Hey dear friends,
I would like to discuss 1st and 2nd section step by step ,in more detail. I will work on 2nd section ,but all processes depend so 1st chapter ; Fractional derivative setting in Hilbert space. As we discussed yesterday, in this paper, Fractional derivatives of Caputo and Riemann-Liouville types generated by Laplace-Fourier transform and coincide since the interval of integration starts from -∞. We can consider these cases on (a,t),t>0 (for time-derivative we can select a=0 ). But, again we will use the same formula , it will be repetition.What do you think about this matter? My suggestion is investigation the Cauchy problem on operator fractional differential equtions of
(D^α x)(t) + λAu(t) = f(t), t ∈ I, α ∈ (0, 1), λ> 0,
(J^(1−α) x)|t=0 = x.0 or x(0)=x.0,
where A is a linear m-accretive operator.
Sum up, please sent us all steps of plans related to 1st and 2nd section in more detail. Thanks in advance.
Jan Meichsner, 2020/04/29 13:35
Dear Ismail,
I think there is a confusion about vocabulary. I would not say that the derivatives coincide. You should always see the full problem which is the differential equation, yes, but also the additional equations we typically refer to as initial conditions (or whatsoever). Just seeing the differential operator, you often simply cannot invert it due to the missing injectivity. The additional conditions enforce this and, therefore, are needed for uniqueness.
As for the paper, I would stick to their vocabulary calling Dαu(t)=f(t,u(t)) with either u(0)=u0 or Dα−1u(0)=u0 the Caputo or Riemann-Liouville problem, respectively. By the way, it does not matter if you would agree with this terminology based on your knowledge. Just accept it for the moment and let us use it.
Due to our lack of time, we decided to just deal with the Caputo problem, i.e., Dαu(t)=f(t,u(t)), u(0)=u0. Let us stick to the general full line setting (−∞,∞). For an initial value problem, I actually would assume (0,∞) to be more suitable (we may have a look at this as well if there is time) and, moreover, to be more or less the same as the full line result. It is pointless to consider the problems on finite intervals. Here, you cannot see the exponential weights. Further, and this is also a problem you have with (0,∞), one would have to know much more about sectorial operators to introduce Dα since Fourier techniques can't be applied either. I know. The operators are still normal. However, you will not be able to determine the corresponding spectral measures (I guess; don't know it for sure, though.).
To sum up, considering finite intervals for the beginning will not give you a better insight but rather makes things more complicated. That is a typical effect when you pass to non-full space problems as the underlining group structure, and therefore the Fourier techniques, get lost.
Maybe one more note on the problem Dαu(t)+Au(t)=f(t). This problem we actually already solved with the general theory. At least for A being skew-selfadjoint. Namely, this can be rewritten as
Dα−1Du+Au=f.
For α∈(0,1), the first term is a valid material law. I am convinced one could also do something for more general A such as (m-)accretive operators. The case α≥1 requires some more thoughts, though.
However, this should not be our business for this project. Please let us concentrate on the fractional derivative with non-linear but Lipschitz continuous right-hand side. Last thing. If you wanted to do the last thing as initial value problem, you may check out Lecture 9 again.
So, to sum up this part as well, let us concentrate for the time being on the setting from the paper.
I am note completely sure about how helpful my comment was but I hope it clarifies some things for you. We may continue this discussion here of course if more clarification is needed.
Best,
Jan
Lars Niedorf, 2020/04/14 19:24
Hi everyone,
I have some questions concerning the proof of Lemma 4.2 (d) where one wants to extend the fractional derivative
∂β0,ϱ:(Hβ+|α|ϱ(R,H),‖⋅‖ϱ,α)→(Hα−βϱ(R,H),‖⋅‖ϱ,α−β)
to a unitary operator
∂β0,ϱ:(Hαϱ(R,H),‖⋅‖ϱ,α)→(Hα−βϱ(R,H),‖⋅‖ϱ,α−β)
for β>0 and α∈R.
