Mathematics - 2025-01-06

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Soham Saha

04:38

Hi everyone. Does anyone know which SE site is best suited for illusions (visual, auditory, etc) ?

leslie townes

05:27

soham: psychology & neuroscience has an optical-illusions tag, psychology.stackexchange.com/questions/tagged/optical-illusion although i can imagine questions about illusions that might not be a good fit for the description of that tag

Jakobian

@psie spaces with this property are called exhaustible by compact sets

names - very important

that sounds a bit crazy if you understand it wrong, but hear me out

if you know names, you can search for it

you can remember it better as well

your brain creates a separate container with a label that is the name
and your brain, like a computer searching for a file, doesn't go out of description alone to search for it, but searches for the name and then finds related stuff

well it doesn't actually create any labels as far as I know how brains work by being a human being

but just the fact that brain has ability to relate things to each other (via connections in your brain) does the job

no one needed me TED-talking (Ted talking, heh), but here you go. A speech for the audience that no one needs, for people who don't care about it, for a topic that everyone thinks is obvious enough that it goes without saying

and no one asked for

Soumik Mukherjee

05:45

This is a new variant of TED talking, this is TED chatting

2 hours later…

Soham Saha

07:54

@leslietownes thanks for the link :)

2 hours later…

nickbros123

09:48

the viewpoint of a field over a subfield as a vector space is more strong than i thought

psie

@Jakobian cool :) I'll remember this name. I agree, it's great when things have a name.

psie

10:37

> Theorem Let $p\in[1,\infty)$.

> (1) If $(E,\mathcal A,\mu)$ is a measure space, the set of all integrable simple functions is dense in $L^p(E,\mathcal A,\mu)$.

> (2) If $(E,d)$ is a metric space and $\mu$ satisfies $\mu(A)=\inf\{\mu(U):U\text{ open},A\subset U\}$ for every $A\in\mathcal B(E)$, then the set of all bounded Lipschitz functions that belong to $L^p(E,\mathcal B(E),\mu)$ are dense in $L^p(E,\mathcal B(E),\mu)$.

> (3) If $(E,d)$ is a separable locally compact metric space, and $\mu$ is finite on compact subsets of $E$, then the set of all Lipschitz functions with compact support is dense $L^p(E,\mathcal B(E),\mu)$.

> Consequences The space $C_c(\mathbb R^d)$ of all continuous functions with compact support on $\mathbb R^d$ is dense in $L^p(\mathbb R^d,\mathcal B(\mathbb R^d),\lambda)$, where $\lambda$ is Lebesgue measure.

I wonder; my guess is that the consequence must follow from (3), but $\sqrt{x}$ on $[0,1]$ is continuous with compact support but not Lipschitz, right? I'm confused.

psie

11:16

I got muddled up. The Lipschitz functions with compact support, which are dense, is a subset of the continuous functions with compact support.

Bml

12:01

@leslietownes Thank you. Isn't there a more explicit proof?

@Jakobian Thank you.

nickbros123

12:27

@Jakobian personally what sticks with me is what I think will be important, or what I find "neat". It could be that I read something rly boring, Ill forget it very easily; but when it gets applied somewhere else (somewhere more interesting) and I read it again from a different context, it sticks way easier and more concretely

handan_toddler

12:53

memory is the residue of thought

and meaning is what you should think about

Soumik Mukherjee

13:12

@handan_toddler why?

handan_toddler

why what?

