Mathematics - 2023-12-06 (page 1 of 2)

Jakobian

00:00

It might seem like I'm jumping onto GMT now, but its actually just that Bartle ordered me to see proof of Banach-Zarecki theorem and I figured out this would be the best method to prove that if $f'$ exists on some measurable set $E$ then $f'$ is measurable (this is needed for the proof).

Ted Shifrin

00:11

@oscarmetalbreak Um, NO.

oscarmetal break

Plausible.

Maybe not in torus. How about circle?

Xander Henderson

@oscarmetalbreak If by "circle" you mean "disk", sure, why not?

oscarmetal break

yes a disk

Xander Henderson

(Assuming you mean "A rectangle is homeomorphic to a disk". Maybe you mean some other form of isomorphism?---I just assumed topology....)

oscarmetal break

yes a homeomorphism from rectangle to a disk

Is it because you can find a continuous function from one point to another so it is true?

Xander Henderson

00:24

Two spaces are homeomorphic if there is a homeomorphism between them. What is the definition of a homeomorphism?

oscarmetal break

I guess I will encounter the same problem in reading dynamical systems latter...Thanks

Akiva Weinberger

05:00

0

A: Is $\sin n/\sin m$ irrational for nonzero integers $n\neq m$?

As shown in the answer you linked, the Lindemann–Weierstrass theorem implies that $\cos 1$ is transcendental. It turns out that $\dfrac{\sin n}{\sin 1}$ is a polynomial in $\cos 1$; in particular, $$\frac{\sin n}{\sin 1}=U_{n-1}(\cos 1)$$ where $U$ is the Chebyshev polynomial of the second kind. ...

@TedShifrin

Ted Shifrin

Why you pinging me, DogAteMy?

Akiva Weinberger

Because you interacted with this question earlier

Ted Shifrin

Not I.

copper.hat

05:52

DogAteMy???

Ted Shifrin

Where have you been? That goes back 6 or 7 years.

Remember the old “the dog at my homework” excuse?

copper.hat

Is that a label for someone in chat or a generic term you use?

"left my copy book at home, Sir"

"3 slaps for you, hold your hand out"

Ted Shifrin

06:12

Akiva had a dog-something name years ago.

leslie townes

if you have a former nickname on this chat, ted knows it and still uses it.

copper.hat

thankfully i don't have a nickname

leslie townes

you are legally copper.hat. it's on your passport.

copper.hat

card carrying member of the electronic frontier foundation

clear & tsa don't even know i'm there

leslie townes

@AkivaWeinberger this is cool.

user 85795

06:55

😎

Koro

07:12

Suppose that $f:X\to \bar R$ is $\scr F$ measurable, $\scr F$ is some sigma field on X. Let $\mu$ be a finite measure on $(X,\scr F)$. Given that $\int_A f d\mu=0$ for all $E\in \scr F$, can it be said that f=0 a.e.?

if f were non negative, then sure.

I think yes one can say that.

I take $E^+=\{x\in X: f(x)\ge 0\}$. Since, $\int_{E^+} f =0,$ it follows that f=0 for a.e. x in E^+.

Similarly, f = 0 on a.e. x in E^-.

$X= E^+\cup E^-$ so f is zero for a.e. x in X.

Mad

08:03

anyone can help me with this ?
https://chat.stackexchange.com/transcript/message/64807719#64807719

Jakobian

10:04

If $f$ is gauge integrable and $G$ is a continuous a.e. differentiable $N$-function (for example an indefinite integral of a gauge integrable function), then $\int_{G(a)}^{G(b)} f = \int_a^b (f\circ G)\cdot G'$ if $(f\circ G)\cdot G'$ is gauge integrable

For example, $G$ can be an absolutely continuous function here

porridgemathematics

12:37

@Jakobian in your proof, when you use Vitali's covering theorem, how do you get that $\lambda( \bigcup \mathcal{A} \setminus \bigcup_{n=1}^{\infty} I_{n}) = 0$?

oh I guess you just keep approximating with finitely many intervals, is that what you're doing?

also, I don't see why $\bigcup_{n=1}^{\infty} \bigcap_{k=1}^{\infty} E^n_{k} \subset E$, I can see why the other inclusion holds though

