Dear all,

here's a late comment on Example 2.4.2. I would simplify the proof of the inclusion V(m)∗⊆V∗(m) as follows: Let u∈dom(V(m)∗). Then for all f∈L2(μ) and n∈N we know that 1nf∈dom(V(m)), so ⟨f,1nV∗u⟩=⟨V(m)(1nf),u⟩=⟨1nf,V(m)∗u⟩=⟨f,1nV(m)∗u⟩, and it follows that 1nV∗u=1nV(m)∗u for all n∈N. Recalling ⋃n∈NΩn=Ω we conclude that V∗u=V(m)∗u∈L2(μ) and therefore u∈dom(V∗(m)), V∗(m)u=V(m)∗u, which proves the desired inclusion.

(Note that I distinguish between the notations V∗u and V∗(m)u: for the latter I need to know that u∈dom(V∗(m)), for the former I don't; V∗u just denotes the product of two functions!)

Best wishes, Hendrik

Dear all,

in Example 2.4.2 we considered the multiplication operator V(m) induced by an L2(μ) function V∈L∞(μ) and show its boundedness. In particular we see that ‖V(m)‖L(L2(μ))≤‖V‖∞.

This raises the question whether we can also obtain equality, i.e. an isometric mapping L∞(μ)→L(L2(μ)),V↦V(m).

This question should have a positive answer in semi-finite measure spaces (I hope the following does not contain too many errors):

For ε>0, set Aε:=[V≥‖V‖∞−ε] and choose Bε⊆Aε with 0<μ(Bε)<∞.

Then ‖V(m)1Bεμ(Bε)‖2=1μ(Bε)‖V1Bε‖2≥‖V‖∞−ε for all ε>0 which shows that ‖V(m)‖L(L2(μ))≥‖V‖∞.

Do you know a counterexample to the isometry of mapping (1) in measure spaces that are not semi-finite?

Best wishes, Fabian

The provided solutions to Exercise 2.7 assume that C is non-empty. I would like to complete the solutions by considering the case if C is an empty set.

Let H=L2[0,1],dom(A)={f∈H:f is absolutely continuous,f(0)=0 and f′∈H},Af=f′ for all f∈dom(A).

Let λ∈C. For g∈H define (Bg)x=−∫x0eλ(x−t)g(t) dt. An easy computation gives (λ−A)B=I. Therefore λ∈ρ(A).

This gives that ρ(A)=C and hence σ(A)=∅.

Dear Isem-Team,

a minor remark:

p.20, l.9: I think the stated equality should instead be an inequality since we have

∑∞k=0‖(λ−μ)(λ−A)−1‖k=‖(λ−μ)(λ−A)−1‖1−‖(λ−μ)(λ−A)−1‖<11−‖(λ−μ)(λ−A)−1‖.

Best wishes,

Fabian

Dear all,

here are some more comments from Bremen.

1. On page 12, line -7 I would just say that A is called linear if it is a linear subspace of X0×X1 (even though this sounds rather tautological).

2. The equality (2.1) follows from the first equality in Theorem 2.2.5, applied to the relation A−1: One obtains (ranA−1)⊥=ker(A−1)∗ and thus (domA)⊥=ker(A∗)−1=A∗[{0}].

3. I find the proof of Lemma 2.2.8 remarkable; the only proof I know uses the Fréchet-Riesz theorem, and it appears that this is not needed here. (Of course the projection theorem is used, which is not too far away from the Fréchet-Riesz theorem, but I still find the proof given here very nice.)

4. For Corollary 2.3.3 I was tempted to use Lemma 2.2.7, but unfortunately this is not possible since 2.2.7 assumes linearity while 2.3.3 does not.

Best wishes, Hendrik

Dear all,

here are a few comments from Bremen.

1. At the beginning of Section 2.1 I get the impression that a “bounded linear operator” from X0 to X1 is always defined on all of X0. This is in conflict with the later usage of “bounded”, e.g. in Lemma 2.1.3 and Corollary 2.1.5, where the operator need not be defined on all of X0.

2. Corollary 2.1.5 in this form is new to me – I like it

3. Is is mentioned anywhere that a bounded linear relation is always an operator? I found this helpful for Corollary 2.1.5.

4. I find it a bit odd that linearity of A∗+B∗ is mentioned in Corollary 2.3.3 but not in Theorem 2.3.2.

5. I would write down the proof of Proposition 2.3.8 like this, with the abbreviation A|D:=A∩(D×H1): D core⟺¯A|D=¯A⟺¯A|D⊥=¯A⊥⟺(A|D)⊥=A⊥⟺(A|D)∗=A∗.

Best wishes, Hendrik

Dear virtual lectures,

here are some minor remarks from our group meeting in Darmstadt.

We thought that for consistency with Lecture 1 the title of the lecture should be “Unbounded Operators” and similarly in the other headings in this lecture.

On p.12 after the definition of L(X0,X1) we wondered whether there are unbounded operators defined on whole of X0 and, using the axiom of choice (to construct a Hamel basis), came up with an example. It was not clear to us whether an example of such an operator can be constructed without choice. Maybe a formulation like: “… the latter only need to be defined on a subset of X0” would clarify that unbounded operators can be defined on the whole space, too.

p.13 l.1: for consistency reasons one might consider to change the order of “composition” and “scalar multiples” so that it fits with the order in the definition.

In Proposition 2.1.1. we were not sure whether there are any additional, reasonable assumptions under which CA would be closed, too.

on p.14 l.6: we think that it should be “as a Banach space in its own right”.

on p.20 l.-3: We think that there got a (m) missing and that it should be “If V≠0 μ-a.e. then V(m) is injective…” (instead of “V is injective”).

Best, Jonas

Dear Nathanael,

I think that the remark on the multivalued part has been alluded to in the notes already, despite not being explicated. When it says that we call a linear relation A a linear operator if

A[{0}]={y∈X1:(0,y)∈A}={0}.

The point is that, if this set coincides not just with {0}, then we would be dealing with the “multivalued part” of the operator.

Despite not being needed for what follows, it's interesting nonetheless to delve into. I really like the presentation in Arens on these ideas.

Cheers,

Michael

I want to mention that it might be a good idea to complement the notion ker, ran, and dom for a linear relation T between X0 and X1 by the multi-value part mulT={y∈X1:(0,y)∈T}. This highlights the duality of ker, ran and mul, dom.