In the lecture 3, one of the main objectives is to give meaning to time-derivation of functions in L_{2,\nu}(\mathbb{R}; H). In fact, the idea, I think, came from distribution approach, already made for function in L^2(\mathbb{R}^n, \mathbb{R}).
My question is as follow. Can we speak about “space”-derivative of H-valued functions? That is the derivation of a function f \in L^2(\mathbb{R}^n, H), or the scalar character of \mathbb{R}, in L_{2,\nu}(\mathbb{R}; H), plays a crucial role.
Omar.
Hendrik Vogt, 2019/11/19 08:23
Dear all,
Jürgen already posted a very nice idea concerning the proof of the denseness of M:={φ′;φ∈C∞c(R)} in L2,ν(R), which is important in the proof of Proposition 3.2.4. I'm a bit late to the party, but I'd like to add my thoughts on this. Moreover, I'll indicate how one can use the vector-valued analogue of this denseness property to avoid Lemma 3.2.5 (and it's somewhat technical proof) and prove Corollary 3.2.6 in a rather different way.
The basic idea is the well-known fact from functional analysis that a linear functional on a normed space is bounded if and only if its kernel is closed. Thus, if one has an unbounded functional, then its kernel (being 1-codimensional) is dense. Now observe that M={φ∈C∞c(R);∫Rφ=0}. Let the space C∞c(R) be provided with the L2,ν-norm. Then one easily sees that the functional C∞c(R)∋φ↦∫Rφ∈K is unbounded, so its kernel M is dense in C∞c(R) and hence in L2,ν(R) as well, for every ν∈R. This proof may not be simpler than Jürgen's, but it offers a different viewpoint, and it works for ν=0 as well (which is not needed in the present context).
Finally, in the very same way one shows that {φ′;φ∈C∞c(R;H)} is dense in L2,ν(R;H). This can be used to prove Corollary 3.2.6! The idea is as follows: we know that ∂t,ν is surjective and that ∂t,ν⊆−∂∗t,ν+2ν. Thus, we obtain equality if we can show that −∂∗t,ν+2ν is injective (because a surjective mapping can't have a proper injective extension). Now ker(−∂∗t,ν+2ν)=ran(−∂t,ν+2ν)⊥, so we are done if ∂t,ν−2ν is surjective. The latter is true since ∂t,−ν is surjective and the operators ∂t,ν−2ν and ∂t,−ν are unitarily equivalent, by Corollary 3.2.7.
Best wishes, Hendrik
Sascha Trostorff, 2019/11/19 09:02
Dear Hendrik,
thanks a lot for your nice proof of Corollary 3.2.6. I really like the idea and I am happy that we can avoid the formula for the adjoint of Iν.
Best regards
Sascha
Hendrik Vogt, 2019/11/19 12:24
Dear Sascha,
thanks for your answer – glad that you like it
Best wishes, Hendrik
Jürgen Voigt, 2019/11/19 11:40
Dear Hendrik,
your proof of the denseness of M by arguing that it is the kernel of an unbounded linear functional doesn't convince me. After all, how do you “easily see” the unboundedness of the integral functional η? You have to find a sequence (fn) of functions tending to 0, such that η(fn) does not tend to 0. For ν≠0, the easiest way is to start with a function f with η(f)≠0 and take as (fn) a sequence of translates of f to the appropriate side. But then you can also take the sequence (f−fn), which lies in M and tends to f, and this shows that M is dense. And how do you show the denseness for ν=0. In this case you use fn:=1nf(⋅n), and again (f−fn) does the job showing denseness of M.
Best wishes, Jürgen
Hendrik Vogt, 2019/11/19 12:23
Dear Jürgen,
well, what do you think was the reason I wrote “This proof may not be simpler than Jürgen's” ? It's because I glossed over the “easily sees” part. I still find this different viewpoint useful since, at least for me, it better explains why the set M is dense. Ultimately our arguments are not that different because the explicit sequences you write down show that the functional is unbounded. (You might say I add an additional (unnecessary) layer to the argument, and I wouldn't disagree, but my brain will insist that it finds that layer useful.)
Best wishes, Hendrik
Jürgen Voigt, 2019/11/19 13:25
Dear Hendrik,
sorry, but I simply have to reply. Sure, the extra observation of the unboundedness of η is interesting and may be helpful for understanding. But I refuse to do the double summersault of first showing the unboundedness of η by an argument showing the denseness of M, and then using the unboundedness of η to show the denseness of M. (For me, this is not a question which of the proofs is “simpler”!)
Best wishes, Jürgen
Hendrik Vogt, 2019/11/15 14:24, 2019/11/15 19:06
Dear all,
I have a few comments on the integral of simple functions. Somehow one is temped to say that for simple functions all the rules of integration are “clear”. When one tries to write it up, there are several options. One way is to say that a simple function is a function f:Ω→X of the form f=∑A∈FxA1A with a finite subset F⊆Σ, xA∈X and μ(A)<∞ for all A∈F. Then one defines ∫fdμ:=∑A∈FxAμ(A); now the main task is to show that the integral is well-defined. This is somewhat tedious, in particular concerning notation, but later on it simplifies the proofs; e.g., linearity becomes clear.
The virtual lectures chose the “opposite” option, in a sense: they take the canonical representation f=∑x∈f[Ω]x1Af,x of a simple function, and thus they don't have to worry about the integral being well-defined. The “price” they have to pay is that later on the proofs become more technical. To be honest, I was not too keen to follow the computation in the proof of linearity (Proposition 3.1.4). (RETRACTED, see below: Moreover, in our seminar session we saw that in the displayed formula at the end of the proof of Proposition 3.1.5(a), the first equality is not “clear”; one would need to argue with the sets AB∘fn,y as far as we saw.)
I would suggest the following: already in Remark 3.1.1 it is worth noting that simple function are all the functions f:Ω→X of the form f=∑A∈FxA1A with a finite partitionF⊆Σ of Ω, xA∈X for all A∈F and xA=0 if μ(A)=∞. (This is easy to see.) Then, if f=∑A∈FxA1A and g=∑B∈GyB1B are simple functions with the representations having the indicated properties, one obtains f+g=∑A∈F,B∈G(xA+yB)1A∩B, which is a representation of f+g with the finite partition {A∩B;A∈F,B∈G} of Ω, and xA+yB=0 if μ(A∩B)=∞. With these considerations one easily confirms Remark 3.1.1.
Now for the Bochner integral I would proceed as follows. First of all, if f is a simple function, then F:={Af,x;x∈f[Ω]} is a finite partition of Ω, and f=∑A∈FxA1A if one sets xA:=f(ωA) with some ωA∈A. If f=∑B∈GyB1B is another representation of f, with a finite partition G and yB∈X, then by definition
∫fdμ=∑A∈FxAμ(A)=∑A∈F∑B∈GxAμ(A∩B)=∑A∈F∑B∈GyBμ(A∩B)=∑B∈GyBμ(B)
since xA=yB if A∩B≠∅. Thus one can compute the integral of f with any such representation of f via a finite partition.
Now linearity of the integral is easy to show: if f and g are simple functions, one just has to observe that there are representations of f and g with one common finite partition (cf. the 3rd paragraph of my deliberations). Likewise, from this viewpoint the displayed formula at the end of the proof of Proposition 3.1.5(a) becomes clear.
