Mathematics - 2025-01-14

one potato two potato

05:12

peep

Vuk Stojiljkovic

08:51

Hello guys, can I ask a question ?

Soumik Mukherjee

yes, you can just ask, you don't need to ask to ask

Vuk Stojiljkovic

Can someone take a look at my question in Mathstackexhange and help me with understanding the concept I am asking, sorry if I am asking too much.

Soumik Mukherjee

which one? the most recent question that you posted?

Vuk Stojiljkovic

Yea, I am confused with some facts, I really hope one could explain to me some things.

Soumik Mukherjee

0

Q: Direct sum of Hilbert spaces, operator matrix

I have a question regarding the understanding of the term in question. I know what a direct sum is for subspaces, that is the intersection of $U$ and $V$ is zero and each element can be written in a unique way. Such example can be $\mathbb{R}=U\oplus V$ where $U=\{(x,0)|x\in \mathbb{R}\}$ and $V=...

There are people in this chatroom who are into functional analysis, when they come online, they may help you.

Vuk Stojiljkovic

08:58

That is it, thank you for posting and for giving time to my question. I understand that it is maybe basic, but I am confused as to how does it work

As I Mentioned in the question, it is trivial to consider a direct sum of spaces as I mentioned, but I am confused as to how that applies for an arbitrary operator acting on H^2=H\oplus H

Soumik Mukherjee

@VukStojiljkovic yeah sorry but I have to go now, I posted the question here so that others can notice easily.

Vuk Stojiljkovic

Right, see you! Have a nice day, thanks for posting!

Vuk Stojiljkovic

10:43

Dear mathemticians, can someone please try to answer my question above. Thank you very much!

leslie townes

11:04

vuk what is the problem exactly? the question on main kind of rambles on a bit, slightly more focus might help. one underlying point (which someone mentions in the comments) is that if you think of an operator on H+H as a 2x2 matrix, the entries of the matrix are not all of the same type. in fact they all have different types.

so it is not quite the abstract algebra situation where considers the set of nxn matrices over a [blah] where blah is something like a ring or a field.

of course in the background it is abstractly possible to identify any hilbert space with any other hilbert space of the same dimension, but conceptually it is not particularly helpful to do so.

in fact it mgiht be helpful to consider the case where that kind of identification is not possible. consider C + H where C is the complex numbers with usual inner product and H is a hilbert space of larger dimension than 1.

nickbros123

11:23

@Jakobian Of course. I also want to study the cool / crazy stuff of set theory sometime in the future. In my curriculum, theres no set theory

I can imagine myself, for eg, reading stuff about CH while sipping tea 10 years down the line

Soumik Mukherjee

@nickbros123 Start reading Jech's book then

@VukStojiljkovic you too have a nice day

leslie townes

modern set theory is one of those things where you can very much get by without knowing anything about it, but it is sometimes helpful to know it is out there. sometimes it is a source of counterexamples, or the truth of something actually depends on your set theory. there is a ton of interesting math where it does not come up. but it is sometimes helpful to know 'about' it even if you don't literally study it.

pie

12:01

Hey guys, what do you think of my three definition of simple closed function, basic elementary function and combinations of two functions in math.stackexchange.com/questions/5022713/…

I feel like my terminology is bad maybe these concepts has a name that I don't know, the other problem is that my definitions doesn't seem professional, can somebody help me rewrite them?

Pizza

12:28

@SineoftheTime yes

Jakobian

@pie it looks fine

Elementary function

In mathematics, an elementary function is a function of a single variable (typically real or complex) that is defined as taking sums, products, roots and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions, and their inverses (e.g., arcsin, log, or x1/n). All elementary functions are continuous on their domains. Elementary functions were introduced by Joseph Liouville in a series of papers from 1833 to 1841. An algebraic treatment of elementary functions was started by Joseph Fels Ritt in the 1930s. Many textbooks and dictionaries do not give...

