I have a late question concerning Exercise 13.7, in connection with Theorem 13.3.1. Under the hypotheses of Theorem 13.3.1, let f∈L2,ν(R;H). Then it is shown that
un:=(∂t,ν+Bn)−1f→u:=∂−1t,νM(∂t,ν)f
weakly in L2,ν(R;H), with the material law
M(z):=∞∑k=0(−1z)kCk.
(This is not quite the same M as defined in the middle of p. 168 of the Lecture Notes: this is because I want to have the extra power of ∂−1t,ν outside.) This means that u will satisfy the equation
∂t,νu=M(∂t,ν)f.
This is almost the equation I am required to show in Exercise 13.7; only the material law is at the wrong place. Unfortunately, I am at a loss to produce a material law such that the equation required in Exercise 13.7 is satisfied.
Can you help me there, please?
Best wishes, Jürgen
Marcus Waurick, 2020/05/20 12:59
Dear Jürgen,
thank you for your question. With the help of a Neumann series expression (choosing ν>0 large enough) you can compute M(∂t,ν)−1, which is also a material law operator N(∂t,ν) for a suitable N.
You would then end up with
∂t,νN(∂t,ν)u=f.
Does this help?
Best regards,
Marcus
Jürgen Voigt, 2020/05/20 13:18
Dear Marcus,
thanks a lot! This is precisely what you discuss in Remark 13.3.2. (Phew!)
Best regards, Jürgen
Gabriel McCracken, 2020/02/06 15:44
Dear virtual lecturers,
In Remark 13.3.2 I missed a reference to section 3.2 for using ||∂−1t,ν||≤1/ν. I asked myself whether ν>2L is optimal, and realized that ν>L instead of ν>2L is actually enough to make the series in k converge, but not necessarily the series in l: The more general qν>L gives the norm estimate ‖∞∑k=1(∂−kt,ν)Ck‖<q1−q which is ≤1 for q≤1/2.
thank you very much for your remarks. It would be really good to have this reference to Section 3.2 there. (We are so used to these concepts that we use it almost without thinking.) You are also right concerning the convergences of the series. This should be worth a comment in the final lectures notes.
Also thank you for your remark concerning the weak star topology. It makes sense to have this recalled, too.
Best regards,
MM
Jürgen Voigt, 2020/02/05 20:30
Dear virtual lecturers,
concerning Theorem 13.2.2: I think you can prove this result – with the same arguments – also in the more general context where (Vn(m)) is replaced by a sequence (Bn) in L(H) converging to B∈L(H) in the strong operator topology. (This would make it more parallel to Theorem 13.3.1.)
Best wishes, Jürgen
Marcus Waurick, 2020/02/05 20:38
Dear Jürgen,
Thanks for this comment. You are completely right, we haven't been very consistent in formulating these results.
Best regards,
MM
Jürgen Voigt, 2020/02/05 20:20
Dear ISemTeam,
I/we need a hint concerning Exercise 13.5. As I understand, the problem amounts to characterising weak-operator-topology convergence of the sequence (B−1n) in terms of the original sequence (Bn) of bounded self-adjoint operators, where these operators are bounded below by a common bound c>0. Is there anything more special, because you formulate it in a special context? And how can one exploit the self-adjointness?
Best wishes, Jürgen
Marcus Waurick, 2020/02/05 20:36
Dear Jürgen,
The only thing we were after was a sufficient condition (not a characterisation) for the weak operator topology limit of B−1n to be an inverse of an operator.
Btw, I haven't seen the assumption of(Bn)n being bounded anywhere in this exercise.
Hope this helps,
Regards,
MM
Jürgen Voigt, 2020/02/05 22:14
Dear Marcus,
then you shouldn't write ``characterise'', in the exercise. (And I wrote `a sequence of bounded operators', not `a bounded sequence'!)
In any case, thanks! This makes it easier. Best wishes, Jürgen
Marcus Waurick, 2020/02/06 08:56, 2020/02/06 08:59
Dear Jürgen,
``characterise'' is one part. And I think you have figured out this part, anyway.
The other part is about finding a sufficient condition. One could also interpret my `Btw'-remark as a hint. (At least I thought it was one yesterday evening.)
Best regards,
MM
Jürgen Voigt, 2020/02/06 16:52
Dear Marcus,
you are quite correct concerning `characterise'; once more I didn't read carefully enough. (In my defense: I didn't understand that `in terms of convergence of (Bn) in a suitable sense' would allow to include `convergence of (B−1n) in wot'.)
What I still do not see is, how to utilise the self-adjointness of the Bn's.
Best wishes, Jürgen
Jan Meichsner, 2020/02/04 09:43
Dear virtual lecturers,
I have a (mostlikely dumb) question concerning Theorem 13.1.1. Namely, in the proof it is said that M∈M(H,ν) which means sb(M)<ν (in my opinion a desirable property) but the pointwise convergence of the sequence (Mn) only holds for z∈CRe>ν. Can you understand, and even more important help, me with my confusion? Thanks in advance and sorry for the very late question.
