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02:32
@geocalc33 h3y
02:43
given $y' + x^{-1}y = 1$, since it is a first order linear DE, we can solve it using the integrating factor $e^{\int x^{-1} \ dx}$. the lecturer says it is $e^{\ln x}$ (and this works). why not $e^{\ln x + C}$? or should i just assume we've set $C = 0$?
answer: you can just cancel out $C$ as soon as you multiply the equation by the integrating factor
$e^c = C$ which will still be just some constant. So you would be multiplying by some constant
just like you said in the answer part
yeah. the way to think about it is that the integrating factor in that method just needs to be some antiderivative of the thing in that method. you don't need the most general one, or any particular specific one.
which is maybe somewhat at odds with how something like "$\int P(x) \, dx$" is usually interpreted, but, thats what it's standing in for, in that formula.
Leslie used dollar signs????? 🤯🤯🤯
Must be having a relaxing day...
don't worry, i'm not rendering them on my end.
i'm procrastinating entering my time for march into my work's timekeeping software. those dollar signs gave me one more half second of procrastination.
this is the stuff AI will be great for
02:55
@DLeftAdjointtoU sup broseidon
I thought you would have a personal assistant do that?
@geocalc33
Want me to find the book for you on google?
The right one
Is this the homological methods in transverse foliational derived functor space text?
Yeah can you find that one for me
dc: i think the only people who do that at my work are very old and sufficiently inactive that their secretary can easily keep track of how they spend time.
@DLeftAdjointtoU Weibel?
03:00
@geocalc33 yes you can preview it on amazon
Okay
price?
@leslietownes They're made and just cruising off of the name and reputation now.
@geocalc33 yes, you'll have to buy it, but I found this for the first page for now:
Please let's create a room
@geocalc33
Let's do this
let's do what
I'ma teach you some HomAlg in that room
03:25
is the determinant of the characteristic polynomial matrix the product of eigenvalues iff we're considering the polynomial at $x = 0$?
nonsense statement
don't read
warning
@D.C.theIII Must be an impostor!
@shintuku Up to sign, depending on your definition, yes. The det is the product of the eigenvalues.
thanks!
 
