{ε} (λ))) = 0$ which is equivalent to the fact that $| u (x) - λ | \geq ε$ for $μ$ -a.e. $x \in X$ (in particular, this implies that $\frac{1}{u - λ}$ belongs to $L^{\infty} (X, μ)$ ). Thus, $M_{\frac{1}{u - λ}}$ is indeed a bounded inverse for $M_{u - λ}$ , in other words $λ \in ρ (M_{u})$ .

Conversely, if $λ \in e s s r a n u$ , for any $ε > 0$ one can consider $f := \frac{1_{u^{- 1} (B_{ε} (λ))}}{μ (u^{- 1} (B_{ε} (λ)))^{\frac{1}{2}}}$ (note that w.l.o.g. we suppose that $μ (u^{- 1} (B_{ε} (λ))) < \infty$ , because otherwise, by $σ$ -finiteness of $(X, μ)$ , we find a measurable set $Y \subseteq X$ such that $0 < μ (u^{- 1} (B_{ε} (λ)) \cap Y) < \infty$ and we can replace $u^{- 1} (B_{ε} (λ))$ by $u^{- 1} (B_{ε} (λ)) \cap Y$ in the definition of $f$ and do the same argumentation afterwards). Then clearly $‖ f ‖ = 1$ as well as $‖ (M_{u} - λ) f ‖ = \frac{1}{μ (u^{- 1} (B_{ε} (λ)))^{\frac{1}{2}}} (\int_{u^{- 1} (B_{ε} (λ))} | u (x) - λ |^{2} d μ (x))^{\frac{1}{2}} < ε .$ If there would exist a bounded inverse to $M_{u} - λ$ , say $(M_{u} - λ)^{- 1}$ , then we can choose for $ε := \frac{1}{2 ‖ (M_{u} - λ)^{- 1} ‖} > 0$ a function $f \in L^{2} (X, μ)$ such that $‖ f ‖ = 1$ and $‖ (M_{u} - λ) f ‖ < ε$ . But this implies $1 = ‖ f ‖ \leq ‖ (M_{u} - λ)^{- 1} ‖ ‖ (M_{u} - λ) f ‖ < ‖ (M_{u} - λ)^{- 1} ‖ ε = \frac{1}{2}$ which is a contradiction. Hence, $λ \in σ (M_{u})$ .

Best wishes, Patrizio

Anna Muranova, 2022/11/27 15:44

Dear Patrizio,

Thank you very much! That's a really helpful answer, so detailed!

Best, Anna

Jochen Glück, 2022/11/12 23:05

Dear all,

Thanks a lot to the virtual lecturers for the very interesting fourth lecture!

Here are a few remarks:

(1) On the bottom of page 48 it is claimed that continuity of the resolvent follows from the resolvent identity (i.e. from the identity in line -7 on page 48). I'm a bit confused by this wording, for the following two reasons:

The wording seems to indicate that the continuity were an immediate consequence of the resolvent identity, but this doesn't seem to be the case, as one doesn't know a priori that the resolvent is locally bounded. (The existence of Exercise 1 also seems to indicate that it's not completely straighforward.)

Isn't the most common way to obtain continuity of the resolvent via a Neumann series argument which does not use the resolvent identity?

(2) Minor nitpick: When “measurable sets” in $R$ are mentioned, does this refer to Borel or Lebesgue measurable sets (or maybe I overlooked this)?

(3) Remark 3.21 shows that positive does not imply positivity presevering. Maybe it's worthwhile to also add a counterexample which shows that the converse implication is false, too. For instance, the self-adjoint matrix $A := (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$ is positivity preserving (even positivity improving), but its spectrum is ${- 1, 3}$ , so $A$ is not positive.

(4) May I make one more suggestion regarding Remark 3.21? To avoid confusion for participants who read up about some topics in other parts of the literature, maybe Remark 3.21 would also be a good opportunity to add a warning that the terminology on positive and positivity preserving operators (and semigroups) is used very inconsistently in the literature. If I remember correctly, the following conflicts in terminology are quite common:

In the theory of self-adjoint operators, I think some authors call $A$ positive if $σ (A) \subseteq [0, \infty)$ (this is the convention used in the current ISEM), others call $A$ positive if $σ (A) \subseteq (0, \infty)$ , and yet others call $A$ positive if $σ (A) \subseteq [0, \infty)$ but $A$ is injective.

In the context of operators on function spaces (or Banach lattices), some authors use the word positive to refer to spectral properties of self-adjoint operators on $L^{2}$ (as is the case in the ISEM) and the notion positivity preserving for operators which map the positive cone into the positive cone (also as in the ISEM). Other parts of the literature (for instance, books that focus on Banach lattice theory, and the classical literature on $C_{0}$ -semigroups on Banach lattices) use the word positive for an operator that is positivity preserving in the sense of the ISEM, and notions such as positive definite or positive semi-definite to refer to spectral properties of self-adjoint operators on Hilbert spaces.

The terminology again seems to be different in the community that works mainly on finite-dimensional linear algebra and Perron-Frobenius types theorems there. There, I think, the word positive is used to describe operators which are called positivity improving in the ISEM, and the word non-negative is used for operators which are positivity preserving in the sense of the ISEM.

So unfortunately, it's apparently quite a mess throughout the literature, and I often noticed that this tends to confuse people. Maybe it would thus be worthwhile to add a comment on this in Remark 3.21 (or to add another remark about it)?

Best wishes, Jochen

Christian Seifert, 2022/11/15 16:25

Dear Jochen,

Many thanks for the comments.

(1) You are absolutely right. We somehow had self-adjoint operators in mind, where we have a corresponding uniform bound. However, we did not write it down in that way. We'll adjust the text accordingly.

(2) We always mean Borel measurable.

(3) Many thanks. We will include the example as well.

(4) We will add a corresponding warning concerning the different use of the terminology.

Best, Christian

Jochen Glück, 2022/11/16 11:18

Thanks for your response, Christian!

Best, Jochen

ISem26

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Discussion on Lecture 04

Discussion on Lecture 04