Mathematics - 2021-07-20 (page 1 of 2)

robjohn

00:08

I went to Balboa Is every summer when I was a kid. My parents rented a different place every year.

We know someone who still rents a place (the same one) there every summer, so we go for a day once a year.

amWhy

00:36

@robjohn Wisconsin is a great place to consider, unless you've ruled out winters. Do you know snow exists some places, experimentally?

;-)

leslie townes

01:23

i like snow, i dont' like the post-snow period where everything is slushy and warming up.

one of my work friends used to sell frozen bananas on balboa. you may have met him.

copper.hat

snow makes it easier to climb some hills.

leslie townes

but why would anyone want to climb hills

you climb the hill, i'll supervise.

my daughter brings me my junk mail every day. i let her open it unless it's important (could not have our car registration shredded by toddler hands). it's always mortgage related with the equal housing logo on it. "it's another house!" my daughter announces.

copper.hat

02:46

i'll take the george mallory on that.
teach her to open them carefully.

leslie townes

03:01

she just spent 20 minutes yelling about getting into pajamas.

yelling and struggling and kicking me in places that are more pleasant not to be kicked.

i'll be happy when this period of development is over.

robjohn

@amWhy Mammoth Lakes is known for skiing, so I have heard that it snows somewhere.

@leslietownes There are two main shops on Marine that sell frozen bananas, Dad's and Sugar & Spice. We go to Dad's for our Balboa Bars and frozen bananas.

leslie townes

i am also partial to Dad's.

robjohn

They both claim to have invented the Balboa Bar.

leslie townes

i think that's funny. they're on the same block. if i get a vote, it's dad's.

robjohn

That's where my loyalty lies.

leslie townes

03:15

they're not quite right next to each other but really close, if memory serves.

robjohn

Yeah, they are a few storefronts apart.

leslie townes

there's always money in the banana stand.

robjohn

We have always taken a walk around the island after dinner, and Dad's is always on the itinerary.

It is always crowded outside both of them.

leslie townes

i like balboa a lot. in the beforetimes when we had a summer program at my work i would take people for a frozen banana at dad's and expense it.

i work at fashion island. you can see my office in the 'show me the money' scene of jerry maguire.

who knows when they'll let us back into it, at this rate.

robjohn

Ah, Fashion Island. We pass by there when we are going down MacArthur.

No, Jamboree. MacArthur is the next exit. We take that sometimes

leslie townes

03:22

we've just been told masks on again. we had masks off for about a week at the office if people were vaccinated, which i am.

i was going to say, jamboree.

robjohn

Yeah, LA is mask on when indoors (except homes).

leslie townes

i have a number of friends with compromised immune systems who i have not been able to see in a great while. i mask up even if it is not formally required.

i live in LA county where compliance seems to be very good, at least in my neighborhood. in OC it is a mixed bag.

robjohn

I have no problem, but I know that there are people who seem to find wearing a mask an imposition. Awww.

leslie townes

yeah, god forbid you take 5 seconds not to kill somebody. poor thing. my neighbhorhood is really good about it.

i go to OC for office stuff or something else and it's all across the map.

robjohn

For the most part, people around here are as well. (compliant)

leslie townes

03:28

i have to be careful because my daughter is too young to be vaccinated and goes to a day care where people come down with covid every week or two. i think my wife and i have both had it, but we aren't sure and don't want to roll dice with other people's lives

robjohn

One of our favorite places to meet our friends from San Diego is the Five Crowns, which is just down Coast Hwy from Balboa.

We meet there for a late Christmas

leslie townes

in CDM? i think i've been there.

robjohn

yep

British looking place

leslie townes

i overlook the colonialist overtones and enjoy the fine steak

robjohn

We have the Tam'O'Shanter here in Los Feliz. Much closer, but Scottish.

leslie townes

03:30

my best friend and i go there. it's one of our mainstays.

robjohn

It is actually in Glendale, which is where I grew up.

