Mathematics - 2022-05-15

leslie townes

00:45

we're running the air conditioner for the first time this year.

trying to get the second floor below 80.

PM 2Ring

I love almost integer stuff. It's interesting that there are lots of examples involving $e$, the most famous ones coming from the Heegner numbers. en.wikipedia.org/wiki/Almost_integer It's pretty easy to make quadratic ones, using Pell's equation.

Also, I hear the 4th root of (9^2 + 19^2/22) is pi.

leslie townes

oh, that's evil.

even by xkcd standards.

PM 2Ring

One of the pi guys (Borwein ?) did a similar thing involving cosh, IIRC. I should be able to find the anecdote on MO...

leslie townes

oh yeah, i've read that anecdote. it was one of the borweins.

PM 2Ring

01:05

John Baez has an article about that stuff: Patterns That Eventually Fail

Ted Shifrin

@leslietownes It's not that bad here.

leslie townes

it's muggy here.

Ted Shifrin

Well, it's because munchkin is so full of hot air.

leslie townes

SD always has better weather. it's the best weather on earth.

PM 2Ring

I think the Borwein example involved the integral of a product of sinc terms, not cosh. More examples here: mathoverflow.net/questions/15444/…

leslie townes

01:15

we have a coastal pine tree nearby that reminds me of SD.

PM 2Ring

We've had a couple of days with no rain (or little rain), but they've been very rare for months. At least the weather is starting to get cooler, so it's merely humid, not muggy.

Found it!

17

A: Integral of the Sinc product

Borwein integrals As Eckhard wrote in comments B[n] is the n-th Borwein integral. (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k])...

As a prank, Jonathan Borwein reported this to Maple, claiming there was a bug in the software. Maple computer scientist Jacques Carette spent 3 days trying to figure out the problem. Then he realized: There was no bug! That's what these integrals really equal!

leslie townes

that's the anecdote. you found it.

sadistic.

PM 2Ring

Perhaps. I like to think of it as an "interesting" teaching technique. ;)

Ted Shifrin

01:38

"Interesting," huh?

PM 2Ring

I'm trying to write a program to calculate the time of sunrise & sunset. Some of the equations I'm using come from this paper from 1989. doi.org/10.1093/mnras/238.4.1529 Unfortunately, their equation for sidereal time has a sign error. It took me a few days to fix that. :( But now I'm wondering if there are other errors in smaller terms that are harder to notice. I can check my results against results from JPL, but that doesn't really help much with the smaller terms.

@TedShifrin Well, I bet Jacques Carette now knows a lot more about those integrals than he otherwise would. :)

robjohn

02:03

@PM2Ring Back in the late 80s or early 90s, I wrote a program that would print out the sky including moon and planets. It was pretty accurate, but the moon position was a bit off since I used the Earth as a sphere.

I had about 880 stars, hand input from the Nautical Almanac and Astronomical Ephemeris.

I was wrong, it was written for the PET, so it was in the early 80s, maybe late 70s

copper.hat

loved the pet

was supposed to go to healdsburg winetasting with friends, but i met another friend yesterday who just texted that he tested +ve, so i had to cancel my end of the trip :-(

robjohn

I got an extra graphics board that had 32K of RAM to use and 2 PROM sockets that I could use for some 3-D graphics software I'd written.

copper.hat

cool!

PM 2Ring

Nice work! This site has some amazing algorithms. neoprogrammics.com/index.html Accurately compensating for the effects of the Moon isn't easy. Here's the IAU2000 stuff for computing the effects of nutation on longitude & the obliquity of the ecliptic: neoprogrammics.com/nutations/nutations_1980_2000b I'm definitely not putting that stuff in my program!

robjohn

@copper.hat It was that software, I think, that got me the job at Apple.

copper.hat

02:11

@robjohn I can see why.