Best,
Jan
Sascha Trostorff, 2020/02/04 10:12
Dear Jan,
I am not sure if I understand your question correctly but let me nevertheless try to give an answer.
The definition of of sb(M) was the infimum over all ν0∈R such that M is holomorphic and bounded on the halfplane CRe>ν0. So, the definition of M(H,ν) ensures that each material law in this set is holomorphic and bounded on at least CRe>ν. So it seems quite natural to state the pointwise convergence in Theorem 13.1.1 on this half-plane, since for each bigger half-plane you may have elements in your sequence, which are not defined there.
I hope this clarifies the matter.
Best regards
Sascha
Jürgen Voigt, 2020/02/04 10:30, 2020/02/04 10:45
Dear Sascha,
I think, Jan's problem is a serious one, only he didn't really formulate his question completely: How can you conclude that the limiting material law M, which you obtain as the limit on CRe>ν, has sb(M)<ν, i.e., belongs to M(H,ν)?
I hope, the remedy should be to replace the `<' in the definition of M(H,ν) by `⩽'.
And then you need a modification in the proof: M(∂t,ν) is only defined for ν>sb(M); so one should work with ν0<ν and M(H,ν0) (with the modified definition).
Best wishes, Jürgen
Sascha Trostorff, 2020/02/04 10:50
Dear Jürgen,
thanks for the clarification. Yes, I think your modification would resolve the problem. Another possibility is to define M(∂t,ν) also for ν=sb(M), which is possible via a limit construction (that is what I did in my habilitation thesis).
Best regards
Sascha
Jürgen Voigt, 2020/02/04 11:06
Dear Sascha,
sounds interesting. But it would make the proof of the convergence more involved, I think. Incidentally, it is not only the proofs that have to be adapted, with my suggestion from above, but also the formulations of Theorems 13.1.1 and 13.1.4.
Best wishes, Jürgen
Jan Meichsner, 2020/02/04 13:55
Dear Jürgen and Sascha,
first of all thanks for the discussion. Jürgen, thanks for helping to explain what I meant.
May I suggest a workaround as well (possibly more or less the same you already suggested)?
I think you should interpret M(H,ν) as a space of germs, i.e., an inductive limit. So let us say Mn→M in M(H,ν) if there is μ<ν and n0∈N with sb(Mn)<μ for all n≥n0 and then of course Mn(z)→M(z) for all z∈CRe>μ. I think that's the natural topology (at least for bounded nets) on the limit.
Best,
Jan
Jürgen Voigt, 2020/01/31 15:17, 2020/01/31 18:49
Dear virtual lecturers,
this is about the proof of Theorem 13.3.1:
At the beginning there appears an f `out of thin air'.
Middle of p. 169, Step 1: I had to think a little, what `unitary equivalence' you allude to. It might be nice to mention Proposition 5.3.2 at this place.
Some lines below: The functional Ψ maps to C; the real case is not possible here.
And two lines below: OK; you can use the uniform boundedness principle, but it is a little far-fetched here, because it also follows from the boundedness of the sequence (Bn).
Line -2 of p.169: How would you get equality for all t? There is no reason, why ˜h should be continuous, a priori. Anyway, I suggest a minimally different version for the argument. This is a typical case of the `sub-subsequence argument': You show that every subsequence of (hn) has a subsequence converging locally uniformly to a function which is equal to ˜h a.e., at the same time showing that without restriction ˜h is continuous and the convergence is everywhere. (Agreed: It is just another twist of your argument!)
Line 6 of p.170: Which `identity theorem' do you refer to, here?
Best wishes, Jürgen
Jürgen Voigt, 2020/02/01 16:47
An additional comment on the beginning of the Proof of Theorem 13.3.1: How you write un and the series in the following line, one gets the impression that it is important to keep ∂−1t,νBn together. This is, however, not the case. I would suggest to extend the line
un=…=∞∑k=0(−∂−1t,ν)kBkn∂−1t,νf,
and similarly in the following line. And for Mn I would find it even more important to have also the expression Mn(z)=∑∞k=0(−1z)kBkn1z, because of the convergence stated some lines later.
For the convergence of the sequence (Mn(z)) to M(z) in the weak operator topology one might drop a word that the reader should convince herself/himself that this holds. (I don't think that it is covered by earlier treatment. Actually, this could be an additional exercice for this lecture.)
Incidentally, it just occurs to me that in the Exercises 13.6 and mainly 13.7 it would be nice to replace the operators T and Tn by B and Bn, in order to have the notation parallel to Theorem 13.3.1.
Marcus Waurick, 2020/02/02 13:26
Dear Jürgen,
yes, rewriting the series in the way you did makes sense and helps to clarify the argument.
In an earlier version of the manuscript we hinted at the dominated convergence theorem (for instance) to deduce the convergence of (Mn(z))n. Thinking this might add confusion rather than clarification, we decided to state the convergence right away. Putting this statement to the exercises section is a good idea, though.
