4:43 AM
Hello everyone, I am unable to access math.se website and I get a page which says the connection between SE servers and cloudflare is timed out. Is anyone facing this issue? I hope this is temporary.
paramanand: the main site loads for me right now. i haven't tried posting in weeks, but commenting seems to work as of a few minutes ago.
Maybe the issue lies with the cloudflare servers in Chennai, India. At least that's what the error page says. I checked on stackstatus.net and it says all the stackexchange sites are working fine.
Yeah, I’ve been on the site off and on all day.
Happy goose to Munchkin.
munchkin is turning 5 soon.
Time doth fleet!
She’ll soon be kicking her parents out.
5:06 AM
yes.
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Let $R$ be a ring with unit element, $R$ not necessarily commutative, such that the only right-ideals of R are $(0)$ and $R.$ Prove that R is a division ring. This was a problem from the book, "Topics in Algebra" by I.N Herstein from Chapter- 3 (Ring Theory) , Page number- 139 (2nd Edition). I tr...

I need some help with this excercise from Herstein's book.
I am able to access the site again. So it was more of a temporary and local problem.
6:14 AM
0

Let $R$ be a ring such that the only right ideals of $R$ are $(0)$ and $R.$ Prove that either $R$ is a division ring or that $R$ is a ring with a prime number of elements in which $ab = 0$ for every $a,b\in R.$ This was a problem from the book, "Topics in Algebra" by I.N Herstein from Chapter- 3 ...

I need some help with this excercise from Herstein's book.
6:46 AM
I would like to ask for (the slightest) hint to prove this: "Let $f_n$ be a sequence of continuous functions such that $f_n \to f$ uniformly on $(a,b)$. Show that $f_n \to f$ uniformly on $[a,b]$." I actually have an idea, but I don't know if it works. My idea is that $\lim_{n \to +\infty} \sup_{x \in [a,b]}|f_n(x)-f(x)|=\lim_{n \to +\infty} \sup \left(|f_n(a)-f(a)|,\sup_{x \in (a,b)}|f_n(x)-f(x)|,|f_n(b)-f(b)|\right)$.
Using that $f_n \to f$ uniformly on $(a,b)$ and that $f_n$ is continuous, I deduce that $f$ is continuous on $(a,b)$ and so $|f_n-f|$ is continuous on $(a,b)$, hence $\lim_{n\to+\infty} |f_n(a)-f(a)|=\lim_{n\to+\infty} \lim_{x \to a^+} |f_n(x)-f(x)|$.
Being $x \to a^+$, I can assume $a<x<b$ and so I can now interchange the limits because of the uniform convergence on $(a,b)$, so $\lim_{n\to+\infty} \lim_{x \to a^+} |f_n(x)-f(x)|=\lim_{x \to a^+} \lim_{n\to+\infty} |f_n(x)-f(x)| \le \lim_{x \to a^+} \lim_{n\to+\infty} \sup_{x \in (a,b)}|f_n(x)-f(x)|=0$ (because $f_n \to f$ uniformly on $(a,b)$).
Similarly, $\lim_{n \to +\infty} |f_n(b)-f(b)|=0$. Hence, $\lim_{n \to +\infty} \sup \left(|f_n(a)-f(a)|,\sup_{x \in (a,b)}|f_n(x)-f(x)|,|f_n(b)-f(b)|\right)=0$ because each one of the three sequences in the supremum tends to $0$. So, $f_n \to f$ uniformly on $[a,b]$.
Oh no, this doesn't work I'm afraid. By the continuity on $(a,b)$ I cannot deduce that $|f_n(a)-f(a)|=\lim_{x \to a^+} |f_n(x)-f(x)|$; I need the continuity at $x=a$ too.
putting aside any question about uniformity, can you show that the limit lim_n f_n(a) exists and is f(a) [and similarly for b]? what are the domains of f_n and where are they assumed continuous?
Sorry for not writing that, the domains are $[a,b]$ and they are continuous on $[a,b]$ for each $n\in\mathbb{N}$.
For the other thing you asked, I think that the reasoning I used above works. Since $f_n$ is continuous on $[a,b]$ for each $n\in\mathbb{N}$, we have $f_n(a)=\lim_{x \to a^+} f_n(x)$. If $x \to a^+$, we have $x>a$ and so by uniform convergence of $f_n \to f$ in $(a,b)$ I can exchange the limits and get $\lim_{n \to +\infty} f_n (a)=\lim_{n \to +\infty} \lim_{x \to a^+} f_n(x)=\lim_{x \to a^+} \lim_{n \to +\infty} f_n(x)=\lim_{x \to a^+} f(x)=f(a)$.
No wait, again the same mistake. I need the continuity of $f$ in $x=a$ now to say that $\lim_{x \to a^+} f(x)=f(a)$.