1 hour later…
04:49
definitely not a nonsense statement, just weirdly written.
not unlike a whole lot of other stuff on here.
Is there an infinite amount of real numbers between any pair of real numbers?
05:07
yes
05:20
Show that for a space X, the following three conditions are equivalent:
(a) Every map $S^1→X$ is homotopic to a constant map, with image a point.
(b) Every map $S^1→X$ extends to a map $D^2→X$.
(c) $\pi_1(X,x_0) = 0$ for all $x_0 ∈ X$.
I didn't know until now that it's holiday here today. :)
So for (a) to (b): Let $f: S^1\to X$ be any map. It is nullhomotopic by (a) so there exists a homotopy $f_t:S^1\to X$ and a constant map c from $S^1$ to X such that $f_0=f, f_1=c$
I define $F:D^2\to X$ as follows: $F(x)=f(x)$ if x is in S^1 else F(x)=c(x).
But I'm not sure how to show continuity of F.
is there a typo there? what is c(x)? how are you using the existence of (f_t)?
c is the constant map from $S^1$ to X. Existence of $f_t$ is by (a).
@leslietownes Because if the interval is from $[a,b]$, we can always find a number that is halfway between each successive interval? $a+\frac{a+b}{2},a+\frac{a+b}{2}+\frac{a+\frac{a+b}{2}+b}{2},...$
So there exists some $x_0\in X$, $c(x)=x_0$ for all x in S^1.
koro: my point is, your formula for F doesn't seem to involve (f_t) at all? just f and the constant map?
05:29
yeah. I don't know how to bring f_t into the mix.
:(
cotton: yeah, that's a slick/simple way to show it, regardless of how you construct the set of real numbers
I have seen this somewhere, probably in Rotmann's but I can't find it rn.
Just making sure my stats professor knows there's an infinite number of outcomes for the responses in the experiment lol
koro: what if you do a kind of polar coordinates thing. identify D with complex numbers of mod <= 1, and C with the unit circle and do F(r e^(it)) = f_r(e^(it)), where i'd choose f_0 to be the constant map (so this is well-defined, i.e., independent of t, when r = 0) and f_1 to be f, or otherwise fiddle with the formula.
Ah, I didn't think about polar coordinates.
thanks.
05:33
and continuity just comes out of whatever the definition of homotopy of paths is.
06:22
The true meaning of Jordan Holder theorem is a well definedness of a finite group $G$ to a power set $\mathcal{H}$ of finite simple groups up to isomorphism.
06:39
In mathematics, a hyperelliptic surface, or bi-elliptic surface, is a surface whose Albanese morphism is an elliptic fibration. Any such surface can be written as the quotient of a product of two elliptic curves by a finite abelian group. Hyperelliptic surfaces form one of the classes of surfaces of Kodaira dimension 0 in the Enriques–Kodaira classification. == Invariants == The Kodaira dimension is 0. Hodge diamond: == Classification == Any hyperelliptic surface is a quotient (E×F)/G, where E = C/Λ and F are elliptic curves, and G is a subgroup of F (acting on F by translations). There are seven...
It's different from the hyperelliptic surface I know in Riemann surface
07:23
We may assume that each $f_j$ is of degree $3$? $f_j$'s are meromorphic functions, in particular degree = 0.
I think it should be the order of poles is 3 counting multiplicities.
meromorphic function with simple pole cannot exists on compact RS so it starts with two poles. If $f_j$ has two poles then $D_j$ can be chosen to be degree $2$ so $M$ is hyperelliptic.
Oh degree here is not a map from divisor class to Z... it's a degree as a map. so the same thing
07:51
What is Peter Weyl theory? What does it have to do with functional analysis?
there's a hyphen in there (peter and weyl are two people). see e.g. en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem
i dunno if anyone calls it a 'theory,' results like the ones mentioned on that page are often what people somewhat ambiguously refer to as 'the peter-weyl theorem'
Weyl appears everywhere
What I don't understand is why our functional analysis teacher is doing this in class and he is also doing representation theory.
i'd suggest asking him, but i've heard enough about how your school works that i have a feeling that it wouldn't be a helpful suggestion
:)
haha
07:58
i can think both of reasons why you might include it, and reasons why you might not include it, alongside other stuff in functional analysis, so, who knows. is it in a textbook that the instructor likes?
So our syllabus on college website includes compact operators and there is not much detail available as to what extent the operators are to be covered.
@leslietownes No, he does not follow any textbook.
Also, please note that this is an introductory functional analysis course so I think that the course structure should follow Kreyszig's or Conway's.
etc.
