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7:24 PM
Good time of day, everyone.
 
QED
Hi
 
@Srivatsan There is no context.
 
QED
uggggg
so many "simplify" questions recently
I suppose that's at least not as bad as "what number comes next"
 
I've been reading the Wikipedia article about mollifiers and I was wondering if someone in here can tell me what a limit in the distributional sense is. (See property (3) in the linked definition). I did a bit of reading and the norm on the Schwartz space looks complicated so I was wondering if one can ignore the word distributional there and replace it with a pointwise limit and get the same.
 
7:49 PM
Anyone here have some knowledge about physics?
 
QED
what physics?
 
I was just wondering what the difference in a few branches of physics was, as I lack any deep understanding of them
Quantum Optics, Quantum field theory, Quantum Mechanics I , Classical Transport Theory etc
 
QED
I think Quantum field theory is a more unified version of Quantum Mechanics, but I don't know about the rest.
 
Just thinking about what Subjects I will be taking in a few years
 
QED
7:56 PM
definitely do QM
 
I guess I will wait and see, my mind will obviously change
 
QED
assuming you've done mechanics already
 
@Matt well, it depends on what you want to do... Pointwise (whatever that should mean) is certainly far too weak for most purposes, but e.g. for proving density of smooth functions in $L^p(\mathbb{R}^n)$ you can do with far less. Convergence in the sense of distributions is the "right" way to phrase it since it gives all that you need and can hope for in most contexts. A decent introduction is given in Felder's notes on MMP chapter 4, but
you should probably read section 2.5 first.
 
8:10 PM
@Matt You can approximate distributions with your test functions :-).
 
It's written in my book "$3.456440 (6D)$" or other numbers as the final answer with (D) and (S) , e.g. 2D , 4S ... What does it mean?
 
@Gigili It means that the end is near.
 
What does the number mean? Why 2 or 3 or 4D? How is it computed?
 
@Gigili I've never seen that. Could you give some more context?
 
Here I come to wreck the daaaaaaaay!
 
8:25 PM
@AsafKaragila So, what's up?
 
Tired.
Yourself?
 
Very tired.
 
I'll get to my point.
Is there a nice reference on known product topologies? (Like product/box/$\kappa$-product/etc...)
 
Well, I don't know of a reference that treats them all at once. There's quite a bit on the box topology in Kunen's handbook of set-theoretic topology. I would look there first.
 
I don't really care about the box topology. I just thought about something which may be useful to someone (perhaps topological vector bundles?)
If we define a topological product. Let $\tau$ be a topology on the index set $I$, and $X_i$ a topological space for $i\in I$. We define a topology on the product $\prod_I X_i$:
$U$ is open if and only if $\{i\in I\mid U_i\neq X_i\}$ is a closed set in $\tau$
Then the usual product is the co-finite topological product; the box product is the discrete topological product...
 
8:35 PM
@tb Thank you, I'm going to have a look. Although Felder gives me wiggings. When we were revising for linear algebra we solved old exams and usually the ones where we didn't even understand the questions would be Felder's.
 
@AsafKaragila So you're describing the basic open sets, not the topology, right?
 
@tb Yeah.
It sounds like an interesting topology, no?
 
@AsafKaragila I admit I haven't seen that so far. I've seen people using filters on the index set to describe various products at the same time.
 
What do you mean?
 
You choose a filter on the index set. As a base on the product you take the products of open sets such that $\{i\in I\mid U_i = X_i\}$ belongs to the filter.
The usual topology is given by the cofinite filter and the box topology by the powerset filter.
 
8:47 PM
Yes, I thought about that. Filters can be treated as open sets in a topology.
That is, the topology itself is a filter + $\varnothing$.
 
Basically, what you need is that the structure defining your product is closed under finite intersections, no?
 
Yeah, I think so.
 
Let me think a moment about where I've seen that. Just a second.
 
Hi, Asaf. How was your trip to Jerusalem?
 
It was fine.
 
8:55 PM
I was thinking of Knight, Box topologies and here's one I stumbled over while Googling: L. Brian Lawrence Paracompact product spaces defined by ultrafilters over the index set
 
Hmmm... I'm not sure if my paywall is down or the second link is bad.
Ah, you forgot the / :-)
 
Ah, thanks, took a moment :)
@Matt Don't be scared, those notes are very nicely written, I think.
 
