Mathematics - 2025-01-08

ZaWarudo

01:30

I must study the continuity in $(0,0)$ of $f$ defined as:
$$f(x,y)=\begin{cases} \frac{x^2+y^2}{\pi+2\arctan(\frac{y}{x})}, && \text{if} \ x\ne0 \\ \frac{y^2}{2\pi}, && \text{if} \ x=0 \end{cases}$$
Using the polar coordinates with $\theta \in [0,2\pi)$, I get:
$$f(x,y)=\begin{cases} \frac{r^2}{\pi+2\arctan(\tan \theta)}, && \text{if} \ \cos \theta \ne0 \\ \frac{r^2 \sin^2 \theta}{2\pi}, && \text{if} \ \cos \theta=0 \end{cases}$$
My doubt: do I have to distinguish cases for the inverse of the tangent? That is, I must use:

Sine of the Time

01:45

Why are you using polar coordinates?

ZaWarudo

They seemed useful because of the $x^2+y^2$ and the $y/x$

And, now that I tried them, I was curious about the inverse tangent situation as well

Sine of the Time

instead of analyzing all the cases you can verify whether $\theta\mapsto \frac{1}{\pi+2\arctan(\tan \theta)}$ is uniformly bounded

ZaWarudo

Thanks for your help! I will try to find a uniform bound right now. Is the cases situation correct anyway?

Sine of the Time

I don't know by heart all these trig formulas

nickbros123

03:52

@SoumikMukherjee idk, I don't want to start an argument

Nirbhay Krishnan

i created a question as a challenge anyone wanna try?

Soham Saha

06:16

@NirbhayKrishnan Yep. What’s the question?

Tapi

09:56

how to show that Evaluation Maps Generate Borel-sigma(C[0,1])?

Jakobian

10:15

@Tapi I don't think they do

Oh maybe I misunderstood you

Tapi

@Jakobian this function: ev_t :C[0,1] to R, takes f to f(t) , (vary t) doesn't generate borel (C[0,1])?

Jakobian

Preimages of Borel sets of R?

Tapi

yes

evaluation maps are continuous (using sup metric)

Jakobian

If you take $C[0, 1]$ with topology of pointwise convergence?

Nirbhay Krishnan

@SohamSaha

$$T_r+1=aT_r+(ab)^r $$

Find the sum of first n terms

it is not challenging but calculation is there this was adapted from jee mains q
also T1=ab in the original question a was 2 and b was 3

Tapi

10:25

@Jakobian no im not sure if we are taking ptwise convergence

Jakobian

As I see it this is the same as saying that $\mathcal{B}(C[0,1]) = \mathcal{B}(C[0,1]_p)$ where the index means we're taking pointwise convergence

Tapi

oh

i don't get what you're saying

Jakobian

What part

Tapi

i was tryin to show $\sigma(ev_t : t \in [0,1]) = \mathcal{B}(C[0,1])$

why convergence is relevant

Jakobian

Pointwise convergence

It means topology as a subspace of product of R's

Tapi

10:29

I see

Jakobian

@Tapi I still think this might be wrong with sup metric

Because I think that the members of Borel sigma-algebra generated from those evaluation maps depend on countably many coordinates

Tapi

ok

Jakobian

While sup metric has open sets such as $\{f : \|f\|<1\}$ which don't

@Jakobian ah sorry that's not true. But they do have a generating set, members of which only depend on countably many coordinates

Tapi

this can be written as countable intersection union of sets of the form f| ev_t < 1-1/n

Jakobian

Ah wait

I'm correct actually because what I said before was not true

The evaluation maps don't generate all pointwise convergence open sets

Because the basic open sets don't generate the Borel sigma-algebra

For example $\{f\}$ where $f$ is any map is not in that Borel sigma-algebra

@Tapi where did you get this from?

Binky

11:24

But to see if a differential form is exact I just need to find the differentiable function f such that the differential of the function coincides with the differential form or df=w?

Jakobian

12:45

I got rudely ignored

Tapi

I had a class

Just came back

@Jakobian I was doing an exercise and I thought this was the way to go

It's about constructing random elements which are functions

Jakobian

so you wanted to show it by showing that certain compositions are measurable?

what does "random continuous function" mean. I guess it just means random variable into $C[0, 1]$?