PM 2Ring

Lunar theory has improved significantly, especially in the last 20 years, thanks to refinements in Lunar Laser Ranging.

copper.hat

plus the 6502 was used in the II

robjohn

@PM2Ring Yeah, I'm not real interested in trying to resurrect and improve that program. It lies on cassette tapes and printed on paper.

copper.hat

in Basic?

robjohn

@copper.hat Yeah, I'd also written an assembler for the 68000 and upgraded to 68030+MMU

copper.hat

02:13

i know its clunky by today's standards, but it was a great leaning vehicle

wow,

robjohn

@copper.hat The PET software was in BASIC

PM 2Ring

JPL uses a fairly sophisticated model for the Moon's interior, but they also fit their ephemeris calculations to a lot of actual data.

robjohn

@PM2Ring I'll leave that to them ;-)

copper.hat

the discovery that you could peek & poke opened up a whole world of exploration for me. we had no documentation at all, and coming across such stuff was practically impossible. byte magasine had some good stuff from time to time

PM 2Ring

It's still fun to compute orbits, but there's no way to compete with JPL. They integrate the equations of motion to the level where they model all the major bodies of the Solar system down to the top 340 asteroids as individual bodies, plus they have access to a huge amount of terrestrial & spacecraft observations for data fitting.

robjohn

02:18

@copper.hat That is how I wrote the 3-D software on the PET 6502 code POKEd in a BASIC program.

@copper.hat Loved BYTE

@PM2Ring If it were anyone but JPL, I'd say "get a life"

PM 2Ring

The best lunar theory from a couple of decades ago, ELP didn't bother trying to model actual observations directly, they took all their data from the then-current JPL ephemerides.

copper.hat

i wrote a little plot program that managed 'double resolution' plots. the pet had 16 characters with each quadrant on/off.

not terribly exciting, but i was so proud of that little routine

all sorts of things suddenly became interesting, scatter plots least squares, etc

PM 2Ring

I thought the PET was wonderful. I never owned one, but I played with them in the shop. A bit later, I had an Osborne

Osborne 1

The Osborne 1 is the first commercially successful portable computer, released on April 3, 1981 by Osborne Computer Corporation. It weighs 24.5 lb (11.1 kg), cost US$1,795, and runs the CP/M 2.2 operating system. It is powered from a wall socket, as it has no on-board battery, but it is still classed as a portable device since it can be hand-carried when the keyboard is closed. The computer shipped with a large bundle of software that was almost equivalent in value to the machine itself, a practice adopted by other CP/M computer vendors. Competitors quickly appeared, such as the Kaypro II. ��2...

A few years later, I got an Amiga 2000. In fact, I still own one (not my original), which I use as a monitor stand. It might still work, (assuming its electolytics haven't dried up), but I no longer have an analog monitor for it. I guess I could get an adapter...

leslie townes

this is the best one.

Sphinx (home automation system)

The Sphinx (Russian: СФИНКС, romanized: SFINKS) was an experimental Soviet project for a home automation system, commissioned by the State Committee for Science and Technology and designed by Dmitry Azrikan, in collaboration with A. Kolotushkin and V. Goessen, in 1987. Sphinx, an acronym for Super Functional Integrated Communication System (Russian: Суперфункциональная интегрированная коммуникативная система), was intended to be an ensemble of modules that would allow consumers to easily interact with information systems. The home environment, as described in a 1987 issue of Soviet magazine Technical...

PM 2Ring

@copper.hat You could do a similar double res trick on the Osborne.

anak

02:37

Alessandro was the set theory fellow, right?

Calvin Khor

@TedShifrin I took a stats class with some math in it and i recall "manifold" and "chi squared test" being uttered in the same sentence ...

anak

the reasons to use the chi square test are manifold

robjohn

03:48

@copper.hat The graphics board I got (with the 32K RAM) had 320x200 pixel resolution, the full resolution of the screen!

I wrote a printer driver that used a three screen width buffer to print on the 960 dot wide dot matrix printer that the company my dad worked for made.

that was a 24K buffer (big on a 64K address space).

Thank heaven that the pixels were only 1 bit (black and white, no shades of gray)

Ted Shifrin

04:08

Is this ancient history being retold? :D

PM 2Ring

Some of those ancient ideas are still in use (eg, GIF is still very common for animations on the Web). Or are operating at a smaller scale on modern hardware, doing fun things with sub-pixels.

Modern high-speed printers are capable of much more accurate ink placement than the old printers. So some techniques that were rejected decades ago are now practical. Also, there are some superior image processing algorithms that were rejected years ago because they were too slow &/or consumed too much memory that are now practical on modern computers.

copper.hat

@robjohn impressive

robjohn

04:23

@TedShifrin yes, it's printed on a cave wall in dot matrix

Ted Shifrin

Versatile printer!

PM 2Ring

04:45

OTOH, there's a vast amount of incorrect image processing software. Even well-known proprietary software like Adobe Photoshop isn't immune.

copper.hat

04:57

what??? software with bugs???

