Discussion on Lecture 03

Yuming Paul Zhang, 2022/12/17 20:58

Dear all,

As Kai and Joachim had pointed out, from closed graph theorem the constant C depends on x. I am wondering if it is still true that we can find a bound independent of x or not?

Thank you!

Best, Paul

Christian Seifert, 2022/12/21 16:55, 2022/12/21 16:55

Dear Paul, the constant $C$ depends on $x$ and in order to get a uniform constant one needs further properties of $b$ (e.g. that $sup_{x \in X} ‖ b (x, \cdot) ‖ < \infty$ ). Best, Christian

Samir BOUJIJANE, 2022/11/15 17:40

Dear all,

First of all many thanks for all the nice lectures and also for the interesting discussions elaborated on here. I have two comments:

In the proof of Lemma 2.5(b). I think there's no need to assume that $x_{i}$ are pairwise different. Indeed, it is only used that $x_{i} \sim x_{i + 1}$ which is given by the definition of a path.
In Theorem 2.10 (Minimal principle). The condition $α > 0$ could be replaced by ( $α > 0$ or $c (x_{0}) > 0$ ) where $x_{0} \in U$ a vertex where the minimum is reached. This can be easily observed from the inequalities displayed at the end of page 36.

Best regards, Samir

Anna Muranova, 2022/11/11 19:36

Dear lecturers,

thank you for the 3rd lecture.

I have the following question: why in Lemma 2.11 $sup_{n} \sum_{y \in X} b (x, y) u_{n} (y) = \sum_{y \in X} b (x, y) u (y)$ or which formulation of the monotone convergence theorem do you use?

Best, Anna

Kai Jennissen, 2022/11/12 18:56, 2022/11/13 10:32

Dear Anna,

I think this follows with the given definition of $m = 1$ (point measure and $X$ countable) from the Lemma of Beppo Levi (monotone convergence) with $f_{n} (y) = b (x, y) u_{n} (y)$ and $f (y) = b (x, y) u (y)$ : $\begin{aligned} lim_{n \to \infty} \sum_{y \in X} b (x, y) u_{n} (y) & = lim_{n \to \infty} \sum_{y \in X} f_{n} (y) \\ = l i m_{n \to \infty} \int_{X} f_{n} (y) d m (y) = \int_{X} f (y) d m (y) \\ = \int_{X} b (x, y) u (y) d m (y) = \sum_{y \in X} b (x, y) u (y) \end{aligned}$

Best regards, Kai

Anna Muranova, 2022/11/14 12:07

Dear Kai,

thank you very much!

Indeed, I have been thinking about the theorem which just says, that monotone sequence converges to its supremum.

Best, Anna

Youssef Hakiki, 2022/11/15 18:46, 2022/11/15 18:51

Dear Anna and Kai,

Here is anopther reformulation of the Lemma 2.11

Indeed, consider the measure space $X$ with the measure $B_{x}$ given by

  $$
  B_x(M):=\sum_{y \in M} b(x,y).
  $$

Since $u_{n} \geq 0$ for all $n \in N$ and $u_{n} ↗ u$ , then by Beppo-Levy theorem in the space $(X, B_{x})$ we infer that

  $$
  \lim_{n\rightarrow \infty}\sum_{y\in X}b(x,y)u_n(y)=\lim_{n\rightarrow \infty}\int_Xu_n(y)dB_x(y)=\int_Xu(y)dB_x(y)=\sum_{y\in X}b(x,y)u(y).
  $$

Now, since $u_{n} \in F$ for all $n$ , we write $\sum_{y \in X} b (x, y) (u_{n} (x) - u_{n} (y)) + (c (x) + α) u_{n} (x) = (L + α) u_{n} (x) .$ Therefore the facts that $\sum_{y \in X} b (x, y) < \infty$ , $(L + α) u_{n} (x) \to f (x)$ and $u_{n} ↗ u$ ensures that $\sum_{y \in X} b (x, y) u (y) < \infty$ .

Best, Youssef

Patrizio Bifulco, 2022/11/11 15:39, 2022/11/11 15:40

Dear all,

many thanks for this nice and interesting third Lecture!

