Mathematics - 2015-12-19 (page 1 of 6)

user174558

00:12

@OFFSHARING Hi Chris.

OFFSHARING

@JohnNash JASPER!!! Hi :-)

user174558

@OFFSHARING Hehe. Have you finished writing the preface?

OFFSHARING

@JohnNash Waiting for the book to be published. :-) Maybe you mean the preface to the second book. :D

user174558

@OFFSHARING Oh OK. The preface to the second book will just say 'This continues our work in the first book.' LOL

user174558

@OFFSHARING Ah, who stole it? Or is it coincidence?

OFFSHARING

00:16

@JohnNash I won't comment more on it. :-)

user174558

@OFFSHARING OK. It is not me, because I cannot understand your work. I can only open my mouth and admire the beauty, LOL.

OFFSHARING

@JohnNash hehe, don't worry.

@JohnNash Perhaps just a coincidence ...

user174558

@OFFSHARING Do you like Star Wars movies?

OFFSHARING

@JohnNash When I was young I liked them, but then I was less interested in them.

These days I rarely watch movies.

@JohnNash I mean when I was kid, I'm still young (my English, sorry).

Agawa001

@skillpatrol right on the mark pal!

skill patrol

00:26

:-)

OFFSHARING

By the way, how is the last Star Wars movie? Did anyone watch it?

Agawa001

dont know what is wrong im sleepless

OFFSHARING

imdb.com/title/tt2488496

user174558

@OFFSHARING I think Huy has watched it.

OFFSHARING

135 min - that's a alot from my life! In 135 min I write some pages in a book.

Agawa001

00:28

maybe its due to maths @skillpatrol

skill patrol

Thinking too much?

Agawa001

need a soporific

user174558

@Agawa001 Eventually you will sleep.

skill patrol

But, will you sleep deeply?

user174558

Eventually, he will.

skill patrol

00:32

How do you know?

user174558

Because the body eventually gets very tired.

user174558

Only an abnormal person will not get very tired.

Agawa001

eventually , but now isnt eventual

user174558

And eventually we would sleep forever.

Agawa001

keep that eventuality for u

skill patrol

00:34

Tired of being awake, or tired of thinking?

user174558

For me, no, for everyone.

user174558

Everyone dies, right?

Agawa001

the eventuality of "sleeping" after dying

skill patrol

Tired of being alive?

user174558

It's just a metaphor, stop picking on words, lol.

skill patrol

00:36

Or tired of living

Agawa001

ok need to try over, maybe counting from 0 to 1000 cyclic way would work now

see ya tomorrow

pjs36

Feels like Christmas cheer in here

skill patrol

later pal

Danu

@BalarkaSen How about... you know... out drinking with friends :D

skill patrol

Kids don't drink.

Balarka Sen

00:40

@Danu Seems like you could win the 5o'clock hat.

skill patrol

Not at 16, usually.

Danu

@skillpatrol Eh... what?

You know I'm a graduate student, right?

@BalarkaSen What's that?

Balarka Sen

There's a new secret hat called 5'oclock in the morning somewhere

skill patrol

@Danu Oops, I thought that was advice for him :-/

user174558

I am now using Linux Mint, Cinnamon, 64-bits.

Balarka Sen

00:45

4 answers with 0 upvotes. Unsung hero of the week!

user174558

If you are unsung, just sing and you will be sung.

skill patrol

Now who is acting childish?

user174558

@skillpatrol Who's, not whose.

skill patrol

Who is?

Danu

@skillpatrol Balarka is 16?

skill patrol

00:53

So he says.