Best wishes, Hendrik
Sascha Trostorff, 2019/11/15 15:56
Dear Hendrik,
thanks for your remark. However, I do not understand, why there should be a problem in the first equality in Proposition 3.1.5 (a). One should not use the definition of the integral, but its linearity. Since B∘f=B(∑x1Af,x)=∑Bx1Af,x you obtain the formula by the (already shown) linearity of the integral and the definition applied to the simple function Bx1Af,x.
Best regards
Sascha
P.S. The idea to define simple functions in that way and to avoid a proof for the well-definedness of the integral is taken from one of your lecture notes
Hendrik Vogt, 2019/11/15 19:14
Dear Sascha,
I did realise that the section on the Bochner integral is adapted from my PDE lecture from 15 years ago If you have a look, then you'll see that I indicated (a version of) my above suggestion already back then!
You're right, there is no actual problem in the proof of Proposition 3.1.5(a). Thanks a lot for pointing this our! It is just not so easy to see what argument is used in that first equality in the displayed formula, in particular since the identity f=∑x1Af,x is not presented as far as I see.
Best wishes, Hendrik
Sascha Trostorff, 2019/11/15 20:42
Dear Hendrik,
yes, indeed we should work out this argument in the final version.
Best regards
Sascha
Jürgen Voigt, 2019/11/16 21:03, 2019/11/16 21:07
Dear Hendrik, dear all,
this is an addition to Hendrik's post from 2019/11/15 19:06. To my taste, the canonical way to introduce simple functions and their integral is the one you mention in your first paragraph. Here is the proof that the integral is well-defined.
It clearly is sufficient to show: if f=∑A∈FxA1A=0, then ∑A∈FxAμ(A)=0. There exists a finite collection G⊆Σ of pairwise disjoint non-empty sets such that each set A is the union of those sets in G which are subsets of A. This can be expressed by saying that there exist αAB∈{0,1}, for (A,B)∈F×G, such that
1A=∑B∈GαAB1B(A∈F).
Then 0=∑AxA1A=∑A,BxAαAB1B=∑B(∑AxAαAB)1B implies
∑AxAαAB=0(B∈G);
hence ∑A∈FxAμ(A)=∑AxA∑BαABμ(B)=∑B(∑AxAαAB)μ(B)=0.
Is this “tedious, in particular concerning notation”?
Best wishes, Jürgen
Hendrik Vogt, 2019/11/16 23:00
Dear Jürgen,
well, what I meant is that it is a bit tedious to explicitly write down G.
Best wishes, Hendrik
Jürgen Voigt, 2019/11/17 00:34, 2019/11/17 11:18
Dear Hendrik,
this is not really true (and somehow below the level of our present discussion). In fact, you know how it can be done: For ∅≠A⊆F one defines
AA:=(⋂A)∖⋃(F∖A);
then AA consists of those points belonging to all sets A∈A that do not belong to any of the sets in F∖A. This shows that the family (AA)∅≠A⊆F is pairwise disjoint, and ⋃∅≠A⊆FAA=⋃F. Discarting the empty sets one obtains the desired G.
Incidentally, the collection {A∩B;A∈F,B∈G} in one of your previous posts will not necessarily be a partition, because the members of a partition should be non-empty.
Best wishes, Jürgen
Hendrik Vogt, 2019/11/17 13:21
Dear Jürgen,
well, now you wrote down that piece of notation that I called “a bit tedious” Of course I knew how it could be done, but you found a concise form that I hadn't thought of before (or that I managed to forget) – nice! (I wouldn't call this “below the level of our present discussion” because my starting point was “When one tries to write it up”, i.e., really write up all the stuff about integration of simple functions that ought to be “clear”.)
Thanks a lot for pointing out that that collection I used is not necessarily a partition! So I should say “collection of pairwise disjoint sets” instead of “partition”.
thank you very much for all the detailed proofs in this lecture. It really helped me to understand the topic. I only noticed some small problems while reading it:
(1) I think that you switched the arguments of the inner product here. I guess that we still want the second argument in the inner product to be linear. However, if you look at the proofs it seems that you switched to the first argument being linear. For example, in the proof of Lemma 3.2.5, you should better write
⟨g,Iνf⟩L2,ν(R;H)=…
Of course, there, it is not really a mistake. However, in the proof of Proposition 3.2.3, we want the inner product acting as a linear functional rather than an anti-linear functional.
(2) Related to this: I also think that in the second part of the proof of Lemma 3.2.5, we need to calculate
⟨g,2νIνI∗νf⟩L2,ν(R;H)=…
to avoid using Fubini-Tonelli for the Bochner integral. Of course, it is the same calculation but one can use the “normal” Fubini-Tonelli theorem then.
Marcus Waurick, 2019/11/12 08:39
Dear Julian,
thank you for your comments.
(1) Your suggestion is closer to the definition of the adjoint as highlighted in Lecture 2. In any case, ⟨Ax,y⟩=⟨x,A∗y⟩ for a bounded linear operator A and all x,y∈H is (evidently) equivalent to ⟨y,Ax⟩=⟨A∗y,x⟩ for all x,y∈H. So, I reckon your remark is rather on didactics.
(2) Thank you for your suggestion. In particular, as we have not said anything on Fubini–Tonelli for the Bochner integral.
thank you for answering. With respect to (1), I still have to say that in the proof of Proposition 3.2.3, it is not just a didactic argument because we want the inner product acting as a linear functional rather than an anti-linear functional.
Best wishes,
Julian
Hendrik Vogt, 2019/11/18 21:13
Dear Julian,
a late answer: you have an issue in Proposition 3.2.3 because one has to pull an anti-linear functional argument out of the integral, right? This is actually not a problem at all since the functional is an R-linear operator! Generally it is often helpful to forget that there are complex scalars, i.e., to consider all the spaces involved as R-vector spaces.
Best wishes, Hendrik
Julian Großmann, 2019/11/19 21:24
Hello Hendrik,
thank you very much for your answer. You are completely right! In fact, I didn't have an actual problem with using anti-linear functionals but just thought that using linear ones would be easier/more straight-forward. Especially since we proved Proposition 3.1.5(a) for this.
Thank you again!
Best wishes,
Julian
Janik Wilhelm, 2019/11/10 20:17
Dear virtual lecturers,
I would like to share some comments from our meeting in Darmstadt last week:
Shouldn't we assume (Ω,Σ,μ) to be complete, as we later define Bochner-measurable functions as pointwise almost everwhere limits of simple functions.
You chose to replace the notation S(μ;X) by S(Rd;X) in the situation where μ is absolutely continuous with respect to the Lebesgue measure λ. We wondered, if this is a well-defined notation. In our opinion, the density ρ(x):=1/|x| defines a σ-finite measure μ on L(R) such that S(λ;X)≠S(μ;X). (For instance, we have f∈S(λ;X) but f∉S(μ;X) for f(x):=χ[0,1].
Throughout the lecture, the norm on X is called ‖⋅‖X. Thus we assume there is an X missing in the proof of Proposition 3.1.2 in the definition of gn as well as in the last line of page 27 after the first equality sign.
