Mathematics - 2023-11-27 (page 1 of 2)

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Jakobian

00:21

@leslietownes Apparently there is a way to equip Henstock-Kurzweil integrable functions with a bornological space structure

But this vector space is bad in a lot of ways (unknown to me yet)

Ted Shifrin

Is that for born-again spaces?

Thorgott

00:37

I misread that as homological and got (very briefly) excited

Ted Shifrin

00:58

Reading comprehension is overrated.

Jakobian

01:14

I think there is some advantage in writing proofs with errors

You don't get stuck overthinking the details

Ted Shifrin

01:27

Not great if it’s your PhD thesis.

2 hours later…

Sha Vuklia

02:58

Is there a clever way to find an explicit $\delta>0$ such that for all $(x,y)\in B(0,\delta)$ with $y\neq 0$ we have $yx^3-y^2<0$? (Such $\delta>0$ exists by degree considerations)

Case distinction: If $y>0$, then since $yx^3-y^2=y(x^3-y)$ it can only go wrong if $x>0$. So for $x,y>0$, I'm trying to find this $\delta$ such that $x^3<y$.

Hm, now I'm confused, because if we are on a circle and $y$ goes arbitrarily close to zero, then $x^3<y$ fails...

oh... I shouldn't have been considering this at all

meaning, I shouldn't have been asking this question

so never mind, it's irrelevant now (to my original problem)

I want to find $\delta>0$ such that for $(x,y)\in B(0,\delta)$: $y(x^3-x-y)<0$ whenever $y\neq 0$

I'm going to try level sets

nvmnvm for all this (again a mistake in original problem)

copper.hat

03:23

there are always mistakes

@ShaVuklia There is no such $\delta>0$.

Sha Vuklia

@copper.hat yes, that's how I realized my mistake

I've corrected it now, and the resulting thing to prove was now easy

2 hours later…

Dannyu NDos

05:29

Is it acceptable to write the square root symbol in modular arithmetic? For example, is it safe to write $\sqrt{-1} = 2 \; (\text{mod} \: 5)$?

leslie townes

i don't know what you mean by 'acceptable.' there are a couple of technical issues that you confront if you try to define something like that, which require making some choices, and AFAIK there is no reasonably widespread 'standard' way of making those choices. in view of that, it probably wouldn't be a good idea to assume any such notation would be self-explanatory, or to use it without saying what it means.

the main technical issues that come to mind are that given n and b, first and foremost, there may be nothing worthy of being called "sqrt(b)" mod n, and secondary to that, when there are such things, there will generally be more than one such thing, requiring you to make a choice as to which one is "the" sqrt( ).

Dannyu NDos

Oh right, there are two square roots...

leslie townes

as one example, with square roots of nonnegative real numbers b it is common to say "okay, take the nonnegative solution to x^2 = b and call it sqrt(b)" and there is nothing like positivity mod n.

there may be more than two. if n is prime, there will be at most two, and kinda like with R, they will differ up to "sign," only unlike R, there will be no "positive" one.

as you see already with both 2 and -2 = 3 being square roots of 4 mod 5.

you could always do something like, okay, sqrt(b) mod n, when it exists, could be defined to be the least x with 0 <= x < n and x^2 = b. that would single out 2 as "the" square root of 4 mod 5, for example, and would even make sense if there were more than two potential choices of square root.

Dannyu NDos

What about writing $\mathbb{F}_5(\sqrt{2})$? There's no square root of 2 mod 5, and the field extension should identify as $\mathbb{F}_5[x] / \langle x^2 - 2 \rangle$.

leslie townes

but for many purposes that would an arbitrary choice.

you can construct fields of characteristic 5 that contain square roots of 2, and if you like you could call one of them sqrt(2). here you run into the same thing people run into generally in abstract algebra, however, in that there is nothing like positivity in real numbers that allows you to disitnguish between roots other than you just picking one and naming it that.

Ted Shifrin

05:46

Sorta like the difficulty with distinguishing $\pm i$.

leslie townes

there used to be a highly-voted Q/A on MSE or MO about abstractly distinguishing between i and -i in complex numbers. i can't find it now, but it is about that.

ted reads my mind once again.