We go there on Mother's Day and Thanksgiving, at least

leslie townes

we go there whenever i can get up to see her.

robjohn

we missed last year, of course

leslie townes

a weird thing about LA county is how it is one of the largest counties on the planet but it's very small in terms of locations and social connections. we probably have a mutual friend.

robjohn

Most likely.

leslie townes

03:32

and you end up knowing everybody. i know people who have directed feature films and i also know their drug dealers.

weird place to live.

robjohn

My mother used to know 70% of everyone in Glendale. She was a member of a slew of social organizations.

leslie townes

my bff's roommate did all of her work up there. there's definitely a connection. wouldn't need six degrees.

robjohn

I had a friend who worked in that industry, I bet he knew some of the people you knew

Dubias

Can someone pls help me with a quick question that I have regarding random variables in statistics?

leslie townes

i can entertain such questions although i do not know my availability to evaluate them until they are posed.

Dubias

03:35

great, thanks, ill post it right ahead

robjohn

ask, don't ask to, oh, just say it...

leslie townes

i thought about tapping the sign.

robjohn

of course, if they are on mobile, you don't see that.

Dubias

The question is as follows: Let $S=\{1,2,3,...\}$ ,and let $X(s)=s^2$ and $Y(s)=$1/s for $s \in S$. For each of the following quantities, determine (with explanation) whether or not it exists. If it does exist, then give its value. (a) $min_{s \in S}X(s)$, (b) $min_{s \in S}X(s)$, (c) $max_{s \in S}Y(s)$, (d) $min_{s \in S}Y(s)$

I know the answers for a and b, but the answers for c and d at the back of the book im using seem to not checkout with what i have

robjohn

a and b look the same

Dubias

03:39

for a I got the value 1 and for b I got that it doesnt exist

robjohn

@Dubias so b must be max

Dubias

oh my apologies, i didnt notice

robjohn

c is 1 and d does not exist?

Dubias

thats what ive also thought but the textbook seems to disagree lol

robjohn

what does the book say?

Dubias

03:41

it says that c doesnt exist while d is 1

im not sure if it has something to do with the fact that X and Y are random variables

robjohn

it's got them backwards then

Dubias

thats a bit odd i checked the page where the author listed all the answer key errors and its not listed there which means nobody had reported that its mixed up

i guess its so basic that people didnt bother reporting it

robjohn

you have c and d copied correctly?

Y(S)={1,1/2,1/3,1/4,...}

Dubias

oh no it was me who mixed them up... lol so sorry

robjohn

ok. It has a max but no min

Dubias

03:46

it should've been (c) $min_{s \in S}Y(s)$ and (d) $max_{s \in S}Y(s)$

robjohn

then I agree wit the book

Dubias

can't we simply plug in 1 for the min though?

and get 1/1 = 1?

robjohn

that is greater than all the other elements of Y(S)

1 min ago, by robjohn

Y(S)={1,1/2,1/3,1/4,...}

inf is 0, but it is not in the set, so no min

Dubias

sorry what do you mean by inf is 0?

@robjohn ahh this makes sense

robjohn

the infimum of Y(S) is 0.

Dubias

03:52

got it, so the value for (d) would be 1 because when we sum everything in the range of Y(s), it will add up to 1?

robjohn

no, the maximum of Y(S) is 1. That is the biggest element.

Dubias

I see, thanks a lot for your help!

robjohn

there is no smallest element

Dubias

Got it! so (a) is 1 simply because the smallest value $s^2$ can take is 1 and max is inf

makes sense, thanks

robjohn

yep

LearningCHelpMe

05:41

Transformations in the complex plane is the bane of my existence...

epsilon-emperor

imgur.com/a/nPX13xk

Could I get some help with parts (b) and (c)?

epsilon-emperor

06:05

(a) uses Baire Category, so I'm guessing (b) and (c) might do so too

For (b) we need an explicit construction. For (c) I have no clue.

PM 2Ring

@AMDG If you can multiply, you can compute reciprocals using Newton's method. Here's a brief demo in Python. Note that this algorithm doesn't even need to use bit shifts, it just accesses the mantissa & exponent values of the floating-point format directly.

LearningCHelpMe

What programs support the graphing of the loci of complex numbers? I would like to find out what this sort of thing would trace out $$arg(\frac{z-a}{z+b}) = \theta$$. It doesn't seem Desmos can do it.