Here are some very minor remarks/questions from my side:

At the bottom of page 30 I would suggest to write $f (x)^{2}$ instead of $f^{2} (x)$ .
There is a typo in the center of page 31 (when recalling the definition of normal contractions): 'a map' instead of 'at map'.
In the proof of Lemma 2.5(d) on page 34 'As the sequence $((f_{n})$ ' there is a bracket to much.
In the proof of Lemma 2.17(b) on page 42 at the end of the third line of the displayed estimate there is a $m (y)$ missing.
On page 43 in the definition of $κ := sup_{x \in X} D e g (x)$ there is a $x$ missing.
In part (ii) of Exercise 2, I guess it should read '[…], then $A f (x) \geq 0$ ' instead of $A f \geq 0$ ? And do you really mean 'local maximum' here?

Best regards, Patrizio

Christian Seifert, 2022/11/14 12:11

Dear Patrizio,

Many thanks for the reamrks.

Concerning your question: “local maximum” is really meant here.

Best, Christian

Kai Jennissen, 2022/11/10 14:11, 2022/11/11 12:18

Dear all,

I also have a question regarding the proof of Theorem 2.16. I found the part on p. 41 a little tricky therefore I'd like to verify whether my understanding is correct.

From the closed graph theorem we conclude that the map $j$ is linear and bounded and thus the existence of $C_{x}$ with $| | j (f) | |_{ℓ^{1} (N_{x}, b (x, \cdot))} \leq C_{x} | | f | |_{ℓ^{2} (N_{x}, m_{N_{x}})}$ follows. Since $f \in ℓ^{2} (X, m)$ is defined on $X$ we constrain the domain of $f$ through composition with $1_{N_{x}}$ such that the inequality reads as follows

$\begin{aligned} \sum_{y \in X} b (x, y) | f (y) | & = \sum_{y \in N_{x}} b (x, y) | f (y) | = \sum_{y \in X} b (x, y) | f (y) 1_{N_{x}} | \\ = | | j (f 1_{N_{x}}) | |_{ℓ^{1} (N_{x}, b (x, \cdot))} = | | f 1_{N_{x}} | |_{ℓ^{1} (N_{x}, b (x, \cdot))} \leq C_{x} | | f 1_{N_{x}} | |_{ℓ^{2} (N_{x}, m_{N_{x}})} \\ \leq C_{x} | | f 1_{N_{x}} | |_{ℓ^{2} (X, m)} \end{aligned}$

Thus the map $ℓ^{2} (X, m) ∋ f \mapsto \sum_{y \in X} φ_{x} (y) | f (y) | m (y) = \sum_{y \in X} b (x, y) | f (y) |$ is a bounded linear functional, aka. an element of the dual of the Hilbert space $ℓ^{2} (X, m)$ . By the Riesz representation theorem there exists a unique element of $ℓ^{2} (X, m)$ such that the functional can be written as inner product. But since $\sum_{y \in X} φ_{x} (y) | f (y) | m (y) = ⟨ φ_{x}, f ⟩$ the uniqueness implies this element is given by $φ_{x}$ and therefore $φ_{x} \in ℓ^{2} (X, m)$ .

Best regards, Kai

Christian Seifert, 2022/11/11 12:19

Dear Kai,

You are right and your understanding is correct.

Best, Christian

Kai Jennissen, 2022/11/09 22:28, 2022/11/10 13:37

Dear all,

first of all I want to thank the organizers.

I have difficulties to understand the details of the implication $i) ⟹ i i i)$ in the proof of Theorem 2.18. Given $i)$ and the definition of $a (x, y)$ it's clear that $\sum_{y \in X} a (x, y) m (y) = D e g (x) \leq sup_{x \in X} D e g (x) =: κ$ and therefore by Lemma 2.17 a) the operator $A f (x) = \sum_{y \in X} a (x, y) f (y) m (y) = \frac{1}{m (x)} \sum_{y \in X} b (x, y) f (y) + \frac{c (x)}{m (x)} f (x)$ is bounded. But how does this imply that $L$ is bounded? Given $L f (x) = \frac{1}{m (x)} \sum_{y \in X} b (x, y) (f (x) - f (y)) + \frac{c (x)}{m (x)} f (x) = \frac{f (x)}{m (x)} \sum_{y \in X} b (x, y) - \frac{1}{m (x)} \sum_{y \in X} b (x, y) f (y) + \frac{c (x)}{m (x)} f (x) = f (x) D e g (x) - A f (x) + \frac{c (x)}{m (x)} f (x)$ I'm not able to verify $| | L f | | \leq κ | | f | |$ .

Furthermore some minor comments.