Tapi

@Jakobian I wanted to show that the map Xn is measurable

@Jakobian yes

typically when it's R we say it's a random variable, when it's R^k we say it's a random vector and when C[0,1] we say it's a random continuous function

Jakobian

13:01

your idea wasn't a bad one, but the issue here is that $C[0, 1]$ is not the subspace of the categorical product of $[0, 1]$ amount of copies of $\mathbb{R}$

Tapi

I basically have to show inverse image of a set from sigma(C[0,1]) under Xn would belong to \mathcal{A}

Jakobian

while for finite products this works, it stops working for infinite ones

Tapi

right

Jakobian

so your argument has to be more explicit

Tapi

so the idea i was going with is finding a generating set and showing the inverse image thing for that set

generating set that forms a pi system

Jakobian

13:04

a generating set that we know of, is the open balls in the sup metric

not sure of any more useful one for this problem at the moment

so you take a function $f\in C([0, 1])$ and want to say that $\{\omega : \sup_{t\in [0, 1]} |f(t)-X_n(t, \omega)| < r\}$ is measurable

where $X_n(t, \omega)$ means $X_n(t)$ at $\omega$

(unfortunate notation)

Tapi

okay

Jakobian

$X_n$ is supposed to be a function on $\Omega$ so writing it as $X_n(t)$ for $t\in [0, 1]$ triggers me a little bit, but what can you do

so now what do we know

we know that $[0, 1]$ has a countable dense set, which is good for all the sigma-algebra considerations

and that both $f$ and $X_n(\cdot, \omega)$ are continuous

@Jakobian let's denote this set as $A$

from this we can say that $A = \{\omega : \sup_{t\in Q} |f(t)-X_n(t, \omega)|<r\}$ where $Q$ denotes some countable dense set of $[0, 1]$

and this can now be restated as $A = \bigcap_{t\in Q} \{\omega : |f(t)-X_n(t, \omega)|<r\}$

Tapi

got it

Jakobian

oh wait

you were actually right, it is generated by those evaluation maps

Tapi

:(

@Jakobian this is true though

Jakobian

13:13

my mistake was to think that $\{f\}$ is does not depend on countably many coordinates

Tapi

it's continuous that's why?

Jakobian

yeah

@Tapi no I think that's false

well anyway this argument shows the same thing

Tapi

oh because we can approximate the function at irrational points using rational points since f is continuous

by using a polynomial approximation

Jakobian

@Jakobian so here $A$ is clearly measurable since $X_n(\cdot, \omega)$ is measurable

@Tapi well no, because a continuous function can be determined by its values on a dense set

and here this dense set is countable

Tapi

okay

Jakobian

13:18

here if you take preimages of Borel sets of $\text{ev}_t$, then those are measurable sets in $C[0, 1]$

and conversely if you take an open set $U = \{g : \|f-g\| < r\}$ then $U = \bigcap_{t\in Q} \{g : |f(t)-g(t)| < r\}$

and so $U$ is in the sigma-algebra generated by $\text{ev}_t$ for $t\in [0, 1]$ (or even for $t\in Q$)

Tapi

@Jakobian why is $X_n(\cdot, \omega)$ measurable?

Jakobian

so yeah you were right, and this argument that $A$ is measurable above basically uses the same argument

@Tapi oh sorry I meant $X_n(t, \cdot)$

it's measurable because it's a linear combination of random variables

Tapi

@Jakobian Let B be a borel set in R; preimage of B= $\{ f \in C[0,1] | f(t) \in B) \}$

Why is this measurable?

Jakobian

14:26

@Tapi because $\text{ev}_t$ is continuous

Soham Saha

@NirbhayKrishnan Did you mean $T_{r+1}=aT_r+(ab)^r$ ?