PM 2Ring

sagecell.sagemath.org/…

@LearningCHelpMe SageMath can probably do what you want. I haven't used it much for complex numbers, but I posted a 3D plot of the branches of the Lambert W function here a little while ago.

LearningCHelpMe

06:20

@PM2Ring Thanks. I will check it out

PM 2Ring

Mar 29 at 23:46, by PM 2Ring

Lambert branches

robjohn

@LearningCHelpMe Mathematica can, if you pose it right: ContourPlot[Arg[(x + I y - 1)/(x + I y - 2)] == Pi/3, {x, 0, 3}, {y, -1, 1}]

PM 2Ring

Here's the Sage docs for standard 2D complex plots: doc.sagemath.org/html/en/reference/plotting/sage/plot/…

robjohn

here is the image

LearningCHelpMe

Thank you all for the help

robjohn

06:35

@LearningCHelpMe it gives the expected circle when you add the option AspectRatio -> Automatic, that forces the x-y units to be equal

@LearningCHelpMe Do you understand why that should be part of a circle? see inscribed Angle

PM 2Ring

06:51

@LearningCHelpMe Here's the basic Sage plot of that function: plot

robjohn

07:16

@PM2Ring that matches the plot above where the hue transitions from orange to yellow.

PM 2Ring

07:27

@robjohn Indeed! Sage has a lot of plotting options, and I'm only familiar with some of them. Here's something kinda equivalent to your version: implicit plot

Lately, I've been exploring its 3D parametric plots. Eg, drawing Bezier curves on an ellipsoid of rotation.

I'm actually trying to draw circles on ellipsoids. (That is, the locus of points at a constant geodesic distance from the centre, rather than a path of constant curvature). I can create geodesic paths ok, but doing those circles is a little trickier.

Unfortunately, my differential geometry skills are fairly limited, although I did write an answer a few years ago that does some differential stuff on an ellipsoid: math.stackexchange.com/a/1340899/207316

Mad Spaces

09:18

For $ A \subset B $ does then $ f^{-1}(A) \subset f^{-1}(B) $ For a function f continues?

For a general funciton, this does not work, true?

Leaky Nun

@MadSpaces ça se vérifie en général

Mad Spaces

i am sorry i do not understand you. I speak English or German.

Leaky Nun

sorry

the fact is true in general

Mad Spaces

This doesnt make sense, i drew two cricles, with the smaller one being inside the bigger one and i call the smaller one A and the bigger one B. Then i drew lines out of the circle (as function building) and i can make all the lines from the smaller one be concentrated on another part of the original set X and the other lines of B concentrate on another part of B

Then it wont work, correct?

Oh. excuse me

I am dumb, yea you are right hahaa. now i get it

Mad Spaces

09:57

Heres another question, we have as example in our lecture for a Function that is continuious and bijective but its reverse function is not continuious.

$ X = [0,1[ \cup [2], Y = [0,1] , f(t) = t, t \in [0,1[ $ and $f(2) = 1$

Why is $f^-1$ not continuious ? i have tried to check with the Open balls definition and the closed balls but i am not reaching a result?

Alessandro Codenotti

10:11

Is $A=(1/2,1)$ open in $X$? Is $f(A)$ open in $Y$?

Mad Spaces

Well $f(A) = A $, and they seem both open to me, but i am guessing that is not the answer you expecting, mind enlightening me?

epsilon-emperor

1

Q: $ M(X) $ is a Banach space if $ || \mu || = | \mu |(X) $. Rudin's book (RCA).

let $ M(X) $ be the space of all complex regular Borel measures on a locally compact hausdorff space $ X $. Prove that is a Banach space if $ || \mu || = | \mu |(X) $. Where $ | \mu |$ is the total variation of the complex measure $\mu $. the idea to solve this exercise is to define an isomorphi...

measure-theory

Is there a way to solve this without that isomorphism?

Alessandro Codenotti

10:34

Ah wait I was not thinking, I went backwards. Anyway f^-1 is mapping a connected space to a disconnected one, hence is not continuous

Mad Spaces

We have not learned that statement. Maybe you could argue with open and closed balls or epsillon delta creiterium ?