Soumik Mukherjee

13:32

@handan_toddler why one should think about meaning?

handan_toddler

It is easier to remember.

handan_toddler

14:18

because there is more to think about

Soumik Mukherjee

14:36

i see

ILikeMathematics

@VladimirLysikov I'm doing it a different way now - the usual $f(x) = f(y) \implies x = y \implies$ injective

So I think "Let $u \otimes \varphi(v) = z \otimes \varphi(w)$. We need to show $u \otimes v = z \otimes w$" should be enough

2 hours later…

Vladimir Lysikov

17:09

@ILikeMathematics Even in this case, you need to consider arbitrary $x$ and $y$, and not only $x = u \otimes v$ and $y = z \otimes w$

Here is an example of what you need to worry about:
Take an $n$-dimensional vector space $V$ with a basis $e_i$, consider the dual basis $e_i^*$ in $V^*$, and define a tensor $I = \sum_{i = 1}^n e_i \otimes e_i^*$.
Consider a map $\psi \colon V \otimes V^* \to V \otimes V^*$ given by $\psi(v \otimes y) = v \otimes y - \frac{1}{n}y(v)\cdot I$
It is not injective: one can check that $\psi(I) = 0$.
But if $n \geq 3$, then it is true that $\psi(u \otimes x) = \psi(v \otimes y) \Rightarrow u \otimes x = v \otimes v$

psie

18:14

In the Riemann-Lebesgue lemma, the author says it suffices to prove $$\hat{f}(\xi)\underset{|\xi|\to\infty}{\to}0$$for step functions on $\mathbb R$ (simple functions where the sets are intervals). This is because the step functions are dense in $L^1(\mathbb R,\mathcal B(\mathbb R),\lambda)$ and so let $(\varphi_n)$ be a sequence of step functions that approximate $f$ in the $L^1$ norm.

Then \begin{align*}\sup_{\xi\in\mathbb R}|\hat{f}(\xi)-\hat{\varphi}_n(\xi)|&=\sup_{\xi\in\mathbb R}\left|\int f(x)e^{\mathrm{i}x\xi}\,\mathrm{d}x-\int \varphi_n(x)e^{\mathrm{i}x\xi}\,\mathrm{d}x\right|\\ &\leq\|f-\varphi_n\|_1\end{align*}

I don't understand what this inequality shows. That $(\hat{\varphi}_n)$ converges uniformly to $\hat{f}$? What is the justification for proving the lemma for step functions only?

ILikeMathematics

@VladimirLysikov Ah, you're right

Thanks

Jakobian

@psie continuity of the Fourier transform as a function from $L^1$ to $L^\infty$

psie

@Jakobian ok, hmm. I have to give this some thought.

Jakobian

18:36

Let's refer to the condition as being in $C_\infty$

(the Fourier transform is in $C_\infty$)

once you show that $C_\infty$ is closed in $L^\infty$, you are done

psie

@Jakobian why is it a function to $L^\infty$ though? The expression $\sup_{\xi\in\mathbb R}|\hat{f}(\xi)-\hat{\varphi}_n(\xi)|$ is the sup-norm, which is not the same as the $L^\infty$ norm. The inequality reminds of a sort of Lipschitz condition.

Jakobian

it's the same once you know that Fourier transform is continuous

psie

ah ok, then we're good, I think I know that the Fourier transform is continuous :)

Jakobian

let me correct myself

it's the same once you know that if $f\in L^1$ then $\hat f\in C$

psie

18:53

The inequality seems to be a very concise way of proving the lemma. I think there are a few key steps that have simply been omitted :(

Sine of the Time

left to the reader

psie

:(

I understand almost uniform convergence is equivalent to convergence in $L^\infty$, and that is what the inequality shows; $(\hat{\varphi}_n)$ converges in $L^\infty$ to $\hat{f}$. But I don't see how this proves the lemma.

Sine of the Time

$\hat \varphi_n$ vanishes at infinity right?

for each $n$

psie

@SineoftheTime yup, it has compact support for each $n$.

Sine of the Time

so also $\hat f$ vanishes

Soumik Mukherjee

19:04

user image

I am triggered by this terminology

Sine of the Time

:|

psie

@SineoftheTime why is compact support preserved under uniform convergence?

