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4:04 PM
@Ilya Sorry for that. Just give it a week and you won't remember a thing.
 
@Gigili Are you still mad at me?
 
No.
 
Great ;-)
 
4:25 PM
For those who haven't done so already: will you star JM's notice on the board? Thanks. -->
 
@robjohn Hi Rob
 
hey there. I am at UCLA, getting ready to proctor an exam.
 
What is the course?
 
Hello @robjohn and @Skullpatrol did you star JM's notice?
 
@Skullpatrol First term sentential logic
@MattN No, that had to be starred by Asaf
 
4:38 PM
@robjohn No it needs stars. It has only 7 and the other two dudes got 11 : /
 
Why would extra stars help?
 
Thanks to whoever starred it just now.
 
His is pinned. Extra stars won't help
That was me
 
@robjohn Show that we miss him just as much as we missed the other two deserters?
@robjohn Nice, thanks : ) 3 to go : )
 
I don't see anything with 11 stars. If JM doesn't know that we will miss him in his absence, he hasn't been paying attention.
deserters? who are they?
 
4:41 PM
@robjohn Srivatsan and teddy. Both got 11 stars on their notices.
 
@MattN They are not deserters!!
 
5:12 PM
@robjohn Like this? $$ \|\mathcal{F}f \|_{\alpha, \beta} = \sup_x \left | x^\alpha D^\beta \mathcal{F}f(x) \right | = \sup_x \left | \mathcal{F} x^\beta D^\alpha f(x) \right | = \| \mathcal{F} f^\prime \|_\infty < \infty $$
 
5:39 PM
Hi
 
Hi @Daniil.
 
Hi @Gigili
Hi @Daniil
Hi @MattN
 
6:05 PM
Yo
 
@Skullpatrol Hello.
Awesome! One more star to go : )
@JonasTeuwen Are you there?
 
@MattN Yeah, just got here.
 
@N3buchadnezzar Yo, Yo what's up?
 
@JonasTeuwen Great. Can you tell me if I have the thing up there^ right?
 
Looks a bit fishy, that would imply that the Schwartz seminorm is independent on $\alpha$ and $\beta$.
Oh, right you still have the Fourier transform.
Let's see.
 
6:18 PM
This is part of the proof that $\mathcal{F}$ is continuous (as an operator on the Schwartz space).
 
One more star to go. We'll have to get someone do the honours! :-)
 
Oh! That was quick : )
Thank you!
 
@MattN I'm trying to figure out what you've done.
 
@JonasTeuwen I renamed the variable and changed $x$ to $D$ when I pulled it into the Fourier transform. What do you think of that?
 
6:21 PM
Let me verify.
 
@N3buchadnezzar Thank you!
 
@JonasTeuwen What I want to do is to show $ \|\mathcal{F}f \|_{\alpha, \beta} < \infty$ from $ \|\mathcal{F}f \|_\infty < \infty$ by replacing $\mathcal{F}f$ with $\mathcal{F}x^\alpha D^\beta f$.
 
@MattN Okay, don't you have $x^\alpha D^\beta \mathcal F f(x) = (-2 \pi i)^\beta \mathcal F x^\beta D^\alpha f(x)$?
 
I'm doing this the wrong way around.
@JonasTeuwen Yes. True.
 
As I'm quite surprised that given $f$ we compute $\mathcal F f$ which is $g$ and $\mathcal F f'$ which is $h$, then we have $\|g\|_{\alpha, \beta}$ constant.
For all $\alpha$, $\beta$. Doesn't sound right, right?
 
6:26 PM
It does not, no.
 
Hm, for some reason, MathJax script is not working for me :(
 
So, okay $| x^\alpha D^\beta \mathcal{F}f(x) | = (2 \pi)^\beta| \mathcal F x^\beta D^\alpha f(x)|$.
 
I'm doing it wrong: I can't write $f^\prime$, I just have to use the inequality I proved.
 