Nirbhay Krishnan

15:24

@SohamSaha yessss

psie

I'm reading about the Radon-Nikodym theorem, and in the first part of the proof, the author assumes the two $\sigma$-finite measures involved are finite and $\nu\leq\mu$ pointwise (in particular, $\nu\ll\mu$). Then, through Hilbert space theory, the author derives that there exists a function $g\in L^2(E,\mathcal A,\mu)$ such that $$\int f\,\mathrm{d}\nu=\int fg\,\mathrm{d}\mu,\quad \forall f\in L^2(E,\mathcal A,\mu).$$

The author goes on to show that $0\leq g\leq 1$ $\mu$-a.e. Now, in the second part of the proof, the assumption is only that $\mu,\nu$ are finite, but the author uses the first part of the proof (by replacing $\mu$ with $\mu+\nu$) to find a measurable $0\leq h\leq 1$ such that for every $f\in L^2(\mu+\nu)$, $\int f\,\mathrm{d}\nu=\int fh\,\mathrm{d}(\mu+\nu)$. In particular he says, for every bounded measurable $f$, $$\int f\,\mathrm{d}\nu=\int fh\,\mathrm{d}\mu+\int fh\,\mathrm{d}\nu.$$

Why is this last formula true, and does it hold for any function in $f\in L^2(\mu+\nu)$?

psie

15:37

I'm guessing the formula is true by measure theoretic induction, where we define $g=fh$, which is measurable if $f$ and $h$ are. But what would one induct over? $g$, $f$ or $h$?

Thorgott

I don't know what measure-theoretic induction is, but this is linearity of the integral in the measure

Jakobian

@Thorgott where you go from simple functions, to non-negative, to integrable

it's an informal name of this method

Thorgott

ok sure, that's how one can prove this

I've just never heard the name before

Jakobian

I get that. I studied in a department full of probability theorists, that's how I know

Jakobian

16:08

(I didn't check what you said, psie)

Claudio

16:19

sorry guys, I have the following solid $E = \{ \mathbf{x} \in \mathbb{R}^3: 2\sqrt{x^2+y^2} \le z \le 1+x^2+y^2 \}$. Well, I can't see how to know what the bounds for $x,y$ are, it seems that this subset is not bounded above

if you take only the $x \le 1$, (e.g a disk of radius 1) then I guess you have a solid

Jakobian

@Claudio yeah it's not bounded

there is no bounds for $x, y$ that you can have

Claudio

thanks @Jakobian. There should therefore be a mistake, let me check the solution

Jakobian

@Claudio you mean $x^2+y^2\leq 1$ I suppose. I don't know what's your definition of a solid

"solid" is not something I have ever defined

Claudio

yeah exactly

I mean there's symmetry wrt. the z-axis

so $x\ge 0 \longmapsto \sqrt{x^2+y^2}$

and u get a solid of revolution

Jakobian

what do you mean?

it's pretty clear that there's rotational symmetry around the $z$-axis

Claudio

16:28

lmao my bad

Jakobian

I don't know why you are apologizing. What you're saying is confusing

Claudio

I'm agreeing with you :p

ignore my messages under yeah exactly

Jakobian

okay

Claudio

I checked the solution: it says to rewrite $E = \{ \text{ same as above }, (x,y) \in D \}, D =\{x^2+y^2 \le 1\}$

Jakobian

solution to what

Claudio

16:31

it's ax exercise about the divergence theorem

Jakobian

@Claudio it's hard to rewrite it as something it isn't

Claudio

yeah there's probably a mistake, and they forgot about adding the bounds for $x,y$

Jakobian

unless perhaps the set $E$ was described in words and you didn't understand it correctly

Claudio

the solid: E = ....

Jakobian

the set of $(x, y, z)$ such that $z = 2, x^2+y^2 = 1$ is where the two bounding surfaces "touch" so it makes sense to take only the lower part

I see. The book is to be blamed

Claudio

16:36

@Jakobian yeah, that was my guess to what it actually meant indeed

psie

@Jakobian well, now you can (if you want). We have $$\int fh\,\mathrm{d}(\mu+\nu)=\int fh\,\mathrm{d}\mu+\int fh\,\mathrm{d}\nu,$$with $f,h\in L^2(\mu+\nu)$ and $0\leq h\leq 1$. First we check that the formula is well-defined. $fh\in L^2(\mu+\nu)\subset L^1(\mu+\nu)$ since $\mu,\nu$ are finite and hence also $\mu+\nu$ is finite ($fh$ is in $L^2(\mu+\nu)$ since $f$ is and $h$ is bounded above by $1$).

