Mathematics - 2021-10-04 (page 1 of 2)

love_sodam

01:17

0

Q: Showing the equality between chain map and chain homotopy

This is one of the exercise problem in Hatcher. Let $C$ and $C'$ be chain complexes and let $I$ be the chain complex consisting of $\Bbb Z$ in dimension $1$ and $\Bbb Z\times\Bbb Z$ in dimension $0$, with the boundary map taking a generator $e$ in dimension $1$ to the difference $v_1-v_0$ of gen...

I have a question on one of the hatcher exercise problem

I wonder if I understood the problem correctly

SmokenSieEinBitteChebaHitBits

@love_sodam

I upvoted your post

coincidentally I'm only one section before that in Lang

*homotopies

I understand tensor product but not what you're doing with it in your question

Could you teach?

@love_sodam for example, what is a chain homotopy

love_sodam

You can google it (just a definition). I don't think this problem requires some high level algebra.

SmokenSieEinBitteChebaHitBits

You're doing HA - that is by definition higher level algebra

lol

Let $f, g : E \to E'$ be two morphisms of complexes

Then $f$ is homotopic to $g$ if there exists a morphism $h : E \to E'$ of degree $- 1$ such that $f_n - g_n = d'^{n-1}h_n + h_{n+1} d^n$.

@love_sodam you should at least draw a picture of that or have a picture of it in your head

first thing

The CD I mean

love_sodam

Yeah I know what that is. What are you trying to say?

SmokenSieEinBitteChebaHitBits

I'm just trying to learn it myself :)

So you have:
https://q.uiver.app/?q=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

@love_sodam click "Show / Hide Grid" on toolbar for a clearer picture

nvm that that diagram doesn't exactly commute

Bob Pen

02:54

$\left\|\left(\boldsymbol{X}^{T} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{T} \boldsymbol{v}\right\|_{2} \leq \frac{1}{\sigma_{\min }(\boldsymbol{X})}\|\boldsymbol{v}\|_{2}$ Does anyone have any hints as to how I can proceed in proving this inequality?

leslie townes

03:18

mm, the LHS is the least squares solution to Xa = v (where a is the unknown). do you know the characterization of this solution in terms of the SVD of X?

SmokenSieEinBitteChebaHitBits

03:59

Test

My name update didn't show here yet

Test again

leslie townes

might take a while. i remember when i updated my profile image it took a while before it went from my profile page to the chat window

BKJA1

Hello World !

SmokenSieEinBitteChebaHitBits

hi

copper.hat

05:04

@BobPen Write $X = U \Sigma V^T$ and expand.

Koro

05:16

Suppose that $a\gt 0$ and let it be known that $a^{m+n}=a^ma^n$ for all $m,n\in \mathbb Q$. Now I want to show that $a^{x+y}=a^xa^y$ for all $x,y\in \mathbb R$. For this, I let $(x_n), (y_n)$ to be sequences of rationals that converge to $x$ and $y$ respectively so $a^{x+y}=a^{\lim x_n+\lim y_n}=\lim a^{x_n}. \lim a^{y_n}=a^xa^y$ (as $a^{m+n}=a^ma^n$ for all $m,n\in Q$)

I think that this is wrong as it nowhere uses the fact that $x\to a^x$ a continuous function. What am I missing?

Ah no, I made a mistake: $a^{x+y}=a^{\lim x_n+\lim y_n}=a^{\lim (x_n+y_n)}=\lim a^{x_n+y_n}=\lim ( a^{x_n} a^{y_n})$ because $x_n, y_n$ are rational numbers.

where the third equality follows by continuity of $x\to a^x$. So it follows that $a^{x+y}= \lim a^{x_n} \lim {a^{y_n}}= a^x a^y$. This proves the result. Is my understanding correct? Thanks.

leslie townes

05:51

yes, that's where continuity is used, assuming you have access to that fact. continuity is often addressed (implicitly) within the definition of a^x itself.