Ted Shifrin

I channel munchkin.

leslie townes

math.stackexchange.com/questions/2368034/is-i-equal-to-sqrt-1 and the posts that people link to in comments under the OP.

munchkin 2 had a very fussy day today. a whole lot of yelling.

Ted Shifrin

And related to the issue of ambiguous roots is this post.

leslie townes

we have enough golf balls to illustrate 7 x 5 now.

Dannyu NDos

05:52

Here's a more practical question: How can I easily identify an isomorphism between $\mathbb{F}_5(\sqrt{2})$ and $\mathbb{F}_5(\sqrt{3})$? After all, they're both Galois fields with 25 elements, right?

Ted Shifrin

I would still be happy with $2\times 3$ and $3\times 2$ :D

We don't even write that sort of thing, @Dannyu. But since $-1$ is a square in $\Bbb F_5$, those are the identically same field.

Dannyu NDos

Weird.

Ted Shifrin

To write $F(\alpha)$, you really need a field extension of $F$ in which $\alpha$ lives in which to do your arithmetic. It's best to write $F[x]/(x^2-2)$.

If you think that's weird, you should go back to basics.

leslie townes

imvho it's best to think of the square roots of 2 and 3 that you get in fields like that as abstract things. not as some result of taking "sqrt(2)" or "sqrt(3)" as things that already exist and just adding them into F_5.

Ted Shifrin

You do get that $3=-2$, and so $3=2a^2$ for some particular $a\in\Bbb F_5$.

To make sense of $F(\alpha)$, as I said, arithmetic needs to take place in some field extension $K$, such as the algebraic closure of $F$ ... unless you define this directly as a quotient ring.

And saying the field extension is $F(\alpha)$ is beyond begging the question.

Dannyu NDos

05:57

So... It's foolish to write $\mathbb{R}(i)$ in place of $\mathbb{C}$?

leslie townes

one of my algebra instructors once said of a diagram, "remember, these two fields are both sitting in some common extension of [the ground field]. they aren't just out there flapping in the breeze."

Ted Shifrin

It's understood that $\Bbb R(i)$ is a subring/subfield of $\Bbb C$, since $\Bbb C$ is the algebraic closure of $\Bbb R$.

leslie townes

since then, not a year goes by that i don't at some point think about fields flapping in the breeze.

Ted Shifrin

Do fields flap? I suppose that depends on whether it's in Florida or not.

Dannyu NDos

And it's too sadge that I cannot reserve "$\mathbb{C}$" to refer to computable real numbers.

Ted Shifrin

06:01

Huh?

Dannyu NDos

I mean... there is no standard notation for them.

Ted Shifrin

I don’t even know what computable means.

Dannyu NDos

Computable number

In mathematics, computable numbers are the real numbers that can be computed to within any desired precision by a finite, terminating algorithm. They are also known as the recursive numbers, effective numbers or the computable reals or recursive reals. The concept of a computable real number was introduced by Emile Borel in 1912, using the intuitive notion of computability available at the time.Equivalent definitions can be given using μ-recursive functions, Turing machines, or λ-calculus as the formal representation of algorithms. The computable numbers form a real closed field and can be used...

A real number is computable if there exists an algorithm that approximates the number with an arbitrary precision.

Soumik Mukherjee

So what is not a computable number?

Dannyu NDos

Examples of non-computable real number: Limit of Specker sequence, and Chaitin's constant.

Ted Shifrin

06:04

This is a topic in recursion theory, not algebra.

Dannyu NDos

Who knows? Personally I am interested whether the extension from the rationals to the computables is of finite degree.

Ted Shifrin

Is $\root n\of 2$ computable?

Dannyu NDos

Yes, for every positive integer $n$.

Ted Shifrin

Then you just answered that question. Are any transcendental numbers computable?

Dannyu NDos

Huh, so the degree is countably infinite.

Ted Shifrin

06:13

At least.

Dannyu NDos

There are countably infinitely many computable real numbers because there are countably infinitely many Turing machines.

Ted Shifrin

You didn’t answer my last question. There could be non-akgebraic computable numbers … or not?

Dannyu NDos

$\pi$ is computable. $e$ is computable. $\log 2$ is computable.

Yet very unfortunately, $\mathbb{Q}(\pi, e, \log 2)$ doesn't have decidable equality.