@tb With pointwise I meant to treat $x$ as a constant and then $\varphi_\varepsilon (x)$ becomes $\varphi (\varepsilon)$ and in my concrete example I can do a pointwise limit $\lim_{\varepsilon \to 0} \frac{1}{\varepsilon} e^{-\frac{1}{1-\left ( \frac{x}{\varepsilon} \right )^2}}$. Of course, I can't treat $x$ as a constant as it ranges over the entirety of $\mathbb{R}$. : )
 
@Matt but that pointwise limit doesn't make sense! you have $\varphi_{\varepsilon}(x) \to 0$ for $x \neq 0$ and $\varphi(0) \to \infty$. On the other hand $(\varphi_{\varepsilon} \ast g)(x) \to g(x)$ for all, say, smooth functions with compact support (so that the limit of $\varphi_\varepsilon$ can't be zero)
 
What is the best programming language?
Waves flag
Start the debate!
 
9:03 PM
Waves flag
Stop the debate!
 
Common Lisp!
 
Pointless discussion as it's all about personal taste. : )
@tb I can't make sense of this but I think if $x$ was a constant then I don't see why the pointwise limit $\varepsilon \to 0$ wouldn't make sense.
 
@Matt Of course you have for $x \neq 0$ fixed that $\varphi_{\varepsilon}(x) \to 0$. But the crucial requirement is that $\varphi_{\varepsilon} \ast g \to g$ for nice enough $g$ in an appropriate sense, so you don't want to the limit of the $\varphi_\varepsilon$ to be zero.
You should think of $\varphi_{\varepsilon}$ as an approximation of the Dirac $\delta$-function at zero. When $\varepsilon \to 0$ the convolution with a mollifier converges to pointwise evaluation at $0$ for nice enough $g$.
 
@tb I think the problem is that when you write limit in this sentence you might be talking about a limit in a distributional sense and I have not read chapter 4 yet so I don't know what it means.
 
Yes, that's what I'm talking about. I'm trying to convince you that the pointwise limit is an inadequate interpretation.
 
9:12 PM
Hey, do anyone know how to solve semi-complex logarithm expressions?
$$ \log^2(x+2) = \log^4(x) $$
 
@tb You can't convince me of something I don't understand. Let's leave it at that for the moment and I'll badger you again when I have thought about it some more. Is that okay?
 
@Matt I'm aware of that, of course. Since I'm not sure what you're trying to do with mollifiers I can't give you a good example. One standard usage for them is that you can prove density of smooth functions in $L^p(\mathbb{R}^n)$ for $1 \leq p \lt \infty$ (and I saw that you had a question on that quite a while ago). The point here is that $\varphi_{\varepsilon} \ast g$ is a family smooth function in $L^p$ which tends to $g$ in $L^p$.
 
The reason why I thought about it was because it came up in this homework question and they only made us verify properties (1) and (2) of mollifiers so we didn't actually prove that the thing is a mollifier.
 
That's what I had in mind. You don't need to prove that your $\varphi_{\varepsilon}$ are a family mollifiers. What you should see however is that the $\varphi_{\varepsilon}$ do not tend to $0$ in $L^1$.
That's because $\int \varphi_{\varepsilon} = 1$ for all $\varepsilon$ and the integral is a continuous linear functional on $L^1$.
 
Holy cow. When you go to another university as a visitor, who usually pays the accommodation?
 
9:23 PM
@JonasTeuwen It depends. Usually someone invites you and then it is the one who invites you who pays.
(every professor has some money for paying visitors, usually)
 
Oh, great. Because at the ANU it is like $445 a week!
 
ANU ?
 
@JonasTeuwen But in what context are you going there?
 
Yes.
 
@tb The homework question doesn't ask me to prove that it's a mollifier but if I wanted to prove it anyway would I have to compute the distributional limit?
 
9:24 PM
@tb As a PhD student, my advisor wants every PhD student of his to spend a year abroad.
 
