Mathematics - 2023-01-16

BAYMAX

02:21

0

Q: On a simple relation between four variables connected to an inequality.

Is there a simple relation between $x,y,z,w \in \Bbb{R}$ for which $\det(AB+A+I) = (x-y)z+(y+1)w+(x+1) = (y-x)(w-z) + (x+1)(w+1) < 0\\$ and $\det(BA+B+I)=(z-w)x+(w+1)y+(z+1) = (y-x)(w-z) + (y+1)(z+1)<0$ simultaneously. This arises from this question: A pen-and-paper proof for a matrix implicati...

Any ideas!

Akiva Weinberger

03:24

@mick This is the same as $\frac34x^2+(y-\frac12x)^2$

Not sure if that helps

Akiva Weinberger

There once was a man named Dobińsky
who could, with a smile and a wink-sy
excitedly tell
you the numbers of Bell
with two infinite series, methinks-y

user4539917

03:41

> He was the oldest of four children, having two younger sisters, Anna Maria and Maria Magdalena, and a younger brother, Johann Heinrich.

Koro

@leslietownes indeed. So I assume $X\times Y$ to have product topology and then I also show that the metric induced by the norm $\|(x,y)\|:=\sqrt{\|x\|^2+\|y\|^2}$ on $X\times Y$ actually generates the product topology.

leslie townes

03:57

koro: yeah, or even the max norm would work and might be easier to prove stuff about? i dunno. just, 'it comes from a norm' is what you seem to be using.

koro: when both (maybe at least one) of the spaces are complete, you can show that separate continuity of the form in each slice is enough to imply continuity in the product topology / aka the boundedness of the form. you need 'actual' functional analysis for that.

user4539917

@leslietownes did ted visit when plato came over to tutor?

leslie townes

user: well, we mostly only have ted's word for that. and a bunch of scrolls and tablets and stuff. i don't know there are any eye witnesses to any of this. except ted.

user4539917

perhaps ted was a part of the dialogues

ted's dialogues: a geometric approach

shintuku

ah yes, when he was referred to as Θεόδωρος

user4539917

Feb 25, 2022 at 18:29, by Ted Shifrin

I screwed up. Instead of my "geometric approach" theme, the subtitles should have been "an ironic approach."

leslie townes

04:10

that page of chat reminds me i need to get working on my functional analysis book.

Ted Shifrin

You mean, Functional Analysis , a Pedestrian Approach?

shintuku

#rekt

Koro

@leslietownes Ohh, you mean continuity in x and y implies the overall continuity under the said hypothesis?

If p is a sublinear functional on a real vector space X, show that there
exists a linear functional f on X such that -p(-x)<=f(x)<=p(x).

Well, the zero functional will work here.

But this is not probably the soul of this exercise. So I assume $X$ to be a non zero space, f to be not necessarily a zero map.

I think this is the situation that was meant in the exercise.

I don't understand what the following means:

we shall now derive another useful result
which, roughly speaking, shows that the dual space X' of a normed
space X consists of sufficiently many bounded linear functionals to
distinguish between the points of X.

what does 'consists of sufficiently many bounded linear functionals to distinguish between points of X' mean?

The theorem after this is: Let X be a normed space, $x_0\in X-\{0\}$. Then there exists a bounded linear functional f such that $f(x_0)=\|x_0\|, \|f\|=1$.

leslie townes

04:32

they mean: if x and y are distinct points of X, then there is a bounded linear functional g on X for which g(x) isn't equal to g(y). (consider the nonzero x - y as your x_0 and take g to be f_0 produced by the theorem)

the reason why you might care specifically about that fact over others might not be clear at this point in their treatment. but that's what it means for there to be 'sufficiently many' functionals to 'distinguish' the points. given distinct points, you can find a functional that takes different values on those points.

i don't know why a textbook would put "the following, roughly speaking, shows that [blah wordy thing that has not been defined]" type language before a statement of a theorem. that seems like bad organization to me. maybe it works in a classroom with some motivation first.

Ted Shifrin

04:51

Try this textbook.

leslie townes

that's one for the errata page.

copper.hat

05:06

@user4539917 it was a jk, playing on the name Leonhard

Yai0Phah

07:33

I just learned that the Hahn–Banach theorem can be proved via the Tychonoff theorem for compact Hausdorff spaces.

copper.hat

surely the hahn banach theorem needs a tvs?

Raffallo

07:55

hello! How can I solve AX=0 to get the correct values? Should I use regularized matrix computations?

Yai0Phah

08:05

The Hahn–Banach theorem only needs a real vector space with a sublinear functional.

Koro

08:17

@Yai0Phah more generally a functional that is subadditive and satisfies $p(ax)=|a|p(x)$ for all x and all scalars a.

@leslietownes thanks Leslie. I understood that now.

does one enounter 'digon' in algebraic topology?

it seems like a very mysterious object.

Koro

08:35

Given normed space X:
Algebraic dual of X: all linear functionals on X.
Dual of X: all bounded linear functionals on X.

one potato two potato

10:42

5

A: Surface groups and subgroups of fundamental groups

Such a covering map always exists. Indeed, suppose more generally that you have two closed surfaces $S$ and $T$ and a monomorphism $\pi_1(S)\to \pi_1(T)$ whose image $H$ is a subgroup of finite index. Then there exists a finite-sheeted covering $T'\to T$ such that $\pi_1(T')=H$. Furthermore, $...

iff condition $\pi_1(\Sigma_n)\hookrightarrow\pi_1(\Sigma_m)$.