Alessandro Codenotti

11:49

@MadSpaces you can use open sets too. Is {2} open in X? Is f({2}) open in Y?

Mad Spaces

12:41

I think the answer is yes to both these questions.

Alessandro Codenotti

Why is f({2}) open?

Mad Spaces

because any radius 0<M<1the ball B(1,M) is contained in Y.

Alessandro Codenotti

But you want the ball to be contained in f({2}) to say that the latter is open

Mad Spaces

the set [f(2)]=[1] is obviously not open in Y . Right

So basically [2] is open in X but f[2] is not in Y, thus f-1 is not cont

Alessandro Codenotti

13:17

Right

Mad Spaces

Alright, got it, thanks!

epsilon-emperor

13:30

@robjohn I'm reading this (math.stackexchange.com/a/61549/425395) answer of yours, and I do not understand what you mean by "apply Riesz Representation Theorem"

I have concluded the existence of $C < \infty$ such that $\|fg\|_1 \le C \|g\|_q$ (not the way you did, but still got till here)

How do you propose applying Riesz Representation theorem now to get the result? and there are so many Riesz Representation theorems, which one do you want to apply?

vitamin d

14:49

@BalarkaSen Hi Balarka

Balarka Sen

@vitamind Up for some plurisubharmonicity?

vitamin d

@BalarkaSen Absolutely

Balarka Sen

This is what we had last time:

yesterday, by Balarka Sen

Let me round the discussion off for today by saying @vitamind has proved f : C -> R is subharmonic if d^2f/dzdzbar >= 0, and a plurisubharmonic function f : C^n -> R is one which restricts to a subharmonic function on every holomorphic curve (= a copy of C) in C^n

Can you combine these two to show that for any plurisubharmonic function $f : \Bbb C^n \to \Bbb R^2$, $\partial^2 f/\partial z_i \partial \overline{z}_j \geq 0$?

For all $1 \leq i, j \leq n$

vitamin d

Not quite sure what you mean by: restricts to a subharmonic function on every holomorphic curve.

Balarka Sen

For any holomorphic map $g : \Bbb C \to \Bbb C^n$, $f \circ g : \Bbb C \to \Bbb R$ is subharmonic yes? You can think of $g$ as a curve inside $\Bbb C^n$. Curve because complex dimension 1.

It's image in $\Bbb C^n$ rather.

vitamin d

15:01

Ok.

Balarka Sen

Sorry, I meant $\partial^2 f/\partial z_i \partial \overline{z}_i \geq 0$. Although it's an interesting question what happens for $i \neq j$.

... and I wrote $\Bbb R^2$, whereas I meant $\Bbb R$.

vitamin d

I think I need a hint. Please note that I'm not too familar with SVC, I only wanted to have an image of a pseudoconvex domain in my head for starters, but probably not possible without understanding (pluri)subharmonicity.

Balarka Sen

Consider $g(z) = (0, \cdots, 0, z_i = z, 0, \cdots, 0)$ in the definition of plurisubharmonicity above.

That is the holomorphic curve which is the $i$-th coordinate axis.

vitamin d

Ok.

Balarka Sen

What does that yield?

vitamin d

15:15

Do you mean what the Wirtinger derivative yields?

Balarka Sen

Yes

vitamin d

\ge0?

Balarka Sen

$f \circ g$ is a subharmonic function, I want you to write that in terms of Wirtinger derivatives and keeping in mind $g$ is the very special function as above.

vitamin d

\del^2\del{z_i}\delbar{z_i}(f(0,0,..,z_i,0,..0))

Balarka Sen

OK. And then you can do this computation at the point (0, ..., 0) to get d^2f/dz_idzbar_i >= 0 at the origin

If this is true at the origin it is true at all points in C^n. Just translate, yes?

vitamin d

15:22

Yes. Should I apply the chain rule?

Balarka Sen

Correct.

But it's OK, I am satisfies with what you did at the origin for now.

I was going to ask you to do the above with the function g(z) = (0, ..., 0, z, 0, ..., 0, z, 0, ..., 0) where it's nonzero at the i-th place and the j-th place

vitamin d

\del^2\del{(z_i,z_j)}\delbar{(z_i,z_j)}(f(0, ..., 0, z, 0, ..., 0, z, 0, ..., 0))?