Next, is $fh\in L^1(m)$ where $m$ equals $\mu$ or $\nu$? (don't know how to reason about that). Assuming we've shown that, then we just have to prove the formula for the cases when $fh$ is a measurable indicator function, a simple function, etc. Indicator function is just the definition of $(\mu+\nu)(A)=\mu(A)+\nu(A)$, linearity gives the formula for simple functions, and then its true for nonnegative functions by monotone convergence.

Jakobian

Ask Thorgott. I'd have to catch up on the other half of what you said

If Thorgott has nothing to say about it them I might chime in

psie

16:59

Ok. Well, it does say in the text, prior to displaying $\int fh\,\mathrm{d}(\mu+\nu)=\int fh\,\mathrm{d}\mu+\int fh\,\mathrm{d}\nu$ that "in particular, if $f$ is a bounded measurable function...", so I'd guess by the word 'in particular' the author means the formula only holds for $f$ being bounded measurable, since then it is in $L^1(m)$ for $m$ equal to $\mu$ or $\nu$. Otherwise, I don't see that it necessarily holds for $f,h$ simply in $L^2(\mu+\nu)$ and $0\leq h\leq 1$.

Anyway, time to move on I think :)

X4J

17:29

Let G be a finite group and g_1,...,g_n are rep. for all distinct conjugacy classes in G. Suppose H is a subgroup of H such that g_1, ..., g_n lie in H, does G=H?

Jakobian

@X4J If $H$ is normal then certainly $H$ contains all the conjugates of $g_1, ..., g_n$ so that $G = H$. So I would try to search for some non-normal, but relatively large subgroup of $G$. Perhaps take $G = S_3$?

Soumik Mukherjee

@X4J This is a nice question, what have you tried?

Jakobian

If the amount of conjugacy classes is $> \frac{1}{2}|G|$ then $G = H$ as well

so it needs to be some group with the number of conjugacy classes $\leq \frac{1}{2}|G|$

Soumik Mukherjee

This is true for finite groups

Jakobian

this is a finite group

Soumik Mukherjee

17:43

yes, I am saying that the statement X4J wrote is true

Jakobian

18:03

I see. It's called Jordan's lemma apparently (or at least follows from it)

Soumik Mukherjee

18:18

I remember this question particularly because it was asked to me in an interview 4 years ago. I couldn't prove it back then.

Homesick Iguana

18:36

Is it possible to define a map that is simultaneously a graph and group automorphism, and an isometry?

Soumik Mukherjee

I don't understand the question

Jakobian

@SoumikMukherjee that's a hard question

Soumik Mukherjee

yeah :(

This was their first question and it was my first interview so I got more nervous at that time.

Vladimir Lysikov

19:04

@HomesickIguana identity preserves any structure you want

Homesick Iguana

@VladimirLysikov ah great :)

thanks

Jakobian

a graph?

Soumik Mukherjee

19:19

I have the same question, a map can be a group automorphism and an isometry if it's on a metrizable topological group but what does it mean to be a graph?

Homesick Iguana

a graph with smooth edges is what I was thinking

like as in "graph theory"

Vladimir Lysikov

I read "graph and group isomorphism" as "graph (isomorphism) and group isomorphism"

Homesick Iguana

I should have also stated that there is not a global smooth manifold structure at play but each stratum is indeed smooth on its own

the graph can be the 1-strata

maybe deleting the graph disconnects the space

Tapi

19:59

Is the set of sequences with finitely many rational non zero entries (rest are 0) not in a bijection with the power set of rational numbers?

I figured

Jakobian

@SoumikMukherjee to be an isometry you do need to have a metric

@Tapi it's injects into finite subsets of $(\mathbb{Q}\setminus \{0\})\times \mathbb{N}$, so is at most countable.

Soumik Mukherjee

20:21

@Jakobian yes, I wrote metrizable for that reason

ILikeMathematics

In analysis, it seems we do a lot over $\mathbb R$ and $\mathbb C$. Why not try to generalize as much as possible like in linear algebra, where we do most things over arbitrary fields $K$?