Thanks also for your remark concerning the notation in 13.6 and 13.7. We shall change the notation in the final version of the manuscript.
Best regards,
MM
Jürgen Voigt, 2020/02/03 13:21
Dear Marcus,
indeed, mentioning the dominated convergence theorem in connection with the convergence of (Mn(z)) wouldn't have helped me. I had rather thought of something like this (as an exercise):
Let H be a Hilbert space. For k∈N let (Ak,n)n be a sequence in L(H) converging to Bk∈L(H) in the weak operator topology. Let (An)n be a sequence in L(H), and assume that the sequence ((Ak,n)n)k converges to (An)n `uniformly in operator norm', i.e., supn‖An−Ak,n‖→0 as k→∞.
Show that (Bk)k is convergent in operator norm to some B∈L(H) and that (An)n converges to B in the weak operator topology.
Best wishes, Jürgen
Marcus Waurick, 2020/02/03 13:44
Dear Jürgen,
fair enough; this looks better than my idea. I was thinking more along the lines of (N,P(N),#) with # the counting measure as the measure space and (Ak,n)n converging in the weak operator topology to some Ak with ∑∞k=0supn‖Ak,n‖<∞ and then consider for ϕ,ψ∈H the sequence
(∞∑k=0⟨ψ,Ak,nϕ⟩)n.
Best regards,
MM
Marcus Waurick, 2020/02/02 13:17, 2020/02/02 13:17
Dear Jürgen,
thank your very mich for your comments!
The mentioning of the f is not really needed anyway. One can start with a Neumann series representation of the inverse in question right away. Sorry for the confusion!
Mentioning Proposition 5.3.2 makes sense; thanks!
We also should replace K by C in the definition of Ψ.
Also, the application of the uniform boundedness principle is not necessary here. Thanks!
I like your argument concerning line -2, page 169 better. Here we decided to carry out the subsequence principle in detail. In fact, this principle is also used in Lecture 14. We are thinking of shuffling this around a bit.
The identity theorem we are referring to is the one that asserts equality of two holomorphic functions on a connected open set, given they coincide on a set with accumulation point.
Best regards,
MM
Jürgen Voigt, 2020/01/28 15:31
Dear virtual lecturers,
concerning Theorem 13.1.4 I have the similar comment as for Thm 13.1.1: Supposing that (Mn(z)) converges, it is silly to state that there existsM(z). One rather should state that the limit M(z) belongs to M(H,ν) and has such and such properties.
On p. 162, line -5 one could say that sb(M)=0 unless V=0.
p.164, line 3: I think that ν≥1+sup… is good enough. And two lines below I think that it would be fair to mention that Theorem 13.1.4 should be applied with Mn(z)=(1+1zVn(z).
Concerning the comment pointing to Exercise 13.3: This, obviously, is because the material law multiplied by z is still a material law. In this connection I was missing a statement of the fact that the set of material laws M(H,ν0) is an algebra, and the mapping M↦M(∂t,ν), for ν>ν0, is an algebra homomorphism. Actually, I needed such a statement already, when studying Lecture 12, and I think it would be noteworthy enough to state it already in the context where the mapping was defined.
Best wishes, Jürgen
Marcus Waurick, 2020/01/28 17:12, 2020/01/28 17:13
Dear Jürgen,
thank you very much for your comments.
You have been right with Theorem 13.1.1 and, thus, you are so, too, for Theorem 13.1.4. Whether this is `silly', I don't know, you're perspective and formulation certainly is the better one.
Concerning p 162, fair enough.
Concerning p 164, yes, ν≥… is sufficient. I think there is a typo in your comment; but you are right, we apply Theorem 13.1.4 to
Mn(z)=1+1zVn(m)
One needs to be slightly careful here as Vn(m) denotes the multiplication operator on L2(μ) and not on L2(R;H).
The multiplication operator in L2(R;H) with H=L2(μ) reads
(Mn(im+ν)f)(x,ω)=f(x,ω)+(1ix+ν)Vn(ω)f(x,ω)for x∈R and ω∈Ω.
Exercise 13.3 is not difficult because THIS material law multiplied by z is still a material law. Note that you need to check boundedness of the analytic function in question. You are completely right with your comments after that. It is most certainly helpful to highlight the properties of the mapping M↦M(∂t,ν) and we shall do so in the final version of the lecture notes. Thank you!
Best regards,
Marcus
Jürgen Voigt, 2020/01/28 21:12
Dear Marcus,
thanks for pointing out the typo! Reading once more the beginning of the section I realised that you had this material law M already at the beginning, in the same context; so this was quite well prepared.
I don't quite understand your `slightly careful', because this material law seems to be an especially simple case Mn(z)=1+1zBn, with bounded operators Bn∈L(H). This probably means that Theorem 13.2.2 could also have been treated analogously to Theorem 13.3.1, but with a sequence (Bn) in L(H) converging to B in the strong operator topology. Correct?