I believe him.

user174558

He is 16 and I am a banana.

skill patrol

Define: banana

user174558

banana

Noun: banana ‎(countable and uncountable, plural bananas)

An elongated curved tropical fruit that grows in bunches and has a creamy flesh and a smooth skin.
The tropical tree-like plant which bears clusters of bananas. The plant, usually of the genus Musa but sometimes also including plants from Ensete, has large, elongated leaves and is related to the plantain.
(mildly pejorative, slang, ethnic slur) A person of Asian descent, especially a Chinese American, who has assimilated into Western culture or married a Caucasian (from the "yellow" outside and "white" inside). Compare coconut ‎(“assimilated Hispanic or Black”) or Oreo ‎(“Black person who is "black outside" and "white inside"”).
banana f ‎(plural bananes)
banana f ‎(plural bananes)

(14 more not shown…)

Adjective: banana ‎(not comparable)

Curved like a banana, especially of a ball in flight.
2002, Andrew Collins, Guild of Honor, page 53, ISBN 1403371490.
2006, Richard Witzig, The Global Art of Soccer, page 247, ISBN 0977668800.
banana ‎(invariable)

Verb: banana

third-person singular past historic of bananer

skill patrol

20 mins ago, by John Nash

It's just a metaphor, stop picking on words, lol.

Perhaps, the 3rd definition? @JohnNash

3. (mildly pejorative, slang, ethnic slur) A person of Asian descent, especially a Chinese American, who has assimilated into Western culture or married a Caucasian (from the "yellow" outside and "white" inside)

Wiki: metaphor

:-/

Agawa001

01:55

someone must teach me how to sleep

or may i read a book about it

skill patrol

Semiclassical

02:15

my version of that: put on something which is sort of interesting but requires consistent attention to follow. if i'm at all tired, then it starts to blur together

and then i'm out

Thomas Andrews

03:47

I kept a blackboard in my dorm room in college, and kept it through grad school. It got damaged in some basement flooding at some point and I had to pitch it.

Rajesh Dachiraju

0

Q: A strategy to prove/disprove a real analysis problem setup?

Given $g(x,\omega)$ and $f(x,\omega)$ are smooth, and as $\omega\to\infty$, $g(x,\omega) \to 0 \forall x$, except $x = x_o$ where $g(x_o,\omega)\to D$, where $D$ is a constant.More conditions/specific details of $g,f$ are given in the problem, but I want to avoid all of them as they are cumbersom...

real-analysis

skill patrol

06:13

@ThomasAndrews did looking at it help to put you to sleep? :P

Though, the sound of finger nails on a blackboard would certainly make a nasty alarm clock sound ;-)

Ted Shifrin

07:01

@anon: I saw your ping from a few days ago. I'm not ignoring you, just haven't been here. I think the answer is yes ... basically you need to remember that when you put the bi-invariant metric on a Lie group $G$, the one-parameter subgroups are the geodesics through the identity element. What you're doing in your example is lifting to $SO(n+1)$ and taking a geodesic there. That generalizes to any symmetric space of compact type, for sure.

anon

I figured something like that

skill patrol

07:21

Nice hat Professor @TedShifrin I take it you had a change of heart?

Dec 15 at 1:54, by Ted Shifrin

Nope, @Jasper. I'm way too old for hats.

Perhaps, still young at heart :-)

Agawa001

09:55

@skillpatrol took ur adviced and worked, i fell asleep lik a daed koala

David Lin

10:14

Just thought of a neat question.

Is there a mathematic model for happiness?

0celo7

@DavidLin $1-10$

iwriteonbananas

10:51

Hey @DanielFischer. I would like to show that the operator $$A:D(A)=\{f\in C_b(\Omega):Af=mf\in C_b(\Omega)\}\to C_b(\Omega), \quad Af=mf$$ is closed, where $m\in C(\Omega)$ and $\Omega\subset \Bbb R^n$.

So we take $(f_n)_n\subset D(A)$ with $\lim_n f_n=f$ and $\lim_nAf_n = g$, and we need to show that $f\in D(A)$ and $Af=g$.

I can't seem to figure out why $f\in D(A)$, i.e. why is $mf$ bounded?

Do you happen to have an idea for me?

wendy.krieger

@DavidLin I used twelfty for a quarter century, makes the maths easier.

Balarka Sen

Hi @iwriteonbananas

iwriteonbananas

Hey Balarka

I spent 20-30 minutes writing an answer to this math.stackexchange.com/questions/1581724/… but I gave up

Balarka Sen

posted lots of answers

@iwriteonbananas It's a very unclear question.

iwriteonbananas

indeedio

Balarka Sen

11:03

if it's asking for intuition of cross product, i'll just post my "picture" ;)

iwriteonbananas

What picture?