@onepotatotwopotato that's because a Riemann surface is a complex curve, where as this is a complex surface
maybe he has a physics background, they often at least pretend to be into representation theory and aspects of functional analysis at the same time
In algebraic geometry, a hyperelliptic curve is an algebraic curve of genus g > 1, given by an equation of the form where f(x) is a polynomial of degree n = 2g + 1 > 4 or n = 2g + 2 > 4 with n distinct roots, and h(x) is a polynomial of degree < g + 2 (if the characteristic of the ground field is not 2, one can take h(x) = 0). A hyperelliptic function is an element of the function field of such a curve, or of the Jacobian variety on the curve; these two concepts are identical for elliptic functions, but different for hyperelliptic functions. == Genus == The degree of the polynomial determines...
I don't know any of AG. But I heard some correspondence between RS and complex curves.
08:02
I'm guessing this is what you mean by hyperelliptic Riemann surface
it's not that unusual for something called "functional analysis" in a course catalog to be a kind of grab bag of topics
He also covered Haar measure and I don't know why.
I think that he 'shows off' in class how much he knows and keeps digressing to things unrelated to course.
I prefer analytic formulation of RS only because I don't know AG
I think I have DG next sem. I bet the teacher teaching that here will be the worst so far.
It's almost exactly the same object, complex curves just have more structure
08:05
Fourier analysis is somehow related to representation theory so maybe his intent was to present some connection between those two fields?
So I wish to study all of it during our summer break.
koro, haar measure comes up all over the place but also gives you a way of constructing representations of things on hilbert space
Ohh
I'm yet to study it. I'll study it soon.
One day he 'taught' something in class and then at the end, he said-ah, I confused this class with another class. This is not related to your syllabus.
😅
@Astyx Good to know. Maybe I need to study AG at some point (maybe this summer?)
There is one food delivery service here, if they delay delivering the food then they pay the whole amount back +the food that was ordered.
This business model seems good to increase revenue.
08:13
@Koro one day, i bet one of these instructors trips, and when they fall, they break into several pieces and a bunch of wires and robot parts fall out
So viewing RS in analytic way is no longer considered these days @Astyx?
I don't know, it probably is by some people
it certainly still is taught
@TedShifrin I should have said it correctly. Instead of saying loop at $\sqrt 2$, I meant to say loop at $(\sqrt 2,\sqrt 2)$.
Instead of rational no., I should have said: a point ( , ) in R^2 with both entries rational.
I think uncountability of the group is because: I take a straight line through $\sqrt 2$ and consider circles $\gamma_a$ through $(\sqrt 2, \sqrt 2)$ and point a on the line. So it can be seen that $[\gamma_a]\ne [\gamma_b]$.
This is because between $\gamma_a$ and $\gamma_b$, there is rational no. hence both are not hpy. equivalent.
So we have {$[\gamma_a]: a$ is on the line and on complement of $Q^2$} uncountable.
08:35
how do you know those circles are paths in R^2 \ Q^2? i don't really care about the underlying problem, it just seems that there's no immediately clear reason why such a circle would have to avoid Q^2, or what you'd need to worry about in order to ensure that
08:50
I should replace circles by loops: So from base point $(\sqrt 2, \sqrt 2)$ to $a$, we have a path constructed as follows: there are only countable elements of $Q^2$ so we avoid all them to join to a.
Then we come back to basepoint using another path.
Concatenation of these two paths is a loop at the basepoint.
I should consider this loop instead of circle.
But I see this requires lot more details. I'll think about this.
09:05
does anyone have any insight into why this ought to be true?
$A$ is an integer matrix
and $p$ is a positive integer prime
then $\text{tr}(A^{p^{n-1}) $ is equal to $\text{tr}(A^{p^n})$ mod $p^{n}$ for $n \geq 1$ a natural number
or at least does anyone see some straightforward way to prove this
sorry $\text{tr}(A^{p^{n-1}})$ is equal to $\text{tr}(A^{p^{n}})$ mod $p^n$
so $n=1$ is basically just a consequence of fermats little theorem
it does not look like there is going to be necessarily some straightforward way to be very careful with congurences and extend the $n=1$ approach for higher $n$ though
(this result is true and now basically considered 'classical')
09:22
there may be a nice graph theoretic way to see this actually
with $A$ of course being the adjacency matrix of some graph (with positive edge weights)
 