"Simple closed form" is not something that anyone defines, really, good that you did

Your question lacks motivation though

pie

@Jakobian $(\Gamma^{-1}(a))^x$ is an elementary function (it is in form of $c^x$) where $\Gamma^{-1}(a)$ is the inverse gamma function of $a$,

that is why I needed to exclude any annoying $c$ that will break my definition

@Jakobian what does motivation mean here? I am just curious to see if there is any a with a simple closed form have a gamma of a simple closed form,

Jakobian

@pie motivation is why you are considering a problem

It's not really sufficient motivation to consider something "because you are curious"

At least not on this site

I also consider things because I am curious about them. But this curiosity didn't come from thin air

pie

12:44

@Jakobian well then, The reason I considered this question is that I had an ODE final exam and other students seem to try to evaluate $\Gamma(1/4)$ in an exercise which I know for sure that would be a waste of time, so I advised them not to try to find $\Gamma(a)$ for any $(0,1)$ except $1/2$ since it is impossible to find, but that made me wonder "really? no other value in $(0,1) has a simple closed form?"

Jakobian

12:58

@pie see, that's a good motivation

It'd be good to include it in your question

ILikeMathematics

Determine the set of accumulation points of $A$

Xander Henderson

@ILikeMathematics No, thank you. I would rather do something else.

ILikeMathematics

I was thinking about simplifying this set

one potato two potato

7

Q: The Gauß map of a minimal surface: is it holomorphic or antiholomorphic?

I'm reading A survey on classical minimal surface theory, by William H. Meeks and Joaquín Pérez. In the early beginning, they start giving eight definitions of minimal surfaces. The last of them is Definition 2.1.8 A surface $M\subset \Bbb R^3$ is minimal if and only if its stereographically ...

differential-geometry definition riemannian-geometry minimal-surfaces

ILikeMathematics

But I'm not sure how we could do that

one potato two potato

13:06

I've never questioned this.

Jakobian

13:34

@ILikeMathematics d_2?

The set of accumulation points in this case is the same as the closure

It consists of the union of closed balls and $\{(0, 0)\}$

Binky

13:58

Hi

Sine of the Time

hi

Binky

14:25

How do I find the dimension of an eigenspace

ILikeMathematics

14:42

@Jakobian Euclidean norm

@Jakobian How can we express that as set?

Jakobian

14:58

@ILikeMathematics what's your favourite notation for a closed ball

ILikeMathematics

@Jakobian We haven't covered 'closed balls' yet as a concept

Jakobian

@ILikeMathematics what

I don't believe you

you must have covered closed and open balls

ILikeMathematics

No

It's probably not necessary to know about them here, I know the definition of an accumulation point

Jakobian

I'm shocked

ILikeMathematics

We can probably still find them

Jakobian

15:03

do you not know how would you write your set $A$ in terms of open balls

ILikeMathematics

I don't know what an open ball is

Jakobian

oh. Well it's really not complicated, this comes when talking about metric spaces

and it's basically why we consider metrics

okay so let's say $B(x, r) = \{y\in X : d(x, y) < r\}$ and $D(x, r) = \{y\in X : d(x, y)\leq r\}$ where $x\in X$ and $r > 0$

you call $B(x, r)$ the open ball with center $x$ and radius $r$, and $D(x, r)$ the closed ball with center $x$ and radius $r$

here $d$ is some metric

now in the plane, if $d = d_2$ is the Euclidean metric, then open balls are literally open disks, and closed balls are closed disks

and you can see how it generalizes to $\mathbb{R}^n$ with Euclidean metric

ILikeMathematics

@Jakobian Ah, so in a plane, they're like two disks

@Jakobian Yeah

Jakobian

sure, one with the boundary and the other without it

ILikeMathematics

Ok maybe then we should think about the accumulation points of one such ball

It should just be the ball itself

With boundary

Jakobian

15:08

so now the reason why you call $B(x, r)$ the "open ball" is because it's an open set

wait, how do you define accumulation points if you don't know about open balls

normally, you would say that $x$ is an accumulation point of the set $A$, if $A\cap (B(x, r)\setminus \{x\})\neq \emptyset$ for all $r > 0$

or more generally, given any open set $U$ with $x\in U$, that $A\cap (U\setminus\{x\})\neq \emptyset$ (this definition works even outside of metric spaces)

although I can imagine you haven't covered open sets

ILikeMathematics

Let $(M, d)$ be a metric space and $A \subset M$. Then $x_* \in M$ is an accumulation point of $A$ if $\forall \varepsilon > 0 \exists y \in A \setminus \{x_{\ast}\} \quad d(y, x_{\ast}) < \varepsilon$