4 hours later…
11:02 AM
I see a lot of symbols in chat
Any idea to prove the quasi triangle inequality for this quasi-norm: $N(x,y,z)=(\sqrt {|x|}+\sqrt {|y|}+\sqrt {|z|})^2$
so the $\ell^{1/2}$ one
I'm not familiar with that, we use $\ell$ for spaces of sequences
its standard to use $\|x\|_p = \sum |x_i|^p$ for the quasi-norm in $\ell^p$ for $0 < p < 1$, just here its squared
here its squared but I don't think it'll be a problem
yes, that's true
11:08 AM
I guess this one isn't homogeneous though so that's why the square
it's homogeneous
but I can't prove the quasi triangle inequality
the one I wrote isn't because I didn't take power of 1/p
you have some inequality of the form $(x+y)^p \leq 2^{p-1}(x^p+y^p)$ for $x, y\geq 0$
it's proven in a chapter after this exercise, but it does the job
11:14 AM
$f(x) = x^p+1 - (x+1)^p$ then $f'(x) = px^{p-1} - p(x+1)^{p-1}$ and $f'(x)\geq 0$ for $0 < p < 1$ so that $f$ is increasing. Moreover, $f(0) = 0$.
thank you
does someone know about affine varieties?
@SineoftheTime yeah I mixed it up because you use $d(x, y) = \sum |x_i-y_i|^p$ to have a metric so it satisfies triangle inequality
what I've wrote proves that $(x+y)^p \leq x^p + y^p$ for $0 < p < 1$ and $x, y\geq 0$
But you still have squares so you need something like... $(x+y)^2 \leq 2x^2+2y^2$
yes, I' m working on the squares
this should go from Jensen's inequality and convexity of $x^p$ for $p > 1$
@ZaWarudo If $f_n\to f$ uniformly on $A$ and on $B$, then $f_n\to f$ uniformly on $A\cup B$. And $f_n(x)\to f(x)$ iff $f_n\to f$ uniformly on $\{x\}$. Thus if we show that $f_n \to f$ pointwise on $[a, b]$, then $f_n\to f$ uniformly on $[a, b]$ as well
@ZaWarudo is $f$ assumed to be continuous as well? If so, then you can prove that $f_n$ converges pointwise to $f$ on $[a, b]$
I assume so since otherwise the question wouldn't make much sense
The essential thing to use here is the following inequality: $$|f(a)-f_n(a)| \leq |f(a)-f(x)| + |f(x)-f_n(x)| + |f_n(x)-f_n(a)|$$
where $x \in (a, b)$
you can pick $N$ large enough so that for $n \geq N$ the middle term is small, and the other two terms will be small from continuity, by picking $x\in (a, b)$ close enough to $a$