A question concerning Corollary 3.1.6: It was left to the reader to show that any continuous function on [a,b] is Bochner-measurable. We assume that a way to approximate a continous function by step functions in this case is to use that its image is compact. For any positive ϵ, we can thus cover the image by finitely many ϵ-balls and define a step function given on the preimage of each such ball by some value inside the ball. As we only need pointwise convergence, a similar procedure should also work for a Borel measure space that can be covered by countably many compact sets. However, we wondered which is the precise condition needed for a Borel measure space in general to ensure that all continous functions are measurable. And why.
We noticed that the last part of the proof for Proposition 3.2.4 does not require specific properties of the operator ∂t,v, but only uses the boundedness of ∂−1t,v which implies the equivalence of the graph norm and the norm x→‖Ax‖L2,v(R;H). This furthered our understanding of the proof.
In the proof of Corollary 3.2.6, we would have considered further explanation helpful as to how dom(∂∗t,v)⊂dom(∂t,v) follows from ran(I∗v)⊂dom(Iv), for instance a reference to Lemma 2.2.2 or even a short computation.
Finally, we suppose you should replace f∗ϕ∈C∞c(R) by f∗ϕ∈C∞(R) in Exercise 3.2.
Thank you for reading our comments and we are curious about your answer.
Best regards
Janik
Marcus Waurick, 2019/11/11 08:19
Dear Janik,
thank you very much for your comments from the Darmstadt team.
Can you further explain, why we need completeness of the measure space? I don't see this being necessary for the definition of Bochner measurable. The definition at hand is used in order to integrate reasonably many functions.
Your are right concerning S(μ;X). Throughout the text, we use S(R;X) in the case of the Lebesgue measure. Note that S(R;X) is dense in L2,ν(R;X); for X a Hilbert space and all ν∈R.
Your question concerning measurability is interesting. For this I refer you to the discussion in
today we had the same discussion about completeness as well in Hamburg. The problem is very simple:
If you have a sequence of measurable functions fn:Ω→R that converge pointwisely to f:Ω→R, then f is also measurable. However, this is not true when the pointwise convergence holds only μ-a.e. In this case, you need that the measure is complete to ensure that f is also measurable.
Now, in the lecture 3, we cannot conclude:
f:Ω→X Bochner-measurable ⇒‖f(⋅)‖:Ω→R measurable.
This means that Remark 3.1.1 (b) is wrong. However, the definition of Lp(μ,X) still works because the measurability of ‖f(⋅)‖ is explicitly given there.
Jürgen Voigt, 2019/11/11 23:44, 2019/11/11 23:46
Dear Julian,
let me ask you why you think that the implication
f Bochner-measurable⇒‖f(⋅)‖ measurable
is wrong. This is proved in the text, so you should say where you think there is a mistake. Note that in this case both functions are pointwise a.e. limits of simple functions.
Maybe you misinterpret what means that
fn(ω)→f(ω)μ-a.e. This does not mean that the set of ω where (fn(ω)) does not converge to f(ω) is a μ-null set, but rather that this set is contained in a μ-null set.
So, if there is a μ-null set N such that fn(ω)→f(ω) on the complement of N, then also ‖fn(ω)‖→‖f(ω)‖ on the complement of N.
Best wishes, Jürgen
Jürgen Voigt, 2019/11/12 14:42
Dear all,
sorry, I have to withdraw part of my comments: somehow I did not understand/realise that Bochner-measurability of R-valued functions is not the same as μ-measurability, if μ is not complete.
maybe I misinterpreted what you mean by “μ-measurable”. I was under the impression that it just means Σ−B(R)-measurable in this context.
Best wishes,
Julian
Sascha Trostorff, 2019/11/12 15:11
Dear Julian,
okay, now I got your point. So, the function ‖f(⋅)‖ is not measurable in the classical sense, but equals a measurable function outside a μ-null set, which then doesn't affect the integral. So, even if we do not require the measurablity of ‖f(⋅)‖ in the definition of Lp, everything makes sense, if we pass to a measurable representative or do I oversee something?
Best regards
Sascha
Julian Großmann, 2019/11/12 16:19
Hello Sascha,
yes, exactly. I also think that there is no problem at all in the lecture when using a non-complete measure. The only confusion was just the formulation in Remark 3.1.1 (b).
Best wishes,
Julian
Jürgen Voigt, 2019/11/12 17:56, 2019/11/12 17:57
Dear Julian,
as I already wrote, I mixed up Bochner-measurability as R-valued functions with μ-measurable (i.e., Σ-B(R)-measurable). And your comment concerning Remark 3.1.1(b) is quite justified.
Best wishes, Jürgen
Jürgen Voigt, 2019/11/12 12:54, 2019/11/12 12:56
Dear Marcus,
I do not see how your reference to wikipedia solves the problem that is brought up by Janik and Julian. As I understand, the problem is the following.
Accepting how measurable functions f:Ω→X are defined: if (fn) is a sequence of measurable functions converging to f a.e., show that then f is measurable, i.e. an a.e.-limit of simple functions. Actually, it seems to me that you neglect this problem generously and uncommented in your proof of the completeness of Lp(μ;X), Proposition 3.1.2. Also, I do not see how completeness of the measure space could help to prove this.
I think I know a proof, with the help of Egorov's theorem, and I will post it if you want. But here I just want to explain what is the problem.
Let B0 be a set of functions f:Ω→X, then form the set B1 consisting of pointwise limits of sequences in B0, and repeat the procedure with sequences in B1, to obtain B2. (Starting with continuous functions in certain situations, the sets B1,B2,… are known as Baire classes, and it is well-known – and not difficult to proof, e.g., for B0:=C[0,1] – that, in general B2 is strictly larger than B1. (And I fear that you run into the same problem starting with B0:=S(Ω;X)!)
Best wishes, Jürgen
Marcus Waurick, 2019/11/12 15:13
Dear Jürgen,
thank your for your remarks. My reference to wikipedia was mentioned in order to clarify what conditions are needed to render continuous functions measurable. Thus, in the reference to Corollary 3.1.6. I did not intend to refer to the wikipedia in order to clarify that measurable functions are point wise a.e. limits of step functions (or some statement of this sort). I have seen that my previous post wasn't very clear in that.
In any case, I apologise if I happened to misunderstand Janik's comments. Please clarify, if need be.
Now, concerning measurability in general (not with regards to continuous functions only): The difference of Baire classes from the notion of measurable functions is that in our case, we ask for the limits to be only almost everywhere rather then everywhere. I think you are right, Jürgen, when starting off with B0=S(Ω;X), that B1, the set of pointwise everywhere limits is strictly smaller than the set of measurable functions.
In the construction of the Bochner spaces in the lecture notes, the important bit is the space Lp(μ;X) (and its scalar-valued analogue).
In particular, in our proof of Lp(μ;X) being complete, we use completeness of Lp(μ). So, if I understand your post correctly, what one needs to show is that every function in Lp(μ) is a.e. pointwise limit of simple functions.
I try to show this with using the following statements:
(1) Lp(μ) is complete and for every convergent sequence (fn)n in Lp(μ) there exists a subsequence, which is majorised and converges pointwise almost everywhere (Theorem of Fischer–Riesz).