I think $C$ should be replaced with $D$ in the $(i i i) ⟹ (i)$ section of the proof of Theorem 2.18.
Shouldn't it be $\sum_{z \in X} b (y, z) + c (y) \leq C m (y)$ instead of $\sum_{z \in X} (b (y, z) + c (z)) \leq C m (y)$ ?

Best regards, Kai

Sahiba Arora, 2022/11/10 00:07

Dear virtual lecturers, dear Kai,

I have the same problem: I get $D = 3 κ$ and don't see how to make the bound sharper.

Regards,

Sahiba

Kai Jennissen, 2022/11/10 13:38, 2022/11/10 13:38

How did you derive $D = 3 κ$ ?

Best regards, Kai

Sahiba Arora, 2022/11/10 14:41, 2022/11/10 14:41

Dear Kai,

For example, from the last equality in the last displayed equation, each of the last three terms is bounded above by $κ$ .

Regards, Sahiba

Kai Jennissen, 2022/11/10 15:48, 2022/11/10 16:06

The bound for the first two terms was clear but obviously it also holds that $\frac{c (x)}{m (x)} \leq D e g (x) \leq κ$ . Thanks.

Regards, Kai

Christian Seifert, 2022/11/11 12:11

Dear Kai and Sahiba,

There is a mistake in this proof and the statement. One needs $D = 2 κ$ . To prove (i) $\Rightarrow$ (iii), set $a (x, y) := \frac{b (x, y)}{m (x) m (y)}$ for $x \neq y$ and $a (x, x) := 0$ . Then $L = D e g - A$ (here, $D e g$ considered as a multiplication operator). Now, a bound for $D e g$ provides also a bound for the norm of $A$ , so one obtains the $2 κ$ then.

We'll update the statement and proof in the final version.

Best, Christian

Sahiba Arora, 2022/11/11 12:24

Dear Christian,

Isn't there also a $\frac{c (x)}{m (x)} f (x)$ term also as in Kai's last displayed formula?

Regards, Sahiba

Christian Seifert, 2022/11/11 13:18

Dear Sahiba,

Yes indeed; this comes from the $D e g$ as well.

Best, Christian

Sahiba Arora, 2022/11/11 13:28, 2022/11/11 14:32

Dear Christian,

So $L = D e g - A + \frac{c}{m}$ and thus the bound should be $3 κ$ instead of $2 κ$ .

Regards, Sahiba

Sascha Trostorff, 2022/11/11 14:18

Dear all,

since we agree on the bound for $Q$ couldn't we argue as follows:

For $f, g \in C_{c}$ we know by Greens formula $| ⟨ L f, g ⟩ | = | Q (f, g) | \leq 2 κ ‖ f ‖ ‖ g ‖,$ which yields that $φ : g \mapsto ⟨ L f, g ⟩$ is a bounded functional which can be uniquely extended to $ℓ_{2}$ . Then by Riesz-Frechet, we know $L f \in ℓ_{2}$ and $‖ L f ‖ = ‖ φ ‖ \leq 2 κ ‖ f ‖$ . Clearly, this inequality extends to $f \in ℓ_{2}$ by density.

Best regards

Sascha

Christian Seifert, 2022/11/11 14:34

Dear Sahiba,

The $D e g$ contains the term $\frac{c}{m}$ already, so one gets $L = D e g - A$ (note that $a$ is zero on the diagonal, see my post above).

Best, Christian

Sahiba Arora, 2022/11/11 14:38

Dear Christian

Yes, sorry I didn't notice that in your comment your choice of $a$ is different from the notes.

Alles Klar!

Regards, Sahiba

Souhadou Diallo, 2022/11/16 22:19

Thank for the clarification

EL-Houcine OUALI, 2022/11/09 18:42

Hello, Thank you very much for this third lecture.

Sahiba Arora, 2022/11/09 15:41, 2022/11/10 00:00

Dear all

First of all, thanks to the virtual lecturers for all the lectures till now and apologies for not participating in the forum sooner.

I have a couple of small doubts:

Am I correct that the second inequality on top of page 33 was obtained using Hölder's?
Proof of Proposition 2.9(a), I don't see how the equality $\sum_{x} φ (x) L f (x) m (x) = \sum_{x} L φ (x) f (x) m (x)$ is obtained as the former is equal to $\sum_{x, y} b (x, y) φ (x) (f (x) - f (y)) + c (x) φ (x) f (x)$ but the latter is equal to is equal to $\sum_{x, y} b (x, y) f (x) (φ (x) - φ (y)) + c (x) φ (x) f (x) .$
Proof of Theorem 2.13: I don't see how the last displayed equation on page 38 is obtained.
Proof of Theorem 2.13: Why does $L u_{t} (x) = 0$ hold for $t = T$ ?
Example 2.14: It is claimed that $L 1_{x}$ vanishes outside $N_{x}$ . I see this for $z \neq x$ but not for $x \notin N_{x}$ , because $L 1_{x} (x) = D e g (x)$ need not be zero?