Jakobian

use \ast instead of *

ILikeMathematics

Thanks

Jakobian

@ILikeMathematics yeah this definition is the same. Basically, in any punctured neighbourhood of $x$ there exists an element of $A$

@ILikeMathematics so your set $A$ here, we could basically just write it as $\bigcup_{n\in\mathbb{N}} B((1/n, 1/n), 1/n)$

the centers are points $(1/n, 1/n)$ and the radius is $1/n$

ILikeMathematics

Yes

And now we can follow that the accumulation points are the balls themselves with borders

So it will be A, but with borders

Jakobian

15:14

@ILikeMathematics yeah, they are

ILikeMathematics

So we just represent it as closed balls

Right?

Jakobian

no that's not correct

ILikeMathematics

Replace < with <=

Why not?

Jakobian

if $A'$ denotes the set of accumulation points of $A$ (derived set as it's called), then $A' = \{(0, 0)\}\cup \bigcup_{n\in\mathbb{N}} D((1/n, 1/n), 1/n)$

@ILikeMathematics well what can I say? It's just not true

if you have a point outside of $(0, 0)$, then sure, if a neighbourhood of it intersects some ball, then only finitely many

ILikeMathematics

@Jakobian Ah, well almost. So what I said, with the point (0, 0)

Jakobian

15:18

right. Note that the limit points of $B(x, r)$, they're not always equal to $D(x, r)$

this is something special because we're in the plane

Xander Henderson

We're on a plane?

Are there snakes there?

Jakobian

@ILikeMathematics yeah, that's the point where "local finiteness" of the family of open balls breaks. That is, the neighbourhoods of it intersect infinitely many of the balls.

by the way, a neighbourhood of the point $x$ is a set which contains $B(x, r)$ for some $r > 0$

ILikeMathematics

Hm, why do we always have $(0, 0)$ in it?

Jakobian

in what?

ILikeMathematics

As accumulation point

in the set

Jakobian

15:21

I don't know what you mean

ILikeMathematics

@Jakobian Why $A' = \{(0, 0)\} \cup \bigcup_{n \in \mathbb N} D((1/n, 1/n), 1/n)$, why not $A' = \bigcup_{n \in \mathbb N} D((1/n, 1/n), 1/n)$?

Why do we have (0, 0) in A'

Jakobian

because every open ball around $(0, 0)$ contains an element of $A$

ILikeMathematics

Oh, right

Jakobian

for instance, if you take the ball $B((0, 0), r)$, then $(1/n, 1/n)$ will be in it for small enough $n$

ILikeMathematics

Thank you!

Jakobian

15:25

👍

Claudio

15:42

If I have an ode of the form $y'(t) = \frac{y(t)}{y(t)+2t}$ I can solve it, as my book says, by dividing both the num and den by $t$ and letting $z(t) = y(t)/t$ to obtain a separable-variable one

ILikeMathematics

@Jakobian Isn't $(0, 0)$ already part of $A' = \bigcup_{n \in \mathbb N} D((1/n, 1/n), 1/n)$? I guess we would need to show it's not

Claudio

now I did so, and after substituting I get that $\frac{|y+t|}{|t|}\frac{1}{\frac{y^2}{t^2}}= e^c|t|$

ILikeMathematics

Probably just $d((1/n, 1/n), (0, 0)) = \sqrt{\frac{2}{n^2}} = \frac{\sqrt 2}{n} < \frac 1n$

And that gives a contradiction

Claudio

now I can rename $e^c = C \in \mathbb{R}$ and obtain for $t \ne 0$: $$|y(t)+t| = Cy^2(t) $$

now my book never mentions imposing $t \ne 0$ anywhere, moreover it says that the costant solutions, namely $y_0 \equiv 0$ can be obtained by imposing $z^2(t)+z(t) =0 $

the latter also gives the solution $z = -1 \mapsto y(t) =-t$

and it says that this one can be obtained by setting $C = 0$ in the general solution $y(t)+t = Cy^2(t)$. My doubts are: why does the book leave out the absolute value? And how can $e^c$ be zero???