2 hours later…
10:58
pozhar v pravom dvigatele
11:15
HI
12:12
If a divisor $D$ on a compact riemann surface of genus $g$ has degree $2g-2$, then $D$ is a canonical divisor? I know that if $D$ is a canonical divisor then its degree is $2g-2$.
12:29
Hi folks! I'm trying to show that $\sum_{i=1}^\infty E |X_i - X|^p < \infty$ for p > 0 implies $X_n$ converges almost surely to $X$

could you please have a look if my proof makes sense?

https://imgur.com/a/8oGv5kE
if $g=1$, i.e. the Riemann surface is a torus, a divisor is canonical iff it is principal. do you know if every degree $0$ divisor on a torus is principal?
13:04
why is that?
I know some criteria for a divisor to be principal
13:16
$\{x_n\}, \{y_n\}$ are converging cauchy sequences. Can I get a bound on $|x_iy_i - x_jy_j|$ from $|x_i-x_j|< \epsilon$ and $|y_i-y_j| < \epsilon$?
it isn't, my point is that it doesn't hold
I mean divisor is canonical iff principal in g=1 case.
13:42
@shintuku $|x_iy_i-x_jy_i+x_jy_i-x_jy_y||$
ah that works, thanks!
13:53
I'm reading Evans Semigroup theory (7.4.1). We have $\mathbf{u}'(t)=A\mathbf{u}(t)$ and $\mathbf{u}(0)=u$. He said $A:D(A)\to X$ a linear operator.
Where $D(A)$ is a subspace of $X$ ($X$ is a Banach space.)
I don't understand few things 1) operator means same space to same space, 2) If domain is not full space then $A\mathbf{u}(t)$ may not be defined.
He said we'll write $\mathbf{u}(t)=S(t)u$ to display explicitly the dependence of $\mathbf{u}(t)$ on the initial value $u\in X$.
How to read this equation? I am asking this because I want to explain myself the meaning of $S(t)S(s)u$.
I don't understand why $S(t+s)u=S(t)S(s)u$.
By definition, $\mathbf{u}(t+s)=S(t+s)u$. Similarly, $\mathbf{u}(s)=S(s)u$ so now $S(t)S(s)u=S(t)\mathbf{u}(s)$. Here $\mathbf{u}(s)$ is our initial function. So we get another solution of the ODE $\mathbf{v}(t)=S(t)\mathbf{u}(s)$.
14:10
Can anyone please help me with this : math.stackexchange.com/questions/4672608/… ?
Sorry my bad
en.wikipedia.org/wiki/Cauchy%27s_functional_equation Somehow we have to get to Cauchy's Functional equation
@Franklin if you assume $f(x) = cx$ the moment you start your proof, you will not have shown that there exists a $c$ that satisfies $f(x) = cx$
@shintuku I dont get what you mean by :"you will not have shown that there exists a $c$ that satisfies $f(x) = cx$" ?🥲
You have to prove that any function which satisfies the equation must linear.
What you are doing is that you're showing linear functions satisfy the equation.
@PNDas Yes, now that's a good one! My proof only shows, that if f is linear, it satisfies the conditions, but the point is, I have to show: Any function that satisfies the conditions, can't be anything apart from being a linear function, right?
14:23
no you have to give a specific $c$ that makes true the statement
@shintuku It's obviously true for any non-zero $c$ ?
so you have to start your proof with this: let $c$ be nonzero in $\mathbb R$. then, we have blahblah
@Franklin why non zero
@shintuku yes, I agree...
@PNDas I said, it just like that. Now, of course for non-zero c it's valid, so I asserted it. Btw, your point is good, but don't you think, you are interpreting in a different way?
$f(x)=\frac1t\int_0^t(f(x+y)-f(y))\,dy\implies\frac1t\int_0^t(f(x+y)-f(y)-f(x))\,dy=0$. So taking derivative both sides, we get f(x+t)=f(t)+f(x).
14:27
I think the interpretations of this question, might vary.
@Franklin you can't assert it is obvious and be done with it, the point is that you have to prove it
Suppose you're given nonzero $c$. prove that you have $f(x) = cx$
if it is valid for nonzero $c$ the proof is complete, but this is what you have to prove
pndas is giving good commentary
In my last comment, the equation is true for any, $x\in\mathbb R,t>0$.
In the Cauchy's equation, the equation is true for any $x,y\in\mathbb R$. So our case is not exactly same as the Cauchy.
but I believe it can be proved in the same manner.
@shintuku I have edited the solution. I hope this fixes the issue you suggest...
you've assumed $f(x) = 2x$, instead of showing, given $2$, we have $f(x) = 2x$
if this is your first time doing an existence proof, try something easier. prove $\forall n \in \mathbb N \exists k \in \mathbb N$ such that $n < k$
@shintuku ok, now I get it. I am just rollbacking it, and and I will add : for a given c\in R, if f(x)=cx, then we see ....
I think this will solve this?
@shintuku no, its just that I am confused with the language of this problem
14:37
then you will prove a different theorem, not existence
martin R has commented valid proofs on your question
Like I said, we want to prove that "any" solution is of "this" form. What you're showing is that functions of "this" form are solutions.
@Franklin try proving $\forall n \in \mathbb N \exists k$ such that $n < k$, and $\forall q \in \mathbb Q \exists m,n \in \mathbb Q$ where $m,n \neq 0$ and such that $mq+p = 1$
What happened to $n$ and where did $p$ come from in the second statement?
14:59
woopsie, $\forall q \in \mathbb Q \exists m,n \in \mathbb Q$ where $m,n \neq 0$ and such that $mq + n = 1$
15:09
any interesting examples of functions that map function spaces to function spaces
15:30
@shintuku What function spaces are you familiar with?
i haven't done any work on function spaces yet, i was just curious about the possibility of a space made out of functions
Well, define some yourself :)
@TedShifrin, do you know semigroup theory?
Oh okay thanks
15:44
@leslietownes I have difficulty showing that F is continuous.
argh i need more topology to make spaces out of things
unless i make it a vector space
We have $F|_{S^1}= f 1_{S^1}, F|_{\{0\}}= c_{x_0}$
@shin Nothing wrong with vector spaces, but they'll be infinite-dimensional in general.
Pasting lemma doesn't apply to directly show F is continuous.
soon i will be done with eigeneverything and will have gotten up to power series in real analysis... soon i move on to the juicy maths
15:52
koro: i will be idle very shortly but i think the idea would be to investigate conditions on the map A: I x [0, 2pi] to X that allow you to meaningfully define a map from D to X given by g(r e^{it}) = A(r,t) and check that maybe those conditions are satisfied in this case
i.e. we're not just randomly mashing together prescribed values on two subsets here
Leslie, I tried to do it like this now. To show continuity of F, I used the definition of coontinuity: take any open set U in X and show that $F^{-1}(U)$ is open.
it isn't clear to me why you'd necessarily want to go right back to the definition of continuity to do that, but OK
use a certain quotient map $S^{n-1}\times I\rightarrow D^n$
15:55
i have to be idle now
geometrically, the point is that the cone over a sphere is a disk
I should also comment on well-definedness of F. But suppose for now that it is well defined.
@PNDas I get it. But, when we arrive at the point $f(x)+f(t)=f(x+t),$ then $t>0$ how do we solve this function $f(x)$ ?
@leslietownes I really wanted to avoid that but couldn't come up with an alternative.
@leslietownes Thank you so much for the help and reviewing the solution also :-).
I'm specially happy that I came up with that last set theoretic bit in the last part of the image.
@shintuku I get what you mean. But then again, how to solve and say, $f(x)+f(t)=f(x+t),$ the only solution is $f(x)=cx.$
15:58
where I use the fact that arbitrary U of open sets is open.
@Franklin you deleted the thread so i don't know what the statement is
@Franklin This is called Cauchy functional equation (not sure if you have to assume continuity of f also) but you may look up this term.
@Koro yes, continuity should be assumed. I should've mentioned it...
See this
3
Q: Proving continuity of a function that satisfies Cauchy functional equation and is bounded on an interval.