Also, a couple of typos:

Theorem 2.13: Second bullet point should have $[0, T] \times U$ instead of the other way round.
Proof of 2.18, first line: I think it should be “(b) and (a) respectively” instead of “(a) and (b) respectively”.
Proof of 2.18, (iii) $\Rightarrow$ (i): The constants $C$ should be replaced by $D$ .

Lastly, maybe a small suggestion: As the function $m : X \to (0, \infty)$ gives rise to the corresponding measure, maybe it would be a good idea to just define $n := d e g$ in Section 2.5.2, instead of defining $n$ on the subsets? It just seems more natural to me (unless there is a reason to define it on the subsets first?).

Regards,

Sahiba

Sascha Trostorff, 2022/11/10 21:31

Hi Sahiba,

I have an answer to your second question:

The two terms are in fact equal. Just note that $\begin{aligned} \sum_{x, y} b (x, y) φ (x) (f (x) - f (y)) & = \sum_{x, y} b (x, y) φ (x) f (x) - \sum_{x, y} b (x, y) φ (x) f (y) \\ = \sum_{x, y} b (x, y) φ (x) f (x) - \sum_{x, y} b (y, x) φ (y) f (x) \\ = \sum_{x, y} b (x, y) f (x) (φ (x) - φ (y)) \end{aligned}$ due to the symmetry of $b$ .

Best regards

Sascha

Sahiba Arora, 2022/11/10 21:33

Dear Sascha,

Ah yes, of course. Thanks!

Regards, Sahiba

Christian Seifert, 2022/11/11 11:57

Dear Sahiba,

Many thanks for your questions and comments. Concerning your questions: 1. Yes this is Cauchy–Schwarz (or Hölder with $p = q = 2$ if you like). Just smuggle in $\frac{b (x_{i}, x_{i + 1})^{1 / 2}}{b (x_{i}, x_{i + 1})^{1 / 2}}$ .

2. See Sascha's answer (@Sascha: Many thanks!)

3. We have that $L u_{t} (x) = 0$ , but looking at the first displayed formula on page 39 we see that $L u_{t} (x)$ is a sum of two non-positive terms, so both terms have to be zero. The first term being zero then gives the equality.

4. We have $(L + \partial_{t}) u_{t} (x) = 0$ and $\partial_{t} u_{t} (x) \leq 0$ , so $L u_{t} (x) \geq 0$ . But the fist displayed formula on page 39 also gives $L u_{t} (x) \leq 0$ .

5. This is a typo (many thanks for spotting it!): $L 1_{x}$ vanishes outside $N_{x} \cup {x}$ .

Best, Christian

Sahiba Arora, 2022/11/13 00:11

Dear Christian,

Many thanks for your response. I'm still confused about 3 and 4.

For 3, how did you start with the observation that $L u_{t} (x) = 0$ ? In fact, this is only concluded much later using the last displayed formula on Page 38. So the argument becomes circular.

Moreover, for 4, I think one needs to use that the positivity of $(L + \partial_{t}) u_{t} \geq 0$ carries over to $t = T$ .

Regards, Sahiba

Christian Seifert, 2022/11/20 14:10

Dear Sahiba,

sorry for the confusion and the long silence with respect to your question. As it turns out, you are completely right. One somewhow needes that $t \mapsto \partial_{t} u_{t} (x)$ is continuous at $t = T$ to obtain the last displayed formula on page 38. Moreover, to reason that $L u_{T} (x) = 0$ one needs to know that $(L + \partial_{t}) u_{T} \geq 0$ . We hav adusted the statement accordingly.

Many thanks for pointing this out!

Best, Christian

Leon Berghoff-Flüel, 2022/11/09 12:06

Hello everyone,

thank you from the Darmstadt group for the nice lecture. We came up with a simpler proof for the converse direction in Lemma 2.5 (d):

The assumtion $\underset{n \to \infty}{lim sup} Q (f_{n}) \leq Q (f_{n})$ together with proposition 2.4 already implies $Q (f_{n}) \to Q (f)$ . And $f_{n} (o) \to f (o)$ is just due to the pointwise convergence.