1. Why can't we just say that we define the desired map ∂β0,ϱ via the concatenation of unitary operators
Hαϱ(R,H)Lϱ→Hα(im+ρ)(im+ρ)β→Hα−β(im+ρ)L∗ϱ→Hα−βϱ(R,H)
where we use the extension of the Fourier-Laplace transform from part (b) of the lemma? Also the restriction β>0 wouldn't be necessary then.
2. In the proof, they write L∗ϱ(im+ϱ)−βφ∈⋂γ∈RHγϱ(R,H). It's not of great importance for the proof since one only needs that L∗ϱ(im+ϱ)−βφ is an element of Hβ+|α|ϱ, but how should this intersection be interpreted? The spaces Hγϱ(R,H) are by definition just completions of some pre Hilbert spaces. So this intersection is meant as an intersection in the space of distributions D(R,H)′, right?
3. What is the exact argument for extending the map
∂β0,ϱ:Hβ+|α|ϱ(R,H)→Hα−βϱ(R,H)? We know that this map is an isometry and has dense range, but why do we know at this stage that the closure of Hβ+|α|ϱ(R,H) with respect to the norm ‖⋅‖ϱ,α is given by Hαϱ(R,H)?
Best regards, Lars
Jan Meichsner, 2020/04/16 12:42
Hey Lars,
I thought a bit about it. Here comes the outcome in the same sequence you asked.
1. Good question. I am not sure if I oversee something but I am tempted to agree with you. First of all, I would assume the statement to be true for all β∈R. Second, I have not checked completely but for the time being I don't see a good reason why one would be interested in Hβ+|α|ϱ as a core. One could also use C∞c. Which brings me to your thought. I believe that works. I was thinking about the first lines in the proof of Lemma 4.2 (d). But I don't see that we need β>0 to show that (im+ϱ)β is unitary. So, I believe you can define the desired extension in the above manner. Afterwards, one probably wanted to show as well that the so defined operator, restricted to C∞c, is really just the derivative.
2. Fair point but I believe this was done in order to help the reader. One should not think about the fact why the desired statement L∗ϱ(im+ϱ)−βφ∈Hβ+|α|ϱ holds but about the easier statement L∗ϱ(im+ϱ)−βφ∈⋂γ∈RHγϱ from which one can directly see what we actually want.
As for the intersection, I think one can give a very intuitive meaning to the objects in H−αϱ for α>0 and I am wondering why this was not done because in my opinion one can also understand the support later on better. However, the point here, at least for me, is that one just wants the elements in all Hγϱ, γ∈R to be functionals on one common function space. I believe H∞ϱ would suffice here but, of course, smaller choices such as D are allowed. On the other hand, formally intersecting sets of completely unrelated objects should not cause to many issues either, should it?
3. Öhm, Lemma 4.2 c) right? β+|α|>α which is why Hβ+|α|ϱ is dense in Hαϱ. Works?
Best,
Jan
Lars Niedorf, 2020/04/16 15:31
Hi Jan,
thanks a lot for your response! 2. and 3. are fine then, especially 3. . Here are some more thoughts on 1.
As far as I understood the idea of the proof, the reason why we assume β>0 is the second line of the proof where one proves that ∂β0,ϱ is an isometry. Since u should be in the domain of ∂β0,ϱ, we need β+|α|≥0 to apply part (a) of the lemma. So in order to get rid of the assumption β>0, we need to take another space where we know that our operator ∂β0,ϱ is defined.
I think taking C∞c(R,H) might cause some trouble, since we need that C∞c(R,H) is dense in Hαϱ(R,H) to extend the operator, but density will be proved in Lemma 4.8 afterwards by using Lemma 4.2 (d): In the proof there we consider ∂α0,ϱf∈L2ϱ(R,H) for a given function f∈Hαϱ(R,H). (By the way, we need Lemma 4.2 (d) there also for β<0, right?)
