Mathematics - 2012-03-24 (page 1 of 5)

« first day (598 days earlier) ← previous day next day → last day (4415 days later) »

12:05 AM

Short q

How do i normalize a function ?

I have V(x) = x^2*(-2*x+(1/4)*a)

But I want V to go from 0 to g, where g is some factor I decide.

So Max(V) = g

12:59 AM

Okay.

I've finished three sections out of the seven planned.

Tomorrow I think I can finish this.

Now I can go to sleep. Goodnight.

Goodnight! @asaf

Does someone know why "the space $M_1$ of probability measures is included in the dual of the space $C_b$ of bounded continuous functions" in Didier's reply? Any reference on the dual of $C_b$?

I will cook dinner and be back soon.

1:43 AM

@robjohn very nice

Hiya

Hi @tb

still awake ?

obviously :)

I thought you had the habbit of chatting in the sleep...like walking in the sleep :-)

yeah, it's like LaTeX, in my vegetative nervous system

I always wondered...How people on M.SE take so much pain in typing up illustrious answers in good Latex formatting.....I wanted to say hats off to them for their patient efforts.....But little did I knew that they had it in their Vegetative nervous system !

@tb

1:54 AM

@Tim It's a version of Riesz representation theorem.

@RajeshD at some point it's rather automatic... :)

@tb : I got a thought when i was reading Do Cramo book, Is there any way we can compute the topological properties (like number of closed loops which are prominent, etc) of online handwritten characters

By Online handwritten characters we have the X and Y co-ordinates of the pen tip sampled evenly in time

Huh, "kernel" isn't French. Another maths word to learn...

Noyau

2:09 AM

French liquor ?

No, kernel.

For both, kernel in the sense of algebra and in the sense of integral kernel.

And, apparently, operating systems too.

or "exploitation systems", hehe.

Kernel is the central (core) part of a system which is responsible for the behaviour of the system....like kernel of an operating system...Linux has a kernel written in 'C'

Oh, if you really want to suffer, read computer stuff in French. Impossible. Octet for byte, for example.

2:12 AM

I see...

@tb : Octet seems more natural than a byte

@RajeshD It is not entirely clear to me what you want to do, so you want to enter a character using the mouse as they do here, for example? And you want to determine the number of holes, self-crossings, etc? I'm sure that it shouldn't be too hard to implement.

Huh, do they not say co-égalisateur in French?

I don't know what they use. I don't think I've ever seen that word in French.

Google gives <1000 results. I suspect they just say conoyau.

2:16 AM

For cokernel, yes. But for co-equalizer, I don't know.

I'm looking at Demazure and Gabriel now, and they say conoyau where I would say coequaliser. But then again, it seems most algebraic geometers do that...

This guy likes exclamation marks

Yes, yes he does.

Oh, did you know that Verdier's thesis is now online?

In PostScript? Someone must have gone to a lot of trouble to retypeset all that...

2:23 AM

Maltsiniotis did. It appeared in the mid-nineties in the astérisque series.

ah

Okay, time for me to leave. Good night!

Goodnight!

5 hours later…

7:30 AM

Hello Boyos.

Kannappan Sampath

Hello!

What time is it there?

Kannappan Sampath

1305 hours

Oh, quite late already.

Kannappan Sampath

Yes, but did not manage to get anything done! : (

7:41 AM

Hey guys

@JonasTeuwen I have an analysis question

I have been thinking about topologies induced from metrics on functions spaces

Okay, yes.

@JonasTeuwen Say we equip $C[0,1]$ with the sup metric

Yes.

what is a basis for the topology here? I mean like what do the open balls look like

Or for example $C[0,1]$ with the $d_2$ metric

$d_2(f,g) = \sqrt{\int_0^1 |f-g|^2}$

Oh, right. Well, for those spaces it is actually better to work with your metric. So your balls are scaled and translated versions of $\{f : \|f\|_{\infty} \leq 1\}$.

7:49 AM

why norm less than 1?

I just take the $1$-ball.

Oh ok

Hmm this may not form a basis for the topology

Sure, you need $\{f : \|f - g\| \leqslant \epsilon\}$.

I have a feeling that the topologies induced by the two metric I stated above are not equivalent....

Yes.

You can make $\int_0^1 |f|$ finite while $\sup |f|$ is infinite.

Oh, on $C[0,1]$, not $C(0,1)$.

7:56 AM

right,

hi

@BenjaminLim You can take for example a Fourier series that converges in $L^2$ norm but not uniformly 8-).

I'm not so advanced jonas

Okay, so what you need to do is find a sequences of functions that converges in one norm but not in the other.

@JonasTeuwen How can sup $|f|$ be infinite?