2 hours later…
1:11 PM
is it true that if $u : \mathbb{C} \rightarrow \mathbb{R}$ is a harmonic function for which $0$ is not a critical value, then $u^{-1}{0} \neq \emptyset$ has finitely many components? I know that there should be at most countably many components.
nvm, that was a silly question
what I mean to ask is this, suppose $u$ is harmonic on some open subset of the Riemann sphere, $a < b$ are non-critical values of $u$. Consider the regular domain $\{ a \leq u \leq b \}$. Is it possible for this things boundary to have infinitely many components?
also we may assume its a compact regular domain.
so the boundary consists of a bunch of analytic jordan curves
2:09 PM
Is $f:\Bbb R\to\Bbb R$ defined by $$f(x)=\begin{cases}\sin(x)&x\in\Bbb R\setminus\Bbb Q\\\frac1{|x|+1}&x\in\Bbb Q\end{cases}$$ measurable?
oh, again the answer is certainly yes.. because the open set $u$ is defined on to begin with could be disconnected..
im guessing even if it was a domain, the answer could still be yes, but an obvious example showing this may be doesnt come to mind
@PNDas yes it is
@porridgemathematics How to show that?
use the definition of measurability
in general things like this will always be measurable, its only when you start constructing things using constructions that require uncountably many intersections do you need to be cautious
and usually for things like that there are ways to approximate your object by sequences of measurable functions
$f^{-1}(O)$ is measurable for all open set $O$ in $\Bbb R$. If $O\subset [-1,1]^c$ then $f^{-1}(O)$ is countable or empty so its measurable but otherwise I'm not able to see it.
@porridgemathematics I agree.
continuous functions are measurable, your function has two branches, and its measurable on both of them
now use the definition of measurability
2:18 PM
Ah yes.
They also asked is it true that $f(E)$ is measurable for every non empty bounded open subet $E$?
My guess is false but how do I show that?
I guess I have to consider $[0,2\pi]$ then it's image is of measure 1. so there exists a non measurable set in the image
and take its preimage?
i think its true
@porridgemathematics Why?
because both of these functions are basically invertible
i.e. they both essentially have continuous inverses (so measurable inverses)
Sorry I wrote it wrong. They also asked is it true that f(E) is measurable for every non empty bounded subset E ?
ah, this is likely to be false
if E doesnt even need to be measurable
2:31 PM
Yes
so I think you could prove this is false as follows: consider [0,pi/2], take a non-measurable irrational subset of it, and take f of that subset, f restricted to [0,pi/2] - Q is bi-measurable, so the image wont be
@porridgemathematics What I said is also correct right? chat.stackexchange.com/transcript/message/64504683#64504683
it can work, once you establish that f(G) is measurable whenever G is
its simpler to just establish that for one of the branches where its invertible imo
g2g
2:49 PM
Morning all! I'm just learning some formal theory about polynomials for the first time and my book alludes to but does not out and out say if the following is true, so I am guessing:
That is the lemma as I have guessed it. Is it correct?
Oops, should be $\leq$ in both
@EE18 with that correct it is correct
well the "if and only if" is a bit vague
I see, thank you. The part I am in particular worried about is the "if and only if". The book only states the "if"
it's true that $R$ has no zero divisors if and only if for all polynomials p,q the inequality holds
I guess you're saying for the "only if" I should have a "for all p,q"
@EE18 yes, exactly
if you put that it's correct with an if and only if
2:52 PM
OK, I will just rewrite and put back here for verification if that's OK
sure
Probably there are nicer ways to write it, which gives me a lot more respect for textbook authors who do so so easily!
there is another variation of this: in (b) you have equality as soon as the leading coefficients are non zero divisors, even if $R$ has some other elements which are zero divisors. But what you wrote is correct
Noted, I will put that in my notes right now. Thank you!
When you say leading coefficients are nonzero divisors, can we further weaken that to say "even if they are zero divisors, as long as the other is not an element which is such that $ab = 0$"?
That is, for any two specific $p,q$ we have equality if and only if the leading coefficients are not such that $p_{\deg{p}} q_{\deg{q}} = 0$.
3:18 PM
Um, check the second one.
Oof, yes I missed a plus sign. That's what you mean right?
Rather important, yes.
I missed it twice!
Thanks for catching that Prof. Shifrin
You could do composition ….
Sounds painful! I wonder if they exclude that just because composition of polynomials doesn't come up so much in analysis proper? But I defer to you, I have no clue if it does or doesn't
3:33 PM
Also, am I right to be suspicious about informality as in the sums without upper bounds in (8.14) and (8.15) here:
That is, they should have bounds in terms of the degree of the given polynomial, no?
I wrote something like that instead for comparison to (8.14)
why deg $p+q$?
It should be max$\{\textbf{deg} p, \textbf{deg} q\}$
3:49 PM
That would not be correct, because it could be strictly less than that maximum right?
Or I guess it would be with leading coefficient equals zero
True true, I see what you mean
4:04 PM
You can cancel out the highest degree stuff, so not equal.
Per Soumik's suggestion, I went with the above. Is that what you mean Ted?
Personally, I don’t write the formulas out like this. I look for the highest degree term. You still haven’t discussed why the degree may be <.
The strict inequality would obtain iff $p,q$ have the same degree and their leading coefficients are additive inverses of each other, right?
I agree that to look for highest degree term is good, but I guess I'm trying to be general and to write something down which holds both for the equality and inequality case, since $p,q$ are arbitrary here
Hopefully that makes sense?
Can anyone give a hint for how you would argue for the equality (the starred * equality) of the following expressions in some commutative ring $R$. That is, how to "prove" the Cauchy product is the correct expression. I'm thinking induction on the degree of one of the two polynomials?
$$\left(\sum_{j=0}^{\deg (p)}p_jx^j\right)\left(\sum_{i=0}^{\deg (q)}q_ix^i\right) = \sum_{j=0}^{\deg (p)}\sum_{i=0}^{\deg (q)}p_jq_ix^{j+i} \stackrel{*}{=} \sum_{n=0}^{\deg (pq)}\left(\sum_{j=0}^n p_jq_{n-j}\right)x^n.$$
Wikipedia and MathSE answers always talk about infinite series and how essentially this is the definition of products in such cases, but I am looking to prove it simply from the ring structure in the "finite sum" case.

2 hours later…
6:03 PM
Is anyone familiar with Teschl's book Ordinary Differential Equations and Dynamical Systems? I have a question about his definition of a sub solution. It's quite short. Let me know if you are interested :)
6:59 PM
I'm doing the exercise above. I managed to show (a), and (c) follows from (b). I've also noticed that (b) <=> (d). I could prove (d) by proving that $\mathbb Z[(1+\sqrt{-3})/2]$ is a PID (by showing it's Euclidean), and then I'm done, however I don't think this is how the exercise is intended. Does anyone see how (b) follows from (a), assuming that the exercise is set up logically?
I have an idea
I should focus on the second part of (a) I believe for the solution
(which I initially got wrong)
7:34 PM
i have no idea but yeah my first question would be does the case of equality in (a) have anything to do with the cases identified in (b)
still working on that x'D

2 hours later…
9:15 PM
@ShaVuklia Do let us know if you work it out.
@F.White sure will do (so far no luck, but I'm going to discuss this exercise with two fellow students tomorrow, so hopefully that will yield something)

1 hour later…
10:19 PM
@ParamanandSingh Hello, I needed some guidance here. If possible, can you leave a comment on my answer to confirm whether my answer to this question is correct conceptually rather than algebraically? How consistent is my solution? Does the solution contain a trivial conceptual error? Thank you .
didn't someone here say something about the fact that the faculty to admin ratio shrank in the past couple of decades?