However, I think taking the space H|β|+|α|ϱ(R,H) instead of Hβ+|α|ϱ(R,H) should solve this problem so that we obtain the statement of the lemma also for β<0.
Concerning my proposal to define the extension ∂β0,ϱ:Hαϱ(R,H)→Hα−βϱ(R,H) via the concatenation of unitary maps, the point why the paper uses a different argument might be (as you mentioned) that one wants of course the extension ∂β0,ϱ:Hαϱ(R,H)→Hα−βϱ(R,H) to coincide with the derivative ∂β0,ϱ:dom(∂β0,ϱ)⊆L2ϱ(R,H)→L2ϱ(R,H) on at least C∞c(R,H). But I think this can then be seen via the diagram
H|β|+|α|ϱ=W|β|+|α|ϱLϱ→H|β|+|α|∩L2(im+ρ)β→H|β|+|α|−β∩L2⊆Hα−β∩L2L∗ϱ→Wα−βϱ⊆Hα−βϱ
since we don't have to take the extension of the Fourier-Laplace transform there. So we know that ∂β0,ϱ:Hαϱ(R,H)→Hα−βϱ(R,H) is an extension of ∂β0,ϱ:H|β|+|α|ϱ(R,H)→Hα−βϱ(R,H).
Best, Lars
Jan Meichsner, 2020/04/17 08:04, 2020/04/17 08:08
Hey Lars,
ah, the density result comes after. Sorry, I did not pay enough attention. Well, one can prove that independently if I am not mistaken. As an exercise (I really hope I am not wrong here) try to show the density of C∞c in Hkϱ, k∈N0 as follows. Let f∈Hkϱ be given. Consider g:=e−tϱf (sorry but I needed to get rid of this m. When you see t, I am using the function as multiplication operator). Show g∈Hk (ordinary Sobolev space). Approximate g by a sequence (φn) in C∞c. Set ψn:=etρφn which is still in C∞c. Show ψn→f in Hkϱ. Together with the density result from Lemma 4.2 c), you can interpolate and get the density in all Hαϱ. Again, if I am not mistaken, you will need the following formula:
∀m∈N,n∈N0,m≥n:m∑l=n(ml)(ln)(−1)n−l=δmn.
I tried this yesterday, and one can give a proof for this by induction.
Have a try if you wish, and let me know if I talked shit.
As for your point that you need Lemma 4.2 (d) also in case β<0, I believe this is not true. If I understand you thought correctly, you would like to have this Lemma also for β<0 to get the mapping property Dα:Hαϱ→L2ϱ (sorry, I also cannot stand ∂0,ϱ), right?. But this follows since Dβ, β>0 is unitary with inverse D−β providing you also with the right mapping properties for negative values.
The last argument seems fine to me. I would just write H|α|+|β| instead of H|α|+|β|∩L2 since we know H|α|+|β|⊆L2. Same comment goes also for the point after applying the multiplication operator. Or do I not understand something here?
Anyway, nice thoughts and good job. Keep going.
Best,
Jan
Lars Niedorf, 2020/04/17 12:02
Hi Jan,
probably I overlook something, but where shall the formula come into play?
Concerning the other two points you mentioned, I completely agree with you (taking the inverse of ∂α0,ϱ, that's insane… ). Yes, the intersection with L2 is maybe a little bit overcautious.
Thanks and regards, Lars
Jan Meichsner, 2020/04/17 12:33
Hey Lars,
I used it when I wanted to show that ψ(m)ntof(m) in L2ϱ but maybe you did that differently, and actually more intelligent, then me. Well, did everything work out? If you want, you can send me your solution so I can learn something as well.