Thorgott

@SoumikMukherjee lol I've never heard of that

psie

@SineoftheTime actually, it is $\varphi_n$ that has compact support for each $n$. I'm not sure about $\hat \varphi_n$.

well, actually I am

that's the point of the proof

we show $\hat \varphi_n$ vanishes at infinity only

Jakobian

19:20

Again, I will repeat myself

If $\hat\varphi_n\to \hat f$ in $L^\infty$ and $\hat\varphi_n\in C_\infty$ then $\hat f\in C_\infty$

in other words, $C_\infty$ is closed in $L^\infty$

well... not in other words

what you want to use is that $C_\infty$ is closed

psie

so one has to show that $C_\infty$ is closed (and that it is a subset of $L^\infty$)

ILikeMathematics

20:06

Let $\beta: V \times V \to K$ be a bilinear form. Then if $1 + 1 \neq 0$ in $K$ and $\beta$ is symmetric, there exists an orthogonal basis of $V$. Why do we need $\beta$ to be symmetric? Assume $\beta$ is not symmetric and $1 + 1 \neq 0$, can't we pick an arbitrary basis of $V$ and apply Gram-Schmidt?

Vladimir Lysikov

In Gram-Schmidt there can be problems with denominators.
For example, for any antisymmetric form $\beta$ we have $\beta(x, x) = 0$ and this messes with Gram-Schmidt
And indeed antisymmetric forms do not have orthogonal bases, we introduce symplectic bases for them

Also, the notion of orthogonality is only nice when it is a symmetric relation, that is, $\beta(x,y) = 0 \Leftrightarrow \beta(y, x) = 0$, so not for arbitrary forms

psie

20:58

@Jakobian do you have a hint how to show $C_\infty\subset L^\infty$ and $C_\infty$ being closed?

Thorgott

what is $C_{\infty}$?

psie

good question, because I'm not so sure myself, it was never really defined explicitly, only stated in words. I think it is the space of all Fourier transforms vanishing at infinity. Like Jakobian said, if $\hat\varphi_n\to \hat f$ in $L^\infty$ and $\hat\varphi_n\in C_\infty$ then $\hat f\in C_\infty$. The logic of this makes sense, but only if $C_\infty\subset L^\infty$ is closed.

here $\hat\varphi_n$ is the Fourier transform of a step function

Thorgott

21:17

@psie you only really need to observe that you can replace the sup by a limsup as $\xi\rightarrow\pm\infty$ on the LHS

Jakobian

@psie continuous functions which vanish at infinity

that is, for any $\varepsilon > 0$ there exists a compact set $K$ such that $|f(x)| \leq \varepsilon$ for $x\notin K$

equivalently, in this context, $\lim_{|\xi|\to \infty} f(\xi) = 0$

To show that $C_\infty$ is closed, one needs to take $g_n\in C_\infty$ with $g_n\to g$

psie

@Thorgott I'm thinking (to show an arbitrary Fourier transform vanishes at infinity given we know that the Fourier transform of a step function vanishes at infinity plus that inequality), we bound $\|\hat f(\xi)\|_\infty\leq \|\hat f(\xi)-\hat\varphi_n(\xi)\|_\infty+\|\hat\varphi_n(\xi)\|_\infty,$, though I'm not sure what happens with the term $\|\hat\varphi_n(\xi)\|_\infty$

Jakobian

while proving that it's closed, preferably don't use any previous notation, the hats and Fourier transform merely obscure the picture

Thorgott

you do know what happens with that term

it's the very premise of your original question

Jakobian

either way, one can fix find $n$ such that $\|g_n-g\|\leq\varepsilon$

we have $\|g\|\leq \|g_n\|+\|g_n-g\|$

now you can find a compact set $K$ such that $|g_n(x)|\leq\varepsilon$ for $x\notin K$

and so on. I won't complete the argument, I guess I already said a lot

there is no difference from any analysis argument whatsoever

psie

21:36

Ok, thank you!

2 hours later…

Thorgott

23:59

(sorry, misclick)

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