Peter John

If two spaces have the same homology group for each degree, then they have the same cohomology group for each degree?

Balarka Sen

What does d^2/d(z_i, z_j) mean?

@PeterJohn Yes. Universal coefficient theorem.

vitamin d

\del^2\del{z}\delbar{z}(f(0, ..., 0, z, 0, ..., 0, z, 0, ..., 0))?

Balarka Sen

15:33

You're not making sense, vitamind.

Peter John

@BalarkaSen You mean theorem in en.wikipedia.org/wiki/Universal_coefficient_theorem ?

Balarka Sen

Yes

vitamin d

I don't know.

leslie townes

balarka you're too young to know about the universal coefficient theorem. please delete that from your mind and try again later.

robjohn

@epsilon-emperor The same theorem that the OP is using. I have just shown that such a $C$ must exist, and the missing $C$ was what they were looking for.

leslie townes

15:43

nothing makes me feel older and less competent than seeing people on here. which is good. i will need the next generation to toil in order to support me in old age.

epsilon-emperor

@robjohn Makes sense. I was able to complete the proof a few hours after I posted on this chat!

Wolgwang

What is the process to convert an answer to a comment?

leslie townes

it begins with a mommy and daddy who love each other very much. that's my copper hat comment of today.

he's not even here to see it. what a shame.

hyper-neutrino

If an answer should be a comment, flag it for mod attention and indicate that it should be a comment. If you flag it as Not An Answer, it goes into the Low Quality review queue and might be deleted by regular users instead.

leslie townes

oddly enough i just saw a few comments that should be answers. :)

hyper-neutrino

15:49

That's a pretty common thing too :) and there is no way for mods to do that change (for good reasons), only vice versa.

but yes, something a lot of people don't know is that NAA and VLQ flags are only shown to moderators after an hour

they are first put into the review queue instead

leslie townes

i have been guilty of this many times. the problem is i click into the stuff and begin what is honestly a comment and then it turns into an answer.

hyper-neutrino

@Wolgwang well, there is no other way to convert an answer to a comment

the other option is to just ask the OP to post it as a comment instead and delete the answer

Wolgwang

Ohk Thanks :-)

epsilon-emperor

The dual space of $\mathbb C$ is $\mathbb C$ itself?

hyper-neutrino

it takes three clicks to convert an answer to a comment (more if you want to access the other options) but IIRC math.SE has a decent flag backlog so if you feel that it is not that important it wouldn't hurt to give the mods here a bit less of a headache :)

Wolgwang

15:56

@hyper-neutrino User's 'last seen' is 2016 :-/

hyper-neutrino

welp :\ guess that's not an option lol

leslie townes

they'll come back.

you gotta believe.

i vanished for a few years and then returned, like a phoenix from the ashes.

epsilon-emperor

The dual space of $X = \mathbb C$ is the set of all bounded linear functionals on $X$, to $\mathbb C$ (which is just $X$). Suppose $f\in X^*$, and $f(x) = ax$ for some $a\in \mathbb C$. Then $f(x)$ is determined on all of $X$, for obvious reasons. Hence, $X^*$ is isomorphic to $\mathbb C = X$.

Sounds right?

robjohn

@leslietownes That is often because the question is really a good question, but posed as a PSQ. People answer in a comment so that they don't get penalized for violating the EoQS.

leslie townes

i have been guilty of that, too. :)

robjohn

16:09

@leslietownes Did is a very high rep user who has not been here since 8 Feb 2019. One never knows if these people will ever come back.

hyper-neutrino

Howard was a very prominent user on Code Golf (and Stack Overflow) and in 2014 he just vanished and has never been seen since.

leslie townes

that's sad. you hope it's good stuff and not bad stuff that kept them from coming back.

hyper-neutrino

Yeah... we can only hope they either just found better things to do with their life and moved away from here. But a user that prominent and active last being seen within 2 minutes on their two sites and never reappearing again ever is... definitely not comforting.

leslie townes

i left because of a career change and came back largely because of a pandemic induced desire not to be going crazy while i'm waiting for work stuff to come in.

hyper-neutrino

His last activity on SO was a year before (and Code Golf about four months before) he never showed up again so I do hope it was just eventual loss of interest and he just logged off and never came back.