(2) The set of simple functions S(μ) is dense in Lp(μ).
Now, let f∈Lp(μ). Then by (2), there exists a sequence (fn)n of simple functions converging to f in Lp(μ). By (1), we can assume that (fn)n converges almost everywhere to f.
So, I think in the end, the difference between the notions of measurability (Bochner measurable vs μ-measurable) is somehow `swallowed' by the equivalence relation of being equal almost everywhere. Please let me know, whether I made a mistake here.
Best regards,
MM
Marcus Waurick, 2019/11/12 16:15
Dear all,
Sascha and I just had a discussion on this. Sascha will write a post that is going to resolve the problem in due course.
Best regards,
MM
Hendrik Vogt, 2019/11/12 23:48, 2019/11/13 11:20
Dear all,
Jürgen made the following important point: in the proof of Proposition 3.1.2 it is used that the a.e. limit of a sequence of Bochner measurable functions is again Bochner measurable. Here's how I would prove this fact:
First of all, if f:Ω→X is Bochner measurable, then it follows from the definition that the set [f≠0] is σ-finite. Likewise, if f is the a.e. limit of a sequence of Bochner measurable functions, then [f≠0] is σ-finite.
Now let fn:Ω→X be Bochner measurable, for n∈N, and fn→f a.e. Let [f≠0]=⋃n∈NAn with measurable sets An of finite measure, A1⊆A2⊆…. Then one easily checks that gn:=1An∩[‖fn(⋅)‖≤n]fn→f a.e. Now comes the main idea: for each n∈N, gn is in L1(μ;X), and thus there exists a simple function hn such that ‖gn−hn‖1≤2−n, by Lemma 3.1.3. Then
∫Ω∞∑n=1‖gn(ω)−hn(ω)‖Xdμ(ω)≤1,
which implies that ∑∞n=1‖gn(ω)−hn(ω)‖X<∞ for a.e. ω∈Ω, and it follows that gn−hn→0 a.e. Thus hn→f a.e., and we conclude that f is Bochner measurable.
thank you very much for this nice proof. In a discussion with Moppi we only had the idea to prove Pettis' Theorem, which then would imply the statement as well. However, your proof is much nicer and more elementary.
This whole discussion shows once again, that measurability (in which sense ever) is a very delicate matter and we apologize for not taking enough care of that subject.
Best regards
Sascha
Hendrik Vogt, 2019/11/16 15:21
Dear Moppi,
you write that use S(R;X) in the case of the Lebesgue measure and that S(R;X) is dense in L2,ν(R;X). Do you mean that S(R;X) is supposed to be read as the space of simple functions that is built w.r.t. Lebesgue measure? Then you have a problem: S(R;X) is not a subspace of L2,ν(R;X) unless ν=0! In fact, it is easy to construct a set A of finite Lebesgue measure such that 1A∉L2,ν(R;X), but of course 1A∈S(R;X). I think the correct solution would be to work with S(A;X) with a ring A⊆Σ that generates Σ; then for Ω=R you can choose as A the collection of all relatively compact measurable subsets of R.
Florian Pannasch suggested a far easier proof of Lemma 3.1.7 in our discussion in Kiel, even for p∈[1,∞) and X an arbitrary Banach space:
Since we already know that S(μ;X) is dense in Lp(μ;X) by Lemma 3.1.3 it suffices to approximate elements of the form 1Ax for some A∈Σ and x∈X. The latter is trivial, since we find a sequence (φn)n in the linear hull of D with φn→1A in Lp(μ). Then
‖1Ax−φnx‖p=‖x‖X‖1A−φn‖p→0
and thus the claim follows.
Best regards
Sascha
Jürgen Voigt, 2019/11/05 13:27
Dear virtual lecturers,
Lemma 3.2.1: Maybe my question is stupid, but shouldn't one say something why the resulting function k∗f is measurable? Even in the expression on line 3 of the proof of the lemma I start asking myself, why the function resulting from the inner integral is measurable.
Proposition 3.2.4: I have a suggestion for an alternative proof. If ν>0, I suggest to define
φn(t):=∫t−∞(f(s)−f(s−n))ds.
(And with f(s+n), if ν<0.)
Two comments on terminology (admittedly, both a matter of taste):
“one-to-one” in Prop. 3.2.3: I prefer the injective-surjective-bijective
terminology. One reason is, how – according to wikipedia – one would have
to express a bijective mapping: a “one-to-one correspondence”! (OK, you can say “one-to-one and onto”.)
“arbitrarily differentiable” (in Prop.3.2.4) is – in a way – better than “infinitely differentiable”, but still is not meaningful by itself, but rather the abbreviation (and thereby the definition) of “arbitrarily often differentiable”. Clearly “infinitely often differentiable” – used by many authors – is simply nonsense; but I have decided – for myself – that “infinitely differentiable” is defined as “arbitrarily often differentiable”.
And a misprint on p.30, line -3: The index at μ should be p,ν, in this order.
Best wishes, Jürgen
Sascha Trostorff, 2019/11/05 14:15
Dear Jürgen,
thank you very much for your comments.
Your proof of Proposition 3.2.4 - at least for me - is nicer than ours. Thanks a lot!
Concerning the question of measurability: You are completely right, we have just overseen this. One way to show it would involve Pettis' Theorem and Fubini to obtain the measurability of the convolution tested with a Hilbert space element. But since we do not want to employ Pettis, I am curious if you or anybody else have a more elementary proof for this fact.
Best regards
Sascha
Jürgen Voigt, 2019/11/05 17:47
Dear Sascha,
I think I might have an answer to the measurability question (without invoking Pettis' theorem). If f is of the form φx, then the measurability of k∗f follows from the scalar case. Then the same holds for the linear hull of these functions, and for functions of the linear hull one can show the inequality as in the lecture notes. Then (cautious!) approximation.
Best wishes, Jürgen
P.s. Writing posts with the offered platform is perfect. Thanks, Christian!
Jürgen Voigt, 2019/11/06 18:18
Dear Sascha and dear all,
concerning the measurability: First concerning the question why the function resulting from the inner integral is measurable. This follows from the scalar case, because it is just the convolution of |k| and ‖f(⋅)‖, and the estimate of the proof shows that there exists a null set N⊆R such that k(⋅)‖f(t−⋅)‖ is integrable for all t∈R∖N.
Now comes the “cautious” approximation. The main observation is that the density lemma, Lemma 3.1.3 deserves to be reinforced by adding “and for each f∈Lp(μ;X) there exists a sequence (fn) in S(μ;X) such that fn→f a.e. and ‖fn(⋅)‖≤2‖f(⋅)‖ for all n∈N. (This is what in fact is proved!)
Let f∈L2,ν(R;H) and (fn) be as above. As mentioned above, there exists a null set N⊆R such that k(⋅)‖f(t−⋅)‖ is integrable for all t∈R∖N. Then
|k(⋅)|‖fn(t−⋅)‖≤2|k(⋅)|‖f(t−⋅)‖(t∈R∖N,n∈N),
but also k(⋅)fn(t−⋅)→k(⋅)f(t−⋅) a.e.; hence the dominated convergence theorem implies that
∫k(s)fn(t−s)ds→∫k(s)f(t−s)ds(n→∞,t∈R∖N).