Best wishes, Jürgen
P.s. Sorry for the “silly”; it was meant in the sense of “albern”, not “stupid”.
Marcus Waurick, 2020/01/29 09:44, 2020/01/29 09:45
Dear Juergen,
the 'slightly careful' was just referring to a clash of notation; if one had written the material law as an operator acting in L2(R;L2(μ)) one ended up with
Mn(im+ν)=1+1im+νVn(m),which does not read well because the first m on the right-hand side and the second one have different interpretations. (Maybe some subscripts could resolve the matter here.)
You are correct with your observation concerning Theorem 13.2.2. Since we had a more general statement already mentioned in Theorem 13.1.4, we wanted to be a bit more specific in Theorem 13.2.2.
Best regards,
Marcus
P.s. “albern” is just fine ;)
Jürgen Voigt, 2020/01/29 13:20
Dear Marcus,
thanks for the clarification! This had indeed escaped my notice, and it is an important point.
Best, Jürgen
Marcus Waurick, 2020/01/23 13:53
Dear participants,
Sebastian Franz found a misprint in the expression of the limit equation in Remark 13.3.2 (and therefore also in Example 13.3.3). The right-hand side expression of the last equality on page 167 misses a time-derivative.
The last expression in Remark 13.3.2 should rather read:
f=∂t,ν˜u+∂t,ν∞∑ℓ=1(−∞∑k=1(−∂−kt,ν)Ck)ℓ˜u
Thus, the limit expression in Example 13.3.3 is
∂t,ν˜u+∂t,ν∞∑ℓ=1(−∞∑m=1(−∂t,ν)−2m(2m)!(m!2m)2)ℓ˜u=f. Note that the highest derivative order of the expressions is still just 1. The derivative in front of the sum term is cancelled out by inverses in the sum, which are at least of order 1.
Thanks again Sebastian for spotting this!
Best regards,
MM
Jürgen Voigt, 2020/01/22 21:43, 2020/01/22 22:05
Dear virtual lecturers,
a comment on the first part of Theorem 13.1.1: I find the way you formulate this a little confusing, in particular in view of the proof. The first statement should be that the weak limit M(z) (which exists by the hypothesis) is a material law, and then Mn(∂t,ν)→M(∂t,ν). And the uniqueness of M is no issue - I think -, because the mapping M↦M(∂t,ν) is injective, by Theorem 8.2.1.
And a comment on Remark 13.1.2: I do not think that you need separability. If you have the countable subset of convergence as you mention, then, for all x,y∈H, the Vitali convergence theorem tells that (⟨x,Mn(z)y⟩)n converges for all z with Rez>ν - call the limit b(z)(x,y) -, and the mapping z↦b(z)(x,y) is holomorphic. Then b(z) is a bounded sesquilinear functional, with the same bound as the common bound of the Mn(z). Then the operators M(z) defined by ⟨x,M(z)y⟩=b(z)(x,y) are the weak limit of (Mn(z))n, and you are in the situation of Theorem 13.1.1.
Best wishes, Jürgen
Marcus Waurick, 2020/01/23 11:56
Dear Juergen,
thank you very much for your comments! Yes, you are right with the reformulation of Theorem 13.1.1. This is better than our version. We shall change it, when we begin the final write up.
Also, concerning Remark 13.1.2, you are absolutely right. When I was writing this remark – stupidly enough – I had a different situation in mind: If you only assume that (Mn)n is bounded, then (given H separable) you can choose a subsequence (Mnk)k such that (Mnk(z))k converges in the weak operator topology for a countable set of z accumulating somewhere in CRe>ν. The sequence (Mnk)k then satisfies the assumptions in Theorem 13.1.1. It is not possible to choose a convergent sub-sequence, if H is not separable. However, one can still show that there exists a sub-net(Mn(ι))ι such that Mn(ι)(z) converges for all z∈Cℜ>ν to some M(z), which is again a material law. Then the net (Mn(ι)(∂t,ν))ι converges to M(∂t,ν). This is what the remark should have read like. Apologies for this.
In order to prove the compactness statement for the non-separable case, I need to invoke Tikhonov's Theorem in combination with Montel's Theorem (see e.g. the sketch of the proof in Theorem 2.24 in the survey paper
Picard, R.; Trostorff, S.; Waurick, M.
Well-posedness via Monotonicity. An Overview.
Operator semigroups meet complex analysis, harmonic analysis and mathematical physics,
Oper. Theory Adv. Appl., 250:397-452, Springer, Cham, 2015
Concerning the compactness and sequential compactness of the weak operator topology of operators on bounded subsets of L(H), I also recommend Exercise 13.6.
Best regards,
MM
Sebastian Franz, 2020/01/22 14:59
Dear Virtual Lecturers,
to do a little nitpicking, your reference to Abramovitz/Stegun gives the definition of the modified Bessel function of the first kind with
I0(z)=1π∫π0exp(z⋅cos(x))dx.
Although, I'm sure your representation of I0(z) is fine (MAPLE says so ): How can you show it?
