Discussion on Lecture 05

Francisco Ezquerra Larrodé, 2022/11/29 15:07, 2022/11/29 15:18

Dear all,

In team Hagen we are struggling with part d) of Exercise 1. We need to compute $Q^{'}$ . Could you elaborate here a bit more? We could not go further than apply the definition or transform it by using the spectral measure. But we are unsure on what to compute

Finally, would be OK to use part c) directly to prove the closedness of the form in a)?

Thank you for your time.

Paco

Christian Seifert, 2022/11/29 16:57

Dear Paco,

Many thanks. Concerning (d), I agree that there is not too much to compute (but rather to state). Concerning (a) (and ©), we only defined the associated operator for closed forms. So in order to work on ©, one would need to know (a) already.

Best, Christian

Sascha Trostorff, 2022/11/28 15:27

Dear ISEM-Team,

in the formula in Exercise 4 there is a $λ$ missing on the left hand side of the equality (just plugging in the operator $T = 0$ yields a wrong statement).

Best regards Sascha

Christian Seifert, 2022/11/29 16:52

Dear Sascha,

Many thanks; you are right (and Sahiba and Peer noticed this already below ).

Best, Christian

Peer Kunstmann, 2022/11/28 15:24

Dear all,

just a comment:

In reading Lecture 05 one might get the impression that the spectral theorem for self-adjoint operators is an essential ingredient in order to obtain associated operators from forms.

This is not the case. Most of the properties can also be shown for sectorial forms in complex Hilbert spaces. Only the relation of the form to the square root of the operator needs self-adjointness of the operator (or symmetry of the form). And here, one might argue whether the construction of self-adjoint square roots of positive self-adjoint operators really needs the full strength of the spectral theorem. It might be a matter of taste.

Best, Peer

Anna Muranova, 2022/11/28 12:24

Dear all,

thanks lecturers for the lecture. I have two questions:

1. Is it shown somewhere that $supp μ_{f} \subset σ (A)$ ? Is it obvious (or maybe it is just preliminary knowledge for this course)?

2. In proof of Theorem 3.33 in the first part, how from $Q_{n} (f) ↗ Q (f)$ follows $Q (f) \leq lim inf Q (f_{n})$ ?

Best, Anna

Peer Kunstmann, 2022/11/28 15:15

Dear Anna,

1) Theorem 3.18 states $σ (A) = s u p p (E)$ . And by the relation of $μ_{f}$ and $E$ in Proposition 3.16 we get $s u p p μ_{f} \subseteq s u p p (E)$ .

2) (ii) means that ${f : Q (f) \leq α}$ is closed for all $α \in R$ . It is shown that $Q (f) = sup_{n} Q_{n} (f)$ . Hence ${f : Q (f) \leq α} = ⋂_{n} {f : Q_{n} (f) \leq α} .$ But, as $Q_{n}$ is continuous for each $n$ , the set ${f : Q_{n} (f) \leq α}$ is closed for each $n$ , and closedness of ${f : Q (f) \leq α}$ follows.

Best, Peer

Dennis Schmeckpeper, 2022/11/28 18:59

Dear Anna,

for 2. they used the result that the supremum of continuous functions (of an arbitrary family) is lower semicontinuous. Even stronger: also the supremum of lower semicontinuous functions is lower semicontinuous.

There is a characterization of lower semicontinuous functions via sublevel sets, i.e. a function $g$ on $X$ (where $X$ is some topological space) into the extendend reals ( $R \cup {- \infty, \infty}$ ) is lower semicontinuous if and only if every sublevel set ${t \in X ∣ g (t) \leq α}$ for $α \in R$ is closed (in X). I think that is what Peer is using in his argument.

Best Dennis

Peer Kunstmann, 2022/11/24 07:42

Dear virtual lecturers,

thank you very much for the lecture.

Two remarks on the proof of Theorem 3.24:

1) In the definition of $ψ_{h}$ , I think it should read $\dots + x e^{- t x}$ .

2) I don't think that the bound $2 x e^{- t x}$ is correct in line 2 of page 60. One clearly has $| (e^{a} - 1) / a | \leq e^{| a |}$ . So, I would prefer a bound like $2 x e^{- (t / 2) x}$ if $| h | \leq t / 2$ .

Best regards, Peer

Christian Seifert, 2022/11/29 16:46

Dear Peer,

Many thanks, you are absolutely right.

Best, Christian

Sahiba Arora, 2022/11/22 13:45

Dear lecturers,

If my calculations are correct, then I believe there is a factor $\frac{1}{λ}$ missing on the right-hand side of the desired formula in Exercise 4.

Regards, Sahiba

Peer Kunstmann, 2022/11/24 15:47

I'd rather put a factor $λ$ to the left hand side .

Best, Peer

Sahiba Arora, 2022/11/24 16:20

Dear Peer

Yes, I'd be happy with that too.

Regards, Sahiba

Christian Seifert, 2022/11/29 16:39

Dear Sahiba and Peer,

Many thanks for spotting the typo.

Best, Christian

Jochen Glück, 2022/11/16 11:28, 2023/01/20 09:34

Dear all,

Thanks a lot to the lecturers for the very interesting fifth lecture!

(And a particular personal thank you for including the characterization of resolvents as minimizers! So far, I only had some very vague idea how resolvents are related to minimizing forms, and this makes the picture much clearer for me.)

I was wondering about the following question regarding Proposition 3.29:

Question (a): Is the continuity assumption on the form really needed, or does it follow automatically from the assumption $D (Q) = H$ together with the positivity?