Every continuous function attains it's maximum and minimum on a closed and bounded interval

8:01 AM

@BenjaminLim It cannot be infinite here as you take it on $[0,1]$ and not $(0, 1)$.

@JonasTeuwen Oh sorry your remark was for $C(0,1)$.

@BenjaminLim $x^{\frac{n}2}$ I think works. Converges in $L^2$ but not uniformly.

I'm trying to think about this in terms of balls containing each other

You shouldn't.

Morning.

8:08 AM

hmmmm

@JonasTeuwen How do you come up with examples?

@MattN Hi. What do you mean?

@BenjaminLim This is an analysis question, if you phrase it in topology language it might actually become harder to answer!

hold on

Well if you have two spaces with two different norms and you think they aren't equivalent.

I'm working on something

I am trying to think of an example

8:15 AM

@MattN Hmm, I was thinking "what sequence does not converge uniformly?"

What does $x^{\frac{n}{2}}$ converge to? It doesn't seem to converge even pointwise so how can it converge uniformly?

On $[0, 1]$.

Doh.

It does not converge uniformly.

Thanks.

8:17 AM

But it does converge in $L^2$.

Well on $[0,1]$ its limit is $0$ everywhere except at $1$ where it's $1$.

Yes. So uniform fail.

So it does actually converge pointwise.

But not in the $\sup$ norm.

What about the $\mathrm{soup}$ norm?

8:19 AM

Yes it can't converge uniformly if the limit function is discontinuous.

@JonasTeuwen What about this

Consider the function $g$ that looks on the interval $[0,1]$ like

flat, then a spike high up to two, say and then flat again

Why take something harder?

For any given $\delta$ I can always find a $g$ such that $\int_0^1 |g|^2 dx <1$

Simply by making the function thinner and thinner

But then if this function has sup of $2$

Then taking $\epsilon = 1$.

Shows that such a $g$ will never be in the ball of radius $1$ about the zero function in the sup metric

@JonasTeuwen Is my last sentence right?

Why are you so obsessed with those balls.

@JonasTeuwen I am more used to topology

Am I right no?

I think it is correct

8:22 AM

I'm not sure why it wouldn't be equivalent with your argument.

Equivalent allows it to be "inequal up to a multiplicative constant".

You probably want your spikes to increase in height and keep the same area.

No I think I want the height to be the same and the area to decrease no?

I want to show that there exists $\epsilon > 0$ such that for all $\delta > 0$,

$B_\delta^{L²} (0) \not\subset B_\epsilon^{\text{sup}}(0)$

You want the $\sup$ norm to blow up and the $L^2$ norm to stay the same for example 8-).

Oh no, the $\epsilon$ $\delta$ stuff.

I only use that when I need to actually prove it.

Not when I think about an example!

So Jonas

But I can understand that you do that, I did that too 8-).

Okay, can you tell me what you're exactly trying to do with that function?

see, please try and see from my viewpoint

8:26 AM

Yes, I understand. Sorry :-).

So I want to produce a function in the left that is not in the right

So a function that is in $C[0, 1]$ with the $L^2$-norm (the left?) and $C[0, 1]$ with the $\sup$-norm?

yeah in the left but not the right

Okay, good. So what is your function?

the spike thing I told you

8:28 AM

Spike Lee?

(Suggestion: What does it mean for vector space norms to be equivalent? They have the same convergent sequences)

@BenjaminLim That's not really... clear.

Can you make a small picture of give me a function description?

^

$_ \Lambda _$

$\Lambda$

@JonasTeuwen It looks like that

Okay. I understand, but from where to where?

what do you mean by where to where?

Where is the peak, where is it $0$.

8:32 AM

The peak is say at 1/2

The function is flat

then the upside down V thing

and then flat again

Meh. That doesn't completely determine the function, right?

Anyway. It is in both spaces.

Why is it in both balls?

snickers

You see I want to show that there is an $\epsilon >0$ such that for all $\delta > 0$, I have a function such that it's $L^2$ norm is less than $\delta$ but that it's sup norm is not less than $\epsilon$

Because the $\sup$-norm is finite and the $L^2$ norm is finite because the function is bounded.

8:35 AM

But Jonas

If say I choose the peak function to be such that in the sup norm it is less than any given delta

That does not give a counter example, you would want $d(f, 0) \leq C d'(f, 0)$ to not hold.

At least, not uniformly in $C$.

That's what equivalent means to me: They give the same Cauchy sequences.

Ello : ) How's the head?

'ello! It is tired but nothing else, as far as I can tell without coffee.

Good. Finally!

Kopfschmerz? Talk me about it...