Bob Pen

06:02

In the textbook I am reading right now, it says:

(Restricted Strong Convexity). The matrix A satisfies the restricted strong convexity (RSC) condition of order $k$, with parameters $\mu>0$, $\alpha \geq 1$, if for every $\mathrm{I}$ of size at most $k$ and for all nonzero $\boldsymbol{h}$ satisfying $\left\|\boldsymbol{h}_{\mathbf{| c}}\right\|_{I} \leq$ $\alpha\left\|\boldsymbol{h}_{I}\right\|_{1}$
$$
\|\boldsymbol{A} \boldsymbol{h}\|_{2}^{2} \geq \mu\|\boldsymbol{h}\|_{2}^{2}
$$

However this is said to satisfy the Nullspace property:

This is the book: book-wright-ma.github.io/Book-WM-20210422.pdf, page 89 - 91

Koro

thanks a lot @Leslie for clarifying. :)

robjohn

07:40

@AbstractSpacecraft I updated your profile.

@leslietownes It was still unchanged here until I updated their profile from their main account.

Alex Robinson

09:56

when doing chi squared, if the hypothesis is for a positive correlation, but the data suggests a negative correlation do you halve the p value?

my working example: revising should increase pass rate of a test, but the data suggests revising decreases pass rate and my p value is ~0.08. do i halve it

Typo

11:00

Is there any symbolic way of showing 2 is a solution of log_2x +x=3? I don't want to involve the lambert W function

Nevermind, I got it.

love_sodam

13:05

3

Q: Showing the equality between chain map and chain homotopy

This is one of the exercise problem in Hatcher. Let $C$ and $C'$ be chain complexes and let $I$ be the chain complex consisting of $\Bbb Z$ in dimension $1$ and $\Bbb Z\times\Bbb Z$ in dimension $0$, with the boundary map taking a generator $e$ in dimension $1$ to the difference $v_1-v_0$ of gen...

algebraic-topology homology-cohomology

It seems the problem has an error

Jam

any lecture on grassmanian subvarieties?

robjohn

13:27

@TedShifrin I commented, and might even post to reopen since the author has improved the post.

@Typo For that equation, simply plugging in some simple values will give a solution, but if $3$ were changed to something else (not $6$ or $11$, etc), you probably need Lambert W.

Thorgott

14:43

@love_sodam what do you mean by "f is the same as h", that doesn't really make sense

Semiclassical

@Typo out of curiosity, what did you come up with? it's easy enough to guess and check that x=2 solves your equation, but I don't know how you'd deduce it otherwise

robjohn

pretty much what I was saying

Semiclassical

yeah, not sure how i missed what you said when it was right there

robjohn

yeah, if there were a simpler solution, then we'd have a simpler expression for W

Semiclassical

i wouldn't be surprised if, aside from the trivial cases (i.e., pick some $n$ so that $\log_2 x + x=n+2^n$ has $x=2^n$ as solution), the roots of $y=x+\log_2 x$ are transcendental

there may even be a way to prove that but uh

i don't actually know or care

o.9

15:08

hello good afternoon

Don Thousand

I'm currently writing up some course notes for a course I am teaching on first order logic and the basic theorems. I'm struggling with where to start, however. I want to start with sets, but I'm tempted to do a breakdown on why each axiom in ZFC is important (and give the axiom of regularity some love), but in doing so, there needs to be some level of understanding of definable relations/logical formulas/set maps, which requires first order logic.

Do you think handwaving over the axioms with a intuitive understanding of what logical formulas are to begin the study of set theory, followed by propositional logic and later first order logic makes sense? Or is there a better way to order and present things? I just don't know.

anakhro

15:57

@DonThousand what are the main theorems you are wanting to cover in the course?

fin

hi @TedShifrin

Ted Shifrin

hi @fin

fin

whats up

good morning

Ted Shifrin

Leaving in five minutes for PT

fin

oh fun

o.9

16:03

what is pt

physical therapy?

Ted Shifrin

physical therapy

fin

poop time

idk if i said this when i was drunk but college has been a really good environment for me

im starting to experience some semblance of a challenge

o.9

glad to hear that

Ted Shifrin

yes, that's to be expected

fin

not to sound like an asshole though its just in the STEM classes

Ted Shifrin

16:04

I got most of my Bs in literature classes

fin

interesting

i love writing

Ted Shifrin

I write good analytical stuff, horrid creative stuff.

o.9

why was there bullshit in the literature classes :(

Ted Shifrin

No, the lit classes were excellent.