Jakobian

13:12

@porridgemathematics This is a corollary to Vitali's covering theorem

For all $\varepsilon > 0$ one can find intervals $I_1, ..., I_n$ such that $A\setminus \bigcup_{i=1}^n I_i\subseteq \bigcup_{i=n+1}^\infty J_i$ and $\sum_{i=n+1}^\infty |J_i|\leq \varepsilon$

The intervals $J_i$ are enlargements of intervals $I_i$

In the proof of Vitali's covering theorem its really that $I_1, I_2, ...$ can be fixed all at once

porridgemathematics

sure, Im okay with that part

I think the other part may not be correct though, I believe I found a counterexample

Jakobian

I think that's impossible

what's the counter-example?

porridgemathematics

it might be, im probably confused

basically why is it enough that if $x$ is in the intersection of $[c,d]$'s (length shrinking to zero) for which $\frac{f(d) - f(c)}{d-c} \geq r + \frac{1}{n}$, that the upper derivative at $x$ is $> r$?

Jakobian

If $x\in \bigcup_n \bigcap_k E_n^k$ (because its $E_n^k$ and not $E_k^n$)

Then we can fix some $q$ such that $\frac{f(d)-f(c)}{d-c}\geq r+\frac{1}{q}$ for some $x\in [c, d]$ where $0 < d-c\leq \frac{1}{k}$

porridgemathematics

right

Jakobian

13:21

Then write $\frac{f(d)-f(c)}{d-c} = \frac{d-x}{d-c}\frac{f(d)-f(x)}{d-x}+\frac{x-c}{d-c}\frac{f(x)-f(c)}{x-c}$

Notice that $\frac{d-x}{d-c}+\frac{x-c}{d-c} = 1$ and both are in $[0, 1]$

If both of these quotients were $< r+\frac{1}{n}$, then so would be $\frac{f(d)-f(c)}{d-c}$

So $\frac{f(x)-f(c)}{x-c}\geq d+\frac{1}{n}$ for some $c$ such that $0 < |x-c| \leq \frac{1}{k}$

porridgemathematics

ahh, okay, thats the part I was missing

nice proof

Jakobian

I've mentioned that this inclusion is tricky and requires to see that this quotient is a convex combination in here

but you probably didn't see that

porridgemathematics

i hadnt seen it

the result is still a little surprising to me

because there arent any assumptions on f

Jakobian

I suppose so

but $G(a) = \sup_{x\in [0, a]} f(x)$ is also measurable I guess, no matter what $f$ is

sometimes things just fit so nicely that we don't need any assumptions

porridgemathematics

thats true

Jakobian

13:42

I find the proof really cool because it means I can talk about this comfortably without some weird arguments about measurability of derivative

I was trying to see that if $f:[a, b]\to\mathbb{R}$ is differentiable on a measurable set $E$, then $|f(E)|\leq \int_E |f'|$

so I can see why Banach-Zarecki theorem is true

(literally differentiable, not almost everywhere)

Then the argument goes that if $f$ is a continuous $N$-function of bounded variation, then $|f(d)-f(c)|\leq \int_c^d |f'|$

Thus being bounded by an absolutely continuous function $F(t) = \int_a^t |f'|$, $f$ must also be absolutely continuous

(this is the tricky implication btw)

This is because of Lebesgue's differentiation theorem applied to $f$ and that $f'$ is Lebesgue integrable

If $A$ is any set then I can talk about $|A|$ comfortably, but integrals can only really be done on measurable functions, thats why my concern

porridgemathematics

13:57

don't you immediately get that $f$ is absolutely continuous if you know $|f(d) - f(c)| \leq \int_{c}^d |f'|$? to me thats just saying $|\mu_{f}| \leq |f'| d\lambda$, and $|f'| d \lambda$ is absolutely continuous w.r.t $d\lambda$, thus so is $\mu_{f}$

here $\mu_{f}$ is the measure induced by $f$

(assuming $f$ is continuous and $f'$ is in $L^1$)

but $f'$ in $L^1$ is immediate from bounded variation

there's a correspondence (for measures on the real line) between BV right continuous functions and finite (complex) borel measures

so $f$ being BV and right continuous means its induced measure's absolutely continuous part is finite (on $[a,b]$), which means $f'$ is integrable

similarly theres a correspondence between sigma finite signed measures and BV right continuous functions with bounded positive or negative variation (on the real line)

oscarmetal break

14:22

Hi, I would need some help to a question. I am trying to prove the symmetry group $S_5$ contains a subgroup of cardinality 20. Right now I have able to prove the set of affine transformations $x\to ax+b$ with $ a\in F^x_5$ and $b\ in F_5$ is a group of cardinality 20. And I just need to prove there exists an injective homomorphism from this set to $S_5$ such that the image would be the desired subgroup. However I have the difficulty to formula this map.