This shows the measurability of k∗f. (Then one can go ahead with the estimate as in the lecture notes. The measurability of the functions k∗fn follows from the scalar case, but there is no need to carry out the long estimate first for the simple functions, as I had insinuated in my previous post.)
A side remark to the second paragraph above: It would not really be necessary to add the sentence to the assertion, because of the following general fact. If (fn) is a Cauchy sequence in Lp(μ;X), then there exist a function g∈Lp(μ)+ and a subsequence (fnj) converging a.e. and in Lp(μ;X), and such that ‖fnj(⋅)‖≤g(⋅) a.e. for all j∈N.
Best wishes, Jürgen
discussion/lecture_03.txt · Last modified: 2019/10/21 16:59 by matcs
Discussion on Lecture 03
Dear all.
In the lecture 3, one of the main objectives is to give meaning to time-derivation of functions in L_{2,\nu}(\mathbb{R}; H). In fact, the idea, I think, came from distribution approach, already made for function in L^2(\mathbb{R}^n, \mathbb{R}). My question is as follow. Can we speak about “space”-derivative of H-valued functions? That is the derivation of a function f \in L^2(\mathbb{R}^n, H), or the scalar character of \mathbb{R}, in L_{2,\nu}(\mathbb{R}; H), plays a crucial role.
Omar.
Dear all,
Jürgen already posted a very nice idea concerning the proof of the denseness of M:={φ′;φ∈C∞c(R)} in L2,ν(R), which is important in the proof of Proposition 3.2.4. I'm a bit late to the party, but I'd like to add my thoughts on this. Moreover, I'll indicate how one can use the vector-valued analogue of this denseness property to avoid Lemma 3.2.5 (and it's somewhat technical proof) and prove Corollary 3.2.6 in a rather different way.
The basic idea is the well-known fact from functional analysis that a linear functional on a normed space is bounded if and only if its kernel is closed. Thus, if one has an unbounded functional, then its kernel (being 1-codimensional) is dense. Now observe that M={φ∈C∞c(R);∫Rφ=0}. Let the space C∞c(R) be provided with the L2,ν-norm. Then one easily sees that the functional C∞c(R)∋φ↦∫Rφ∈K is unbounded, so its kernel M is dense in C∞c(R) and hence in L2,ν(R) as well, for every ν∈R. This proof may not be simpler than Jürgen's, but it offers a different viewpoint, and it works for ν=0 as well (which is not needed in the present context).
Finally, in the very same way one shows that {φ′;φ∈C∞c(R;H)} is dense in L2,ν(R;H). This can be used to prove Corollary 3.2.6! The idea is as follows: we know that ∂t,ν is surjective and that ∂t,ν⊆−∂∗t,ν+2ν. Thus, we obtain equality if we can show that −∂∗t,ν+2ν is injective (because a surjective mapping can't have a proper injective extension). Now ker(−∂∗t,ν+2ν)=ran(−∂t,ν+2ν)⊥, so we are done if ∂t,ν−2ν is surjective. The latter is true since ∂t,−ν is surjective and the operators ∂t,ν−2ν and ∂t,−ν are unitarily equivalent, by Corollary 3.2.7.
Best wishes, Hendrik
Dear Hendrik,
thanks a lot for your nice proof of Corollary 3.2.6. I really like the idea and I am happy that we can avoid the formula for the adjoint of Iν.
Best regards
Sascha
Dear Sascha,
thanks for your answer – glad that you like it
Best wishes, Hendrik
Dear Hendrik,
your proof of the denseness of M by arguing that it is the kernel of an unbounded linear functional doesn't convince me. After all, how do you “easily see” the unboundedness of the integral functional η? You have to find a sequence (fn) of functions tending to 0, such that η(fn) does not tend to 0. For ν≠0, the easiest way is to start with a function f with η(f)≠0 and take as (fn) a sequence of translates of f to the appropriate side. But then you can also take the sequence (f−fn), which lies in M and tends to f, and this shows that M is dense. And how do you show the denseness for ν=0. In this case you use fn:=1nf(⋅n), and again (f−fn) does the job showing denseness of M.
Best wishes, Jürgen
Dear Jürgen,
well, what do you think was the reason I wrote “This proof may not be simpler than Jürgen's”
? It's because I glossed over the “easily sees” part. I still find this different viewpoint useful since, at least for me, it better explains why the set M is dense. Ultimately our arguments are not that different because the explicit sequences you write down show that the functional is unbounded. (You might say I add an additional (unnecessary) layer to the argument, and I wouldn't disagree, but my brain will insist that it finds that layer useful.)
Best wishes, Hendrik
Dear Hendrik,
sorry, but I simply have to reply. Sure, the extra observation of the unboundedness of η is interesting and may be helpful for understanding. But I refuse to do the double summersault of first showing the unboundedness of η by an argument showing the denseness of M, and then using the unboundedness of η to show the denseness of M. (For me, this is not a question which of the proofs is “simpler”!)
Best wishes, Jürgen
Dear all,
I have a few comments on the integral of simple functions. Somehow one is temped to say that for simple functions all the rules of integration are “clear”. When one tries to write it up, there are several options. One way is to say that a simple function is a function f:Ω→X of the form f=∑A∈FxA1A with a finite subset F⊆Σ, xA∈X and μ(A)<∞ for all A∈F. Then one defines ∫fdμ:=∑A∈FxAμ(A); now the main task is to show that the integral is well-defined. This is somewhat tedious, in particular concerning notation, but later on it simplifies the proofs; e.g., linearity becomes clear.
The virtual lectures chose the “opposite” option, in a sense: they take the canonical representation f=∑x∈f[Ω]x1Af,x of a simple function, and thus they don't have to worry about the integral being well-defined. The “price” they have to pay is that later on the proofs become more technical. To be honest, I was not too keen to follow the computation in the proof of linearity (Proposition 3.1.4). (RETRACTED, see below: Moreover, in our seminar session we saw that in the displayed formula at the end of the proof of Proposition 3.1.5(a), the first equality is not “clear”; one would need to argue with the sets AB∘fn,y as far as we saw.)
I would suggest the following: already in Remark 3.1.1 it is worth noting that simple function are all the functions f:Ω→X of the form f=∑A∈FxA1A with a finite partition F⊆Σ of Ω, xA∈X for all A∈F and xA=0 if μ(A)=∞. (This is easy to see.) Then, if f=∑A∈FxA1A and g=∑B∈GyB1B are simple functions with the representations having the indicated properties, one obtains f+g=∑A∈F,B∈G(xA+yB)1A∩B, which is a representation of f+g with the finite partition {A∩B;A∈F, B∈G} of Ω, and xA+yB=0 if μ(A∩B)=∞. With these considerations one easily confirms Remark 3.1.1.
Now for the Bochner integral I would proceed as follows. First of all, if f is a simple function, then F:={Af,x;x∈f[Ω]} is a finite partition of Ω, and f=∑A∈FxA1A if one sets xA:=f(ωA) with some ωA∈A. If f=∑B∈GyB1B is another representation of f, with a finite partition G and yB∈X, then by definition ∫fdμ=∑A∈FxAμ(A)=∑A∈F∑B∈GxAμ(A∩B)=∑A∈F∑B∈GyBμ(A∩B)=∑B∈GyBμ(B) since xA=yB if A∩B≠∅. Thus one can compute the integral of f with any such representation of f via a finite partition.