KoroSuppose that $f: \mathbb R\to \mathbb R$ is a function that satisfies Cauchy’s functional equation that is, $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb R$. It can be shown that there exists some constant $c$ such that $f(x)=cx$ for all $x\in \mathbb Q$. Now, one theorem says that if $f$ is bounded...

4
Q: Show that $f(x)=cx$, $\forall x\in\mathbb{R}$.

Sanket BiswasLet $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Suppose $$f(x)=\frac{1}{t}\int_0^t(f(x+y)-f(y))dy$$ for all $x\in\mathbb{R}$ and all $t>0$. Then show that there exists a constant $c$ such that $f(x)=cx$ for all $x$. My approach: It is given that $f:\mathbb{R}\to\mathbb{R}$ is continuo...

@shintuku I was talking about this link given by Martin R
16:03
what is the step you're asking about
@shintuku how to conclude that the only solution of $f(x+y)=f(x)+f(y)$ in this thread, is f(x)=cx
@shintuku @Koro Can we use this as a standard known fact?
.
An additive function is rational homogenous i.e a linear map on the linear space $\Bbb{R}$ over $\Bbb{Q}$.
@Franklin where is this happening, what step are you asking about
you quote the thread, what step in this thread are you asking about
Then continuity or continuity at a point or continuity at $0$ , monotonicity forces a linear extension over R.
In fact an additive function which is lebesgue measurable is linear i.e f(x) =ax.
17
Q: Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$?

Sourav GhoshLet $F\subset\Bbb{R} $ intersect every uncountable $\mathcal{F}_{\sigma}$ set. $B\subset \Bbb{R}$ is said to have the property of Baire if $B=U\triangle M$ where $U$ is open and $M$ is meager. Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the ...