Several amazing teachers.

o.9

so why did you get most of your bullshit there

Ted Shifrin

16:06

And in math the one B I got (I deserved more) was in the course which is most my expertise

You're the one saying bullshit. I never did.

It shows your bias.

o.9

What is Bs

Ted Shifrin

plural of B

o.9

oh

Ted Shifrin

rolls eyes

We usually put B's, but that's actually wrong.

o.9

I C

Ted Shifrin

16:07

Now I understand where you got it. :)

anakhro

@DonThousand You aren't around anymore to elaborate, but my suggestion is to come up with some interesting soft questions in logic to pique interest. e.g. "Are we justified in using infinity in mathematics, despite physical evidence that suggests it isn't an empirical concept?", or "Is there a list of axioms that summarizes all of mathematical reasoning?".

fin

i have my first college exam today

anakhro

Congrats, great fun!

Don Thousand

16:27

@anakhro Sorry, I'm back. Planning to cover up to Lowenheim Skolem and Henkin Models. It's a long course.

@TedShifrin Makes me feel good about my bad grades in uni. Loved the subjects, but still always did bad :(

@anakhro And yea! That was part of my plan for the lecture. It's around 20 students, so it should be fun. But I need to write up course notes for students with disabilities and others who want to prestudy, but I finding it a lot more difficult for the reasons I mentioned.

anakhro

@DonThousand What are your major concerns about the ordering? It's fine to give hand wavy stuff at the start for like a bit, but don't depend on it too much. Catching their interest with big questions that will be answered over the course of the lectures might be one of the more accessible things.

@leslietownes are you familiar with distribution theory at all?

You are the only analysis person I recognize right now. :(

user10478

This might be trivial and I'm just missing it, but can someone explain in more words what the bottom comment here (math.stackexchange.com/questions/4249567/…) is suggesting? It seems to be introducing a condition that isn't part of the problem to relate $C$ and $D$.

leslie townes

anakh: somewhat but it has been a long time

anakhro

@leslietownes maybe worth a try then: are you at all familiar with trying to do something like distributions except instead of test functions in $C^\infty_c$, use $C^0_c$ or something significantly larger?

leslie townes

i guess. i don't see 'distributions' as tied to any particular set of test functions, although maybe some people reserve that term for whatever schwartz was doing.

in my limited experience, the notation in that area is full of baroque notation conventions and superscripts and subscripts to denote exactly what choices are being made in the setup

anakhro

16:41

Heh.

leslie townes

maybe that isn't classical or the most useful thing but it's what ive seen

anakhro

To put it in words, I just wonder if it is something already considered to have the space of test functions be continuous functions, rather than smooth.

leslie townes

i don't recall seeing something with that little regularity being discussed in the context of distribution theory, but people do study that. the space of continuous functionals on C_b(X) is not a very nice space.

it's realizable as a space of finitely additive measures on X, or perhaps slightly more regular measures on a really goofy compactification of X. where measures are signed measures, i think.

people doing distributions tend not to want all of that stuff in there. but who knows,.

anakhro

Ah, I see. It does seem a bit wacky.

Thanks!

epsilon-emperor

Hello

(Infinite dimensional complex Hilbert spaces)

Riesz Rep theorem gives a natural map from X to its dual

but this is not an isomorphism due to the conjugate linearity issue

So do we say that X and its dual are isomorphic, or not?

leslie townes

16:53

people often say 'anti-isomorphic'

epsilon-emperor

Yep, seen that!

but in general if someone says they are isomorphic do I correct them?

and maybe the natural map is not an isomorphism, but some isomorphism exists?