One should post a huge diagram, including the Eilenberg-Zilber map etc.

:P

Balarka Sen

This picture.

iwriteonbananas

Haha, right

Balarka Sen

iwriteonbananas

You love those comics, don't you

Balarka Sen

11:10

Who doesn't like Tintin?

It's great.

@iwriteonbananas Why? That's all there is to it. You have two piecewise linear functions on triangulated $X, Y$ respectively. Pull them back to $X \times Y$ by projection maps $X \times Y \to X$ and $X \times Y \to Y$. Multiply the two functions on $X \times Y$. This is cross product.

:P

Agawa001

i just made a comment, but tend to think that it fits an answer

dunno; im just backhurt with potentially downvoted posts

iwriteonbananas

@BalarkaSen Cross-product is just the map on cohomology induced by the EZ -map :P

Balarka Sen

Do you want me to post another Haddock picture? I have lots.

iwriteonbananas

lol

Balarka Sen

I mean, they're the best way to depict anger through internet.

iwriteonbananas

11:18

Not sure I would agree

Daniel Fischer

@iwriteonbananas We know that $mf_n$ converges pointwise (even uniformly, but we only need pointwise at the moment) to $g$. Where $m(x) \neq 0$, that implies that $f_n$ converges pointwise to $\frac{g}{m}$. But also $f_n \to f$ pointwise, so $g = mf$ where $m(x) \neq 0$. And where $m(x) = 0$ we have $m(x)f_n(x) = 0$ for all $n$, so $g(x) = 0 = m(x)f(x)$. Thus we have $g = mf$.

Balarka Sen

By the way, do you think my cup product question is worth asking on HSM?

iwriteonbananas

@DanielFischer Ah, thanks, makes sense. I should have found that myself.

@BalarkaSen Yeah, go ahead and ask it.

Balarka Sen

iwriteonbananas

lol

user174558

11:22

The Tintin pics look good.

user174558

Should I watch the Tintin movie @BalarkaSen?

Balarka Sen

Spielberg's? It's good, but nothing compared to Herge's comics.

Too much action.

user174558

I hate Batman by the way.

Balarka Sen

You may watch it if you want. I enjoyed it.

user174558

I never liked comics books.

user174558

11:26

I can't believe there is HSM SE.

Balarka Sen

*comic books.

Agawa001

bandes dessinés*

robjohn

12:01

@JohnNash heh. Nice blue square Mr Nash...

Agawa001

12:13

@robjohn thats called "the moody square"

or "chameleonic square"

Balarka Sen

@iwriteonbananas Posted.

robjohn

12:36

@BalarkaSen I really want to post an answer to the subject line, but it would be inappropriate...

OFFSHARING

To see my book published and I'm out of mathematics, completely. I let @BalarkaSen @TedShifrin @CarlMummert @DanielFischer @anon to bring their contribution to mathematics.

Balarka Sen

@robjohn Not sure if I get the joke.

OFFSHARING

And the rest.

Balarka Sen

Ted Shifrin has already contributed a lot to mathematics.

OFFSHARING

Please to never contact me, for all I mentioned (never means never).

I don't need anyone's of these opinions on my book (I'll consider you respect me doing).

user174558

12:43

@robjohn You mean 'Professor Nash', lol.

user174558

@OFFSHARING Of course, I am not on the list, lol.

OFFSHARING

@JohnNash I can always contact me, you always respected my work.

Balarka Sen

Tautologically, yes, you can always contact yourself.

user174558

@BalarkaSen Me too, I have 800 rep on MSE, lol.

OFFSHARING

@JohnNash How about having a website about cooking? No math at all, just cooking. ;)

user174558

12:48

Notice that my last three lines end with ", lol."

Huy

13:16

@Clarinetist you here by any chance?

OFFSHARING

@BalarkaSen tired enough to make a lot of mistakes. Wish you ever in this life attain in your area the level I attained in my area.

No bragging, just a wishing to you.

iwriteonbananas

@BalarkaSen Nice, looks like a lot of effort went into that question.

@DanielFischer I'm trying to modify the argument you gave above to the following scenario: I want to conclude from $(f_n)_n\subset L^p$ with $\lim_nf_n=f\in L^p$ and $\lim_n mf_n=g\in L^p$ that $mf=g$ holds almost everywhere. Here $m$ is a measurable function.