Ted Shifrin

So this field extension has some transcendental degree. Who knows what.

Probably countably infinite.

Dannyu NDos

If the degree of extension is uncountable, the extension field is uncountable, no?

Ted Shifrin

06:21

Yes, but you need to understand the difference between algebraic and transcendental extensions.

Dannyu NDos

Aw shoot, that's where I mixed up.

1 hour later…

Lucky Chouhan

07:34

@SoumikMukherjee Ugh...Being a socially awkward is the worst thing you can do to yourself :/

@SoumikMukherjee I wish... I am already a fan of Sherlock Holmes and Hercule Poirot so it would be fun to be a *spy*😎🧐🧐

user image

@copper.hat see this :))

2 hours later…

one potato two potato

09:29

yesterday, by one potato two potato

I just learned I've never questioned before when I took real analysis course: Let $\mu$ be a finite measure and $\nu$ be a probability measure on $(\Omega,\mathcal{F})$. Let $\mathcal{F}_n\uparrow\mathcal{F}$ be a $\sigma$-algebra filtration of $\mathcal{F}$. Let $\mu_n = \mu|_{\mathcal{F}_n}$ and $\nu_n = \nu|_{\mathcal{F}_n}$. If $\mu_n\ll \nu_n$ for all $n$ and $X = \limsup d\mu_n/d\nu_n$ then $\mu(A) = \int_AX d\nu+\mu(A\cap\{X = \infty\})$.

@Jakobian did you know this before?

I guess people in analysis like Leslie or robjohn already knew.

Jakobian

09:44

Not this specific result

Seeker

10:10

Could someone check if my proof is correct? I did add an answer but it seems have disappeared from the front page. I would really appreciate if someone could give it a look. It’s really short. The link is math.stackexchange.com/a/4815222/1050393

psie

10:56

I'm befuddled about this answer. If $A$ is not singular, we have $$\pmatrix{I&0\\ -CA^{-1}&I}\pmatrix{A&0\\ C&D}=\pmatrix{A&0\\ 0&D}$$ I see how the determinant of the first matrix is just $1$, but why is the determinant of the RHS $\det (A)\det(D)$. I feel like this is just what the original question asked about.

Jakobian

11:50

To establish the ultimate convergence theorems, I shall replace sequences by nets

one potato two potato

I've seen the notion of nets during the functional analysis class and nowhere else.

Jakobian

In your real analysis class

Limits like $\lim_{x\to 0} \frac{\sin(x)}{x}$ are an example of convergence of nets

Mad

12:48

i am a newb to concepts of differential geometry,
given a lie group homomorphism $f: G \rightarrow H$ the push forward $df$ is said to be a lie algebra homomorphism. $g \mapsto h$
i know that Lie Algebras are isomorphic to the tangent space at the identity $T_e G , T_eH$ so shouldnt we speak of $df_e$ as map between those

Mad

13:14

1

Q: Abelian Lie group implies abelian Lie algebra

Here is the exercise from Lee's Introduction to smooth manifold 8-25 Prove that if $G$ is an abelian Lie group, then $Lie(G)$ is abelian. [Hint: show that the inversion map $i:G\rightarrow G$ is a group homomorphism, and use $di_e: T_eG\rightarrow T_eG$ is given by $di_e(X)=-X$.] where $Lie(G)$...

differential-geometry lie-groups lie-algebras smooth-manifolds

this confuses me here, obviously, the differential of the inverse building at unity delievers the - X for a tangent vector X
in the reply, he refers to the push forward in general (not evaluated at Unity)
Yet he uses the identity of -X (which applies if evaluated at unity)

Same issues basicaly, i am missing something?

Thomas Finley

13:27

A continuous function $f:[0,1] \to \mathbb{R}$ attains each of its values finitely many times and $f(0) \neq f(1)$. Show that $f$ attains at least one of its values an odd number of times

4

A: Continous function to be solved by using intermediate value property

Hint. Assume wlog that $f(0)<f(1)$. Start by showing that in the interval $f([0,1])$ there are at most a countable set $E$ of local maximum/minimum values of $f$. Since $(f(0),f(1))$ is uncountable there exists $y\in (f(0),f(1))\setminus E$. By using intermediate value property, show that the car...

I found this answer but this answer appears less detailed to me.