All the best
Sebastian
Sascha Trostorff, 2020/01/22 20:35
Dear Sebastian,
taking the form of the Bessel function in the lecture, we first can do a simple change of variables to obtain
J(s)=12π∫2π0exp(ssin(x))dx.
Then we use that over the whole period sin and cos attain the same values, so you get
J(s)=12π∫2π0exp(scos(x))dx.
Finally, you use the periodicity of cos and that it is even to get
J(s)=12π∫π−πexp(scos(x))dx=1π∫π0exp(scos(x))dx
which is the formula stated in Abramowitz/Stegun.
Best regards
Sascha
Sebastian Franz, 2020/01/23 07:55
Dear Sascha,
thanks! The trick of replacing sin by cos was new to me.
Best regards,
Sebastian
Johann Beurich, 2020/01/22 14:26
Dear all,
the set D={[n+1/2,n+1];n∈Z} in Exercise 13.4 is meant as the union
⋃n∈Z[n+1/2,n+1],
right?
Best regards,
Johann
Marcus Waurick, 2020/01/22 15:58
Dear Johann,
oh! Yes, of course!
Sorry for the irritation!
Best regards,
moppi
discussion/lecture_13.txt · Last modified: 2019/10/21 17:02 by matcs
Discussion on Lecture 13
Dear virtual lecturers,
I have a late question concerning Exercise 13.7, in connection with Theorem 13.3.1. Under the hypotheses of Theorem 13.3.1, let f∈L2,ν(R;H). Then it is shown that un:=(∂t,ν+Bn)−1f→u:=∂−1t,νM(∂t,ν)f weakly in L2,ν(R;H), with the material law M(z):=∞∑k=0(−1z)kCk. (This is not quite the same M as defined in the middle of p. 168 of the Lecture Notes: this is because I want to have the extra power of ∂−1t,ν outside.) This means that u will satisfy the equation ∂t,νu=M(∂t,ν)f. This is almost the equation I am required to show in Exercise 13.7; only the material law is at the wrong place. Unfortunately, I am at a loss to produce a material law such that the equation required in Exercise 13.7 is satisfied.
Can you help me there, please?
Best wishes, Jürgen
Dear Jürgen,
thank you for your question. With the help of a Neumann series expression (choosing ν>0 large enough) you can compute M(∂t,ν)−1, which is also a material law operator N(∂t,ν) for a suitable N.
You would then end up with ∂t,νN(∂t,ν)u=f.
Does this help?
Best regards,
Marcus
Dear Marcus,
thanks a lot! This is precisely what you discuss in Remark 13.3.2. (Phew!)
Best regards, Jürgen
Dear virtual lecturers,
Best,
Gabriel
Dear Gabriel,
thank you very much for your remarks. It would be really good to have this reference to Section 3.2 there. (We are so used to these concepts that we use it almost without thinking.) You are also right concerning the convergences of the series. This should be worth a comment in the final lectures notes.
Also thank you for your remark concerning the weak star topology. It makes sense to have this recalled, too.
Best regards,
MM
Dear virtual lecturers,
concerning Theorem 13.2.2: I think you can prove this result – with the same arguments – also in the more general context where (Vn(m)) is replaced by a sequence (Bn) in L(H) converging to B∈L(H) in the strong operator topology. (This would make it more parallel to Theorem 13.3.1.)
Best wishes, Jürgen
Dear Jürgen,
Thanks for this comment. You are completely right, we haven't been very consistent in formulating these results.
Best regards,
MM
Dear ISemTeam,
I/we need a hint concerning Exercise 13.5. As I understand, the problem amounts to characterising weak-operator-topology convergence of the sequence (B−1n) in terms of the original sequence (Bn) of bounded self-adjoint operators, where these operators are bounded below by a common bound c>0. Is there anything more special, because you formulate it in a special context? And how can one exploit the self-adjointness?
Best wishes, Jürgen
Dear Jürgen,
The only thing we were after was a sufficient condition (not a characterisation) for the weak operator topology limit of B−1n to be an inverse of an operator.
Btw, I haven't seen the assumption of(Bn)n being bounded anywhere in this exercise.
Hope this helps,
Regards,
MM
Dear Marcus,
then you shouldn't write ``characterise'', in the exercise. (And I wrote `a sequence of bounded operators', not `a bounded sequence'!)
In any case, thanks! This makes it easier. Best wishes, Jürgen
Dear Jürgen,
``characterise'' is one part. And I think you have figured out this part, anyway.
The other part is about finding a sufficient condition. One could also interpret my `Btw'-remark as a hint. (At least I thought it was one yesterday evening.)
Best regards,
MM
Dear Marcus,
you are quite correct concerning `characterise'; once more I didn't read carefully enough. (In my defense: I didn't understand that `in terms of convergence of (Bn) in a suitable sense' would allow to include `convergence of (B−1n) in wot'.)
What I still do not see is, how to utilise the self-adjointness of the Bn's.