We don't know that right

leslie townes

i wouldn't, unless i were grading them on finer points of terminology, or unless it seemed like it was interfering with something they were trying to say

if you realize the dual as something with an obvious conjugation on it (e.g. a set of complex valued functions) then composing with that would give you a linear isomorphism

sometimes people are only interested in properties that are preserved under anti-isomorphism, where just saying 'isomorphic' wouldn't seem that bad and is maybe actually right in whatever category they are implicitly working in

but YMMV

epsilon-emperor

@leslietownes yes fair

Yep cool

thank u

do you know if an actual isomorphism (not natural) exists tho?

maybe some magic using orthonormal basis?

leslie townes

17:25

yeah, just choose orthonormal bases for each (by riesz they have the same cardinality) and write down the unitary

epsilon-emperor

yep, thanks!

umm "unitary"?

anakhro

unitary transformation I assume he means

leslie townes

yeah. the norm on the dual satisfies the parallelogram identity, so it comes from an inner product. it's a hilbert space with that inner product

if H and K are hilbert spaces with orthognormal bases of the same cardinality, (h_i)_{i in I} and (k_i)_{i in I}, there is a unique unitary U from H to K satisfying U h_i = k_i, i in I

epsilon-emperor

ohhh okay didn't know this

thanks!

robjohn

17:40

@TedShifrin Using Theorem 5.4 (quoted by the author) makes that problem pretty simple.

leslie townes

all of this somehow gets more confusing if you want to identify the dual with H itself. i guess because the "natural" thing you want to do is antilinear and not linear

leslie townes

17:51

my daughter now wants to be a lemon or a strawberry for halloween. not sure how to make this happen. i might be able to provide a glass of diluted strawberry lemonade.

robjohn

@leslietownes I take it a yellow sheet or a red sheet with black polka dots will not do for the ghost of a lemon or strawberry?

even if it has "Sunkist" printed on it?

Koro

I want to find an invertible matrix M such that $A^T=M^{-1}AM$.

where A is a square matrix.

leslie townes

ooh, good idea

koro, math.stackexchange.com/questions/62497/…

Koro

I tried with a 2 by 2 matrix first. I took A as matrix of a linear transformation T with respect to standard basis in R^2. Then I want to find a base change that will give me $A^T$ and M

Am I in the right direction?

copper.hat

unfortunately we gave away our younger halloween costumes many years ago.

leslie townes

18:04

her first choice was this genuinely frightening jack-o-lantern mask she saw me looking at online. it was for older children and probably would have gotten her kicked out of day care. then she changed to ghost, which i was rooting for because it's the easiest to make.

who knows what it'll be next week

Semiclassical

the answer by Jose (at least the diagonalization argument) is the most elementary approach i see in that link

namely, argue that if both $A$ and $A^T$ are diagonalizable, then they'll both be conjugate to the same diagonal matrix of eigenvalues

and thus conjugate to each other

the issues arise once you have to worry about matrices which aren't diagonalizable

for that you have to prove the case of $A$ in Jordan normal form

which isn't too bad

leslie townes

the comments on mathoverflow.net/questions/122345/… are interesting

Semiclassical

18:25

it reminds me of another question but blast if i can remember which one

leslie townes

18:43

loud fighter jet again

copper.hat

must be kamala on some secret meeting with fox news.

Semiclassical

i want to say the question i'm forgetting was like this. suppose you start with the system of equations $y=Ax$ for vectors x,y and matrix $A$. then the fact that $A$ and $A^t$ are conjugate means there exist $x',y'$ such that $y'=A^T x'$

now, if you know how $A$ and $A^t$ are conjugate, i.e., $A=MA^t M^{-1}$

then one just picks $y'=M^{-1}y$ and $x'=M^{-1}x$

but these two system of equations should also be equivalent by elementary row ops

so "how many" row ops do you need to do to establish this?

but blast if i can find that question again

leslie townes

18:59

looks like kamala is flying from here to las vegas. secret meeting with the craps table, i'll wager.

copper.hat

playing a little golf on the side too, i'll bet

Ted Shifrin

@robjohn Hell, I never even knew this. Shrug.

@anakhro Hard to work with distributional derivatives if you do that!!

Wow, FB is down globally. Good for the whistleblower!

Koro

and so is whatsapp

Ted Shifrin

That's part of the "global."

Koro

19:15

apparently Google Pay is down too

Ted Shifrin

I wouldn't know. I use only Apple Pay.

Koro

Apple pay is not available here :(

1 hour ago, by Koro

Am I in the right direction?

So the proof is easy for real symmetric matrices (in fact any diagonalisable matrices)

I think proving (in general case) $A$ and $A^T$ similar is going to be very difficult using elementary methods.