8:42 AM

I think Ben's example shows what he wants if I understand correctly... He has $\|f_n\|_2 \to 0$ while $\|f_n\|_\infty = 1$, so there can't be $C$ such that $\|f\|_\infty \leq C \|f\|_2$.

But he has no sequence.

I do know that example, yes...

Oh. I haven't read attentively enough. I automatically shrank the base :)

@JonasTeuwen -> tell me about it

Everything else would've surprised me.

Yes.

Plus I don't really understand why there must be balls.

8:44 AM

Why is there such a spate of questions on the Hausdorff distance recently?

Speaking of balls: you certainly know that the unit sphere in Hilbert space is contractible, right?

In fact something much stronger holds: it is diffeomorphic to the Hilbert space itself!

Yes.

I sense excitement :)

Sorry, the unit sphere is diffeomorphic to the Hilbert space?

Hmm, so how would the diffeomorphism look in $L^2$?

Oh, it's not very explicit. You start with a kind of stereographic projection and perturb it to take care of the missing point. It's quite non-trivial.

Oh. :-). I'm not surprised.

8:49 AM

@JonasTeuwen BOO!!!!

Is this an old result?

Bessaga, somewhere in the sixties.

Ah. Let's see!

I guess it's one of those things that you can prove by thinking very hard once you know that it is true, but how on earth should you have the idea?

8-). I often have that question...

8:51 AM

at.yorku.ca/cgi-bin/…

Okay, the next thing I was wondering: Does it hold for Banach spaces? Apparently: almost.

Hmm, MathSciNet does not give me the PDF 8-).

"Every infinite-dimensional Hilbert space is real-analytically isomorphic with its unit sphere"

I'd have linked you to it if it were available online.

Oh right 8-).

Let's see what the title I quite shows.

What do you want to see?

By the way: Maybe! You! Want! To! Go! After! A! Bounty!?

A bounty! Maybe I do, but I'm still in the process of awakening. Let's see 8-).

@tb How the proof goes.

9:10 AM

@tb Mmmm.... bounties...

Hi @tb : Morning I had too leave abruptly...thats why i couldn't reply you

9:24 AM

some annihilation?

Huh... I didn't know that Theo was a Chinese guy, always thought of him as a Swiss (also known as "Zi Gut German").

Bakery time. See you in a bit.

@AsafKaragila Chinese?

@robjohn Chinese.

Heya =)

Kannappan Sampath

9:44 AM

Group Theorists, Can you help me talk about the conjugacy classes of subgroup generated by $\langle (12345),(2345) \rangle$ in $S_5$.

There was question where answers from GAP exist while not from Group theory.

And, Hi all of you!

One of the guys in my office is in algebraic graph theory. What they do often is calculate things in GAP then try to derive the results from the theory.

We often mock him for that. :-P

Kannappan Sampath

Yes, even I don't like doing things with GAP.

Groups are not so computationally taxing, like the finite dimensional projective planes or $t$-designs.

Finite things are often boring.

Kannappan Sampath

Hmm... No, not for me atleast!

I started out in geometric group theory. There we consider finite groups as trivial...

At least virtually so.

9:49 AM

Salem to all mathematician.

Kannappan Sampath

Well, finite groups are interesting in their own right. We don't still know many things about them!

I didn't say otherwise :)

Hi Dharmendra

Hello

Kannappan Sampath

@Dharmendra Hi

I also don't know if the empty set exists or not, which is the most finite thing possible.

9:51 AM

Good day all

@AsafKaragila I think it has 27 elements.

@tb Maybe so.

one for each dimension in string theory and then one for me.

Kannappan Sampath

Geometric group theory, I think is the focus of PCMI this summer.

user image

9:53 AM

It's one of the topics that are quite underrepresented here and on MO, less so on MO

@Dharmendra ???

@tb But... geometric things... eww.

Focus on Step1
I know only angle of the camera
How to find the distance and the height ?
looks like impossible

Kannappan Sampath

@tb It is quite interesting, I heard my supervisor tell me. He says, techniques are quite advanced, asked me to defer attending PCMI, although they wanted only one semester group theory course.

@KannappanSampath the group theory itself that is used is pretty simple. But there's a lot of metric geometry and (algebraic) topology being used. Also, the interesting examples come from many different sources, that's probably why he described it this way.

Kannappan Sampath

@tb Quite likely. But I trust my mentor in these aspects. : )

9:58 AM

@Dharmendra it is impossible to determine height and the distance from the angle only. but don't you know the height in which the phone is held?

Kannappan Sampath

@tb So, I know the prerequisites, now! Thank You for telling me about it!

« first day (598 days earlier) ← previous day next day → last day (4415 days later) »

Transcript for

Mar '1224

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile