Mathematics

Hi

@Srivatsan There is no context.

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?

Daniil

cut-the-knot.org/arithmetic/anEquation.shtml this is quite nice.

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

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

ntnu.edu/physics/courses

7:56 PM

definitely do QM

I guess I will wait and see, my mind will obviously change

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 :-).

Gigili

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.

Gigili

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.

Daniil

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!