@JonasTeuwen well, then you'd have to look for some grant and support. Otherwise you can see if you can do some TA'ing or something to that extent. You should ask your advisor what kinds of possibilities you have. I assume he has something specific in mind.
 
Right. He would arrange it. But he's not here the next week/month, so I wondered if I need to start saving some money...
 
@JonasTeuwen 445 seems too expensive. Did you look this up on the internet? Or did the student exchange office give you this information?
 
I have looked it up on the website of the ANU and have asked a former student of the ANU.
 
Obviously there will be some deal for exchange students from your uni. I'd check with the exchange office.
 
9:28 PM
@Matt Well, you'd have to verify that $\varphi_{\varepsilon} \to \delta_0$ in the sense of distributions. Felder does this in Satz 4.7 in his notes.
 
Okay, I'll just ask my advisor. I thought it would be something common.
Maybe the guy where I'm going to is going to be my co-advisor. We shall see!
@Matt Still with the distributions? :D.
Duistermaat (RIP) has a great book about those.
 
I won't read much more about it, I don't really have any time as I'm already behind with revising stuff that's actually relevant for the exam.
 
There is no such thing as having not enough time to study distributions.
 
I think I have about 4 weeks to do the whole term.
It always ends up like this. Except for this term I at least attempted some of the exercises.
 
That's like eternity.
 
9:33 PM
This year I'll also read some of the notes. Preferably before attempting any homework. : )
 
Matt, then I recommend that you postpone learning about distributions. It takes some time to get used to them and if they weren't treated in the course, it won't help you much for the exam I think (except for some general culture).
 
@tb Thank you, I think that's exactly what I was looking for. Looks complicated. I'll postpone : )
@tb Actually, I think it might be relevant. It reminds me of what I've seen of Soboloev spaces.
But for now I'll postpone anyway, there is more basic stuff I want to do first.
 
Which course is this? :-).
 
Oh, sure it is relevant for Sobolev spaces (and it helps a lot there). But if you don't know about the Schwartz space, yet, stick to the way it was done in the course.
 
@JonasTeuwen Futile Attempts.
 
9:38 PM
The Schwartz space as in that particular decaying function space? Isn't that like basic in any course on Fourier analysis? :-).
 
Replace Fourier by Functional
 
Hmm, where do we need that in functional analysis? The Fourier transform maps Schwartz functions to Schwartz functions (and tempered distributions to tempered distributions of course) and that was, I thought a good reason to introduce them.
 
I think the Schwartz space is an example of a Fréchet space (which also came up in the lecture).
 
@JonasTeuwen Fourier Analysis $\neq$ Futile Attempts $=$ Functional analysis, as far as I understood.
@Matt yes, it is. It is the mother of all Fréchet spaces...
 
Yes, what tb says.
Knowing the definitions would now come in handy. : )
I'm going to go for a little while, see you later maybe.
And thank you!
 
9:44 PM
The only place where I have used general Fréchet spaces was in the course on functional analysis.
 
@JonasTeuwen Futile Attempts = Functional Analysis = what I'm revising.
 
See you later, Matt!
 
@Matt Bye.
 
10:11 PM
Mr. Square, how are you?
 
10:25 PM
Pretty good. We have finished watching the Rose Parade.
We are now watching the Rose Bowl.
 
Roses are flowers...
 
They are. Very good :-)
 
You are throwing parades for flowers??
 
The Tournament of Roses Parade, better known as the Rose Parade, is "America's New Year Celebration" held in Pasadena, California, a festival of flower-covered floats, marching bands, equestrians and the Rose Bowl college football game on New Year's Day (but moved to Monday if New Year's Day falls on a Sunday), produced by the non-profit Pasadena Tournament of Roses Association. Originally started on January 1, 1890, the Rose Parade is watched in person by hundreds of thousands of spectators on the parade route, and is broadcast on multiple television networks in the United States (A...
The Rose Bowl (officially the Rose Bowl Game presented by Vizio for sponsorship purposes) is an annual American college football bowl game, usually played on January 1 (New Year's Day) at the Rose Bowl in Pasadena, California. It is a part of the Tournament of Roses "America's New Year Celebration", which also includes the historic Tournament of Roses Parade. In 2002 and 2006 (2001 and 2005 football seasons), the Rose Bowl game was also the BCS National Championship Game. In the current BCS alignment, the Rose Bowl will host the Big Ten and Pacific-12 conference champions unless they a...
 