16:32
@shintuku Look at the first line
That's what I am talking about...
what's wrong with it
multiply $f$ by $t$ and differentiate
@shintuku Nothing's wrong
The thing I am saying, is
how to conclude that the only solution of $f(x+t)=f(x)+f(t)$ in this thread, is f(x)=cx
@Franklin Subject to what hypotheses?
@XanderHenderson All the hypothesis, given in the thread I posted
This dangling thread just gets more and more snagged.
16:41
@Franklin Either my question has a simple answer, which you could just give, or it does not, in which case, I have no interest in attempting to read through the two or three questions and images posted above which might give me the context I requested. Nevermind.
@Franklin First find f on 0. Then on all integers, then on rationals. Use continuity now.
@TedShifrin I know whats dangling, but I dunno what's snagged!
$f$ on $0$? What kind of English is that?
I'm here to teach English vocabulary, not math, @Franklin.
@TedShifrin Queen's English.
@XanderHenderson Simple answer: f is continuous. I suggested you to read the question in the thread cause I thought it would be convenient for you...
16:43
@Franklin Oh, continuity makes this relatively simple. Can you show that $f(x)= cx$ on $\mathbb{Q}$?
I will only remark that when I came in ages ago and looked at the posted question, I noted that the necessary hypothesis of continuity was missing ... and I moved on.
@TedShifrin Really?!?😂😂😂😂
at
Erp... not constant
constant times $x$... X(
@XanderHenderson f(x)=cx
16:44
Typing faster than I'm thinking.
@XanderHenderson That's becoz ur a pro!!!!( I am not joking)
The question remains: can you show that it is true on the rationals?
@Koro Can I directly conclude, that $f(x)=cx$ ?
@Franklin what is directly?
@Koro as a standard fact?
It appears to me as a popular result!
16:50
@Franklin f' is constant
With the name: cauchy functional equation
that's what you use to conclude $f(x) = xc$
@shintuku but I dont understand, this part...
work it out, draw a picture
I mean, how is it so?
@shintuku ok
16:51
@shintuku $f'$? The only hypothesis I see given is that $f$ is continuous. $f'$ is not a priori guaranteed to exist. However, if $f$ is differentiable, then this makes life a bit easier.
the quoted picture includes proof $f$ is differentiable
i haven't checked it personally, just layed out the structure of the proof for franklin
@Franklin once you understand that you can go from $f'$ constant to there is $c$ such that $f(x) = xc$, you'll understand that you need to prove $f$ is differentiable to get $f'$ constant
I am very confused about what is actually being asked.
My understanding of the question is that @Franklin is trying to show that if $f(x+y) = f(x) + f(y)$ for all $x,y\in\mathbb{R}$ and $f$ is continuous, then $f(x) = cx$ for some constant $c$. Is this not the case?
@shintuku I understand it
@Franklin good, so do you understand the picture now?
@XanderHenderson dunno, i'm explaining the quoted thread and picture
@shintuku yes, but not the proof of differentiability in the picture I posted
16:58
@Franklin alright, what step
in any case, if you understood the last part, you should be able to figure this out yourself
@shintuku Yes, that is why I pinged @Franklin, and not you. I am trying to get him to clarify the question. But most of my input here has been ignored, so maybe it is time for me to bow out.
@XanderHenderson yeah they don't understand what they don't understand so i'm only answering when they give a very definite question
@Xander We old guys have to get used to being ignored.
Last try: @Franklin as I understand it, you are trying to show that if $f$ is a continuous function on $\mathbb{R}$ and $f(x+y) = f(x) + f(y)$ for all $x,y\in \mathbb{R}$, then $f(x) = cx$ for some constant $c$. Is this correct?
@TedShifrin Bah! Kids these days! GET OFFA MAH LAWN!
@XanderHenderson soooo true!!!!
17:05
@Franklin Okay, that seems to be a "yes". So, can you show that if a function $f$ has the property that $f(x+y) = f(x) + f(y)$ for any two rational numbers $x$ and $y$, then there is a constant so that $f(x) = cx$ for any rational number $x$?
did anyone post the master list related to this yet?
140
Q: Overview of basic facts about Cauchy functional equation

Martin SleziakThe Cauchy functional equation asks about functions $f \colon \mathbb R \to \mathbb R$ such that $$f(x+y)=f(x)+f(y).$$ It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for s...