For example, I don't really know how to show, "in the interval $f([0,1])$ there are at most a countable set $E$ of local maximum/minimum values of $f$. "

30

Q: Does there exist a continuous function from [0,1] to R that has uncountably many local maxima?

Does there exist a continuous function from $[0,1]$ to $R$ that has uncountably many strict local maxima?

real-analysis functions

I found this post but it seems to talk about, "strict local maxima" and not local maxima in general.

I seem to be stuck in here for a long time.

Alessandro Codenotti

14:17

@ThomasFinley What is your definition of nonstrict local maxima?

Thorgott

15:08

@Mad the Lie algebra is the tangent space at the identity

alternatively, the Lie algebra is the space of left-invariant vector fields

these are canonically identified cause a left-invariant vector field is uniquely determined by its value at the identity

the Lie bracket is defined in terms of left-invariant vector fields, but the pushforward is defined in terms of tangent vectors at the identity, so you have to keep those identifications in mind

nothing more sinister is going on, however, you just have to look at the identification to see that the left-invariant vector field whose value at the identity is $-X$ (where $X$ is a tangent vector at the identity) is the additive inverse of the left-invariant vector field whose value at the identity is $X$

(more clearly, the canonical identification is a group isomorphism)

copper.hat

@LuckyChouhan why did you tag me for that?

Secret

**Lemma**
Proving that by the fundamental theorem of algebra, any product decomposition of integers is finite
- We have n, Suppose there is a m > n, n can said to be finite relative to m. Then
- By fundamental theorem of algebra, n can decompose into many factors
- We know that (-1)(-1) = 1
- The product of positive factors is a multivariable monotonically increasing function f, so f(x1,x2,x3,x4, ...) = n for given n exists
- 0n = 0, so to have nonzero n, all terms has to be positive or negative

- It follows that the first infinity is an even prime number bigger than any number, and there are obviously infinitely many of them

Mad

16:14

@Thorgott i am a bit confused by your answer.

I understand the isomorphie, but, how would this tranlate into Symbols?

Jakobian

Uh... no

Ted Shifrin

16:34

@copper.hat That seems self-evident.

Mad

@TedShifrin assist me master!

user image

How is he making that step?

Lucky Chouhan

16:57

@copper.hat as Ted said "That seems self-evident." Actually, in that photo I sent real copper hat haha

@Koro koro koro Hello

Thorgott

@Mad Idk what I'm supposed to translate here. Language is much clearer than symbols in this case.

@Mad The exponential map is a homomorphism

Lucky Chouhan

@Thorgott what does Thorgott mean? Can I pronounce it as "Thor-got"?

Koro

@LuckyChouhan hello

Lucky Chouhan

@Koro Is Koro your real name?

Koro

on mse, yes.

Jakobian

17:02

@Koro Nice. Mine is Jack

Lucky Chouhan

Oh @Koro Are you a teacher cuz I just saw you're on MSE from 8yrs?

@Jakobian I gave you this name. You once said your name is "Jacob".

Xander Henderson

@LuckyChouhan You ask a lot of personal questions about people here. This is not really appropriate, and I would ask that you please back off a bit.

3

Jakobian

Sorry, I forgot

my name is Jacob

Lucky Chouhan

Anyway it sounds cool.

Koro

Jakobian: There was a question in my exam that apparently no answered.

Here it is:

Lucky Chouhan

17:04

@XanderHenderson oh sorry. I just ask out of curiosity.

Jakobian

I'm not surprised that someone didn't answer a question on an exam

Koro

yesterday, by Koro

and one was to show that f=0 if given that f is even function supported on [-1,1] and that $|\hat f(\xi)|\le e^{-\xi^2}$ for all $\xi\in \mathbb R$.

Lucky Chouhan

@Jakobian Exams suck.

Jakobian

@Koro my name on mse is Jacob

Koro

there is a lemma called 'sampling lemma' but it is not applicable here as f is not L^2+continuous.

Lucky Chouhan

17:05

@Jakobian No you're still not right. Actually you're name on MSE is Jakobian.

Jakobian

No

Lucky Chouhan

I don't know why people don't put their real name. It shows how vulnerable we are

Jakobian

please respect my MSE name (applies to Lucky Chouhan only)

Aurel-BG

Hey chat

Lucky Chouhan

@Jakobian yeah sure, but I will call you "Jack".