Ted Shifrin

Spray for termites.

PM 2Ring

@copper.hat I wasn't talking about bugs. I'm talking about code that correctly performs incorrect calculations. ;)

Calvin Khor

and not a deliberate design choice? any examples easy to explain?

PM 2Ring

In some cases, yes, it's a deliberate design choice, sacrificing precision for speed. But in some cases, the bad algorithms result from inadequate understanding.

Admittedly, the mathematics of precision image processing is hard, and software developers may not have the necessary maths background. Also, important info related to colour hardware standards can be buried in technical articles &/or behind paywalls.

A simple example is the topic of gamma correction. Pixel colour values, eg in a PNG file use a non-linear encoding. But most image processing treats those values as linear. So any processes that rely on or modify the brightness of pixels end up with wrong values.

The wrong algorithm ends up becoming a "standard", and people using the software end up tweaking images by hand until they get something that looks good, not even realising that they're using bad software.

Calvin Khor

05:20

hm. why are standards even behind a paywall.

PM 2Ring

Good question. Tradition! You put standards in the appropriate industry journal, which all professionals will have access to.

Calvin Khor

:(

i tried to check some iso thing at some point but didnt want to pay 100 swedish krona or whatever

was it iso....anyway.

PM 2Ring

But if you're a software developer, not a colour TV engineer, you probably don't read that journal. And you probably don't even realise that you're missing important info about how pixels on a screen actually behave.

IIRC, there's some thing related to gamma that recent versions of Photoshop can do properly. But by default, they do it the old way, and you have to go into an obscure menu to make it use the correct algorithm.

Calvin Khor

whoever had the bright idea to clearly label definitions and theorems was a genius

PM 2Ring

I guess that goes back to Euclid. We call him a geometer / mathematician, but he was actually employed as a librarian at Alexandria.

Here's an excellent article on gamma, and the errors associated with incorrect gamma handling: blog.johnnovak.net/2016/09/21/…

Calvin Khor

05:33

> The popular graphics library <insert name here> I’m using handles gamma correctly.

lol

i guess they don't patch the older libraries for backwards compatibility?

or have i entered a cult of sacred knowledge

i think this explains why gimp's soft brush was uglier than photoshop's

or their gradient fills

PM 2Ring

Here's a great example from near the end of that page: incorrect gamma handling in 3D rendering is one of the main culprits behind the “fake plasticky CGI look”

Calvin Khor

eyyy theres a gradients section. totes called it

PM 2Ring

@CalvinKhor You generally can't fix this with a simple patch. You don't want old software to suddenly behave differently. But you could add new methods to supplement the old ones.

Calvin Khor

tho apparently ps is still wrong

PM 2Ring

> you have to go into an obscure menu to make it use the correct algorithm.

Calvin Khor

05:42

right

thanks i hate image processing now

PM 2Ring

This is kind of a legacy thing. If we used 16 bits per channel instead of 8 bits, we could have gotten away with linear encoding instead of gamma-adjusted encoding (although gamma would still have advantages of not wasting bits on very dark & very bright colours).

Image processing is fun, if you do it in a colour space that's perceptually linear (eg Luv) and you use enough bits. Using algorithms that were devised when 32k bytes was a lot of RAM is madness.

Calvin Khor

haha im joking i have lots of fun with image editors and this explained a lot of things

PM 2Ring

05:59

Oh, good. :) One of the refs at the end of that article is to poynton.com/GammaFAQ.html It's a bit old, with lots of stuff about analog TV & CRTs, but it's still mostly good, since modern display technology was designed to (sort of) mimic CRTs. Poynton's articles on gamma & colour are the de facto standards on this stuff for coders into image processing.

Calvin Khor

> Your connection is not private
Attackers might be trying to steal your information from poynton.ca (for example, passwords, messages or credit cards). Learn more
NET::ERR_CERT_DATE_INVALID

PM 2Ring

Oops. Try this one: poynton.ca/notes/colour_and_gamma/GammaFAQ.html

Calvin Khor

its still complaining. https such a mood killer

against my browsers better judgement im in

PM 2Ring

Yeah, the certificate expired 8 months ago. Oh well.

S.M.T

06:36

Q is to represent a displacement of. 40km west of north and east Of north

When the angle is turned clockwise , it’s value and direction is said to be positive

In 1st case , it’s positive

min 2nd case , the direction of angle is anti-clockwise.