Jakobian

15:58

@ILikeMathematics clearly not: $d((1/n, 1/n), (0, 0)) = \sqrt{2}/n > 1/n$

leslie townes

claudio without getting too heavily into it i would think of all of those methods as ways of arriving at families of solutions (that, depending on your level of care, might not be exhaustive). for example "e^c" isn't an arbitrary real number, it is only a positive real number, but presumably if you pay more attention with absolute values and signs that won't matter.

in the background, you can often end up convincing yourself that you have found all solutions by finding a big enough set of solutions and appealing to a uniqueness theorem.

for finding things like constant solutions i would just work with the original equation, and not try to confuse yourself by trying to get some other method to find those too

it is very common (in the US at least) for books not to work with any level of rigor at all in finding solutions, and not caring about sign issues is very consistent with that.

you can sometimes phrase these tricks more rigorously than they appear in books, but the work you have to do to do that is somewhat at odds with the idea of arriving at these things via symbolic calculation

Claudio

@leslietownes I see

this exercise is a bit weird indeed, there's no theory in it, like most of the previous exercises were more difficult

like they require you to find the maximal interval and how the solution behaves at $t \to \pm \infty$ without explicitly solving the d.e.

which has turned out to be very difficult, at least for me

leslie townes

16:20

yes, stuff like that can be hard. it is easier to 'just solve' if you are somehow able to do that

Binky

17:41

\[
\begin{pmatrix}
-1 & 0 & 0 \\
2 & 0 & 3 \\
-1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 0 \\ 0
\end{pmatrix}.
\]

I don't understand why I did the exercise wrong

Since the rank is 1 I have two free variables which are those not present in the chosen minor, in this case I took the minor -1 at the top left

So I have the system x=0, y=t, z=s

ILikeMathematics

@Jakobian Oh, and one more thing: Do you have an idea on this? Let $f: \mathbb R \to \mathbb R$ with $f(x + y) = f(x) + f(y)$ and $f(rx) = rf(x)$ for all $r \in \mathbb Q$ and $x \in \mathbb R$. Assume there is an open, bounded interval with $|f(x)| \leq M$ for all $x$ in it. Show that $f$ is continuous and $f(x) = ax$ for some $a \in \mathbb R$.

Xander Henderson

@ILikeMathematics It is often helpful, in dealing with problems like that, to start with some simple cases and see what happens. For example, can you determine $f(0)$?

ILikeMathematics

@XanderHenderson $f(0) = f(0 + 0) = f(0) + f(0) \implies f(0) = 0$

Xander Henderson

Okay, great.

ILikeMathematics

So atleast if it is linear, there won't be a constant term and we are done

Xander Henderson

17:54

Now try another easy case: can you pin down $f(1)$? or, alternatively, if you cannot pin down $f(1)$, what happens if you assume that $f(1) = a$ for some $a\in \mathbb{R}$?

ILikeMathematics

Then $f(2) = f(1 + 1) = f(1) + f(1) = 2a$

@XanderHenderson So we can continue like that

$f(n) = an$

Xander Henderson

Sure, or you can go the other way: $a= f(1) = f(\frac{1}{2} + \frac{1}{2}) = f(1/2) + f(1/2)$, which implies that $f(1/2) = a\cdot \frac{1}{2}$.

ILikeMathematics

So generally $f(r) = ra$ for $r \in \mathbb Q$

Ok, actually this would've directly followed from our $f(rx) = rf(x)$ condition

Xander Henderson

Indeed, which makes things easier.

But I was just following your lead (and, in general, it isn't a terrible idea to push one condition as far as you can before looking at other conditions).

Thorgott

the second condition is actually redundant

and this is part of the argument as to why it is redundant

Xander Henderson

18:01

@Thorgott I was getting to that.

:D

ILikeMathematics

@Thorgott Yep, the previous exercise was to show it, which I did, so I included it like this

Xander Henderson

@ILikeMathematics Notice that once you have $f(1/2) = a/2$, you can then work out values for $f(1/4)$ and $f(3/4)$.

ILikeMathematics

Yes

Xander Henderson

And if you keep doing that, you end up determining the value of $f(x)$ for all dyadic rationals $x$.

Vladimir Lysikov

@Binky The rank is actually 2

ILikeMathematics

18:02

@XanderHenderson Yeah

Xander Henderson

And then you can apply the same reasoning to $f(1/n)$, and work out $f$ for all rational numbers with a denominator that is a power of $n$, and then you can mix-and-match to get all rationals.

But the extra condition saves some time.