OFFSHARING

13:33

The best place to do mathematics is in perfect isolation.

@iwriteonbananas You are a Romanian professor, isn't it? If so, it's not a shame to ask stuff with your real name and say you don't know your stuff.

iwriteonbananas

@OFFSHARING Yes. I enjoy my anonymity.

OFFSHARING

@iwriteonbananas You enjoy it, yes, but you know what? If you ever publish anything using the stuff you learn here it would be good to mention it, not to present yourself to the world as someone you actually aren't.

iwriteonbananas

Alright, thanks for that piece of unsolicited advice.

OFFSHARING

@iwriteonbananas If I didn't have ideas in mathematics I'd leave it in no time. There are many other jobs one can do.

iwriteonbananas

That's great.

OFFSHARING

13:41

Like a DJ in some exotic clubs from Ibiza, Spain (those famous clubs if you know what I mean).

iwriteonbananas

@DanielFischer So by $L^p$ convergence, we can extract subsequences of $(f_n)$ and $(mf_n)$ on which we have pointwise convergence, say $f_{n_k}(x)\to f(x)$ and $mf_{l_j}(x)\to g(x)$. Is it then true that $\lim_k f_{n_k}(x)=\lim_j f_{l_j}(x)$?

Daniel Fischer

13:53

@iwriteonbananas You forgot the "almost everywhere". It is of course true that we have $\lim f_{n_k} = \lim f_{l_j}$ in $L^p$, and that implies a.e. equality of the pointwise limit functions. But it's more convenient to successively extract subsequences. First take a subsequence $f_{n_k}$ converging pointwise a.e. to $f$, then extract a further subsequence $f_{n_{k_r}}$ so that $m\cdot f_{n_{k_r}}$ converges pointwise a.e. to $g$.

iwriteonbananas

@DanielFischer Right, ok. How do we justify extracing a further subsequence $f_{n_{k_r}}$ with $m\cdot f_{n_{k_r}}\to g$?

Oh, nevermind

Thanks.

Balarka Sen

@iwriteonbananas Yeah, it sure did.

I linked stuff to nlab this time in case KCd finds something in wikipedia and picks on me.

:P

iwriteonbananas

@BalarkaSen Haha. Surely he won't bother reading through that nlab article.

Balarka Sen

Yep, he won't.

user159870

Hi!

Balarka Sen

13:59

@OFFSHARING lol, "Romanian professor"

@OFFSHARING Thanks. I simply wish I can do mathematics.

I find quench for learning is a better motivation than "being great" or "attaining the highest level" or "being Ramanujan".

Or whatever those mean.

user159870

We have the ellipsoid $\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1 , p, q, r \neq 0$.

Is $r(u,v)=(p\sin{u} \cos{v}, q\sin{u} \sin{v} , r\cos{u})$ a parametrization of the ellipsoid for each $u$ and $v$ ?

Or do $u$ and $v$ have to satisfy a condition?

gansub

I am unable to get an answer for my question. I am asking myself if I should edit my question to get an answer

Balarka Sen

@iwriteonbananas lol!

gansub

if I have a surface of a sphere expressed as latitude and longitude and I transform this into Cartesian coordinates I would have to deal with only two dimensions am I right ?

iwriteonbananas

@BalarkaSen The Romanian professor thing cracked me up, too :P

Balarka Sen

14:04

Huy

@BalarkaSen: I thought you didn't like comic books.

evinda

Oh yes, right... So:
For k=1: $A_n^1 x=(x_2, \dots, x_n,0)$.
For k=n-1: $A_n^{n-1} x=(x_n, 0, \dots, 0)$
For k=n: $A_n^n x=(0,0, \dots, 0)$
Thus for every $k \geq n$: $A_n^k x=0$, i.e. $A_n^k$ is equal to the zero matrix.
Therefore, $\lim_{k \to +\infty} ||A_n^k||=0$.
But we don't use inductionin this way... So is it enough like that?

gansub

anybody can clarify my doubt ?

Balarka Sen

@Huy That's not true.