@leslietownes Visiting SE is not what I would first recommend if you do not want to go crazy :P

leslie townes

16:18

yeah, i have bad judgment.

there is a lot of Hurry Up And Wait in my field and SE is a balm for that.

how is it 87F at 9 in the morning.

Sayan Chattopadhyay

@leslietownes Isn't it in every algebraic topology course? I just took one in my 3rd year which went over it in quite detail

robjohn

@leslietownes it was only 82ºF at 9 AM here.

leslie townes

it's rarely this hot, we're about a mile from the water. should not be happening.

i don't want to hear about undergraduates taking algebraic topology courses. :)

we got the classification of surfaces only. no algebra and topology at the same time.

it is absolutely boiling right now and some landscapers are at work outside. i can't imagine.

Sayan Chattopadhyay

Oh I see. I had it(classification) in my topology course and the next on altop did all this.

robjohn

@leslietownes seeing people, or seeing young people?

. o O ( hopefully, not dead people )

leslie townes

16:29

not seeing dead people. but people of all sorts. seeing you makes me feel like i lost a lot of mental capacity before i should have.

whatever. i'll never run as fast as a cheetah. i'm fine with that and most of my issues in this area are ultimately about that.

robjohn

@leslietownes you never have to run faster than a cheetah, just the others that are being chased by that cheetah.

the advantage to living in a herd.

Question: If I am being chased by a cheetah, and they are less than 6 feet behind me, do I need to put on my mask?

hyper-neutrino

Only if the cheetah hasn't gotten both shots of the vaccine yet.

robjohn

Ah, thanks!

leslie townes

if you're in LA county i think the mask needs to be on if this is happening indoors.

epsilon-emperor

https://math.stackexchange.com/questions/1183163/fourier-coefficient-of-complex-measure-exercise-in-rudin

So this is a problem from Rudin, and I think it's trivial. I'm probably wrong, so hear me out.

$$\int e^{-int}d\mu=\int\cos(nt)d\mu-i\int\sin(nt)d\mu \to 0$$ as $n\to\infty$ means that the real and imaginary parts go to zero separately. Using $\cos nt = \cos (-nt)$ and $\sin nt = -\sin (-nt)$ we have the result immediately, don't we?

Koro

16:35

at a dense set problem

robjohn

@leslietownes well, yeah, I know that. I'm not stupid ;-)

epsilon-emperor

I don't understand why Rudin even gave this exercise, and why the OP and answerer have done so much? I'm surely missing something

Koro

I am stuck again

4

Q: Positive integer multiples of an irrational mod 1 are dense

I'm not sure how to solve this one. Thank you! $2.$ For any $\alpha\in \mathbb R$ we define $$\lfloor \alpha \rfloor = \max_{n\in\mathbb Z}\{\,n\mid n\leq \alpha\,\}$$ and $$\alpha\bmod 1 = \alpha - \lfloor \alpha \rfloor$$ Let $\alpha$ be irrational. (a) Given $n\in\mathbb N$ show that $\{\,k\a...

real-analysis irrational-numbers

I think the accepted answer is wrong

leslie townes

why cheetahs are chasing you indoors is your business and not mine. i won't pry into your personal life.

Koro

because $\{k_2\alpha\}-\{k_1\alpha\}=\{(k_2-k_1)\alpha\}$ but $k_2-k_1$ can be positive or negative also. This case was not considered in the answer.

robjohn

16:38

@Koro Didn't we go over this in chat recently?

Koro

Hi professor @robjohn, that was when we were discussing on $\mathbb Z$.

Oliver Diaz

@epsilon-emperor: Consider the real an imaginary parts of the measure. Convince yourself that it is enough to assume eta $\mu$ is a (finite) real measure, and then take conjugates.

Koro

But here we have set of naturals $\mathbb N$

That's why I am confused.

again

robjohn

@Koro is there a big difference?