Now linearity of the integral is easy to show: if f and g are simple functions, one just has to observe that there are representations of f and g with one common finite partition (cf. the 3rd paragraph of my deliberations). Likewise, from this viewpoint the displayed formula at the end of the proof of Proposition 3.1.5(a) becomes clear.
Best wishes, Hendrik
Dear Hendrik,
thanks for your remark. However, I do not understand, why there should be a problem in the first equality in Proposition 3.1.5 (a). One should not use the definition of the integral, but its linearity. Since B∘f=B(∑x1Af,x)=∑Bx1Af,x you obtain the formula by the (already shown) linearity of the integral and the definition applied to the simple function Bx1Af,x.
Best regards
Sascha
P.S. The idea to define simple functions in that way and to avoid a proof for the well-definedness of the integral is taken from one of your lecture notes
Dear Sascha,
I did realise that the section on the Bochner integral is adapted from my PDE lecture from 15 years ago
If you have a look, then you'll see that I indicated (a version of) my above suggestion already back then!
You're right, there is no actual problem in the proof of Proposition 3.1.5(a). Thanks a lot for pointing this our! It is just not so easy to see what argument is used in that first equality in the displayed formula, in particular since the identity f=∑x1Af,x is not presented as far as I see.
Best wishes, Hendrik
Dear Hendrik,
yes, indeed we should work out this argument in the final version.
Best regards
Sascha
Dear Hendrik, dear all,
this is an addition to Hendrik's post from 2019/11/15 19:06. To my taste, the canonical way to introduce simple functions and their integral is the one you mention in your first paragraph. Here is the proof that the integral is well-defined.
It clearly is sufficient to show: if f=∑A∈FxA1A=0, then ∑A∈FxAμ(A)=0. There exists a finite collection G⊆Σ of pairwise disjoint non-empty sets such that each set A is the union of those sets in G which are subsets of A. This can be expressed by saying that there exist αAB∈{0,1}, for (A,B)∈F×G, such that 1A=∑B∈GαAB1B(A∈F). Then 0=∑AxA1A=∑A,BxAαAB1B=∑B(∑AxAαAB)1B implies ∑AxAαAB=0(B∈G); hence ∑A∈FxAμ(A)=∑AxA∑BαABμ(B)=∑B(∑AxAαAB)μ(B)=0.
Is this “tedious, in particular concerning notation”?
Best wishes, Jürgen
Dear Jürgen,
well, what I meant is that it is a bit tedious to explicitly write down G.
Best wishes, Hendrik
Dear Hendrik,
this is not really true (and somehow below the level of our present discussion). In fact, you know how it can be done: For ∅≠A⊆F one defines AA:=(⋂A)∖⋃(F∖A); then AA consists of those points belonging to all sets A∈A that do not belong to any of the sets in F∖A. This shows that the family (AA)∅≠A⊆F is pairwise disjoint, and ⋃∅≠A⊆FAA=⋃F. Discarting the empty sets one obtains the desired G.
Incidentally, the collection {A∩B; A∈F,B∈G} in one of your previous posts will not necessarily be a partition, because the members of a partition should be non-empty.
Best wishes, Jürgen
Dear Jürgen,
well, now you wrote down that piece of notation that I called “a bit tedious”
Of course I knew how it could be done, but you found a concise form that I hadn't thought of before (or that I managed to forget) – nice! (I wouldn't call this “below the level of our present discussion” because my starting point was “When one tries to write it up”, i.e., really write up all the stuff about integration of simple functions that ought to be “clear”.)
Thanks a lot for pointing out that that collection I used is not necessarily a partition! So I should say “collection of pairwise disjoint sets” instead of “partition”.
Best wishes, Hendrik
Dear lecturers,
thank you very much for all the detailed proofs in this lecture. It really helped me to understand the topic. I only noticed some small problems while reading it:
(1) I think that you switched the arguments of the inner product here. I guess that we still want the second argument in the inner product to be linear. However, if you look at the proofs it seems that you switched to the first argument being linear. For example, in the proof of Lemma 3.2.5, you should better write ⟨g,Iνf⟩L2,ν(R;H)=… Of course, there, it is not really a mistake. However, in the proof of Proposition 3.2.3, we want the inner product acting as a linear functional rather than an anti-linear functional.
(2) Related to this: I also think that in the second part of the proof of Lemma 3.2.5, we need to calculate ⟨g,2νIνI∗νf⟩L2,ν(R;H)=… to avoid using Fubini-Tonelli for the Bochner integral. Of course, it is the same calculation but one can use the “normal” Fubini-Tonelli theorem then.
Dear Julian,
thank you for your comments.
(1) Your suggestion is closer to the definition of the adjoint as highlighted in Lecture 2. In any case, ⟨Ax,y⟩=⟨x,A∗y⟩ for a bounded linear operator A and all x,y∈H is (evidently) equivalent to ⟨y,Ax⟩=⟨A∗y,x⟩ for all x,y∈H. So, I reckon your remark is rather on didactics.
(2) Thank you for your suggestion. In particular, as we have not said anything on Fubini–Tonelli for the Bochner integral.
Best regards,
MM
Hello Marcus,
thank you for answering. With respect to (1), I still have to say that in the proof of Proposition 3.2.3, it is not just a didactic argument because we want the inner product acting as a linear functional rather than an anti-linear functional.
Best wishes, Julian
Dear Julian,
a late answer: you have an issue in Proposition 3.2.3 because one has to pull an anti-linear functional argument out of the integral, right? This is actually not a problem at all since the functional is an R-linear operator! Generally it is often helpful to forget that there are complex scalars, i.e., to consider all the spaces involved as R-vector spaces.
Best wishes, Hendrik
Hello Hendrik,
thank you very much for your answer. You are completely right! In fact, I didn't have an actual problem with using anti-linear functionals but just thought that using linear ones would be easier/more straight-forward. Especially since we proved Proposition 3.1.5(a) for this.
Thank you again! Best wishes, Julian
Dear virtual lecturers,
I would like to share some comments from our meeting in Darmstadt last week:
Shouldn't we assume (Ω,Σ,μ) to be complete, as we later define Bochner-measurable functions as pointwise almost everwhere limits of simple functions.
You chose to replace the notation S(μ;X) by S(Rd;X) in the situation where μ is absolutely continuous with respect to the Lebesgue measure λ. We wondered, if this is a well-defined notation. In our opinion, the density ρ(x):=1/|x| defines a σ-finite measure μ on L(R) such that S(λ;X)≠S(μ;X). (For instance, we have f∈S(λ;X) but f∉S(μ;X) for f(x):=χ[0,1].
Throughout the lecture, the norm on X is called ‖⋅‖X. Thus we assume there is an X missing in the proof of Proposition 3.1.2 in the definition of gn as well as in the last line of page 27 after the first equality sign.