Regards, Leon

Sascha Trostorff, 2022/11/10 21:24

Dear Leon,

that is true, but that is not what Lemma 2.5 (d) claims. You have to show that $Q (f_{n} - f) \to 0$ for the convergence in $D$ , which does not follow from your argumentation.

Best regards Sascha

Joachim Hofmann, 2022/11/07 14:14

Dear all,

first of all thanks to the organizers for the new lecture. I have some remarks respectively questions:

In the proof of Theorem 2.16 the identity $l^{1} (N_{x}, b (x, \cdot)) = {f \in F | supp f \subset N_{x}}$ is not obvious to me and I am not sure it holds true. So if anyone knows the reason please let me know. The inclusion $l^{1} (N_{x}, b (x, \cdot)) \supset {f \in F | supp f \subset N_{x}}$ however is easy and sufficient. So maybe it would be an option to present it like that.

Immediately afterwards the constant $C \geq 0$ derived from the closed graph theorem should depend on $x$ , as the respective spaces depend on $x$ . So for all $x \in X$ we find such a constant $C_{x} \geq 0$ but not a constant $C \geq 0$ that works for all $x$ .

Regards Joachim.

Leon Berghoff-Flüel, 2022/11/08 19:22

Dear Joachim,

concerning your first question: First of all, one needs to understand that the equality is somewhat formal (maybe $≅$ would have been better than $=$ ) in the sense that $supp f \subset N_{x}$ means we consider the function $f$ to be effectively defined only on $N_{x}$ . On the right hand side of the equation, we could as well have written $F / \sim with f \sim g : ⟺ f = g on N_{x} .$ $supp \subset N_{x}$ then in some sense selects the most natural representant.

Now to why the equality holds: Be aware that $ℓ^{1} (N_{x}, b (x, \cdot))$ is defined analogously to $ℓ^{2} (X, m)$ , i.e. $ℓ^{1} (N_{x}, b (x, \cdot)) = {f | \sum_{y \in N_{x}} | f (y) | | b (x, y) | < \infty} .$ So $b (x, \cdot)$ defines somehow the weight. From this, the equality is quite obvious.

Concerning your second remark, I agree with your observation. Thanks for pointing it out.

Best regards, Leon

Joachim Hofmann, 2022/11/10 12:55

Dear Leon,

thanks for your reply but my problem was not with the definition or how to understand is. The problem is that even though $f$ is supported on $N_{x}$ we don't know anything about the behaviour of $b (z, \cdot)$ on $N_{x}$ for $z \neq x$ . I came up with the following example:

Let $X = N \cup {0} \cup {- 1}$ and let $b (0, n) = n^{- 3}$ , $b (- 1, n) = n^{- 2}$ , $b (0, - 1) = 0$ , $b (n, m) = 0$ , $n, m \in N$ . Then we have $N_{0} = N_{- 1} = N$ . If we now take $f (n) = n$ on $N$ , then $supp (f) \subset N_{0}$ and $\sum_{y \in N_{0}} | f (y) | b (0, y) = \sum_{n \in N} \frac{1}{n^{2}} < \infty,$ so $f \in l^{1} (N_{0}, b (0, \cdot))$ . On the other hand, $\sum_{y \in X} | f (y) | b (- 1, y) = \sum_{n \in N} \frac{1}{n} = \infty,$ so $f \notin F$ .

Best regards, Joachim.

Leon Berghoff-Flüel, 2022/11/12 11:03

Dear Joachim,

ah, okay, now I see what you mean. Thanks for clarifying your remark.

- Leon

Christian Seifert, 2022/11/18 12:30

Dear Joachim and Leon,

Many thanks. You are indeed right: the equality may not be true in general (as your example shows). However, we only need one inclusion, namely ${u \in F ∣ s u p p (f) \subseteq N_{x}} \subseteq ℓ^{1} (N_{x}, b (x, \cdot))$ , which is true.

Thanks for pointing this out!

Best, Christian

Christian Seifert, 2022/11/18 12:32

Dear Joachim and Leon,

Concerning the constant $C$ . We have fixed $x \in X$ beforehand, so we actually only need a $C$ which may depend on $x$ . The additional “for all $x \in X$ ” is wrong as stated and not needed; we hay just erased it.

Best, Christian

ISem26

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Discussion on Lecture 03

Discussion on Lecture 03