Koro

I think that was $\{\sqrt {n}\}$

epsilon-emperor

16:41

@OliverDiaz Hmm, but why doesn't my argument work?

Koro

@robjohn yes, I think. Because $\{k_2\alpha\}-\{k_1\alpha\}\gt 0$ does not imply that $k_2\gt k_1$

and it is crucial here for the above linked question. :'(

Oliver Diaz

@epsilon-emperor: then you have to deal with the measure $\mu'(A)=\mu(-A)$ and then you get into circular arguments.

Koro

if $k_2\lt k_1$ then $0\lt 1-\{(k_1-k_2)\alpha\}\lt \frac 1N$ and this doesn't give me an element of given set in $[0,\frac 1N\rbrack$ and I want this because first I want to show 0 is a limit point of the given set

epsilon-emperor

$\hat\mu(n) \xrightarrow{n\to\infty} 0$ implies that $\int \cos nt \, dt \xrightarrow{n\to\infty} 0$ and $\int \sin nt\, dt \xrightarrow{n\to\infty} 0$. Since $\cos nt = \cos (-nt)$, $\int \cos nt\, dt \xrightarrow{n\to-\infty} 0$. Since $\sin nt = -\sin (-nt)$, $\int \sin nt\, dt \xrightarrow{n\to -\infty} 0$.

Isn't this correct, and gives the conclusion directly?

Ah, shit. Typos corrected: $\hat\mu(n) \xrightarrow{n\to\infty} 0$ implies that $\int \cos nt \, d\mu(t) \xrightarrow{n\to\infty} 0$ and $\int \sin nt\,d\mu(t) \xrightarrow{n\to\infty} 0$. Since $\cos nt = \cos (-nt)$, $\int \cos nt\, d\mu(t) \xrightarrow{n\to-\infty} 0$. Since $\sin nt = -\sin (-nt)$, $\int \sin nt\,d\mu(t) \xrightarrow{n\to -\infty} 0$. Thus, $\hat\mu(n) \xrightarrow{n\to -\infty} 0$.

Oliver Diaz

17:01

@epsilon-emperor: the problem with your argument is that $\mu$ is complex. If you can show the following, then your argument will work: $(a_n+ib_n)(u+iv)\xrightarrow{n\rightarrow\infty}0$ )$a_n,b_n,u,v$ all real) does that imply that $(a_n+ib_n)u\xrightarrow{n\rightarrow\infty}0$?

@epsilon-emperor: the hint by Rudi is to show you that it is enough to consider the case where $\mu$ is a real measure. It is tricky to prove that the hint works, but it is one of those exercise that allow you to practice your understanding of the Stone-Weierstrass theorem (complex version) and the fact that continuous functions are dense in the $L_p(\mathbb{T})$.

epsilon-emperor

@OliverDiaz This is not necessarily true, right?

Oliver Diaz

exactly. You may cook-up some examples where the relation does not hold. On the other hand, the explanation of the OP on how he prove the facts of the hint is not entirely towards the end, that is that $d\mu$ can be substituted by $f\,d\mu$, where $f$ is bounded. I mean, the fact is true, but the argument is not entirely correct. I think that calls for the density of continuous functions in $L_1(\mathbb{T})$. In any vent, once that part (the hint) is worked out, the rest is easy.

epsilon-emperor

17:19

@OliverDiaz So I have showed it for when $f$ is trigonometric, and when $f$ is continuous, i.e. $f\in C(T)$

Is there an obvious way to extend from $f\in C(T)$ to $f$ being a bounded borel function, and then replacing $d\mu$ by $d|\mu|$?

Oliver Diaz

@epsilon-emperor: yes, the Stone-Weierstrass theorem (complex version). In this setting, it says that trigonometric polynomials (polynomials in $e^{it}$ and $e^{-it}$) are dense in $\mathcal{C}(\mathbb{T})$ (under the uniform norm)

epsilon-emperor

@OliverDiaz I have already shown that

I need to show that $f$ can be a bounded borel function, and then we can replace $d\mu$ by $d|\mu|$

I'm thinking of using: A function is measurable if and only if it is the a.e. limit of continuous functions.

epsilon-emperor

17:37

imgur.com/a/rxjTfpL seems helpful too.