Jakobian

@SoumikMukherjee I know, I am saying that metrizable is not enough

@ILikeMathematics some parts of it do get generalized. But not in real/complex analysis

the problem isn't that we don't do that, the problem is that you don't know we do that

now why does there exist fields such as complex and real analysis? Because those fields in particular have some nice properties

ILikeMathematics

@Jakobian I guess $\mathbb R$ is simply more important for analysis then, since in LA we don't make that distinction. We could still consider fields where sup and inf exist and I guess prove a lot of it the same way

Jakobian

and we care about them in particular for various reasons. That's all

@ILikeMathematics it's not only that it's more important. Linear algebra is awfully simple. So much that changing field doesn't matter much. But for analysis, it's not only about algebraic properties of real or complex numbers

people do care about fields other than real or complex numbers, but at this point it can get complicated

no one wants to burden themselves with unnecessary complexity

Xander Henderson

20:37

@ILikeMathematics The field structure is not the most important structure in analysis. It is important, but it is not as important as the smooth structure.

ILikeMathematics

Alright, thanks. So generally in algebra, we don't get a lot of problems when dealing with arbitrary fields, but in analysis, we do

@XanderHenderson Oh, alright

Jakobian

it's just that those things have a lot more structure

Soumik Mukherjee

@Jakobian why? Doesn't metrizable mean that the space has a metric that generates the topology?

Jakobian

@SoumikMukherjee no

metrizable means that there exists such a metric, not that it's given one

Xander Henderson

@SoumikMukherjee No. It means that a metric can be imposed on the space in a way that is compatible with the topology.

Jakobian

20:39

you need to choose a metric to be able to speak about isometries

@ILikeMathematics well in linear algebra, some things are similar, we do restrict to real/complex numbers occasionally even then

Soumik Mukherjee

Okay I get it

Jakobian

for analysis it's not always obvious in which way you want to generalize real/complex numbers

usually you want a field with an absolute value, and you do study things over such fields

X4J

@SoumikMukherjee You mean that this property of H "enforces" H to generate all the conjugates?

Jakobian

it's not really important to do things in this generosity as you say, at least not as far as I know. Unless perhaps you are working over a non-archimedean field, and this turns to its own study

there is such thing as non-archimedean analysis, which I know nothing of, but it's useful for some fields

Soumik Mukherjee

@X4J I don't get what you mean by generate all the conjugates. What I mean is that if a subgroup H of a finite group G intersects all the conjugacy classes then it's the whole group itself, i.e. H=G

This is not true for infinite groups though

X4J

20:50

Yes I have just read some nice proof using group actions

The way a group can be decomposed by group actions fascinates me

psie

21:31

> Consider $[0,1)\subset \mathbb R$ and a partition $[(i-1)/2^n,i/2^n)$ for $i\in\{1,2,\ldots,2^n\}$. Define $\mathcal F_n=\sigma([(i-1)/2^n,i/2^n);i\in\{1,2,\ldots,2^n\})$. Let $\lambda$ and $\nu$ be Lebesgue measure and a finite measure respectively on $([0,1),\mathcal B([0,1)))$. By restricting $\lambda$ and $\nu$ to sets in $\mathcal F_n$, we can view both $\lambda$ and $\nu$ as measures on $([0,1),\mathcal F_n)$. Then $\nu\ll\lambda$ and ...

I'm paraphrasing from my book above. Is $\nu\ll\lambda$ because any $A\in \mathcal F_n$ is a finite union of sets $[(i-1)/2^n,i/2^n)$, so the only set where $\lambda$ equals $0$ is the empty set?

The example above is the beginning of the construction of a Radon-Nikodym derivative $f_n$ that one can show (with the use of martingale theory) converges to a Borel measurable function $f$ $\lambda$-a.e., and that the Lebesgue decomposition of $\nu$ with respect to $\lambda$ is $f\,d\lambda$. Looks harmless at first sight, but maybe not so simple :)

Thorgott

22:18

to do analysis reasonably, what you probably want is a field $K$ together with a valuation $|-|$ (Xander might disagree with this choice of terminology), which is a function $K\rightarrow\mathbb{R}_{\ge0}$ satisfying the axioms you might expect. in particular, this valuation induces a metric and a topology. now if we are slightly more decadent, we also want to demand that $K$ is *complete* with respect to this metric (recall how often you use that property in analysis).
it turns out then that you can distinguish two cases. if the metric satisfies the Archimedean property, the field is neces…

psie

22:31

@psie I found this, but with a few typos in it.

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