Ted Shifrin

Symmetric? Pretty stooopid. Same matrix.

Koro

I take that back and replace symmetric by "diagonalisable matrix"

Ted Shifrin

It’s a bit muddling, as the transpose is naturally the dual map on dual spaces.

anakhro

19:27

@TedShifrin I wasn't planning on exploiting that aspect of the distributions. :P

Ted Shifrin

19:40

That sort of is the whole point, though.

Otherwise just work with $L^1_{\text{loc}}$ or something.

Krijn

'ello

Hey Ted, how are things going

Ted Shifrin

Heya @Krijn

Krijn

Karma can be quite funny

When I was in my bachelor's degree, I really skipped over most of statiscics and probability theory because I enjoyed algebra and number theory much more

Turns out you can't get away with not knowing that stuff in a phd

Xander Henderson

19:55

@Krijn At many institutions, "Prob and Stats" is its own department, separate from mathematics.

So many folk pursuing PhDs in mathematics never do much with prob and stats.

Most people who are interested in analysis end up picking up a fair amount of probability (since probability is "just" measure theory with a couple of new definitions thrown in for funsies), but even they might not ever do much with statistics.

Krijn

I guess, what you need you'll pick up along the way

Xander Henderson

I mean, I completed my undergraduate and masters degrees at an institution with a very strong prob and stats program within the mathematics department (my undergraduate emphasis is even "prob and stats"), but I am a terrible statistician. I have a collaborator who does the stats for us when we collect student data.

Krijn

What kind of reasearch do you do?

Xander Henderson

@Krijn At this moment in time? Very little, unfortunately (yay for teaching 18+ load!).

Semiclassical

@TedShifrin here's a silly question for you. do you know an elementary derivation for the volume of a spherical cap?

the calculus proof is simple ofc

Xander Henderson

20:02

But my masters thesis was on the Assouad dimension of sets which arise as the attractors of iterated function systems satisfying some kind of separation condition, and my PhD thesis is on the complex dimensions of homogeneous metric spaces.

Ostensibly, I know a little something about fractal geometry and analysis on fractals.

Krijn

Whats an Assouad dimension

Semiclassical

Assouad dimension

In mathematics — specifically, in fractal geometry — the Assouad dimension is a definition of fractal dimension for subsets of a metric space. It was introduced by Patrice Assouad in his 1977 PhD thesis and later published in 1979. It was defined earlier by Georges Bouligand (1928). As well as being used to study fractals, the Assouad dimension has also been used to study quasiconformal mappings and embeddability problems. == Definition == The Assouad dimension of X , d A ( X...

leslie townes

first you compute the usa soda dimension, then you permute it

Semiclassical

basically, a definition of dimension that makes sense for fractals

leslie townes

or that more complicated thing

Krijn

20:04

Ahhhh, the Bouligand dimension

Joking, never heard of it

leslie townes

great example of that law of naming. the third thing you learn on that page is that assouad was not the first to introduce the assouad dimension. :)

Semiclassical

i think i've only ever seen box-counting dimension

Xander Henderson

@Krijn Roughly speaking, different notions of "fractal dimension" seek to quantify how the volume of a ball scales with the diameter of that ball. The Assouad dimension compares the volume of a big ball with the volumes of small balls contained within the big ball, which gives a "finer" notion of scaling (it is not just "how does a ball scale?" but "how does a little ball scale when compared to a big ball?").

It is kind of a generalization of the Minkowski-Bouligand dimension, or the Hausdorff dimension.

Krijn

check

Xander Henderson

@leslietownes Baez's Law of Misattribution strikes again!

leslie townes

20:07

can we start calling it Leslie's law of misattribution?

Xander Henderson

@leslietownes No. :P

ncatlab.org/nlab/show/Baez%27s+law

leslie townes

ok. well you happen to have just announced leslie's law of baez's law of misattribution

Xander Henderson

:D

leslie townes

i block ncatlab locally. my computer doesn't even know it exists, except as a server that it should never connect to. i don't need its filth in my life

Xander Henderson

@leslietownes Ha! That is an entirely reasonable approach to life.