Koro

11:28

What is the meaning of $\|f\|_\infty$ where $f\in C([0,1],\mathbb R)$, the space of continuous maps from [0,1] to $\mathbb R$?

I know its definition from measure theory but probably that's not the meaning in the above context.

is it correct to say that $\|f\|_\infty= \sup_{x\in [0,1]} |f(x)|$?

Alessandro Codenotti

@Koro yes (and you can replace sup with max if you prefer since the domain is compact)

Koro

11:48

@AlessandroCodenotti thanks. My concern is what if I replace [0,1] with any compact Hausdorff space.

If it is [0,1], then I can prove sup norm to be infinity norm by assuming Lebesgue measure on [0,1].

But not sure what measure I should consider to define $\|f\|_\infty$ in case of general compact Hausdorff space.

Alessandro Codenotti

Why do you need a measure?

Thorgott

@Koro sounds like a pokemon

Koro

Here is one definition of $\|f\|_\infty:=\inf\{t>0: \mu\{x\in X: |f(x)|>t\})=0\}$.

Martin Sleziak

Maybe related: Definition of $L^\infty$

Koro

with this definition we call $\|f\|_\infty$ to be the 'essential supremum' of f.

If X is compact subset of R, then we can show that indeed $\|f\|_\infty =\sup |f(x)|$.

@Thorgott LOL

@MartinSleziak It seems that we do need measure to define that norm.

@AlessandroCodenotti It is because of the above stated definition.

So in functional analysis, what is usually meant by $\|f\|_\infty$? I think sup norm only.

Thorgott

12:05

yes, sup norm

but of course, only on compact spaces where that sup is actually a max

Alessandro Codenotti

For $f\in C(X,\Bbb R)$, $\|f\|_\infty=\max f$, for $f\in L^\infty(X,\mu)$, $\|f\|_\infty$ is the thing you wrote above

Koro

@Thorgott it seems true. Meanwhile I looked it up and found this mast.queensu.ca/~speicher/Section3.pdf

They also use this notation to mean the sup norm and not the essential sup norm.

Thorgott

@AlessandroCodenotti oh yeah, or that

Koro

@AlessandroCodenotti Is it true for any measure $\mu$?

Thorgott

Koro, this is simply one of those times where two things are similar and related, but different, and which is used depends on the context

in measure theory, you want the essential sup, in functional analysis, you want the sup

Alessandro Codenotti

12:11

@Koro It's weird to call a definition "true", but the definition works for any measure, yes

Koro

and in winter I want soup. It's all related.

Thorgott

also, fwiw, essential sup and sup are the same for continuous functions

that's something to keep in mind

Koro

From $R^n$ to R, yes.

Thorgott

for any Borel space or w/e you call them

Koro

but from compact Hausdorff X to R: it seems from the above comment by Alessandro Codenotti that it is true as well.

But then the comment also seems to say that: $\|f\|_\infty=\max f$ for any measure $\mu$ on X.

Because for different $\mu$'s the sets $\{|f(x)|>t\}$ will be different in the sense that null w.r.t. one measure may not mean null w.r.t. the other.

Alessandro Codenotti

12:16

No that's not what I meant, I'm saying that the symbol $\|f\|_\infty$ is used to denote two different (although somewhat related) things, depending on whether $f$ is a continuous function on a compact space or a measurable function on a measure space

Thorgott

and if a space is both a topological space and a measure space in a compatible way (meaning its sigma-algebra contains the Borel algebra) and we have a function that's both continuous and measurable (this is equivalent to just continuous, actually), then the two notions of $\lVert f\rVert_{\infty}$ agree

Koro

Ahh I see. So the comment is: For $f\in C(X,\mathbb R), \|f\|_\infty: =\max f$ and when $f\in L^\infty (X,\mu)$ then $\|f\|_\infty$ is ...

Thorgott

in either case, there is a notation of $L^{\infty}$ as those continuous/measurable functions for which $\lVert f\rVert_{\infty}$ in the respective sense is finite

so now let $X$ be a topological space and a measure space with measure $\mu$ defined on a $\sigma$-algebra containing the Borel algebra, let $B(X,\mathbb{R})$ denote the set of bounded continuous functions $X\rightarrow\mathbb{R}$, i.e. those for which $\lVert f\rVert_{\infty}\coloneqq\sup_{x\in X}|f(x)|$ is finite.
let $L^{\infty}(X,\mu)$ denote the set of measurable functions $X\rightarrow\mathbb{R}$ for which $\lVert f\rVert_{\infty}^{ess}$ (the essential supremum, I'm introducing new notation to distinguish it from the sup) is finite. Then, essentially, $B(X,\mathbb{R})=C(X,\mathbb{R})\…

this last claim is morally true, but not literally true for technical reasons

it's probably best understood as a pullback, but I haven't thought it through

or, rather, it is literally true, but I actually lied in my definition of $L^{\infty}$ a bit*

Koro

@Thorgott the lie was of measure 0

Thorgott

you want to identify functions that only differ on a set of a measure 0, if that's what you mean

Koro