Oliver Diaz

@epsilon-emperor: The Radom-Nikodym theorem gives you a function $h$ with $|h|=1$ such that $d\mu=h d|\mu|$. Since $\mathbb{T}$ has finite measure, $h$ is also integrable. Then, the density of $C(T)$ on $L_1$ gives you what you need.

epsilon-emperor

@OliverDiaz I see! Let me write this out. So we skipped the step of "bounded borel functions"?

Oliver Diaz

@epsilon-emperor: not exactly. The point is that bounded mesurable functions in $\mathbb{T}$ are also integrable.

@epsilon-emperor: Take a sequence $f_n\in C(T)$ such that $f_n$ converges to $h$ in $L_1(\mathbb{T})$. Then $\Big|\int_Te^{-int}(h-f_n)d|\mu|\Big|\leq\int_T|f_n-h|\,d|\mu|\xrightarrow{n\rightarrow\infty}0$ et voilà!

leslie townes

i just responded to an email with "I don't think that's illegal but maybe don't do it."

epsilon-emperor

17:57

@OliverDiaz Perfect. So that tells me I can replace $d\mu$ by $|h|\ d|\mu|$, and the assumption still holds

copper.hat

18:10

@leslietownes was that regarding adding ice cubes to a perfectly good pinot grigio?

Oliver Diaz

@epsilon-emperor: The real and imaginary part of $h$ are also bounded Borel functions and so, the statement of the problem holds if $d\mu$ is replace by $\phi d|\mu|$ for any bounded real function on $\mathbb{T}$. That is $\int_T e^{-int}h_{real}d|\mu|$ and $\inft_T e^{-int}h_{im}\,d|\mu$ also converge to $0$ as $n\rightarrow\infty$. The conclusion follows by taking conjugates.

epsilon-emperor

@OliverDiaz That makes sense, but we eventually want to conclude that $\int_T e^{-int} d|\mu| \to 0$, right? How are we doing that?

It makes sense for the specific $h$, that we can replace $d\mu$ by $|h| d|\mu|$, and the argument extends to the real and imaginary parts (since $\Re h \le |h|$ and $\Im h \le |h|$)

Also, @OliverDiaz, just wondering, how are you so good at Analysis? It's pretty cool.

Oliver Diaz

@epsilon-emperor: your too kind... I just so happens to have more years (possible one or two decades) of experience at this than you, that is all.

copper.hat

18:29

analysis requires a twisted mind

i mean that is the nicest possible way :-)

epsilon-emperor

Looks like it

copper.hat

i meant "in the nicest..." always seem to have typos.

leslie townes

copper for your benefit chat.stackexchange.com/transcript/message/58671318#58671318

i try to maintain the right tone even when you are not around.

epsilon-emperor

19:06

46 mins ago, by epsilon-emperor

@OliverDiaz That makes sense, but we eventually want to conclude that $\int_T e^{-int} d|\mu| \to 0$, right? How are we doing that?

Oliver, did you understand my question?

copper.hat

@leslietownes :-). i did explain the mechanics to my kids fairly early on. my daughter's response was silence, my son's was "can i go to sleep now?"

looks like Riemann Lebesgue lemma

epsilon-emperor

@copper.hat Yep, looks like it

leslie townes

i remember hearing it fairly early on too. my daughter has yet to ask how mama ducks produce baby ducks. i won't rush it.

she is, however, obsessed with the notion that animals can be babies. we've seen it mostly with birds.

she can recognize juvenile scrub jays even when they aren't blue yet. truly my daughter.

schn

19:49

@robjohn , for the $m$th degree Taylor polynomial it is. Anyway, sloppy wording...

copper.hat

@leslietownes when i was growing up there was an incredible nature series (often focusing on birds) called Amuigh Faoin Spéir (Out Under the Sky). unfortunately there seems to be no record of the shows that i can find. the show certainly provoked my interest in nature (and sketching). it was way ahead of its time.

leslie townes

irish state tv has a really good kids show. we sometimes have my daughter watch it.

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$Mathematics$ Mathematics