I'm not too far off, I see.
 
10:30 PM
hi @robjohn
hi Asaf
 
Srivatsan, what's up?
 
@Srivatsan Hey there. What's up today?
 
robjohn: do you want a puzzle sequences-and-series problem?
 
Sure. I am still working on filling in details for Didier.
A break might be good :-)
 
@robjohn Do you mean break from MSE/math, or break from Stirling?
 
10:34 PM
@Srivatsan A break from Stirling.
 
Discuss the convergence of $\sum \limits_{m, n \geqslant 1} \frac{1}{m^a + n^b}$.
There you go; your break from Stirling.
 
Soon... I'll be out of beer.
 
Drink arak.
Wait, I had the impression you didn't drink beer.
 
Nah, I have to wake up in less than seven hours.
No, I enjoy beer very much.
I also enjoy harder drinks on occasion.
 
Drink a glass of milk and sleep. :-)
 
10:38 PM
$\frac{2}{m^a+n^b}\le\frac{1}{m^{a/2}n^{b/2}}$ so we get convergence if $a,b>2$.
 
@robjohn sure.
 
God how I despise drinking milk. At most I can handle is cocoa. I can't even put milk in my coffee.
To quote some horrible emo-punk lyrics - I like my coffee black, just like my metal.
 
probably if $\frac{1}{a}+\frac{1}{b}\le1$
 
Although I bet he drinks milk based latte.
I drink black coffee, and bitter too.
 
I like my music and coffee, just like the workers in my cottonfarm.
 
10:40 PM
I'd delete that message.
 
leaving out the edge cases of $1$ and $\infty$
 
@robjohn =) ; that's right. At least that's what I got.
 
@AsafKaragila Maybe you are allergic to milk?
 
Not really.
I eat tons of cheese. I just hate milk in its base form.
 
For real, a large percent of people develop that when they get older.
 
10:41 PM
@robjohn Oh, actually I think you need strict inequality
 
@Srivatsan a=b=2 works.
 
@robjohn Diverges for a=b=2.
 
oops, you're right.
 
@Srivatsan: robjohn is a constant, he cannot diverge.
 
as I said first $a,b>2$ :-)
Can we show that this is necessary?
 
10:43 PM
@robjohn Are you revising your guess to "converges iff both a, b > 2"?
 
No...
I was talking about the inequality
 
OK, fine. :)
 
$\frac{1}{a}+\frac{1}{b}<1$
 
I was preparing for the "Ha, got you!" moment. :/
 
Sorry :-)
 
10:45 PM
@robjohn That's correct. But proof needed obviously. =)
 
My question was: is that necessary?
The proof of the sufficiency I have already given. The necessity I am thinking about.
 
@robjohn Are you asking about how to prove that if 1/a+1/b >1, then diverges?
 
@Srivatsan $\ge$
 
@robjohn As Asaf might say: your mother. :=) Where did you give a proof of sufficiency?
[Apologies; not meant to be rude.]
 
Well, the idea, but close to $\frac{2}{m^a+n^b}\le\frac{1}{m^{a/2}n^{b/2}}$
 
10:49 PM
Muahhaha. I have infected Srivatsan. Soon he'll eat meat and drink booze like there is no Buddha!
4
 
Gimme a bit
 
Sure, idea is fine. But AM-GM does not work. Now that you have a good guess of the answer, the fix is quite obvious.
@AsafKaragila chanting Long live arak! There is no Buddha! Asaf is god!
 
I am like Walter E. Kurtz. I got off the boat.
 
$a^{1/a}b^{1/b}mn\le m^a+n^b\text{ when }1/a+1/b=1$
That should prove the sufficiency
 
@robjohn I don't have mathjax installed in this stupid IE. Can you state what you are doing, in words? I can fill the details myself.
// I will fix the other problem in the meanwhile.
 
11:02 PM
Don't you have a brain LaTeX compiler module??
 