@XanderHenderson i agree
@Franklin I wasn't asking you to agree or disagree. I was asking if you could prove this fact.
not to derail any attempts at live interaction, but when something has been done in so many variations on MSE as much as this has
However, if by "i agree" you mean "yes, I can prove this fact", then apply continuity. Suppose that $x \in \mathbb{R}$. Choose a sequence $(q_j) \subseteq \mathbb{Q}$ with $q_j \to x$. What can you say about $\lim_{j\to\infty} f(q_j)$? What does this have to do with continuity?
17:11
I rarely spend time searching the site.
Although now if I'm confident that something's been answered already (especially by me), I do search so as not to be thrown in MSE jail.
@TedShifrin With my moderator hat on, I appreciate that. I really don't enjoy sending moderator messages to people. :/
What I think is: he needs to try to prove it by himself. Instead of looking at others answers. He should take some time. It'll automatically get cleared up.
The math world is full of people who look for answers in books/on-line rather than struggle themselves to learn.
i.e. look inwards, the external world is fake
everything is an illusion
I prefer to think of it as allusions.
17:22
@TedShifrin I do this myself, what can I say about others. When I study for semester exams, if a problem takes too much time just look for solutions online. (This gives their pov, and different ways to the problems). Otherwise, I take my time.
I grew up in the dark ages (or even earlier, if you listen to @leslie). We had no internet to pilfer from. And searching the library for non-obvious journal articles or books was beyond an art form. So I grew up figuring things out for myself (or with my colleagues).
@TedShifrin that had it's own benifits, and disadvantages as you admit, if I am not mistaken...
Overall, I think the modern era has led to plagiarism and a lack of hard work.
no there are no disadvantage. the polio vaccine made people weak
Certainly that's true on MSE.
17:26
as ted says, the polio vaccine made people on MSE weak
No.
How many times do I have to tell people in here to stop and think and actually read and understand what I've already said two or three times? Ugh.
@TedShifrin Till 11th-12th, we didn't have access to internet. So we also experienced same situation.
@TedShifrin A good friend of mine studied under VS Varadarajan. He describes his phd program as being one day after another of being screamed at in an Indian accent: "YOU ARE NOT THINKING! WHY AREN'T YOU THINKING!?"
I will admit that it has changed the framework of research. It's much easier to find out what's known about certain questions. Years ago a graduate student who was auditing my undergraduate differential geometry class asked me a cool question to which I didn't know the answer. I asked a colleague. He didn't know. I emailed a knowledgeable expert. He didn't know. I then googled (duh) and found a paper by Arnold on exactly that topic.
I am now imagining you yelling in an Indian accent.
17:28
i agree with ted, although the existence of a resource like MSE also allows conversations, when they do occur, to be more focused. sometimes instead of working all of the context up from first principles in a long back and forth, one can just link to a question/answer, and say "where in this do you get stuck." which puts a little bit of work on the asker to engage on their own the way they might with a textbook or clay tablet like ted used to use.
I have no Indian accent.
@TedShifrin I don't expect that you do. But it feels right.
My one Ph.D. student would tell you that I was actually surprisingly encouraging and gentle. Perhaps the thousands of undergrads will say something else.
@TedShifrin Yeah, you seem fairly patient here. I can't imagine that you would treat a grad student differently.
Even when you are obviously frustrated. :D
That one Ph.D. student has taught for her career at a 2-year (which then merged with 4-year) college. Even as a grad student, she was a phenomenal teacher.
17:31
@TedShifrin The thing is: Probably you'll admit, life grew much faster and faster. Now, everyone is drowned with exams and time-frames. Everyone, including me, prefers things to work out with some hours at max.
Good for her. We are set to admit our first cohort of 4-year students in the fall.
No, Franklin, we were always drowned. You are too self-indulgent.
You can make all sorts of excuses.
@TedShifrin ugh...maybe...I am depressed...
@XanderHenderson Btw, I proved it. Yes, the part to show, the derivative is constant, just needs some manipulations. That's it... but as I gather, this functional equation is a very standard one
We had plenty of depression, too. I am not minimizing stress, but each generation thinks it's uniquely plagued.
Apparently, chat doesn't want to post ...
I will admit that Covid and imminent world war are things that I didn't deal with as a student, although I almost got drafted to go to the Vietnam war and would have fled the country had I been.
@Franklin You haven't answered my question, which has nothing to do with the derivative.
@Franklin When I was an undergraduate, I was told that a 12 credit hour load was considered a "full load", and that a "full load" should be thought of as the equivalent of a full time job, i.e. 40 hours per week. So every hour of lecture was supposed to correspond to at least two hours of work outside of class (homework, office hours, study groups, etc).
I typically took at least 15 hours per semester. My worst semester as an undergrad was 23 hours.
I was always drowning in exams, papers, and problem sets.
That is the life of an undergrad. But you have to do that work if you are going to learn.
17:38
You both are awesome.
@TedShifrin ohh 😮😮
In fairness, many of today's students do tend to try to (or have to!) work full-time jobs while they go to school. So that just is untenable.
I wish I could get my students to understand that they aren't going to just passively absorb material in lecture, and that they have to actually work outside of class. :/
I thought that many citizens won't even have known about the war.
@TedShifrin Yeah, I don't envy them.
17:39
considering that US military has such a huge budget
Trying to work a full-time job and go to school is the equivalent of working at least two full time jobs. I don't know how anyone does it.
I was one of those radical Democrat war protestors.
@XanderHenderson if you are talking about the rationals portion....Yeah, I proved it
31 mins ago, by Xander Henderson
However, if by "i agree" you mean "yes, I can prove this fact", then apply continuity. Suppose that $x \in \mathbb{R}$. Choose a sequence $(q_j) \subseteq \mathbb{Q}$ with $q_j \to x$. What can you say about $\lim_{j\to\infty} f(q_j)$? What does this have to do with continuity?
and with that why would they draft you (around that time you would be probably an undergrad student).
17:40
@Koro when the US had an active draft, people of drafting age definitely did know about the war :)
maybe these days, it's more of a TV thing for a lot of people, but not then
@XanderHenderson ok, ok, that's an approach. But just a second
ted successfully avoided the draft by reminding his local draft board that he had taught most of their parents calculus and had not been of draftable age since before the US even existed
Can anyone tell me the learning road map of functional analysis? Real analysis (including measure theory) --linear algebra-->complex analysis-->metric space---> general topology--->normed space---->topological vector space.
1
Q: To prove the existence of a constant $c$ in a function

Iti ShreeLet $f: R \to R$ be a continuous function.Suppose $$ f(x) = \frac{1}{t}\int_{0}^{t} (f(x+y)-f(y))dy$$ for all $x \in R $ and $t>0$. Then show that there exists a constant $c$ such that $f(x)=cx$ for all $x$ At first I want to ask how shall I integrate the function give above? And also how am I s...