Best wishes, Jürgen
Dear virtual lecturers,
I have a (mostlikely dumb) question concerning Theorem 13.1.1. Namely, in the proof it is said that M∈M(H,ν) which means sb(M)<ν (in my opinion a desirable property) but the pointwise convergence of the sequence (Mn) only holds for z∈CRe>ν. Can you understand, and even more important help, me with my confusion? Thanks in advance and sorry for the very late question.
Best,
Jan
Dear Jan,
I am not sure if I understand your question correctly but let me nevertheless try to give an answer.
The definition of of sb(M) was the infimum over all ν0∈R such that M is holomorphic and bounded on the halfplane CRe>ν0. So, the definition of M(H,ν) ensures that each material law in this set is holomorphic and bounded on at least CRe>ν. So it seems quite natural to state the pointwise convergence in Theorem 13.1.1 on this half-plane, since for each bigger half-plane you may have elements in your sequence, which are not defined there.
I hope this clarifies the matter.
Best regards
Sascha
Dear Sascha,
I think, Jan's problem is a serious one, only he didn't really formulate his question completely: How can you conclude that the limiting material law M, which you obtain as the limit on CRe>ν, has sb(M)<ν, i.e., belongs to M(H,ν)?
I hope, the remedy should be to replace the `<' in the definition of M(H,ν) by `⩽'. And then you need a modification in the proof: M(∂t,ν) is only defined for ν>sb(M); so one should work with ν0<ν and M(H,ν0) (with the modified definition).
Best wishes, Jürgen
Dear Jürgen,
thanks for the clarification. Yes, I think your modification would resolve the problem. Another possibility is to define M(∂t,ν) also for ν=sb(M), which is possible via a limit construction (that is what I did in my habilitation thesis).
Best regards
Sascha
Dear Sascha,
sounds interesting. But it would make the proof of the convergence more involved, I think. Incidentally, it is not only the proofs that have to be adapted, with my suggestion from above, but also the formulations of Theorems 13.1.1 and 13.1.4.
Best wishes, Jürgen
Dear Jürgen and Sascha,
first of all thanks for the discussion. Jürgen, thanks for helping to explain what I meant. May I suggest a workaround as well (possibly more or less the same you already suggested)? I think you should interpret M(H,ν) as a space of germs, i.e., an inductive limit. So let us say Mn→M in M(H,ν) if there is μ<ν and n0∈N with sb(Mn)<μ for all n≥n0 and then of course Mn(z)→M(z) for all z∈CRe>μ. I think that's the natural topology (at least for bounded nets) on the limit.
Best, Jan
Dear virtual lecturers,
this is about the proof of Theorem 13.3.1:
At the beginning there appears an f `out of thin air'.
Middle of p. 169, Step 1: I had to think a little, what `unitary equivalence' you allude to. It might be nice to mention Proposition 5.3.2 at this place.
Some lines below: The functional Ψ maps to C; the real case is not possible here.
And two lines below: OK; you can use the uniform boundedness principle, but it is a little far-fetched here, because it also follows from the boundedness of the sequence (Bn).
Line -2 of p.169: How would you get equality for all t? There is no reason, why ˜h should be continuous, a priori. Anyway, I suggest a minimally different version for the argument. This is a typical case of the `sub-subsequence argument': You show that every subsequence of (hn) has a subsequence converging locally uniformly to a function which is equal to ˜h a.e., at the same time showing that without restriction ˜h is continuous and the convergence is everywhere. (Agreed: It is just another twist of your argument!)
Line 6 of p.170: Which `identity theorem' do you refer to, here?
Best wishes, Jürgen
An additional comment on the beginning of the Proof of Theorem 13.3.1: How you write un and the series in the following line, one gets the impression that it is important to keep ∂−1t,νBn together. This is, however, not the case. I would suggest to extend the line un=…=∞∑k=0(−∂−1t,ν)kBkn∂−1t,νf, and similarly in the following line. And for Mn I would find it even more important to have also the expression Mn(z)=∑∞k=0(−1z)kBkn1z, because of the convergence stated some lines later.
For the convergence of the sequence (Mn(z)) to M(z) in the weak operator topology one might drop a word that the reader should convince herself/himself that this holds. (I don't think that it is covered by earlier treatment. Actually, this could be an additional exercice for this lecture.)
Incidentally, it just occurs to me that in the Exercises 13.6 and mainly 13.7 it would be nice to replace the operators T and Tn by B and Bn, in order to have the notation parallel to Theorem 13.3.1.
Dear Jürgen,
yes, rewriting the series in the way you did makes sense and helps to clarify the argument.
In an earlier version of the manuscript we hinted at the dominated convergence theorem (for instance) to deduce the convergence of (Mn(z))n. Thinking this might add confusion rather than clarification, we decided to state the convergence right away. Putting this statement to the exercises section is a good idea, though.
Thanks also for your remark concerning the notation in 13.6 and 13.7. We shall change the notation in the final version of the manuscript.