Tintin and Asterix are both great. Tintin is incomparable, of course.

user159870

Is anyone familiar with differential geometry?

gansub

14:08

@user159870 - are you familiar with spherical trigonometry ? :-)

user159870

@gansub What is your question?

gansub

@user159870 scroll up

if I have a surface of a sphere expressed as latitude and longitude and I transform this into Cartesian coordinates I would have to deal with only two dimensions am I right ?

Daniel Fischer

@evinda Sure it's enough. If you wish, you can make an induction proof from it, but why should one?

user159870

@gansub I don't know. Sorry.

gansub

that's the problem

OFFSHARING

14:17

@BalarkaSen some time ago I was tested with an device that measures amongst other things the global energy of a human being (this is not recognized by the medical science, but by the alternative science), and according to the test I have 40 times more global energy than a usual person. I always had a special intuition, and I can easily recognize some stuff although an in unexplainable way.

Balarka Sen

good for you

OFFSHARING

@BalarkaSen Just ask @iwriteonbananas if I'm right.

Balarka Sen

right about what?

OFFSHARING

If he says not, please look at his ip, and then we talk later.

evinda

@DanielFischer Ok... If I would want to make induction would I have to make it on n?
i.e. we would pick an arbitrary $k \geq n+1$, where $n \geq 2$ fixed. Then at the base case we would say that $A_2^k=0$, since $A_2^2=0$, then we would suppose that $A_n^{k}=0$ and then we would have to show that $A_{n+1}^k=0$.

OFFSHARING

14:18

Ah, he leaved ... (of course)

Balarka Sen

sorry, he's not from Romania. you're wrong. he's a German student of math.

I know him.

OFFSHARING

@BalarkaSen Did you see him? Met him in the real life?

Balarka Sen

Yes.

OFFSHARING

Anyway, it's not my business (and surely I won't reveal anything about his identity - I respect the privacy).

Clarinetist

@Huy On my phone. Will be on the computer in a bit

user159870

14:20

@Huy are you familiar with differential geometry?

OFFSHARING

@BalarkaSen I don't even remember the last time my intuition lied to me.

Balarka Sen

You're mistaken this time.

Huy

@user159870: Yes, but I am busy at the moment, sorry.

user159870

oh ok.

Huy

@Clarinetist: ping me when you are on PC and have some time

Balarka Sen

14:22

@Huy: Struggling with the exam? :P

Agawa001

ip cant be revealed by anyone who isnt part of the site moderation staff of technical team

Huy

@BalarkaSen: which one do you mean ?

Agawa001

as far as i know

OFFSHARING

@Agawa001 I know, but at least some might see that I'm perfectly right.

Agawa001

chat server isnt a peer-to-peer system

i think it works as mim

Balarka Sen

14:23

You were creating a linear algebra exam, yes, @Huy?

Or are you already done with that?

Huy

@BalarkaSen: it's a general exam for HS graduation, and I have to create the questions for probability and vector geometry

Balarka Sen

Ah.

Huy

@BalarkaSen: the bad thing is that the solutions of the vector geometry exercise are supposed to be integers, so I'm still struggling with that.

Agawa001

@OFFSHARING dont know where are most of chatters from and i dont give a fly wing

Balarka Sen

@Huy I figured, thus the message above.

Creating problems with nice looking answers can be a tough job.

Huy

14:25

but right now I'm doing some geometric topology which I'll maybe ask you or Mike about soon :P

Agawa001

dont know the point of askin someone where is he from

Mats Granvik

Why does mathematics stack exchange look like this in my browser?

Balarka Sen

@Huy One thing you can do though. Pose whatever question you have, and if the answer is x, tell them to find the nearest integer to x.

Easily done.

Huy

@MatsGranvik: because you use Internet Explorer

Mats Granvik

ok

OFFSHARING

14:26

@Agawa001 I also respect the privacy, but some things annoy me so much.

Agawa001

i dont lie when they ask me where m i from, but i dont like that either

Balarka Sen

@Huy Mike has taken a leave from SE, or so he told me via e-mail.

Huy

ah ok

then I'll ask him via email if you can't help :P

OFFSHARING

@Agawa001 For me it's OK no matter where you come from, in general.