(In my head, $n$ is prime. I probably should have written $p$. If you don't assume it is prime, then you can skip the mix-and-match step.)

ILikeMathematics

@XanderHenderson Yep, I already proved the second condition as a previous exercise from the first

Binky

@VladimirLysikov why

ILikeMathematics

Ok so we know $f(r) = ar$ for all rational $r$ and $a = f(1) \in \mathbb R$

We need to extend it to reals

Xander Henderson

@ILikeMathematics What you have written there is a little bit of nonsense---if $f(r) = ar$, then why do you need to specify that $a = f(1)$?

Vladimir Lysikov

18:06

@Binky The first two rows are clearly linearly independent; in terms of minors you can look at the minor formed by rows 1,2 and columns 1,3

Xander Henderson

I'd rather write "There is some $a\in\mathbb{R}$ such that $f(r) = ar$ for all $r\in\mathbb{Q}$."

ILikeMathematics

@XanderHenderson True, well let's just stay with $f(r) = ar$ and $a \in \mathbb R$, $r \in \mathbb Q$ then

Xander Henderson

(or something like that)

@ILikeMathematics That still doesn't make sense. This seems to imply that $f$ is a multivalued function of some sort---for any $r$, choose an $a$, and then $f(r) = ar$. But that choice of $a$ could vary! and it doesn't even seem to depend on $r$!

You need to fix $a$ first, before you can use it to define $f$. There is some constant $a\in\mathbb{R}$ such that $f(r) = ar$ for all rational $r$.

ILikeMathematics

@XanderHenderson Yep

Xander Henderson

If you really want to write $a$ later, the sentence "$f(r) = ar$ for all rational $r$, where $a$ is a fixed constant," could work, but this is awkward.

ILikeMathematics

18:10

Alright

Ok, so we need to extend r from Q to R now, somehow

Thorgott

and what might be a strategy of extending statements from Q to R?

ILikeMathematics

Well, this is surely just equivalent to $f$ being continuous

Binky

@VladimirLysikov you’ll lose the 3rd variable

ILikeMathematics

@XanderHenderson So really what we're after is continuity

Vladimir Lysikov

@Binky Yes; The solution is x_1 = 0, x_2 = t, x_3 = 0

ILikeMathematics

18:14

Any ideas on showing that?

Xander Henderson

@ILikeMathematics Lots. Why don't you start by assuming that your open interval contains zero. I think that if you can figure it out there, you might be able to figure out the more general result.

ILikeMathematics

So we have $|f(0)| \leq M$

And we figured out that $f(0) = 0$

Ok, that's not very helpful

Thorgott

another thing to note is that the additivity of $f$ implies that $f$ is everywhere continuous as soon as it is continuous at one point (e.g. $0$)

Xander Henderson

@ILikeMathematics No... I suggested that you think about an interval containing zero. Not just zero itself.

ILikeMathematics

Let $(-c, c)$ be such an interval.

Xander Henderson

18:18

That is bad notation, since you are already using $a$.

ILikeMathematics

Then $|f(x)| \leq M$ for all $x \in (-c, c)$

@XanderHenderson I know of the following lemma: let $f : [a,b) \to \mathbb R$ be a function. If for all monotonically increasing $(x_n)_{n \in \mathbb N}$, the sequence $f(x_n)$ converges, then $\lim_{x \to b} f(x)$ exists

Maybe this could help

Xander Henderson

@ILikeMathematics That doesn't look right...

Is $x_n$ converging to $x$?

Or $b$?

Something is missing there....

ILikeMathematics

Let $f : [a,b) \to \mathbb R$ be a function. If for all monotonically increasing $(x_n)_{n \in \mathbb N}$ with limit $b$, the sequence $f(x_n)$ converges in $\mathbb R$, then $\lim_{x \to b} f(x)$ exists

Jakobian

@ILikeMathematics the condition $f(rx) = rf(x)$ for $r\in\mathbb{Q}$ is redundant. This is known as Cauchy functional equation, and under a lot of different conditions, it's known that $f(x) = ax$ for some $a\in\mathbb{R}$, although there do exist different solutions, albeit quite "pathological"

ILikeMathematics

@XanderHenderson Should've been like this

Xander Henderson

18:28

That being said, my intuition is to focus on continuity at zero, in terms of epsilons and deltas (i.e. don't worry about sequential continuity). What would it mean if $f$ were not continuous at zero?