Best,
Jan
Lars Niedorf, 2020/04/17 14:10, 2020/04/17 14:53
Hi Jan,
I hope I don't miss your point. My argument then would have been the following:
If φn→g in Hk (classical Sobolev space), then we have ξℓˆφn(ξ)→ξℓˆg(ξ) in L2 for any ℓ≤k. In particular we have (im+ρ)k(ˆφn−ˆg)→0 in L2. Then
‖ψn−f‖Hkϱ=‖Lϱ(ψn−f)‖Hk(im+ϱ)=‖Fe−mϱ(ψn−f)‖Hk(im+ϱ)=‖ˆφn−ˆg‖Hk(im+ϱ)=‖(im+ϱ)k(ˆφn−ˆg)‖L2→0.
For approximating g in Hk (classical Sobolev space), i would choose ηn:=g∗θn where θn(x)=nθ(nx) is a Dirac sequence and θ∈C∞c, say. Then ˆηn=ˆg⋅ˆθ(⋅n). Since ˆθ is bounded and ˆθ(⋅n)→1 pointwise, we obtain ξℓˆηn(ξ)→ξℓˆg(ξ) in L2 for any ℓ≤k. This implies ηn→g in Hk. Altogether, we know that the Schwartz space S is dense in Hk. In order to obtain density for C∞c, I would approximate a given Schwartz function g∈S via gn:=γng where γn is a sequence of bump functions with support in [−n−1,n+1] and γn|[−n,n]=1 and bounded derivatives up to order k. Via Leibniz rule, we obtain gn→g in Hk.
Best regards, Lars
EDIT: I'm afraid that my claim that ηn lies in the Schwartz space is wrong, but if we approximate ηn via multiplication with cut off functions, the argument should still work I think.
Jan Meichsner, 2020/04/17 22:34
Hey Lars,
Nope, you don't miss anything but indeed acted much smarter than I did. I established the convergence result by using two times the product rule which led to the need of the above formula. Your approach is much better.
As for the classical approximation result, this I would take for granted. The functions ηn are indeed no Schwartz functions, that's right. They are in H∞, though. And the above proof sketches correctly how one would do this. Well done.
This question is completely clarified, is it not?
Best,
Jan
Lars Niedorf, 2020/04/18 00:10
Hi Jan,
that's great, then my question is indeed completely clarified.
Thanks a lot!
Best, Lars
Sebastian Bechtel, 2020/04/14 13:59
Dear all,
if you post something in this forum (it can also be a reply to a previous post) and tick the box “Subscribe to comments” then it seems to be a subscription for the whole discussion board of our project and not merely to replies to that specific post, so it seems to be a good plan that everyone of us uses this feature to not miss the discussions here anymore on the future :)
Best, Sebastian
Jan Meichsner, 2020/04/14 14:36
Good idea. Thanks for the information.
Best,
Jan
Sam Rametse, 2020/04/12 15:46
Hi Everyone,
Hope you are all keeping safe.
I am about 40% through the paper posted by Jan. although I understand parts of it , I am finding other sections a bit technical. Is anyone open to having a Zoom discussion on their understanding of the paper maybe on the 14th April (I think I'll be done reading then)? - are we allowed?
Looking forward to hearing from you,
Sam - University of Johannesburg , South Africa.
Jan Meichsner, 2020/04/13 12:35
Hey there,
Sorry, I just read what you wrote.
I like the idea and would be up for it. And of course, this is allowed and highly appreciated.
Not sure if the others read this here though. Could you write an email to everybody? In this mail, you could also point out that we should regularly check the discussion board. At this point shame on me as well because I didn't check either. Please also suggest a time for the meeting. I have a licensed zoom version and could take care of everything. In particular, we don't have to care for the time limit (which is not a big problem anyway).
Thanks for taking care of everything! See you tomorrow.
Best,
Jan
Sebastian Bechtel, 2020/04/14 13:55, 2020/04/14 13:56
I also missed that post and promise to observe the forum in the future more carefully. I agree with the opinions from the mail that we should stick to our original plan and postpone the general discussion till April 20th. If you have specific questions before that just post them here and we'll have a look!