Balarka Sen

Apparently from shame of spending too much time on hats.

Huy

14:27

I would hope so

Balarka Sen

Well, he said it.

@Huy Good idea.

I won't be able to help most of the time.

Huy

it's just very basic stuff, don't worry

Clarinetist

I'm on @Huy

user159870

@robjohn can I ask you a question about differential geometry?

OFFSHARING

@BalarkaSen when do you plan to start the learning process of calculating some elementary integrals?

Balarka Sen

14:33

Not anytime soon.

I want to learn the implicit and inverse function theorem first.

OFFSHARING

At least you're honest (that I appreciate).

Balarka Sen

Well, there's no point in disagreeing that I don't know anything about your math.

I have no issue with your integrals.

OFFSHARING

@BalarkaSen They come always together, integrals, series and limits (well, after all you can view all like limits).

@BalarkaSen Never let the series out of the picture.

Balarka Sen

fair enough.

OFFSHARING

OK, I try to finish some other proofs.

BBL

Daniel Fischer

14:39

@evinda You can do an induction on $k$ for fixed $n$. "For every $n-k < m \leqslant n$ and all $x$ we have $(A_n^k x)_m = 0$." It's more convenient to make $k = 1$ the base case than $k = 0$, but no big thing.

Agawa001

my posts are voted in binary, thu, i pushed an important effort on em

dunno wats wrong

evinda

@DanielFischer What do you mean with $(A_n^k x)_m = 0$? What does the subindex mean in this case?

Soham Chowdhury

@BalarkaSen eh. check the room.

robjohn

15:02

@user159870 just ask. If someone knows the answer, they will probably help.

Daniel Fischer

@evinda The $m$-th component of $A_n^k x$.

user159870

We have the ellipsoid $\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1 , p, q, r \neq 0$.

I want to show that it is a smooth surface.

$r(u,v)=(p\sin{u} \cos{v}, q\sin{u} \sin{v} , r\cos{u})$ a parametrization of the ellipsoid.

I calculated that $r_u \times r_v=(qr\sin^2{u} \cos{v}, pr\sin^2{u} \sin{v}, pq\cos{u} \sin{u})$.

So that it is a smooth surface, one has to show that $r_u \times r_v \neq 0$, or not?

This doesn't stand. For $u=0$ : $r_u \times r_v=(0,0,0)$, or?

Does one not show in this way that it is a smooth surface?

Huy

15:27

Why did you create a different account @evinda @user159870?

evinda

I am not user159870 @Huy
Why do you think so?

Huy

You talk precisely the same way and the topics also intersect.

evinda

It's definitely not me... I am studying Optimization right now... And I don't occupy with the two subjects simultaneously... @Huy

Huy

also, you ask me and robjohn specifically when there are many other people in the chat (and also some who have been more active in the past hour), which is rather suspicious. but I won't comment on it anymore.

evinda

@DanielFischer So at the induction hypothesis we will say that we suppose that for k=n: $(A_n^n x)_m=0$ and then $(A_n^{n+1} x)_m=(A_n(A_n^n x))_m=0$ since $(A_n^n x)_m=0$, right?

Daniel Fischer

15:43

@evinda No. The induction hypothesis will be that $(A_n^k x)_m = 0$ for all $n - k < m \leqslant n$. Then we use that to see that $(A_n^{k+1} x)_m = 0$ for $n - k \leqslant m \leqslant n$. That step is trivial when $k \geqslant n$ - because the induction hypothesis then says $A_n^k x = 0$ - and for $k < n$, we verify it by showing that if $y_m = 0$ for $a < m \leqslant n$, then $(A_n y)_m = 0$ for $a \leqslant m \leqslant n$.

evinda

15:54

@DanielFischer I am a little confused right now. Isn't it obvious for $k+1 \leq n \Rightarrow k<n$ , using the induction hypothesis?

Daniel Fischer

16:14

@evinda Is or isn't what obvious? In some sense, what we want to prove is obvious in toto.

iwriteonbananas

@DanielFischer Let $\lambda\in \Bbb C$ and $m$ be a measurable function as before. I would like to show that the operator $\lambda \, I - B$ is not bijective if $\{|\lambda-m|<\delta\}$ has positive measure for all $\delta>0$.