Best regards,
MM
Dear Marcus,
indeed, mentioning the dominated convergence theorem in connection with the convergence of (Mn(z)) wouldn't have helped me. I had rather thought of something like this (as an exercise):
Let H be a Hilbert space. For k∈N let (Ak,n)n be a sequence in L(H) converging to Bk∈L(H) in the weak operator topology. Let (An)n be a sequence in L(H), and assume that the sequence ((Ak,n)n)k converges to (An)n `uniformly in operator norm', i.e., supn‖An−Ak,n‖→0 as k→∞.
Show that (Bk)k is convergent in operator norm to some B∈L(H) and that (An)n converges to B in the weak operator topology.
Best wishes, Jürgen
Dear Jürgen,
fair enough; this looks better than my idea. I was thinking more along the lines of (N,P(N),#) with # the counting measure as the measure space and (Ak,n)n converging in the weak operator topology to some Ak with ∑∞k=0supn‖Ak,n‖<∞ and then consider for ϕ,ψ∈H the sequence (∞∑k=0⟨ψ,Ak,nϕ⟩)n. Best regards,
MM
Dear Jürgen,
thank your very mich for your comments!
The mentioning of the f is not really needed anyway. One can start with a Neumann series representation of the inverse in question right away. Sorry for the confusion!
Mentioning Proposition 5.3.2 makes sense; thanks!
We also should replace K by C in the definition of Ψ.
Also, the application of the uniform boundedness principle is not necessary here. Thanks!
I like your argument concerning line -2, page 169 better. Here we decided to carry out the subsequence principle in detail. In fact, this principle is also used in Lecture 14. We are thinking of shuffling this around a bit.
The identity theorem we are referring to is the one that asserts equality of two holomorphic functions on a connected open set, given they coincide on a set with accumulation point.
Best regards,
MM
Dear virtual lecturers,
concerning Theorem 13.1.4 I have the similar comment as for Thm 13.1.1: Supposing that (Mn(z)) converges, it is silly to state that there exists M(z). One rather should state that the limit M(z) belongs to M(H,ν) and has such and such properties.
On p. 162, line -5 one could say that sb(M)=0 unless V=0.
p.164, line 3: I think that ν≥1+sup… is good enough. And two lines below I think that it would be fair to mention that Theorem 13.1.4 should be applied with Mn(z)=(1+1zVn(z).
Concerning the comment pointing to Exercise 13.3: This, obviously, is because the material law multiplied by z is still a material law. In this connection I was missing a statement of the fact that the set of material laws M(H,ν0) is an algebra, and the mapping M↦M(∂t,ν), for ν>ν0, is an algebra homomorphism. Actually, I needed such a statement already, when studying Lecture 12, and I think it would be noteworthy enough to state it already in the context where the mapping was defined.
Best wishes, Jürgen
Dear Jürgen,
thank you very much for your comments.
You have been right with Theorem 13.1.1 and, thus, you are so, too, for Theorem 13.1.4. Whether this is `silly', I don't know, you're perspective and formulation certainly is the better one.
Concerning p 162, fair enough.
Concerning p 164, yes, ν≥… is sufficient. I think there is a typo in your comment; but you are right, we apply Theorem 13.1.4 to Mn(z)=1+1zVn(m) One needs to be slightly careful here as Vn(m) denotes the multiplication operator on L2(μ) and not on L2(R;H).
The multiplication operator in L2(R;H) with H=L2(μ) reads (Mn(im+ν)f)(x,ω)=f(x,ω)+(1ix+ν)Vn(ω)f(x,ω)for x∈R and ω∈Ω.
Exercise 13.3 is not difficult because THIS material law multiplied by z is still a material law. Note that you need to check boundedness of the analytic function in question. You are completely right with your comments after that. It is most certainly helpful to highlight the properties of the mapping M↦M(∂t,ν) and we shall do so in the final version of the lecture notes. Thank you!
Best regards, Marcus
Dear Marcus,
thanks for pointing out the typo! Reading once more the beginning of the section I realised that you had this material law M already at the beginning, in the same context; so this was quite well prepared.
I don't quite understand your `slightly careful', because this material law seems to be an especially simple case Mn(z)=1+1zBn, with bounded operators Bn∈L(H). This probably means that Theorem 13.2.2 could also have been treated analogously to Theorem 13.3.1, but with a sequence (Bn) in L(H) converging to B in the strong operator topology. Correct?
Best wishes, Jürgen
P.s. Sorry for the “silly”; it was meant in the sense of “albern”, not “stupid”.
Dear Juergen,
the 'slightly careful' was just referring to a clash of notation; if one had written the material law as an operator acting in L2(R;L2(μ)) one ended up with Mn(im+ν)=1+1im+νVn(m),which does not read well because the first m on the right-hand side and the second one have different interpretations. (Maybe some subscripts could resolve the matter here.)
You are correct with your observation concerning Theorem 13.2.2. Since we had a more general statement already mentioned in Theorem 13.1.4, we wanted to be a bit more specific in Theorem 13.2.2.