@XanderHenderson If you take a look at this answer, that exactly what I meant about the derivative thing earlier
Simple question....
The second part is to prove that: assuming that every map $S^1\to X$ can be extended to $D^2\to X$, we must have $\pi_1(X,x_0)=0$ for every $x_0\in X$.
17:45
Add on: @TedShifrin and @XanderHenderson you both have become my ideals now....My god! Gifted souls(Not joking or being sarcastic)
@Franklin Please stop linking to other things, and do some work for yourself. I asked you if you were trying to prove that if $f(x)+f(y) = f(x+y)$ and $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}$. You said "Yes". I asked if you could prove the result for $x \in \mathbb{Q}$. You said "Yes". So, given that the result is true in $\mathbb{Q}$, how do you extend to $\mathbb{R}$?
The approach that I would take is outlined in the question I asked you: chat.stackexchange.com/transcript/message/63312417#63312417 .
Why aren't you answering that question?
Or is there some other place in this conversation where we have miscommunicated?
I don't see how the linked question is relevant to the approach that I have been trying to guide you down, given the answers you have provided to my previous questions.
@XanderHenderson this is the case. Alas!
So I take [f] in $\pi_1$ at $x_0$. Here, we have $f: I\to X$. We want to change the domain to $S^1$. I wrongly thought earlier that $I\cong S^1$ considering compactness. But then I realized that it is not so and recalled that $S^1\cong I/$~, where ~ identifies the end points.
@XanderHenderson ok, to answer this
@SouravGhosh Definitely Lebesgue integration, not just measure theory. Complex analysis probably not so much at the beginning.
17:48
@XanderHenderson it's just f(x), correct me if I am mistaken?
With that and using property of quotient maps, there exists $\tilde f: S^1\to X$ such that $\tilde f\circ q=f$, where $q$ is the quotient map $I\to S^1$.
sourav: i don't know that there is a single road map that would work for everybody, but all of those are helpful. i think the most central prerequisites would be real/complex analysis and linear algebra. not so much general topology (you can learn as you go). "normed space and topological vector space" is so close to functional analysis that it's maybe indistinguishable from it (FA contains a lot of the key examples)
This $\tilde f$ can be extended to a map on $D^2$ as per hypothesis. Let's denote this extension also by $\tilde f$.
@Franklin Why? And what does that have to do with $f(q_j)$?
I'm not sure how to go from here.
17:50
Compute the limit two ways.
@XanderHenderson it follows from continuity isn't it?
Atleast I did it like that
f(q_j)=cq_j \to cx as q_j\to x
This is the jist of it
The answer here does not explain a thing:
5
Q: Problem 1.1.5 from Hatcher's

MathematicalPhysicistI want your advice on my solution of this problem. problem Show that for a space $X$, the following three are equivalent: (a) Every map $S^1 \rightarrow X$ is homotopic to a constant map, with image a point. (b) Every map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$. (c) $\pi_1(X...

It makes me want to downvote it.
succumb to your darkest desires
@Franklin There you go.
So, $f(x) = c x$, no?
@XanderHenderson yes!
To be honest: you did half of the work
17:56
What are the current research areas of functional analysis?
@XanderHenderson The thing is: A set of rationals converge to a real , is possible, can be seen more clearly, if he has done a fair amount of real analysis course. But I have not done it so much. True, I know this as a fact up until now. But yes, this actually makes tge argument lot simpler
Maybe, as the real analysis course progresses in my college, I will grow more easy going with these
I don't know what definition of the real numbers you're using in your course, but I would take it as almost axiomatic that every real number is the limit of a sequence of rational numbers.
@XanderHenderson that's just another approach, which I cooked up initially...
I don't think of this as far along in a real analysis course at all.
@TedShifrin in our course, up until now, we are taught to define natural numbers using peano's axiom
17:59
Oh good grief.
So what are the real numbers?
@Franklin I suggested the same thing earlier. I asked you to find f on integers, then extend it rationals and then use continuity. There is a general result: If f and g are continuous real valued functions on R and if they agree on a dense set (Q is dense in R and so is R-Q for example) then they are equal.
And then we are creating integers
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