It works, but not for new messages.
 
@Daniil Same for everybody.
 
6:32 PM
Like this (I only have to fix all the n+1's, they should be gammas): $$\begin{aligned}
& \|\mathcal{F}f \|_{\alpha, \beta} = \sup_x \left | x^\alpha D^\beta \mathcal{F}f(x) \right | = \sup_x \left | (-2 \pi i)^\beta \mathcal{F} x^\beta D^\alpha f(x) \right | \leq \\
& \max(K, K^\prime) \left ( \| x^\beta D^\alpha f(x)\|_\infty + \| x^{n+1} x^\beta D^\alpha f(x) \|_\infty \right ) < \infty
\end{aligned}$$
 
@JonasTeuwen Hm, but I swear it was working some time ago
 
@Daniil Yes it was. But no more. Meta search should help!
 
Right, thanks.
 
@JonasTeuwen Do you agree or did you go to sleep? : )
 
@MattN I didn't get pinged.
I don't know what $K$ or $K'$ are. Or how you get the next inequality.
 
6:48 PM
@JonasTeuwen That's a longer proof : ) (10 lines or so). I have a pdf. Would you like to see it? Or are you not in the mood?
 
@MattN Sure, give it to me. Maybe also the .tex, then I can improve it if you'd like that.
 
@JonasTeuwen That would be awesome.
@JonasTeuwen Sent.
 
@MattN Fine!
 
@JonasTeuwen The thing you want to be looking at is "Theorem 1" and the proof following it.
 
@MattN Okay, I will do it, but first I need to do something else :-).
 
6:59 PM
@JonasTeuwen Sure. : ) (Pour yourself a drink? : ))
 
7:12 PM
@MattN Your phrasing is inaccurate of the continuity of the seminorm.
 
@JonasTeuwen You mean the continuity with respect to the seminorms?
 
Yes, I'll correct it.
 
Can you post the correction here?
 
Yes, first I'll check what you're doing.
 
Thank you.
 
7:19 PM
And install emacs.
 
: )
 
@MattN Do you really want to add the $\mathcal F^{-1} \to \mathcal F$? It is quite obvious once you think about it. The Fourier transform and its inverse are basically the same except for a sign flip and a flip around $x = 0$. No way that that is going to change the norm.
@robjohn Do you have some intuitive explanation for me why applying the Fourier transform twice basically gives us the function back? I was thinking in terms of frequency spectrum, but I have trouble grasping what the frequency spectrum of the frequency spectrum should be.
 
@JonasTeuwen Yes. I really want to spell it out.
 
Okay.
 
I really like that you can type iint and get a double integral : )
 
7:39 PM
$$\iint$$
 
@MattN I have replied!
 
Looking at it now.
@JonasTeuwen But isn't the set that is the union of all neighbourhood bases a basis?
And why is it not good to mention $B$?
 
leo
hi there
 
Hello.
 
@MattN Yes.
@MattN Because it is useless.
And uncommon to denote it by $B$ I think.
 
8:08 PM
@MattN Do you agree with other comments?
 
@JonasTeuwen I don't know yet : )
 
You should apply the same level of verbosity everywhere is my opinion.
 
Yes, I agree.
 
8:44 PM
@JonasTeuwen So this is wrong?
$$\begin{aligned}
& \|\mathcal{F}f \|_{\alpha, \beta} = \sup_x \left | x^\alpha D^\beta \mathcal{F}f(x) \right | = \sup_x \left | (-2 \pi i)^\beta \mathcal{F} x^\beta D^\alpha f(x) \right | \leq \\
& \max(K, K^\prime) \left ( \| x^\beta D^\alpha f(x)\|_\infty + \| x^{n+1} x^\beta D^\alpha f(x) \|_\infty \right ) < \infty
\end{aligned} $$
 
Well, the $K$'s depend on $\beta$, right? For the rest it is okay, I think.
 