Semiclassical

20:11

i find ncatlab occasionally useful

but only occasionally

leslie townes

i'm sure that via the "right" way of looking at something, every mathematical object i've ever seen is a pushout of a 5-fibration of a cohomology of the homotopy of the cohomologies of an infinity-functor. i prefer to live in the matrix

2

Semiclassical

for instance, the ncatlab page on hyperfunctions is surprisingly reasonable: ncatlab.org/nlab/show/hyperfunction

Krijn

Whenever I search for simple topics and ncatlab shows up I panic and close my browser

Semiclassical

it does situate hyperfunctions in the context of sheafs, but that's a natural enough thing to do as i understand it

the other example that comes to mind is states on a star-algebra

Krijn

this one is fun: ncatlab.org/nlab/show/quotient+object#examples

Semiclassical

20:16

see, that's the part of nlab i'm not talking about :P

leslie townes

they serve great steak in the matrix

Xander Henderson

@Semiclassical Sato's book is more useful. :P

@Semiclassical Yes, hyperfunctions come very naturally out of the theory of sheaves. Honestly, I am not sure that there is an alternative framing which makes any sense (but everything I know about sheaves was originally motivated by a desire to understand Sato's book, so what do I know?).

Semiclassical

Sato has a book? i knew of Imai's

Xander Henderson

@Semiclassical I have, somewhere, a scan of a book written by Sato---but it is on my personal computer at home, and not my work computer. It might be just a long article...(?).

Semiclassical

kk

i have no real desire to look into it at the moment

Xander Henderson

20:22

There is also a book by Graf which is very approachable.

Semiclassical

yeah

leslie townes

it's also very avoidable

Semiclassical

one place where i have seen hyperfunctions show up naturally without much context is the following

Xander Henderson

One place where I have seen hyperfunctions show up naturally without much context is the following.

2

Semiclassical

consider the function $1/(1-z^{-N})$ for integer $N$. if we start with some $z$ inside the unit circle and take $N\to\infty$, then the function converges to $0$

if we instead start with $z$ outside, it converges to $1$

Ted Shifrin

20:33

@Semiclassical I assume Archimedes’ approach works for that … remove a cone from a cylinder and it’s just Cavalieri’s principle.

Semiclassical

right

i'm bad at seeing that but i've worked it out before

Rithaniel

Hey folks

Ted Shifrin

It’s easy.

Hi Rithaniel.

Rithaniel

How goes it in here, today?

@XanderHenderson A useful context to keep in mind

Don Thousand

@XanderHenderson Why do you choose to be this way.

Rithaniel

20:45

So, I'm stuck on this problem. I have loose ideas on how to tackle it, but I can't seem to get traction with it

The problem is "Let $y=\{y_n\}$ be such that $\sum\limits_{n=1}^\infty x_ny_n$ converges for every $x=\{x_n\}\in c_0$, where $c_0\subset l^\infty$ denotes the space of all sequences that converge to zero. Show that $\sum\limits_{n=1}^\infty\vert y_n\vert<\infty$."

Don Thousand

This isn't really the right place to ask this, we have a Q/A site for a reason

Xander Henderson

@Rithaniel I am quite certain that this question has already been asked and answered on the main site.

Semiclassical

"Associated with Math.SE; for both general discussion & math questions alike. Just ask; don't ask to ask."

Rithaniel

I generally talk in here, but if the policy has changed to only ever ask questions on the main site, I can do that

Don Thousand

@Rithaniel It isn't like you can't ever ask questions here, but for questions that are askable on the main site, they shouldn't be asked here.

leslie townes

20:47

no policy change.

rithaniel consider math.stackexchange.com/questions/678911/… and the ideas in the arguments.

2

Xander Henderson

@leslietownes You got there first. :P

Rithaniel

Yeah, I was searching around for a good few hours, couldn't find anything good

This is why I ask

Don Thousand

Yes, this is what the main site is for, Rithaniel.

leslie townes

rith this is a special case of a more general problem of 'identifying' (in some sense) the dual of a banach space. c_0 is a standard name for your space and the dual is often written ell^1 (for reasons you discover). the q i linked to may not make exactly the same hypotheses as your problem but it is the same idea.