@Jakobian We've already been over that.

Jakobian

@XanderHenderson oh, sure. I'm late to the party

Xander Henderson

:P

ILikeMathematics

@XanderHenderson It would mean that $\lim_{x \searrow 0} f(x) \neq \lim_{x \nearrow 0} f(x)$

In terms of $\epsilon$s,

Xander Henderson

@ILikeMathematics I don't see any epsilons or deltas there....

Binky

@VladimirLysikov Thanks very much

I didn't know that the minor could also be taken like this

Xander Henderson

18:31

There is a definition of continuity at a point which starts "For every $\varepsilon > 0$, there exists a $\delta > 0$ such that ..." What is that statement, and what is it's negation?

ILikeMathematics

$\neg \left(\forall_{\varepsilon > 0} \exists_{\delta > 0} \forall_{x \in \mathbb R}: |x| < \varepsilon\right)$, so $\exists_{\varepsilon > 0} \forall_{\delta > 0} \exists_{x \in \mathbb R}: |x| \geq \varepsilon$

Xander Henderson

Jeebus... I don't want to parse all that notation. And it doesn't quite look right to me...

ILikeMathematics

Ah, you don't have ChatTeX enabled?

Let me put it into words

Xander Henderson

@ILikeMathematics I do, but I still don't want to parse all that notation. It's gross.

ILikeMathematics

Forgot a condition

Xander Henderson

18:36

In plain English, no matter how small $\delta$ is, we can always find some $x$ such that $|x| < \delta$ and $|x| > \varepsilon$.

Do you know the Archimedean property?

Jakobian

I'll mention that if you treat $\mathbb{R}$ as a $\mathbb{Q}$-vector space, then such maps $f:\mathbb{R}\to \mathbb{R}$ are precisely the $\mathbb{Q}$-linear maps, and you are basically saying that if $f$ is bounded on an open interval, then $f$ is $\mathbb{R}$-linear.

ILikeMathematics

There exists a $\varepsilon > 0$ such that for all $\delta > 0$, there exists an $x \in \mathbb R$ with $|x| < \delta$ but $|f(x)| \geq \varepsilon$

Jakobian

so it can be interpreted as a problem in linear algebra in this way

Xander Henderson

@Jakobian Sure, but that is fairly beyond the scope of the current discussion.

ILikeMathematics

@XanderHenderson Yes

Xander Henderson

18:38

@ILikeMathematics Well, then, consider that a hint. The Archimedean property can play a role here.

Jakobian

It doesn't hurt to give some higher level interpretation

ILikeMathematics

Thanks

Xander Henderson

@Jakobian I disagree. When someone is in the middle of trying to solve a problem, I think that it is pedagogically inappropriate to say "Well, of course, you could use this high powered theorem to solve the problem in a heartbeat!" That isn't the point of exercises like this. The problem wasn't assigned because anyone was unsure about the reality of the result, but because the person who assigned it wanted their students to learn a particular set of tools.

Jakobian

@XanderHenderson when did I say something remotely similar to "Well, of course, you could use this high powered theorem to solve the problem in a heartbeat!"

That's not an interpretation

Thorgott

I don't think linear algebra explains why boundedness on an interval implies continuity here

Jakobian

18:47

for sure. But it does describe the nature of this theorem, and how could it be possible generalized or considered in different settings

Xander Henderson

@Jakobian Which, again, just isn't helpful for a student who is in the middle of trying to solve the problem in the first place. In the usual American educational system, there is a very good chance that someone taking real analysis for the first time won't even understand what it means to view $\mathbb{R}$ as a $\mathbb{Q}$-vector space. It just isn't helpful to bring it up in the current context.

Jakobian

@XanderHenderson I wasn't trying to be particularly helpful. This is nonsense

psie

19:01

> Proposition Let $\nu$ be a positive measure on $(E,\mathcal A)$. Let $g\in L^1(E,\mathcal A,\nu)$ and let $d\mu=g\,d\nu$ be a signed measure. Then $d|\mu|=|g|\,d\nu$.