Are you using Young's inequality?
 
mn <=C(m^a+m^b) when 1/a+1/b=1
 
@AsafKaragila I do. That shuddered at the \text{ and } =)
 
xy <= x^p/p + y^q/q
 
@robjohn Young, no?
 
11:03 PM
@Srivatsan I am going to allow that.
 
That is used to prove Holder
Of course most convexity results are related.
 
@robjohn But isn't this inequality called Young's inequality? Of course, this one is one step if you apply jensen.
 
The horror... The horror...
2
 
yeah that is Jensen , Young, etc
Perhaps it is known as Young, but that is not what I have called it.
 
Sure. OK, that establishes convergence?
 
11:05 PM
Yes when 1/a+1/b<1
 
@robjohn OK. Nevermind the name, yes.
 
I think I'll hit the hay. See you folks on the other side.
 
@AsafKaragila sleep well
good wheat hitting :-)
trying to make it sound like good will hunting and failing
 
Right. Let's diverge now. [This one is easy.]
 
If 1/a+1/b=1
 
11:10 PM
@robjohn Sorry, what is " wheat hitting"?
 
@steveO "hit the hay"
 
4 mins ago, by robjohn
trying to make it sound like good will hunting and failing
@steveO Or as Srivatsan says, hitting the hay (which is not wheat, but I was taking liberties)
 
Thanks, @Srivatsan and @robjohn.
 
@Srivatsan divergence is easy? what am I missing?
 
Does hit the hay mean that sleeping on the hay?
literally
 
11:14 PM
@steveO Hit the hay means going to bed. I have not researched the etymology.
 
@robjohn OK, not hard. :)
 
If we consider the series sum_n 1/ (x + n^b) where x and b are fixed, then the first x^{1/b} terms are more or less equal.
 
@Srivatsan Okay, I see it now.
 
@robjohn Thanks, :-)
 
11:16 PM
@Srivatsan take a "diagonal" of the sum and it diverges.
 
@robjohn Oh, even that works, is it?
@robjohn What is the diagonal you take?
 
oh, hmmm... maybe not.
 
That's what I thought. :)
Let's focus on a=b=2 case. Then for a fixed m, sum_n 1/(m^2 + n^2) is at least m/2m^2, taking the first m terms alone.
 
shh :-)
oh well, go ahead, you've said too much already.
 
@robjohn Sorry. :)
 
11:19 PM
Too much to consider that I solved it myself :-) no, don't worry
 
You're right. Something spurred me on...
To repeat a previous observation: If we consider the series $\sum \limits_n \frac{1}{ (x + n^b)}$ where $x$ and $b$ are fixed, then the first $x^{1/b}$ terms are more or less equal. From these terms alone, we get a contribution of $\Omega(x^{1/b} / x)$
This is the inner sum. For the outer sum, we plug in $x = m^a$ and sum over all $m$.
 
I think the constants in the Omega might contribute something. Can we bound them somehow?
 
@robjohn In general you are right, but the constant I hid is just one-half. :)
 
that bounds it :-)
 
Ignoring the aspect that you cannot take x^{1/b} terms, but floor of that. Throw in another half for that, perhaps.
 
11:25 PM
That is understood
 
@robjohn Yes, just making sure that everything works. Sometimes I get vexed about the constants hidden. It usually turns out ok, but that's not always clear.
To complete the answer, we have 1 / x^{a(1 - 1/b)} >= 1/x. So it diverges.
Is the last step going in the right direction?
$1/a + 1/b > 1 \implies 1/a > 1 - 1/b \implies 1 > a(1-1/b)$, so that's correct.
Anyway, that completes the thing I guess. Down to all the details.
It's quite fascinating where all these Holder conjugates come up.
I said divergence is easy because I did this part first. Solving it gave me the guess on the convergence region. From that, it was easy to guess that I should apply Holder/Young/Jensen and get convergence as well.
@robjohn, There?
:)
 
@Srivatsan Sorry, I had to go afk for a while
 
11:45 PM
@robjohn: I have a general high-level question about non-Euclidean geometry. You're familiar with them, right? [This is a completely a newbie question.]
 

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