@Aurel-BG Hey dear!

Aurel-BG

17:07

hello

Lucky Chouhan

@Aurel-BG how is life?

Aurel-BG

fine and you ? I'm looking for some probable prime numbers

Jakobian

@Koro what other assumptions do we have

Lucky Chouhan

@Aurel-BG pretty cool. For Probable prime see this t5k.org/glossary/page.php?sort=PRP

Koro

@Jakobian that f is L^1.

Lucky Chouhan

17:10

Hey Jack, does chatjax work in your PC?

Koro

but that's obvious from the given info.

Jakobian

Thats not obvious

Koro

it is supported on [-1,1]

ohh right.

1/x on (0,1) and 0 otherwise is not L^1

@Jakobian given: f is L^1

Jakobian

The assumptions given make me think that they want you to consider periodic function on $[-1, 1]$

Koro

I was thinking about that.

and then transfer the business to circles (1-toruses).

that should work. That's how sampling lemma is also proven.

I'll try with that.

Jakobian

17:14

So we have that the Fourier series for $f$ converges in $L^2$ to $f$

Koro

yes

Xander Henderson

Okay... the first part of my Precalc I final exam is written. It is the laziest job I have ever done in writing such an exam---I looked over all of the homework problems I assigned this semester, and simply copy-pasted one problem for each learning objective into a TeX document. I didn't even change any of the numbers for a significant number of them. There is literally nothing which hasn't already been seen on that part of the exam.

I have even told the students that this is exactly what I have done.

shintuku

bastards better pass

Xander Henderson

And I'm willing to bet that the average score will still be less than 75%. :(

Sine of the Time

@XanderHenderson will they trust you?

Xander Henderson

17:23

@SineoftheTime No idea.

Jakobian

@koro math.stackexchange.com/a/10854/476484

this suggests that your function is actually smooth

up to measure zero maybe

Koro

I don't see how you have a rapid decay here but nonetheless I'll think about it some other time.

Lucky Chouhan

@XanderHenderson let's do a bet.

@XanderHenderson don't you get bored from teaching the same stuff for years?

@SineoftheTime long time no see.

Sine of the Time

sup

Lucky Chouhan

@SineoftheTime I haven't got it. :((

Are you safe @SineoftheTime ? Cuz you're from Palestine...

Sine of the Time

17:38

yes I'm safe

Is there a difference between inner and dot produt in general?

Xander Henderson

@SineoftheTime When I hear "dot product", I mentally translate this to "the standard inner product on $\mathbb{R}^n$". So, in my mind, yes, there is a difference. An inner product is a generalization of the dot product.

Jakobian

@koro What if you were to write $f = f^+ - f^-$ thus assuming $f\geq 0$, and write some lower bound on the Fejer kernel?

shintuku

at SineoftheTime: i've been told the inner product gives meaning to the idea of direction in a general space, and the dot product is a specification of this in euclidean space, being an operation which gives meaning to direction in euclidean space

Sine of the Time

@XanderHenderson thank you. I think I need more advanced math to grasp the difference

@shintuku I've been told different things by different professors, but they use both terms as synonyms

Jakobian

@SineoftheTime Do you know what an inner product space is?

Sine of the Time

17:42

no :(

shintuku

the way I understand the inner product is a measure of "sameness of direction", which can have different meanings in different sorts of spaces

Xander Henderson

@SineoftheTime In what context?

Sine of the Time

Hilbert spaces

Jakobian

Its a vector space equipped with inner product, which is supposed to mimic the properties of the dot product

@SineoftheTime if you hear about Hilbert spaces then you heard about inner product spaces

Xander Henderson

@SineoftheTime In what context? What class are you taking? what book are you reading?

Sine of the Time

17:44

@Jakobian we did the introduction defining the scalar product and the professor said that inner and scalar product are the same

shintuku

oh yeah, i recall now. the way I understood inner product is a way to give meaning to the notion of orthogonality (it's in that sense that it can measure "sameness of direction")

Xander Henderson

Do you know what $\ell^p(X)$ and/or $L^p(X)$ denote (where $X$ is typically $\mathbb{R}$, $\mathbb{C}$, or something similar)?