> Proof Let $B$ be a set in the Jordan decomposition of $\mu$ such that $\mu^+(B^c)=0$ and $\mu^-(B)=0$. Then $$|\mu|(A)=\mu(A\cap B)-\mu(A\cap B^c)=\int_{A\cap B}g\,\mathrm{d}\nu-\int_{A\cap B^c}g\,\mathrm{d}\nu=\int_Agh\,\mathrm{d}\nu,$$where $h=\mathbf1_B-\mathbf1_{B^c}$. Taking $A=\{x\in E:g(x)h(x)<0\}$, we infer from the above equality that $gh\geq0$ $\nu$-a.e. ...

I don't understand their inference. We have $|\mu|(A)=\int_A gh\,\mathrm{d}\nu$. How does setting $A=\{x\in E:g(x)h(x)<0\}$ imply $gh\geq0$ $\nu$-a.e.? I get that $\int_Agh\,\mathrm{d}\nu$ is bounded above by $0$, so it seems to be $|\mu|$-null set, but I don't see if this is useful.

Thorgott

19:13

if you integrate a strictly negative function over a set of positive measure, the integral is negative

this involves a little trick, but I'm sure you've seen something like this before

psie

@Thorgott ah, I think I see what you're saying. Since $|\mu|(A)=\int_A gh\,\mathrm{d}\nu$ and $|\mu|$ is a positive measure, $A$ has to be a $\nu$-null set. Thank you :)

Homesick Iguana

19:29

Consider the loop space of the suspension of $\mathcal I$, given by $\Omega \Sigma \mathcal I$. Now $\mathcal I$ itself is a union of four suspensions of $S^1$ under the condition that the basepoints of all suspensions form the 0-skeleton of a unit cube. My question is: What is the weak homotopy type of $\Omega \Sigma \mathcal I$?

ILikeMathematics

@Thorgott I proved continuity at $0$ now

Now I guess just the argument to generalize it is left

Thorgott

that should be easy now

Homesick Iguana

I opened hatcher and immediately cried

Homesick (referring to the loops space question above) this is just completely ill posed

ILikeMathematics

@Thorgott Well, we can't really pull it apart properly, somehow

How would you re-write $f(a)$ with $a \in \mathbb R$ so it becomes clear this boilds down to the continuity of $f(0)$

Homesick Iguana

there are a plethora of weak homotopy types here - you need to be more specific in what you are saying

Jakobian

19:41

@HomesickIguana from happiness

Thorgott

@ILikeMathematics you have $f(x)-f(y)=f(x-y)$

the rest is spelling everything out in terms of epsilons and deltas

ILikeMathematics

Thanks

Homesick Iguana

@Jakobian there is so much to play with in this book - it's awesome.

I can see myself missing important obligations because of this book

its interesting how Hatcher's theme is to take suspensions and coax them into a smooth foliation

just kidding that is my theme :P

GroveRover

20:19

Does every complex representation of the spin group come with a unique representation of the whole Clifford algebra that induces it?

Xander Henderson

@Jakobian Okay, so we agree. You weren't trying to be helpful.

Jakobian

@XanderHenderson let's put it differently. That wasn't my top priority.

and I don't think we agree, since you were claiming e.g. "It just isn't helpful to bring it up in the current context.", even though what I was saying can be helpful for other people in this chat, that are not ILikeMathematics that are just learning the subject

leslie townes

the existence or not of nonlinear solutions to the cauchy problem might be one of those things that depends on your set theory :)

Jakobian

or maybe even for ILikeMathematics if they understand what those things mean, it can be (on a minuscule scale) helpful

putting something in easy to understand context is helpful to those who can understand it, and for those don't understand it, they can simply ignore it

"not particularly helpful" does not mean unhelpful

you wrongly assumed that this is teacher-student situation, while this is a chat that everyone can read, and possibly benefit from, and you also wrongly assumed my intentions

Xander Henderson

@Jakobian The problem with that attitude is that someone who is just learning a topic often does not have a good sense of what can be "just ignored". It is a totally reasonable point to bring up when the problem has been solved, i.e. "there is a more general context for this", but it is actively harmful to insert it into the middle of a process in which the student is confused.

@Jakobian I made no assumption at all about your intentions.

Jakobian

20:30

cool, then it shouldn't bother you

Xander Henderson

I pointed out that your point was out of the scope of the current conversation. You said it didn't hurt, and I disagreed because I think it added confusion for the person who was most actively confused.