Jakobian

14:36

@porridgemathematics Yes. That's why you show that

To prove its absolutely continuous

porridgemathematics

i was just wondering why we need the lebesgue diff. theorem here

or I guess in 99% of what I was saying, the lebesgue diff. theorem is lurking behind in some justifications

like the correspondence between BV functions and radon measures on the line

Jakobian

Lebesgue differentiation theorem says that if $f$ is a function of bounded variation then $f'$ exists a.e.

porridgemathematics

ah okay

Jakobian

yeah its the different one of the same name

porridgemathematics

I see

Jakobian

14:40

I've seen the name only in Bartle but I'm following that book so I'll be repeating it

porridgemathematics

thats the book on the heinstock kurzweil integral right

well, one of them

Jakobian

yeah

Modern theory of integration

The theorem is in chapter 14 and in the appendix

(which I've realized that the proof in the appendix is actually pretty easy)

and by easy I mean digestible by me

porridgemathematics

lol, i end up reading the appendix of a lot of textbooks out of some sort of OCD, its gotten better now but I used to not be able to keep studying out of a book if I couldn't digest the proof of some result cited in the appendix

I would have to wait till I digest the proof of the result cited in the appendix and then continue with the main narrative

occasionally the results in the appendix , or the proofs of them, turn out to be pretty useful to internalise - a lot of other proofs in the main narrative often build on tricks used in proofs for appendix results

Jakobian

I do the same, I have to know why things are true. That's why I've opened another book just for the proof of Banach-Zarecki

And now I'm doing the same thing for integration by substitution for absolutely continuous functions

porridgemathematics

fair enough

Jakobian

14:46

I've used to be more stuck with one book, but I'm more open to exploring and reaching out to other sources now

so I think I'm improving... (also in terms of searching through articles etc.)

Xander Henderson

@porridgemathematics My experience is that the appendices of books summarize results which the author presumes the reader already knows. Hence I tend to skim appendices before I start reading a book, but I don't sweat them too much after that.

Jakobian

Its also where they push theorems that aren't part of the main discussion (which I think, what you're saying above is also a part of)

i.e. things the author doesn't want to focus on

to think of it, thats precisely what appendix is

Xander Henderson

@Jakobian Right. Typically, these theorems are kind of understood to be prerequisites. They aren't part of the main narrative, you probably learned them somewhere else, don't worry about it.

Jakobian

In the example of Lebesgue differentiation theorem above, it was more that the result needs the definition of outer measure and some trickery that's few pages long

so it was more that the author didn't want to focus on it, but the proof wasn't particularly easy (or assumed to be learned before)

porridgemathematics

@XanderHenderson yeah thats true, there are times when things in the appendix aren't prerequisites though (e.g. in hatcher)

or if you're reading a book on geometric analysis, and the chapter is on the hodge decomposition theorem, then a bunch of the sobolev space stuff might be relegated to the appendix (sobolev spaces on manifolds)

also not necessarily prereqs there, just not the main focus of the book , i.e. can be blackboxed and the author intended them to be if they weren't prereqs

Jakobian

14:55

sometimes its also a place to put stuff the author thought was cool

porridgemathematics

yeah

Xander Henderson

@porridgemathematics I mean, to-may-to, to-mah-to. I feel that if an author is "black boxing" something, that means that you are supposed to already know it (and it isn't vital if you don't, but you really should).

@Jakobian I've seen that a few times, but I think that most books tend to use the last three chapters for "here's some cool stuff that I am not an expert in, but it's cool, right?!" :P

Mad

@XanderHenderson are you well versed in differential geometry?

porridgemathematics

yeah, maybe should know of it, I dont think its really necessary to know the exact proofs of certain things, as long as you have a sense of why the result is true this is usually enough

Xander Henderson

@Mad Nope.

porridgemathematics