Sine of the Time

yes

Jakobian

@SineoftheTime I don't see how that relates

Xander Henderson

@SineoftheTime Well, that's where you want a more general notion of an inner product.

There is no "dot product" on $L^2$. But there is an inner product.

Sine of the Time

17:46

make sense

Xander Henderson

I would understand "scalar product" to be a direct synonym of "dot product".

Jakobian

more generally we talk about the concept of dual pair

shintuku

for instance, if you're trying to do optimization, and want two optimize for two parameters that mean opposite things, it might make sense to define an inner product which makes them orthogonal

Jakobian

There is no inner product on $L^p$ for $p\neq 2$, but you can talk about the function $L^p\times L^q\to\mathbb{R}$ defined as $(f, g)\mapsto \int fg$, where $1 < p < \infty$, $1/p+1/q = 1$, and this acts in similar way.

Sine of the Time

we have seen that $L^p$ is Hilbert iff $p=2$

with the usual norm I mean

Jakobian

17:49

Hilbert spaces are a special case of inner product spaces. So if you talked about Hilbert spaces you had to talk about inner product spaces

There is no real way around this

Sine of the Time

no, we haven't

Xander Henderson

@SineoftheTime Huh?

Sine of the Time

we only defined what a dot product is

Xander Henderson

This is an abuse of language.

There is no "dot product" on a general Hilbert space. There is a generalization of the dot product, called an "inner product".

It is very possible that the person teaching you is being sloppy with the language, or does not make a distinction between an "inner product" and the "dot product".

Sine of the Time

ok, maybe there is a language barrier but now I understand

Xander Henderson

17:51

Is this class being taught in a mathematics department? or physics?

Sine of the Time

maths

Xander Henderson

Weird. A mathematician should know better (in my opinion).

Jakobian

Hilbert spaces are inner product spaces which are complete in the sense that with the norm induced from inner product they are complete

There is no real way to define them otherwise

Ted Shifrin

@XanderHenderson I suspect this is a translation issue. Not worth getting up in arms over.

Sine of the Time

yes, so our def of "dot product" is the definition of inner product

Koro

17:53

@Jakobian I'll think about it.

Xander Henderson

@TedShifrin I'm not "up in arms" about it.

But it feels sloppy.

Sine of the Time

and this is why the professor use the terms as synonyms

Xander Henderson

@SineoftheTime Well, there ya go.

"dot product" = "scalar product" = "inner product".

Ted Shifrin

Well, maybe Xander is right, after all. Even I, as sloppy as I can be, use dot product for the Euclidean case and not in Hilbert spaces.

Koro

when I make coffee, either suger is too much or too low. It's never balanced.

Xander Henderson

17:54

@Koro What do you mean "too low"?

Sine of the Time

from a philosophycal point of view, if two things have different names they must differ in something

Xander Henderson

Are you somehow adding negative sugar?

@SineoftheTime That's not always how language works. :P

Koro

adding much less than the right amount of it.

Xander Henderson

For example, I don't distinguish between a "dot product" and a "scalar product". I understand both to mean "the usual inner product on a finite dimensional vector space over $\mathbb{R}$ (or maybe $\mathbb{C}$)."

In any event, time for class.

@Koro But isn't the right amount "zero"? :P

Koro

that's indeed the right amount. It'll take me some time getting used to that one. :)

Soumik Mukherjee

17:56

@XanderHenderson My friend( who is a physics student) once told me that they use these words interchangeably.

Xander Henderson

6 mins ago, by Xander Henderson

Is this class being taught in a mathematics department? or physics?

Koro

in political science department

Jakobian

@Koro If you add sugar to coffee, its too much sugar

Xander has it right

Koro

indeed

user 85795

Context is king.

Jakobian

18:09

@SoumikMukherjee I sometimes do that too

Lucky Chouhan

@Koro WHAT WHAT?

@Jakobian are you free whole day?

user 85795

@robjohn are you toying with my feelings 🤔

copper.hat

@TedShifrin @LuckyChouhan I just responded on my phone, I didn't look back. It is supposedly Lincoln's hat. Copper, or course, turns green when exposed to air.