A question concerning Corollary 3.1.6: It was left to the reader to show that any continuous function on [a,b] is Bochner-measurable. We assume that a way to approximate a continous function by step functions in this case is to use that its image is compact. For any positive ϵ, we can thus cover the image by finitely many ϵ-balls and define a step function given on the preimage of each such ball by some value inside the ball. As we only need pointwise convergence, a similar procedure should also work for a Borel measure space that can be covered by countably many compact sets. However, we wondered which is the precise condition needed for a Borel measure space in general to ensure that all continous functions are measurable. And why.
We noticed that the last part of the proof for Proposition 3.2.4 does not require specific properties of the operator ∂t,v, but only uses the boundedness of ∂−1t,v which implies the equivalence of the graph norm and the norm x→‖Ax‖L2,v(R;H). This furthered our understanding of the proof.
In the proof of Corollary 3.2.6, we would have considered further explanation helpful as to how dom(∂∗t,v)⊂dom(∂t,v) follows from ran(I∗v)⊂dom(Iv), for instance a reference to Lemma 2.2.2 or even a short computation.
Finally, we suppose you should replace f∗ϕ∈C∞c(R) by f∗ϕ∈C∞(R) in Exercise 3.2.
Thank you for reading our comments and we are curious about your answer.
Best regards
Janik
Dear Janik,
thank you very much for your comments from the Darmstadt team.
Can you further explain, why we need completeness of the measure space? I don't see this being necessary for the definition of Bochner measurable. The definition at hand is used in order to integrate reasonably many functions.
Your are right concerning S(μ;X). Throughout the text, we use S(R;X) in the case of the Lebesgue measure. Note that S(R;X) is dense in L2,ν(R;X); for X a Hilbert space and all ν∈R.
Your question concerning measurability is interesting. For this I refer you to the discussion in
https://en.wikipedia.org/wiki/Baire_set
That deals precisely with this question.
You are also correct concerning Proposition 3.2.4; as well as concerning Corollary 3.2.6. Referring to Lemma 2.2.2 would indeed be helpful for this.
Thanks for pointing out the misprint in Exercise 3.2.
Best regards,
moppi
Hello Marcus,
today we had the same discussion about completeness as well in Hamburg. The problem is very simple:
If you have a sequence of measurable functions fn:Ω→R that converge pointwisely to f:Ω→R, then f is also measurable. However, this is not true when the pointwise convergence holds only μ-a.e. In this case, you need that the measure is complete to ensure that f is also measurable.
Now, in the lecture 3, we cannot conclude:
f:Ω→X Bochner-measurable ⇒ ‖f(⋅)‖:Ω→R measurable.
This means that Remark 3.1.1 (b) is wrong. However, the definition of Lp(μ,X) still works because the measurability of ‖f(⋅)‖ is explicitly given there.
Dear Julian,
let me ask you why you think that the implication f Bochner-measurable⇒‖f(⋅)‖ measurable is wrong. This is proved in the text, so you should say where you think there is a mistake. Note that in this case both functions are pointwise a.e. limits of simple functions.
Maybe you misinterpret what means that fn(ω)→f(ω) μ-a.e. This does not mean that the set of ω where (fn(ω)) does not converge to f(ω) is a μ-null set, but rather that this set is contained in a μ-null set.
So, if there is a μ-null set N such that fn(ω)→f(ω) on the complement of N, then also ‖fn(ω)‖→‖f(ω)‖ on the complement of N.
Best wishes, Jürgen
Dear all,
sorry, I have to withdraw part of my comments: somehow I did not understand/realise that Bochner-measurability of R-valued functions is not the same as μ-measurability, if μ is not complete.
Best wishes, Jürgen
Dear Jürgen,
maybe I misinterpreted what you mean by “μ-measurable”. I was under the impression that it just means Σ−B(R)-measurable in this context.
Best wishes, Julian
Dear Julian,
okay, now I got your point. So, the function ‖f(⋅)‖ is not measurable in the classical sense, but equals a measurable function outside a μ-null set, which then doesn't affect the integral. So, even if we do not require the measurablity of ‖f(⋅)‖ in the definition of Lp, everything makes sense, if we pass to a measurable representative or do I oversee something?
Best regards
Sascha
Hello Sascha,
yes, exactly. I also think that there is no problem at all in the lecture when using a non-complete measure. The only confusion was just the formulation in Remark 3.1.1 (b).
Best wishes, Julian
Dear Julian,
as I already wrote, I mixed up Bochner-measurability as R-valued functions with μ-measurable (i.e., Σ-B(R)-measurable). And your comment concerning Remark 3.1.1(b) is quite justified.
Best wishes, Jürgen
Dear Marcus,
I do not see how your reference to wikipedia solves the problem that is brought up by Janik and Julian. As I understand, the problem is the following.
Accepting how measurable functions f:Ω→X are defined: if (fn) is a sequence of measurable functions converging to f a.e., show that then f is measurable, i.e. an a.e.-limit of simple functions. Actually, it seems to me that you neglect this problem generously and uncommented in your proof of the completeness of Lp(μ;X), Proposition 3.1.2. Also, I do not see how completeness of the measure space could help to prove this.
I think I know a proof, with the help of Egorov's theorem, and I will post it if you want. But here I just want to explain what is the problem. Let B0 be a set of functions f:Ω→X, then form the set B1 consisting of pointwise limits of sequences in B0, and repeat the procedure with sequences in B1, to obtain B2. (Starting with continuous functions in certain situations, the sets B1,B2,… are known as Baire classes, and it is well-known – and not difficult to proof, e.g., for B0:=C[0,1] – that, in general B2 is strictly larger than B1. (And I fear that you run into the same problem starting with B0:=S(Ω;X)!)
Best wishes, Jürgen
Dear Jürgen,
thank your for your remarks. My reference to wikipedia was mentioned in order to clarify what conditions are needed to render continuous functions measurable. Thus, in the reference to Corollary 3.1.6. I did not intend to refer to the wikipedia in order to clarify that measurable functions are point wise a.e. limits of step functions (or some statement of this sort). I have seen that my previous post wasn't very clear in that.
In any case, I apologise if I happened to misunderstand Janik's comments. Please clarify, if need be.
Now, concerning measurability in general (not with regards to continuous functions only): The difference of Baire classes from the notion of measurable functions is that in our case, we ask for the limits to be only almost everywhere rather then everywhere. I think you are right, Jürgen, when starting off with B0=S(Ω;X), that B1, the set of pointwise everywhere limits is strictly smaller than the set of measurable functions.
In the construction of the Bochner spaces in the lecture notes, the important bit is the space Lp(μ;X) (and its scalar-valued analogue).
In particular, in our proof of Lp(μ;X) being complete, we use completeness of Lp(μ). So, if I understand your post correctly, what one needs to show is that every function in Lp(μ) is a.e. pointwise limit of simple functions.
I try to show this with using the following statements:
(1) Lp(μ) is complete and for every convergent sequence (fn)n in Lp(μ) there exists a subsequence, which is majorised and converges pointwise almost everywhere (Theorem of Fischer–Riesz).
(2) The set of simple functions S(μ) is dense in Lp(μ).
Now, let f∈Lp(μ). Then by (2), there exists a sequence (fn)n of simple functions converging to f in Lp(μ). By (1), we can assume that (fn)n converges almost everywhere to f.