@Jakobian I have no idea what this is supposed to me. I made no assumption about your intentions. I merely pointed out that what you were saying was out of scope (as much to warn ILikeMathematics that they could ignore it, which is what you recommend that they do, anyway).

Jakobian

@XanderHenderson you are exaggerating, but I guess it could possibly lead to some confusion

still, this won't stop me from mentioning it

Xander Henderson

@Jakobian I don't think that I am exaggerating, because I have honestly seen students go on tilt when someone "helpfully" throws out something beyond their depth, but which is related to whatever problem they are currently studying.

Jakobian

21:02

@XanderHenderson which is not what happened, so I don't know what you are talking about

Xander Henderson

@Jakobian I know you don't, and I don't know why I keep trying to explain pedagogy to you. You clearly have no interest in teaching.

Jakobian

correct

leslie townes

the pedagogue of seville only explains pedagogy to people who don't explain pedagogy to themselves

Jakobian

not teaching, but I do have interest in spreading knowledge

Vuk Stojiljkovic

22:04

Is here someone I can ask something regarding functional analysis?

Xander Henderson

@VukStojiljkovic Just ask the question. You don't need to ask to ask.

Jakobian

@VukStojiljkovic make sure to have a plugin for LaTeX in chat as well (see chat description), otherwise you won't see LaTeX

Sine of the Time

22:26

11 hours ago, by leslie townes

vuk what is the problem exactly? the question on main kind of rambles on a bit, slightly more focus might help. one underlying point (which someone mentions in the comments) is that if you think of an operator on H+H as a 2x2 matrix, the entries of the matrix are not all of the same type. in fact they all have different types.

Leslie replied to your question but maybe you did not receive a notification @VukStojiljkovic

leslie townes

we should all be talking about functional analysis all of the time in here

Jakobian

@leslietownes I'll let you know when I finally begin learning about bornological spaces

leslie townes

when you "get born" as they call it

Jakobian

lol

Claudio

22:55

and it happens again

it goes on discussing what happens when $K>/</= 0$

I can't wrap my head around ignoring the absolute value and replacing $e^c$ with a constant $k \in \mathbb{R}$, but again this seems to be the correct way to do it

Sine of the Time

@Claudio where is this problem from?

Claudio

In some exercises, where initial data are given, one can determine a priori where the solution will be (like t>0 and y>0) and so one can reasonably ignore the absolute values

@SineoftheTime our usual set of exercises

but it's just the ode, not a cauchy problem

I did everthing right, but again the part of the solution above does not convince me

it does so pretty much everywhere :p

leslie townes

claudio again you may just be overthinking it :) note that the formula with absolute value shows that the LHS is never zero (because the right isn't) so if it is a continuous function it is going to have constant sign, so the | | will either do nothing or flip the sign

with flipping the sign giving you those negative scalars 'e^c' and the case where the thing is identically 0 being just slightly missed by this analysis

Sine of the Time

If $u>1$ or $u<0$, then you get $u(t)=\frac{1}{1-K_1\exp(t^2/2)}$; if $0<u<1$ then $u(t)=\frac{1}{1+K_2\exp(t^2/2)}$ where $K_i>0$

So $u=\frac{1}{1-K\exp(t^2/2)}$ for some $K \in \Bbb R$

Claudio

23:11

a continuous function has costant sign?

Oh I see

leslie townes

a continuous function on an interval can't change sign without being zero, which is inconsistent with other aspects of the analysis

Claudio

Yeah you're right, $f \in C^1 \Rightarrow u \in C^2$

yeah, I grasped the meaning of the sentence a bit later, my bad :p

leslie townes

well, it's subtle stuff :) a big reason why a lot of books don't put a lot of effort into explaining it

the assumption of continuous "on an interval" is also being used, you certainly can change sign without being 0 if you don't mind having a domain in more than one piece (like 1/x)

Claudio

I do mind for now hahah :)

Jakobian

23:36

do you guys also get really frustrated when stuck on a problem

A. P.

Hi, take u as a continuous function on the closed ball B[0,r] y x_0 is a point in the boundary \partial B(0,r), then if for all \delta in (0,r) we assume u(x_0) <= max{u(x): x\in \partial B(0,\delta)} then the continuity of u at 0 implies u(x_0)<=u(0)? I cannot see it by using an \epsilon-\eta argument..

Jakobian

see description for LaTeX in chat

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