Don Thousand

@XanderHenderson You know the rules of hyperfunctions and so do I

Rithaniel

20:50

@DonThousand Good to know, Don. You were quite helpful

leslie townes

common exercise type in this area is to show that some type of recipe can be used to define linear functionals on a banach space, and then to try to show that you get all of the functionals via that recipe (or learn about what functionals might not have that form).

if you have a measure theory book you might see this stuff in its sections on duality and L^p spaces. depending on the book.

Rithaniel

@leslietownes Anything helps, really. I don't need an answer, I just need ideas. To talk to someone about it, maybe

I will read what you linked and try to see what comes together

Xander Henderson

@DonThousand :D

Time to teach!

:'(

Rithaniel

I did notice that $\{y_n\}$ defines a functional that is probably in the image of the evaluation map (from $\ell^1$), but didn't have a good argument for why it had to be one

@XanderHenderson Try to enjoy yourself

leslie townes

rith, looking over that link (which i did not read too carefully because i needed to beat xander) i see it's making a boundedness assumption where you do not have one, so you will need to add some ideas in. think about a proof by contradiction. it may help to assume the y_k's are all nonnegative.

Rithaniel

21:04

Yeah, boundedness looks like it'll be a headache

Though, if I could say that an arbitrary functional on $c_0$ must be in $\ell^1$, I think that should work

After all, the dual space contains all functionals, not just bounded ones, right?

Maybe I could show that the functional is continuous

Ted Shifrin

@DonThousand Why are you making rules for this chat? That is not your place.

Rithaniel

(Honestly, I feel bad that I usually don't come around to hang out unless I'm stuck on a math problem. I should try to swing by more often)

leslie townes

rith, usually when i say 'dual' with banach spaces i mean continuous duals. you can't 'write down' unbounded linear functionals on infinite dimensional banach spaces but they do exist (using axiom of choice anyway). here the domain has an order structure and the functionals you're discussing can be compared to a positive linear functional (the one you get by replacing y_k with |y_k|), and positive linear functionals on c_0 do have to be bounded.

off the top of my head i'm not sure if that's generally true in C_0(X) spaces but it's certainly true in C_0(N with counting measure). unless a part of my brain fell out

Rithaniel

21:19

Actually, I was thinking about arguing that summations are sequentially continuous, and since a functional is bounded iff it is sequentially continuous, then this functional must be bounded.

@leslietownes Brain parts going missing is a menace

Happens to me all the time

leslie townes

well, i don't recall it ever happening to me, whatever it was. what were we talking about? where am i?

Rithaniel

Yeah, I think I have the answer. Thank you, leslie. I've been looking all day for some stackexchange question that would have relevant information. You're search-fu is clearly much stronger than mine. You got a hit almost instantly

Now I know that $c_0'=\ell^1$, which is something I didn't know before

leslie townes

i cheated by going back in time and studying a lot of this stuff for a long time

Rithaniel

@leslietownes WHAT? Can I do that?

leslie townes

no. my time machine broke

Rithaniel

21:31

Darn

Ted Shifrin

You can go forward in time and be a bum like me ? :)

Don Thousand

@TedShifrin What rules are you referring to?

Rithaniel

If it means I can cook as well as you, Ted, absolutely

Ted Shifrin

22:23

Lessons are available :)

Rithaniel

22:34

I have a 4 day weekend next week. I might hit you up for a recipe to try out.

leslie townes

few drops of rain just now.

Ted Shifrin

Not here yet, @leslie, although one threatens to drizzle.

robjohn

@TedShifrin That if two diagonals are perpendicular, the area of the quadrilateral is half the product of the diagonals? In general, half the product of the diagonals times the sine of the angle between them is the area.

Ted Shifrin

Ah, that makes more sense. Nah, I've never known that. I just knew the formula from turning it into a rectangle with the mean base.

But this is surely why we want OPs to give us efforts and full information.

I was sort of pleased with my vector algebra proof, too :)

robjohn

@TedShifrin Indeed

Ted Shifrin

22:50

This was a cool differential geometry query.

@DonThousand The ones you were throwing in Rithaniel's direction. We don't need that. Especially from someone who's far from a chat regular or room-owner.

(I didn't post that to get votes, but thanks :P)

Transcript for

$Mathematics$ Mathematics