Jakobian

@LuckyChouhan are you annoying whole day

4

Lucky Chouhan

@user85795 yeah actually when I saw that message of RobJohn then I thought you changed your profile photo after that :))

@copper.hat How is life sir?

user 85795

18:22

:)

robjohn

18:35

@Koro if $f$ is not given as continuous, then all you can say is that it is $0$ almost everywhere. As for $L^2$, both $f$ and $\hat f$ are in $L^2$.

@user85795 you changed your avatar to the image I posted in a reply to Ted, so I asked.

But maybe you were making a joke, and I missed it at first.

user 85795

I was making a joke Dr @robjohn

robjohn

Yeah, I noticed after i replied. Sorry.

user 85795

np pal

:)

Koro

@robjohn indeed f is in L^2

:-)

and so I can apply 'sampling lemma'.

it seems doable now. Thank you so much.

apparently the whole class skipped this in the exam.

LSpice

18:56

@BillDubuque, re, if it is only the @ in my comment that bothers you, then I would be happy to delete and re-post without the @.

I am reluctant to leave off the link entirely, because I think these sites benefit from a rich network of links even when it is quite easy to find their referents oneself; but, if you really feel that a comment below your post ruins its pedagogical effect, then I guess I had rather delete it than make you feel your post was ruined.

Koro

user image

@Thorgott hello! E_f here is the pullback of a fibration with fibre loop space of Y. I understand why $E_f\to Y$ is a fibration. But I think that $E_f=\{(x,\gamma):...\}$ is not correct here. I think it should be $\{((x,y), \gamma): (f(x),y)= \gamma(0)\}$. Right?

or I think this space that I wrote is homeo. to the space E_f in the image.

Ted Shifrin

19:33

@LSpice What are you talking about? There's no need to link to an answer, just above, in the very same question. And, yes, I think you missed Bill's point.

Sine of the Time

Any idea to prove that $V=\{f\in C^1[0,1] | \, f(0)=0\}$ with the inner product $\langle f,g\rangle= \int_0^1f'(t)\overline{g'(t)}dt$ is not Hilbert?

Ted Shifrin

@Sine The hint is that you're missing the usual $L^2$ part of the norm. The immediate counterexample ($f'=0$) is ruled out by the additional condition that $f(0)=0$, so a constant function must be $0$.

LSpice

@TedShifrin The point as I see it is that "just above", or even "on the very same page", is subject to change on MSE, and so it is better not to rely on such time-dependent cues. As to missing the mathematical point, I commented additionally, but was asked to delete the comment, that my original link had no editorial content, and was meant only as a pointer.

Ted Shifrin

I totally disagree with your judgment.

Sine of the Time

@ted wait, I have an idea

I found a counterexample using the Parallelogram law

Ted Shifrin

19:46

How is that? You're starting with an inner product.

Jakobian

LOL

Sine of the Time

oh sorry, I did not mention that we consider the norm $\|f\|=\left(\int_0^1 |f'(t)|^2dt\right)^{1/2}$

Jakobian

What Ted Shifrin said still holds

Koro

aiamq?

Jakobian

aiamq?

Koro

19:50

any idea about my question

Sine of the Time

@Jakobian why?

Koro

here

Jakobian

@SineoftheTime If you're starting from an inner product then you can't establish its not a Hilbert space by using that it doesn't satisfy parallelogram law

because the parallelogram law will hold

you need to somehow establish its not complete instead

Sine of the Time

so I can't use the parallelogram law to determine whether a space is Hilbert or not?

Ted Shifrin

@Sine You're being silly. You have an inner product defined already.

Read carefully. The parallelogram law is required for a Banach space to be Hilbert.

Jakobian

19:54

@SineoftheTime how would you do that?

I can't really answer that if I don't know how you're trying to do it

then I can objectively refute it

Ted Shifrin

(Or indeed for a finite-dimensional normed linear space to be an inner product space.)

Sine of the Time

I get it now

So I have to find a Cauchy sequence that does not converge in norm

to prove it's not Hilbert

Lucky Chouhan

Rearrangement inequality is very intriguing. But I am not getting how to prove it.

Ted Shifrin

What do you mean by Cauchy sequence?

Don't do that.

Don't insult me like that.

Sine of the Time

ok

Ted Shifrin

19:57

So you're using the given inner product to define convergence. What can go wrong?

Sine of the Time

It can converge to something that is not $C^1$

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