Best regards,
Marcus
P.s. “albern” is just fine ;)
Dear Marcus,
thanks for the clarification! This had indeed escaped my notice, and it is an important point.
Best, Jürgen
Dear participants,
Sebastian Franz found a misprint in the expression of the limit equation in Remark 13.3.2 (and therefore also in Example 13.3.3). The right-hand side expression of the last equality on page 167 misses a time-derivative.
The last expression in Remark 13.3.2 should rather read: f=∂t,ν˜u+∂t,ν∞∑ℓ=1(−∞∑k=1(−∂−kt,ν)Ck)ℓ˜u
Thus, the limit expression in Example 13.3.3 is
∂t,ν˜u+∂t,ν∞∑ℓ=1(−∞∑m=1(−∂t,ν)−2m(2m)!(m!2m)2)ℓ˜u=f. Note that the highest derivative order of the expressions is still just 1. The derivative in front of the sum term is cancelled out by inverses in the sum, which are at least of order 1.
Thanks again Sebastian for spotting this!
Best regards,
MM
Dear virtual lecturers,
a comment on the first part of Theorem 13.1.1: I find the way you formulate this a little confusing, in particular in view of the proof. The first statement should be that the weak limit M(z) (which exists by the hypothesis) is a material law, and then Mn(∂t,ν)→M(∂t,ν). And the uniqueness of M is no issue - I think -, because the mapping M↦M(∂t,ν) is injective, by Theorem 8.2.1.
And a comment on Remark 13.1.2: I do not think that you need separability. If you have the countable subset of convergence as you mention, then, for all x,y∈H, the Vitali convergence theorem tells that (⟨x,Mn(z)y⟩)n converges for all z with Rez>ν - call the limit b(z)(x,y) -, and the mapping z↦b(z)(x,y) is holomorphic. Then b(z) is a bounded sesquilinear functional, with the same bound as the common bound of the Mn(z). Then the operators M(z) defined by ⟨x,M(z)y⟩=b(z)(x,y) are the weak limit of (Mn(z))n, and you are in the situation of Theorem 13.1.1.
Best wishes, Jürgen
Dear Juergen,
thank you very much for your comments! Yes, you are right with the reformulation of Theorem 13.1.1. This is better than our version. We shall change it, when we begin the final write up.
Also, concerning Remark 13.1.2, you are absolutely right. When I was writing this remark – stupidly enough – I had a different situation in mind: If you only assume that (Mn)n is bounded, then (given H separable) you can choose a subsequence (Mnk)k such that (Mnk(z))k converges in the weak operator topology for a countable set of z accumulating somewhere in CRe>ν. The sequence (Mnk)k then satisfies the assumptions in Theorem 13.1.1. It is not possible to choose a convergent sub-sequence, if H is not separable. However, one can still show that there exists a sub-net (Mn(ι))ι such that Mn(ι)(z) converges for all z∈Cℜ>ν to some M(z), which is again a material law. Then the net (Mn(ι)(∂t,ν))ι converges to M(∂t,ν). This is what the remark should have read like. Apologies for this.
In order to prove the compactness statement for the non-separable case, I need to invoke Tikhonov's Theorem in combination with Montel's Theorem (see e.g. the sketch of the proof in Theorem 2.24 in the survey paper
Picard, R.; Trostorff, S.; Waurick, M. Well-posedness via Monotonicity. An Overview. Operator semigroups meet complex analysis, harmonic analysis and mathematical physics, Oper. Theory Adv. Appl., 250:397-452, Springer, Cham, 2015
or in the arXiv:
https://arxiv.org/pdf/1401.5294.pdf,
where also more detailed proofs are referenced).
Concerning the compactness and sequential compactness of the weak operator topology of operators on bounded subsets of L(H), I also recommend Exercise 13.6.
Best regards,
MM
Dear Virtual Lecturers,
): How can you show it?
to do a little nitpicking, your reference to Abramovitz/Stegun gives the definition of the modified Bessel function of the first kind with I0(z)=1π∫π0exp(z⋅cos(x))dx. Although, I'm sure your representation of I0(z) is fine (MAPLE says so
All the best Sebastian
Dear Sebastian,
taking the form of the Bessel function in the lecture, we first can do a simple change of variables to obtain J(s)=12π∫2π0exp(ssin(x))dx. Then we use that over the whole period sin and cos attain the same values, so you get J(s)=12π∫2π0exp(scos(x))dx. Finally, you use the periodicity of cos and that it is even to get J(s)=12π∫π−πexp(scos(x))dx=1π∫π0exp(scos(x))dx which is the formula stated in Abramowitz/Stegun.
Best regards
Sascha
Dear Sascha,
thanks! The trick of replacing sin by cos was new to me.
Best regards,
Sebastian
Dear all,
the set D={[n+1/2,n+1];n∈Z} in Exercise 13.4 is meant as the union ⋃n∈Z[n+1/2,n+1], right?
Best regards,
Johann
Dear Johann,
oh! Yes, of course!
Sorry for the irritation!
Best regards,
moppi