~~Equivalently~~ On a related note, one might ask:

Qeustion (b): If $A : H \to H$ is a linear mapping such that $⟨ A u, u ⟩ \geq 0$ for all $u \in H$ , does it follow that $A$ is continuous?

I thought this might possibly follow from the closed graph theorem, but I couldn't find an argument that actually works. Maybe another participant (or one of the lecturers) has some insight into this question?

Edit: Bálint Farkas kindly reminded me of the Hellinger-Toeplitz theorem, which says that an everywhere defined symmetric operator on a Hilbert space is automatically continuous. For complex Hilbert spaces positivity implies symmetry due to the polarization identity, so the answer ~~to the question above~~ to question (b) above is: Yes.

This leaves the following question open, though:

Question ( c): Is the same conclusion still true over the real field (where positivity does not imply symmetry)?

Edit 2 (added much later): In an earlier version of this post I claimed that questions (a) and (b) were equivalent. However, as Dennis kindly pointed out in a response below, this is not true, and the answer to question (a) is no.

Best, Jochen

Dennis Schmeckpeper, 2022/11/22 21:09, 2022/12/13 11:54

Dear Jochen,

I thought a bit about this, and I think it is still true.

Cave: The proof below is not corret. Thanks to Christian (Seifert) for pointing out to me, that as it is written I would need to know that the operator is closed in order to conclude that the range is closed.

However, I think the statement is still correct. See here.

The not correct proof:

I think $⟨ A u, u ⟩ \geq 0$ for all $u \in H$ is in any case strong enough to guarantee that $- 1 \in ρ (A)$ . So $A$ must be closed and by the closed graph theorem bounded.

Heavily inspired by this, here is an argument why $- 1 \notin σ (A)$ . Frist, we notice that $A + I$ must be injective (otherwise -1 would be an eigenvalue). $A + I$ also must have dense range. Suppose not, then there exists $x \in {\bar{r a n (A + I)}}^{⊥}$ such that $⟨ (A + I) x, x ⟩ = 0 \Rightarrow ⟨ A x, x ⟩ + ⟨ x, x ⟩ = 0 \Rightarrow ⟨ A x, x ⟩ = - ‖ x ‖^{2} < 0.$ Furthermore, from $⟨ (A + I) x, (A + I) x ⟩ = ‖ A x ‖^{2} + 2 ⟨ A x, x ⟩ + ‖ x ‖^{2} \geq ‖ x ‖^{2}$ we deduce that the range of $A + I$ is also closed, i.e. $r a n (A + I) = H$ .

Finally, we show that $A_{1} := (A + I)^{- 1}$ is bounded, again by contradiction. Thus, for every $C > 0$ there is a $y \in H$ with $‖ y ‖ = 1$ such that $‖ A_{1} y ‖ > C ‖ y ‖ .$ Let now $C = 1$ and $x \in H$ such that $(A + I) x = y$ . Then we have $‖ (A + I) x ‖ < ‖ x ‖,$ in other words, there exits $x \in H$ with norm 1 such that $‖ (A + I) x ‖ < 1.$ Finally, we deduce a desired contradiction $0 > ‖ (A + I) x ‖^{2} - 1 = ‖ A x ‖^{2} + 2 ⟨ A x, x ⟩ \geq 2 ⟨ A x, x ⟩ .$

Best Dennis

Dennis Schmeckpeper, 2022/12/05 18:39

Dear Jochen,

I think the following argument works: https://cloud.tuhh.de/index.php/s/96frgjpjbsEMYFi

Best Dennis

Dennis Schmeckpeper, 2022/12/13 11:50

Dear Jochen,

you write above that the boundedness assumption in Proposition 3.29 can be dropped due to some application of Hellinger-Topelitz theorem. However, I don't think that this theorem makes a claim on symmetric positive forms.

Consider for instance the delta-distribution at point 0, i.e. $δ_{0} : C_{0}^{\infty} (R) \subseteq L^{2} (R) \to L^{2} (R), f \mapsto δ_{0} (f) = f (0) .$ As a linear operator we can extend it linearly (via a Hamel basis for instance and purely algebraically) to $L^{2} (R)$ by zero. Then we can consider the bilinear form $Q : L^{2} (R) \times L^{2} (R) \to R, (f, g) \mapsto δ_{0} (f) δ_{0} (g) .$

This form is symmetric and positive but not bounded.

I hope this makes sense?

Best Dennis

Jochen Glück, 2022/12/20 16:20

Dear Dennis,

Thanks a lot for your replies, and my apologies for my late reply! (Apparently I had forgotten to activate the email notification for new messages, and thus I only saw your posts right now).

I'm a bit in a hurry right now, but I'll reply in a few days.

Best wishes Jochen

Jochen Glück, 2023/01/20 09:24, 2023/01/20 09:25

Dear Dennis,

Thanks again for your comments, and for your proof in the file you linked!

This is a wonderful argument, thanks a lot for answering my question!

You're also making a very good point in your other comment, where you point out that one actually can't drop the boundedness assumption in Proposition 3.29 (in contrast to what I had claimed).

My mistaken claim was apparently due to the following problem: A symmetric form that is everywhere defined does not yield a mapping $A : H \to H$ , in general, unless we know a priori that the form is bounded. Thus one can't apply the Hellinger-Toeplitz theorem because one doesn't even have an operator to start with. Thanks for pointing this out! (I'll add an edit to my original post where I mention this.)

Best Jochen

ISem26

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Discussion on Lecture 05

Discussion on Lecture 05