So, I think in the end, the difference between the notions of measurability (Bochner measurable vs μ-measurable) is somehow `swallowed' by the equivalence relation of being equal almost everywhere. Please let me know, whether I made a mistake here.
Best regards,
MM
Dear all,
Sascha and I just had a discussion on this. Sascha will write a post that is going to resolve the problem in due course.
Best regards,
MM
Dear all,
Jürgen made the following important point: in the proof of Proposition 3.1.2 it is used that the a.e. limit of a sequence of Bochner measurable functions is again Bochner measurable. Here's how I would prove this fact:
First of all, if f:Ω→X is Bochner measurable, then it follows from the definition that the set [f≠0] is σ-finite. Likewise, if f is the a.e. limit of a sequence of Bochner measurable functions, then [f≠0] is σ-finite.
Now let fn:Ω→X be Bochner measurable, for n∈N, and fn→f a.e. Let [f≠0]=⋃n∈NAn with measurable sets An of finite measure, A1⊆A2⊆…. Then one easily checks that gn:=1An∩[‖fn(⋅)‖≤n]fn→f a.e. Now comes the main idea: for each n∈N, gn is in L1(μ;X), and thus there exists a simple function hn such that ‖gn−hn‖1≤2−n, by Lemma 3.1.3. Then ∫Ω∞∑n=1‖gn(ω)−hn(ω)‖Xdμ(ω)≤1, which implies that ∑∞n=1‖gn(ω)−hn(ω)‖X<∞ for a.e. ω∈Ω, and it follows that gn−hn→0 a.e. Thus hn→f a.e., and we conclude that f is Bochner measurable.
Best wishes, Hendrik
Dear Hendrik,
thank you very much for this nice proof. In a discussion with Moppi we only had the idea to prove Pettis' Theorem, which then would imply the statement as well. However, your proof is much nicer and more elementary.
This whole discussion shows once again, that measurability (in which sense ever) is a very delicate matter and we apologize for not taking enough care of that subject.
Best regards
Sascha
Dear Moppi,
you write that use S(R;X) in the case of the Lebesgue measure and that S(R;X) is dense in L2,ν(R;X). Do you mean that S(R;X) is supposed to be read as the space of simple functions that is built w.r.t. Lebesgue measure? Then you have a problem: S(R;X) is not a subspace of L2,ν(R;X) unless ν=0! In fact, it is easy to construct a set A of finite Lebesgue measure such that 1A∉L2,ν(R;X), but of course 1A∈S(R;X). I think the correct solution would be to work with S(A;X) with a ring A⊆Σ that generates Σ; then for Ω=R you can choose as A the collection of all relatively compact measurable subsets of R.
Best wishes, Hendrik
Dear Participants,
Florian Pannasch suggested a far easier proof of Lemma 3.1.7 in our discussion in Kiel, even for p∈[1,∞) and X an arbitrary Banach space:
Since we already know that S(μ;X) is dense in Lp(μ;X) by Lemma 3.1.3 it suffices to approximate elements of the form 1Ax for some A∈Σ and x∈X. The latter is trivial, since we find a sequence (φn)n in the linear hull of D with φn→1A in Lp(μ). Then ‖1Ax−φnx‖p=‖x‖X‖1A−φn‖p→0 and thus the claim follows.
Best regards
Sascha
Dear virtual lecturers,
Lemma 3.2.1: Maybe my question is stupid, but shouldn't one say something why the resulting function k∗f is measurable? Even in the expression on line 3 of the proof of the lemma I start asking myself, why the function resulting from the inner integral is measurable.
Proposition 3.2.4: I have a suggestion for an alternative proof. If ν>0, I suggest to define φn(t):=∫t−∞(f(s)−f(s−n))ds. (And with f(s+n), if ν<0.)
Two comments on terminology (admittedly, both a matter of taste):
“one-to-one” in Prop. 3.2.3: I prefer the injective-surjective-bijective terminology. One reason is, how – according to wikipedia – one would have to express a bijective mapping: a “one-to-one correspondence”! (OK, you can say “one-to-one and onto”.)
“arbitrarily differentiable” (in Prop.3.2.4) is – in a way – better than “infinitely differentiable”, but still is not meaningful by itself, but rather the abbreviation (and thereby the definition) of “arbitrarily often differentiable”. Clearly “infinitely often differentiable” – used by many authors – is simply nonsense; but I have decided – for myself – that “infinitely differentiable” is defined as “arbitrarily often differentiable”.
And a misprint on p.30, line -3: The index at μ should be p,ν, in this order.
Best wishes, Jürgen
Dear Jürgen,
thank you very much for your comments.
Your proof of Proposition 3.2.4 - at least for me - is nicer than ours. Thanks a lot!
Concerning the question of measurability: You are completely right, we have just overseen this. One way to show it would involve Pettis' Theorem and Fubini to obtain the measurability of the convolution tested with a Hilbert space element. But since we do not want to employ Pettis, I am curious if you or anybody else have a more elementary proof for this fact.
Best regards
Sascha
Dear Sascha,
I think I might have an answer to the measurability question (without invoking Pettis' theorem). If f is of the form φx, then the measurability of k∗f follows from the scalar case. Then the same holds for the linear hull of these functions, and for functions of the linear hull one can show the inequality as in the lecture notes. Then (cautious!) approximation.
Best wishes, Jürgen
P.s. Writing posts with the offered platform is perfect. Thanks, Christian!
Dear Sascha and dear all,
concerning the measurability: First concerning the question why the function resulting from the inner integral is measurable. This follows from the scalar case, because it is just the convolution of |k| and ‖f(⋅)‖, and the estimate of the proof shows that there exists a null set N⊆R such that k(⋅)‖f(t−⋅)‖ is integrable for all t∈R∖N.
Now comes the “cautious” approximation. The main observation is that the density lemma, Lemma 3.1.3 deserves to be reinforced by adding “and for each f∈Lp(μ;X) there exists a sequence (fn) in S(μ;X) such that fn→f a.e. and ‖fn(⋅)‖≤2‖f(⋅)‖ for all n∈N. (This is what in fact is proved!)
Let f∈L2,ν(R;H) and (fn) be as above. As mentioned above, there exists a null set N⊆R such that k(⋅)‖f(t−⋅)‖ is integrable for all t∈R∖N. Then |k(⋅)|‖fn(t−⋅)‖≤2|k(⋅)|‖f(t−⋅)‖(t∈R∖N, n∈N), but also k(⋅)fn(t−⋅)→k(⋅)f(t−⋅) a.e.; hence the dominated convergence theorem implies that ∫k(s)fn(t−s)ds→∫k(s)f(t−s)ds(n→∞, t∈R∖N). This shows the measurability of k∗f. (Then one can go ahead with the estimate as in the lecture notes. The measurability of the functions k∗fn follows from the scalar case, but there is no need to carry out the long estimate first for the simple functions, as I had insinuated in my previous post.)
A side remark to the second paragraph above: It would not really be necessary to add the sentence to the assertion, because of the following general fact. If (fn) is a Cauchy sequence in Lp(μ;X), then there exist a function g∈Lp(μ)+ and a subsequence (fnj) converging a.e. and in Lp(μ;X), and such that ‖fnj(⋅)‖≤g(⋅) a.e. for all j∈N.
Best wishes, Jürgen