I read your comment as saying that the second equality is wrong.
 
No, I was confused by your "substitution".
 
So you're saying you want me to write $K(\beta)$ instead of $K$?
Crap, I should've sent this to the lecturer on Monday.
 
No, no :-). You should just be Aware.
 
8:47 PM
I'm about to send it now.
I'll work on it again tomorrow.
 
@MattN I mean your first equality. How does that follow from your substitution? Your Fourier transform is on the wrong place.
 
So it is wrong!
 
Okay, you can just do that, was just saying :-).
No, you just do differentiation under the integral sign.
 
Yes, I know. I'm worried about that. I don't really know how to justify it.
@JonasTeuwen What would be right?
 
@MattN It is right, but your substitution gives $\mathcal F x^\alpha D^\beta f$.
 
8:52 PM
@JonasTeuwen Wot? But $\mathcal{F}(D^\beta f) = x^\beta \hat{f} (\xi)$, no?
 
@MattN Yes, but if you are that verbose... Was just saying.
 
So $x^\alpha D^\beta \mathcal{F} f $ should be $\mathcal{F}(x^\beta D^\alpha f)$, no?
I need to go over it again tomorrow.
 
@MattN Anyway, you've got the idea right. :-).
 
I don't want to get the idea right, I want to get it all right : )
 
Sure, but those are just details now.
 
9:06 PM
This proof was really painful.
I wish I had a book.
My frustration tolerance is just too low to study without a book.
 
@MattN You did it fine :-).
 
Wednesday was incredibly painful : (
 
Why?
 
@JonasTeuwen you weren't there when I was being taught by robjohn.
 
Mathematics is a synonym for mental pain 8-).
 
9:07 PM
I didn't understand him at all.
 
Sure, but you lack a lot of prerequisites I think.
 
@JonasTeuwen Not so much. If I have the right book I can do it without pain.
 
This is the right book.
 
LOL?
 
Just, when you want to get somewhere you need to suffer.
 
9:08 PM
This book wants you to know everything.
 
Yes. Very good!
 
@JonasTeuwen No. That's what I'm saying. That's a false belief of yours.
 
No intellectual gain without mental pain. 8-).
Well, my experience is different.
You don't understand that stuff for a long time and at once everything is clear!
Instead of someone chewing it out for you, that doesn't learn you how to think or learn your own way of thinking about those things.
You have to think by yourself, you don't have to think like that author, there is already someone that does it and that's the author.
I know it is a bit bluntly said, but that's how I have experienced it.
 
Hm. I think we disagree. : )
 
Yes, I remember that :-).
But there is not really a different book. This one is the easiest. Good luck!
 
9:12 PM
I guess I've read much worse books.
But maybe I just think that because I made it through chapter 0 more or less.
 
If it all were easy, you wouldn't really learn much, you would be just memorizing facts.
 
But I really need to look into finding books that suit me to fill in my missing knowledge. I can't take any more of this otherwise. Wednesday was a nightmare. The additional problem is that once I get frustrated my system shuts down.
 
Okay. What do you miss?
 
I don't know off the top of my head. : /
 
I'll tell you. Yes, gently things.
Okay, what do you want to learn?
 
9:15 PM
Distributions certainly. That's why I ordered Strichartz. I'm not sure it's too casual but even if that turns out to be the case it won't do any harm either.
 
Yes, that's a good book for that, but will not explain you the subject totally.
 
The next thing I was thinking about was getting Stein to learn Fourier analysis.
 
Hah.
So you do like to suffer 8-).
 
Is that a gut stabber book?
 
Very.
 
9:18 PM
Then I'll have to find an alternative.
 
Do yourself a favor and check out Grafakos "Classical Fourier Analysis", you do need to know a bit of functional analysis but not much. (And certainly measure theory).
 
Thanks for the tip.
 
That is very gentle.
What do you know about functional analysis?
 