Felix Schwenninger, 2020/04/06 10:38
Hi Sebastian, hi Jan,
Sorry for spamming, but if I am not mistaken your name, Jan, is mi's'pelled on the page where the projects are listed ;)
Best, Felix
Jan Meichsner, 2020/04/08 08:14
You better be sorry!
But thanks for pointing out. You are right.
Best,
Jan
Jan Meichsner, 2020/04/03 08:29
Hey guys (and everybody else reading this),
This is our nice discussion board. Have fun using it.
Best,
Sebastian & Jan
discussion/project_a.txt · Last modified: 2020/03/07 15:20 by matcs
Discussion on Project A
Hey dear friends, I would like to discuss 1st and 2nd section step by step ,in more detail. I will work on 2nd section ,but all processes depend so 1st chapter ; Fractional derivative setting in Hilbert space. As we discussed yesterday, in this paper, Fractional derivatives of Caputo and Riemann-Liouville types generated by Laplace-Fourier transform and coincide since the interval of integration starts from -∞. We can consider these cases on (a,t),t>0 (for time-derivative we can select a=0 ). But, again we will use the same formula , it will be repetition.What do you think about this matter? My suggestion is investigation the Cauchy problem on operator fractional differential equtions of (D^α x)(t) + λAu(t) = f(t), t ∈ I, α ∈ (0, 1), λ> 0, (J^(1−α) x)|t=0 = x.0 or x(0)=x.0, where A is a linear m-accretive operator. Sum up, please sent us all steps of plans related to 1st and 2nd section in more detail. Thanks in advance.
Dear Ismail,
I think there is a confusion about vocabulary. I would not say that the derivatives coincide. You should always see the full problem which is the differential equation, yes, but also the additional equations we typically refer to as initial conditions (or whatsoever). Just seeing the differential operator, you often simply cannot invert it due to the missing injectivity. The additional conditions enforce this and, therefore, are needed for uniqueness. As for the paper, I would stick to their vocabulary calling Dαu(t)=f(t,u(t)) with either u(0)=u0 or Dα−1u(0)=u0 the Caputo or Riemann-Liouville problem, respectively. By the way, it does not matter if you would agree with this terminology based on your knowledge. Just accept it for the moment and let us use it. Due to our lack of time, we decided to just deal with the Caputo problem, i.e., Dαu(t)=f(t,u(t)), u(0)=u0. Let us stick to the general full line setting (−∞,∞). For an initial value problem, I actually would assume (0,∞) to be more suitable (we may have a look at this as well if there is time) and, moreover, to be more or less the same as the full line result. It is pointless to consider the problems on finite intervals. Here, you cannot see the exponential weights. Further, and this is also a problem you have with (0,∞), one would have to know much more about sectorial operators to introduce Dα since Fourier techniques can't be applied either. I know. The operators are still normal. However, you will not be able to determine the corresponding spectral measures (I guess; don't know it for sure, though.). To sum up, considering finite intervals for the beginning will not give you a better insight but rather makes things more complicated. That is a typical effect when you pass to non-full space problems as the underlining group structure, and therefore the Fourier techniques, get lost. Maybe one more note on the problem Dαu(t)+Au(t)=f(t). This problem we actually already solved with the general theory. At least for A being skew-selfadjoint. Namely, this can be rewritten as Dα−1Du+Au=f. For α∈(0,1), the first term is a valid material law. I am convinced one could also do something for more general A such as (m-)accretive operators. The case α≥1 requires some more thoughts, though. However, this should not be our business for this project. Please let us concentrate on the fractional derivative with non-linear but Lipschitz continuous right-hand side. Last thing. If you wanted to do the last thing as initial value problem, you may check out Lecture 9 again. So, to sum up this part as well, let us concentrate for the time being on the setting from the paper. I am note completely sure about how helpful my comment was but I hope it clarifies some things for you. We may continue this discussion here of course if more clarification is needed.
Best, Jan
Hi everyone,
I have some questions concerning the proof of Lemma 4.2 (d) where one wants to extend the fractional derivative ∂β0,ϱ:(Hβ+|α|ϱ(R,H),‖⋅‖ϱ,α)→(Hα−βϱ(R,H),‖⋅‖ϱ,α−β) to a unitary operator ∂β0,ϱ:(Hαϱ(R,H),‖⋅‖ϱ,α)→(Hα−βϱ(R,H),‖⋅‖ϱ,α−β) for β>0 and α∈R.
1. Why can't we just say that we define the desired map ∂β0,ϱ via the concatenation of unitary operators Hαϱ(R,H)Lϱ→Hα(im+ρ)(im+ρ)β→Hα−β(im+ρ)L∗ϱ→Hα−βϱ(R,H) where we use the extension of the Fourier-Laplace transform from part (b) of the lemma? Also the restriction β>0 wouldn't be necessary then.
2. In the proof, they write L∗ϱ(im+ϱ)−βφ∈⋂γ∈RHγϱ(R,H). It's not of great importance for the proof since one only needs that L∗ϱ(im+ϱ)−βφ is an element of Hβ+|α|ϱ, but how should this intersection be interpreted? The spaces Hγϱ(R,H) are by definition just completions of some pre Hilbert spaces. So this intersection is meant as an intersection in the space of distributions D(R,H)′, right?
3. What is the exact argument for extending the map ∂β0,ϱ:Hβ+|α|ϱ(R,H)→Hα−βϱ(R,H)? We know that this map is an isometry and has dense range, but why do we know at this stage that the closure of Hβ+|α|ϱ(R,H) with respect to the norm ‖⋅‖ϱ,α is given by Hαϱ(R,H)?
Best regards, Lars
Hey Lars,
I thought a bit about it. Here comes the outcome in the same sequence you asked.
1. Good question. I am not sure if I oversee something but I am tempted to agree with you. First of all, I would assume the statement to be true for all β∈R. Second, I have not checked completely but for the time being I don't see a good reason why one would be interested in Hβ+|α|ϱ as a core. One could also use C∞c. Which brings me to your thought. I believe that works. I was thinking about the first lines in the proof of Lemma 4.2 (d). But I don't see that we need β>0 to show that (im+ϱ)β is unitary. So, I believe you can define the desired extension in the above manner. Afterwards, one probably wanted to show as well that the so defined operator, restricted to C∞c, is really just the derivative.
2. Fair point but I believe this was done in order to help the reader. One should not think about the fact why the desired statement L∗ϱ(im+ϱ)−βφ∈Hβ+|α|ϱ holds but about the easier statement L∗ϱ(im+ϱ)−βφ∈⋂γ∈RHγϱ from which one can directly see what we actually want. As for the intersection, I think one can give a very intuitive meaning to the objects in H−αϱ for α>0 and I am wondering why this was not done because in my opinion one can also understand the support later on better. However, the point here, at least for me, is that one just wants the elements in all Hγϱ, γ∈R to be functionals on one common function space. I believe H∞ϱ would suffice here but, of course, smaller choices such as D are allowed. On the other hand, formally intersecting sets of completely unrelated objects should not cause to many issues either, should it?
3. Öhm, Lemma 4.2 c) right? β+|α|>α which is why Hβ+|α|ϱ is dense in Hαϱ. Works?
Best, Jan
Hi Jan,
thanks a lot for your response! 2. and 3. are fine then, especially 3.
. Here are some more thoughts on 1.
As far as I understood the idea of the proof, the reason why we assume β>0 is the second line of the proof where one proves that ∂β0,ϱ is an isometry. Since u should be in the domain of ∂β0,ϱ, we need β+|α|≥0 to apply part (a) of the lemma. So in order to get rid of the assumption β>0, we need to take another space where we know that our operator ∂β0,ϱ is defined.
I think taking C∞c(R,H) might cause some trouble, since we need that C∞c(R,H) is dense in Hαϱ(R,H) to extend the operator, but density will be proved in Lemma 4.8 afterwards by using Lemma 4.2 (d): In the proof there we consider ∂α0,ϱf∈L2ϱ(R,H) for a given function f∈Hαϱ(R,H). (By the way, we need Lemma 4.2 (d) there also for β<0, right?)
However, I think taking the space H|β|+|α|ϱ(R,H) instead of Hβ+|α|ϱ(R,H) should solve this problem so that we obtain the statement of the lemma also for β<0.
Concerning my proposal to define the extension ∂β0,ϱ:Hαϱ(R,H)→Hα−βϱ(R,H) via the concatenation of unitary maps, the point why the paper uses a different argument might be (as you mentioned) that one wants of course the extension ∂β0,ϱ:Hαϱ(R,H)→Hα−βϱ(R,H) to coincide with the derivative ∂β0,ϱ:dom(∂β0,ϱ)⊆L2ϱ(R,H)→L2ϱ(R,H) on at least C∞c(R,H). But I think this can then be seen via the diagram H|β|+|α|ϱ=W|β|+|α|ϱLϱ→H|β|+|α|∩L2(im+ρ)β→H|β|+|α|−β∩L2⊆Hα−β∩L2L∗ϱ→Wα−βϱ⊆Hα−βϱ since we don't have to take the extension of the Fourier-Laplace transform there. So we know that ∂β0,ϱ:Hαϱ(R,H)→Hα−βϱ(R,H) is an extension of ∂β0,ϱ:H|β|+|α|ϱ(R,H)→Hα−βϱ(R,H).
Best, Lars
Hey Lars,
ah, the density result comes after. Sorry, I did not pay enough attention. Well, one can prove that independently if I am not mistaken. As an exercise (I really hope I am not wrong here) try to show the density of C∞c in Hkϱ, k∈N0 as follows. Let f∈Hkϱ be given. Consider g:=e−tϱf (sorry but I needed to get rid of this m. When you see t, I am using the function as multiplication operator). Show g∈Hk (ordinary Sobolev space). Approximate g by a sequence (φn) in C∞c. Set ψn:=etρφn which is still in C∞c. Show ψn→f in Hkϱ. Together with the density result from Lemma 4.2 c), you can interpolate and get the density in all Hαϱ. Again, if I am not mistaken, you will need the following formula: ∀m∈N,n∈N0,m≥n:m∑l=n(ml)(ln)(−1)n−l=δmn. I tried this yesterday, and one can give a proof for this by induction. Have a try if you wish, and let me know if I talked shit.
As for your point that you need Lemma 4.2 (d) also in case β<0, I believe this is not true. If I understand you thought correctly, you would like to have this Lemma also for β<0 to get the mapping property Dα:Hαϱ→L2ϱ (sorry, I also cannot stand ∂0,ϱ), right?. But this follows since Dβ, β>0 is unitary with inverse D−β providing you also with the right mapping properties for negative values.
The last argument seems fine to me. I would just write H|α|+|β| instead of H|α|+|β|∩L2 since we know H|α|+|β|⊆L2. Same comment goes also for the point after applying the multiplication operator. Or do I not understand something here?
Anyway, nice thoughts and good job. Keep going.
Best, Jan
Hi Jan,
probably I overlook something, but where shall the formula come into play?
Concerning the other two points you mentioned, I completely agree with you (taking the inverse of ∂α0,ϱ, that's insane…
). Yes, the intersection with L2 is maybe a little bit overcautious.
Thanks and regards, Lars
Hey Lars,
I used it when I wanted to show that ψ(m)ntof(m) in L2ϱ but maybe you did that differently, and actually more intelligent, then me. Well, did everything work out? If you want, you can send me your solution so I can learn something as well.