My idea was as follows:

Sorry, I forgot to mention: $B:f\mapsto mf$

$f=\mathbb 1\{|\lambda-m|=0\}$ is in the kernel of $\lambda\,I-B$, so it would suffice to prove that $\|f\|_{L^p}>0$

We can write $f$ as the limit of the indicator functions of the sets $A_n=\{|\lambda-m|<1/n\}$

and by monotone convergence $\int |f|^p = \lim_n \mu(A_n)$

Each $\mu(A_n)>0$, but the limit need not be $>0$, right?

Is there a remedy to this debacle? Or a better approach?

Brennan.Tobias

16:31

hello

Daniel Fischer

@iwriteonbananas Yes, the limit can be a null set. The operator is injective whenever $\{ x : m(x) = \lambda\}$ is a null set, but when $A_n$ has positive measure for all $n$, then it isn't surjective. Try to show that there is an $f \in L^p$ such that $\frac{f(x)}{\lambda - m(x)}$ is not in $L^p$ under that condition.

evinda

Here: "That step is trivial when $k \geqslant n$" doesn't this hold for $k+1 \geq n$ ? Also could you explain me further the following part:

" we verify it by showing that if $y_m = 0$ for $a < m \leqslant n$, then $(A_n y)_m = 0$ for $a \leqslant m \leqslant n$. "

Daniel Fischer

@evinda Well, when $k \geqslant n$, we already have $A_n^k x = 0$, so deducing $A_n^{k+1} x = A_n(A_n^k x) = 0$ follows without looking at $A_n$. When $k+1 = n$, then $A_n^k x$ is not necessarily $0$, the first component can be nonzero. Hence that step isn't trivial (it's very easy, though), one has to use the structure of $A_n$ for the conclusion.

We write $y = A_n^k x$. By the induction hypothesis, the last $k$ components of $y$ are $0$. The induction step is done by showing "if $y$ has the last $r$ components $0$, then $A_n y$ has the last $r+1$ components $0$".

iwriteonbananas

@DanielFischer Ok, how about this choice of $f$: Choose $n_0\in \Bbb N$ with $\mu(A_{n_0})\geqslant \frac{1}{\sqrt{n}}$ for all but finitely many $n$. Then put $f$ as the indicator of $A_{n_0}$. Then we have: $$\int |\lambda-m|^{-1} f \geqslant n \int |f| \geq n\cdot \frac{1}{\sqrt n} = \sqrt n$$

Daniel Fischer

16:54

@iwriteonbananas You only know that $\lvert \lambda - m\rvert^{-1} > n_0$, not that it is larger than $n$. Let $B_n = A_n \setminus A_{n+1}$ and do something with the characteristic functions of the $B_n$. You will need to use infinitely many of them.

iwriteonbananas

17:05

@DanielFischer Hmm, I can't seem to figure out what to do with those $B_n$.

Does taking $f$ as indicator of $\bigcup_n B_n$ work?

I'm not sure why that would be in $L^p$ though..

Daniel Fischer

@iwriteonbananas You need $f = \sum c_n\cdot \chi_{B_n}$, where the constants must be chosen so that $f\in L^p$ - so $\sum \lvert c_n\rvert^p\cdot \mu(B_n) < +\infty$ - but $\frac{f}{\lambda - m} \notin L^p$. Use what you know about $\lambda - m$ on $B_n$ to guide you.

iwriteonbananas

17:21

@DanielFischer Right, will $$c_n=\frac{1}{n^2\mu(B_n)^{1/p}}$$ work?

Then $\int |f|^p = \sum 1/n^{2p}<\infty$

And $$\int \frac{|f|^p}{|\lambda - m|^p} = \sum |c_n|^p \cdot \int_{B_n} \frac{1}{|\lambda - m|^p}\geqslant \sum \frac{1}{n^{2p}\mu(B_n)}\cdot \mu(B_n)\cdot n^p=\sum \frac{1}{n}=\infty$$

Johan Larsson

17:50

in Lounge<C++> on Stack Overflow Chat, 2 mins ago, by sehe

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$Mathematics$ Mathematics