Nothing. : )
 
Check out the Springer UTM Linear Functional Analysis by Rynne and Youngson.
Does not really contain everything you should know to be able to do harmonic analysis but it surely will help you gain that knowledge much easier.
 
9:20 PM
Grafakos seems to have two books, one "Classical" and one "Modern".
 
Yes, you need the classical one.
The modern is a followup.
 
@JonasTeuwen I have Kreyszig. I want to read that first. Seems to be the right level for me. : )
 
Yes, that's fine too.
 
But it almost contains nothing.
 
Bogachev's measure theory is in the YELLOW SALE!!!
 
9:22 PM
Have you read Tao's Additive combinatorics?
 
@MattN Yes, but you don't need to know much ;-).
@MattN No.
 
@JonasTeuwen Well there is also an exam looming in the near future where I do need to know something...
@JonasTeuwen I wonder if I can make it through the whole Fourier analytic methods chapter without pain.
 
@MattN And Duistermaat's distributions is in the Yellow Sale!!
 
@JonasTeuwen This? What's the Yellow Sale?
I prefer paperback...
 
@MattN www.springer.com
 
9:25 PM
Is Bogachev painfree?
I'm asking because I can't make my favourite notes into a book (wrong size and I don't have the tex : ( ) so I was thinking a book would be handy.
 
No, certainly not. It is a reference book :-).
Good book: Bauer (he also has a German version).
 
@JonasTeuwen I really don't like German.
 
@MattN No problem! There is also an English one.
 
Ah yes, and my most important question: do you know a painfree book for algebraic topology? I have one but it doesn't contain everything.
 
@MattN Such a thing does not exist. Check out the starred message by Brian.
 
9:33 PM
@JonasTeuwen I think Brian and I disagree on that : )
 
Brian is a very wise man.
 
I can't stand the noise my laptop is making : ( This means: new fans : (
 
No problem!
 
Yes because I have to bring it in and leave it there so I'll be without laptop for 3 days.
Never mind : )
I think I should go to sleep. If I'm unlucky they make me present the rest of last Friday's presentation. I'm really not keen.
Good night!
And thanks for today!
 
Can't you do it yourself?
@MattN Good night! Bye.
 
9:51 PM
Hello, would anyone like to give me a little guidance on an assignment question. Its about real analysis, specifically metric spaces and nest rectangles.
 
Maybe.
 
Please? :)
 
10:36 PM
heya =)
 
11:20 PM
Can anyone here answer an (elementary) question about topological quotient spaces?
 
Just ask.
@Kannappan: PLC also has notes on commutative algebra. I didn't get very far in them, myself.
 
Didn't think anyone was here... Consider the equivalence relation on $\mathbb{R}^{2}$ defined by $(x_{1},x_{2})\sim
(w_{1},w_{2})$ if $x_{1}^{2}+x_{2}^{2}=w_{1}^{2}+w_{2}^{2}$. I'm trying to describe the resulting quotient space. I know that the equivalence classes are circles, but that's about it...
 
Consider a ray emanating out from the origin. This will hit every circle exactly once, so we can identify them all with the point they hit on the ray. Something like that?
 
@anon Right. I was originally thinking that I can identify each equivalence class with a particular real number, say $r$, that is the radius of the circle formed by the points in the equivalence class. But that only gives positive real numbers...
 
11:35 PM
No, nonnegative real numbers. And what of it?
 
@anon Yes. nonnegative. I guess I'm not really sure what the topology on the nonnegative real numbers would "look" like.
 
11:56 PM
hi.. I have a question is anyone in?
 
I am here. I am an idiot though.
Just warning you.
 
no problem, so am I
I just read a blog post which gained top spot on Hacker News
it said that to single out a human from the entire population, one needs 33 bits of information
which is mathematically convincing
 
Well 2^33 is about 8 billion.
 
yes,
 
Since we only have about 7 billion people living nowadays...
 

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