Best, Jan
Hi Jan,
I hope I don't miss your point. My argument then would have been the following:
If φn→g in Hk (classical Sobolev space), then we have ξℓˆφn(ξ)→ξℓˆg(ξ) in L2 for any ℓ≤k. In particular we have (im+ρ)k(ˆφn−ˆg)→0 in L2. Then ‖ψn−f‖Hkϱ=‖Lϱ(ψn−f)‖Hk(im+ϱ)=‖Fe−mϱ(ψn−f)‖Hk(im+ϱ)=‖ˆφn−ˆg‖Hk(im+ϱ)=‖(im+ϱ)k(ˆφn−ˆg)‖L2→0.
For approximating g in Hk (classical Sobolev space), i would choose ηn:=g∗θn where θn(x)=nθ(nx) is a Dirac sequence and θ∈C∞c, say. Then ˆηn=ˆg⋅ˆθ(⋅n). Since ˆθ is bounded and ˆθ(⋅n)→1 pointwise, we obtain ξℓˆηn(ξ)→ξℓˆg(ξ) in L2 for any ℓ≤k. This implies ηn→g in Hk. Altogether, we know that the Schwartz space S is dense in Hk. In order to obtain density for C∞c, I would approximate a given Schwartz function g∈S via gn:=γng where γn is a sequence of bump functions with support in [−n−1,n+1] and γn|[−n,n]=1 and bounded derivatives up to order k. Via Leibniz rule, we obtain gn→g in Hk.
Best regards, Lars
EDIT: I'm afraid that my claim that ηn lies in the Schwartz space is wrong, but if we approximate ηn via multiplication with cut off functions, the argument should still work I think.
Hey Lars,
Nope, you don't miss anything but indeed acted much smarter than I did. I established the convergence result by using two times the product rule which led to the need of the above formula. Your approach is much better. As for the classical approximation result, this I would take for granted. The functions ηn are indeed no Schwartz functions, that's right. They are in H∞, though. And the above proof sketches correctly how one would do this. Well done. This question is completely clarified, is it not?
Best, Jan
Hi Jan,
that's great, then my question is indeed completely clarified.
Thanks a lot!
Best, Lars
Dear all,
if you post something in this forum (it can also be a reply to a previous post) and tick the box “Subscribe to comments” then it seems to be a subscription for the whole discussion board of our project and not merely to replies to that specific post, so it seems to be a good plan that everyone of us uses this feature to not miss the discussions here anymore on the future :)
Best, Sebastian
Good idea. Thanks for the information.
Best, Jan
Hi Everyone,
Hope you are all keeping safe.
I am about 40% through the paper posted by Jan. although I understand parts of it , I am finding other sections a bit technical. Is anyone open to having a Zoom discussion on their understanding of the paper maybe on the 14th April (I think I'll be done reading then)? - are we allowed?
Looking forward to hearing from you, Sam - University of Johannesburg , South Africa.
Hey there,
Sorry, I just read what you wrote. I like the idea and would be up for it. And of course, this is allowed and highly appreciated. Not sure if the others read this here though. Could you write an email to everybody? In this mail, you could also point out that we should regularly check the discussion board. At this point shame on me as well because I didn't check either. Please also suggest a time for the meeting. I have a licensed zoom version and could take care of everything. In particular, we don't have to care for the time limit (which is not a big problem anyway). Thanks for taking care of everything! See you tomorrow.
Best, Jan
I also missed that post and promise to observe the forum in the future more carefully. I agree with the opinions from the mail that we should stick to our original plan and postpone the general discussion till April 20th. If you have specific questions before that just post them here and we'll have a look!
Hi Sebastian, hi Jan,
Sorry for spamming, but if I am not mistaken your name, Jan, is mi's'pelled on the page where the projects are listed ;)
Best, Felix
You better be sorry! But thanks for pointing out. You are right.
Best, Jan
Hey guys (and everybody else reading this),
This is our nice discussion board. Have fun using it.
Best, Sebastian & Jan