Should it value be -30?

Calvin Khor

in the 2nd case, the angle is clockwise

sure, though giving the direction of the arrow is the same as putting a minus sign IMO

$\Huge⏰$

S.M.T

Oh sorry

I did opposite

@CalvinKhor is my 1st case wrong ?

I have drawn according to the question but it’s direction of angle is opposite to that of 2nd one.

Calvin Khor

the choice of the minus sign is not right or wrong[ outside an exam environment ], as long as you're consistent. If anti-clockwise is positive then the diagram for the first case is correct but what you said is wrong

S.M.T

Yes

Anti is positive

Clock is negative

Should I consider to think about about +30 or-30?

Calvin Khor

What do you mean?

S.M.T

06:46

In 1st case , direction is anti clockwise

We know that is positive direction

For 2nd case , direction is clockwise

Have I drawn a diagram of displacement of 40km , -30 or +30 east of north

Calvin Khor

you should follow what your text and teachers say. its clear enough to me but i dont know how they mark you

S.M.T

My text say nothing about the direction of angle

Calvin Khor

no solved examples?

S.M.T

Yes, there is but it does not have a drawn direction in it

Calvin Khor

add it if you want, then see what your teacher says when you recieve the work back. or ask directly. I can't help you decide what will ultimately be right for you

Koro

07:28

$\Huge {(\ddot\smile)}$

Calvin Khor

$\Huge\vcenter{\stackrel{⦿{\large\triangle} ⦿}{\smile}}$

Koro

hahaha

$\Huge {(\ddot\smile)}^2+\Huge {(\ddot\smile)}+1=0\implies \Huge {(\ddot\smile)}=\omega,\omega^2$

Calvin Khor

08:07

Is there anyone keeping track of what mathematical books are entering public domain?

S.M.T

08:25

My teacher says degree is a scalar quantity

porridgemathematics

@TedShifrin thanks for the reply, right I see that, but what I was troubled by was that the second definition the OP states does not even seem to agree with the first one up to a constant factor

also, i have a burning question, this sounds weird to me and I'd like to know if its actually true.. so I seem to have a 'proof' that $H^2(\Omega)$ is finite dimensional, where $\Omega \Subset \mathbb{C}^n$ is a bounded domain, and the former symbol means the square integrable holomorphic functions $f : \Omega \rightarrow \mathbb{C}$

Calvin Khor

@porridgemathematics I know crazy things happen in $\mathbb C^n$ but are you claiming $n=1$ as well? that sounds very off

porridgemathematics

yeah that is my claim

Calvin Khor

but every polynomial should be in $H^2$ as $\Omega$ is bounded

I don't forsee a universe in which multivariate polys aren't "holomorphic", but I am reducing to $n=1$ out of pure fear of multivariate complex analysis

porridgemathematics

yeah that was exactly why it was why I think its wrong

Calvin Khor

08:38

:) whats the proof?

porridgemathematics

so basically my proof went like this: Let $B$ denote the unit ball in $H^2(\Omega)$ (which is a hilbert subspace of $L^2$), then we claim $B$ is a locally bounded family, the reason is because for $f \in B$, $|f(x)| \leq \frac{||f||_{2}}{\sigma(x)}$ is satisfied, where $\sigma(x)$ is the volume of $B(x;d(x,\partial \Omega))$, so this should mean that $B$ is locally bounded

but then a locally bounded family of holomorphic functions is normal in the sense that any sequence has a locally uniformly convergent subsequence

in particular we can take some sequence $f_n \in B$, then take a subsequence $f_{n_k}$ converging locally uniformly to a holomorphic $f$. This $f$ is not apriori the $L^2$ limit of $f_{n_k}$, but I thought it should be because we can exhaust $\Omega$ with compact subsets, and then take limits

Calvin Khor

$B$ isnt a subspace though?

porridgemathematics

I mean $H^2(\Omega)$ is hilbert

Calvin Khor

oh

ill keep reading

ok, i think this is more or less alright, how do you get the dimension?

porridgemathematics

i thought this shows that $B$ is compact in $H^2(\Omega)$

and if $H^2(\Omega)$ was infinite dimensional, that would not be true

Calvin Khor

08:46

hmmmmmmmmm

porridgemathematics

because we could embed $\mathcal{l}^2$ into it if it were infinite dimensional, and the unit ball in the former space is certainly non compact

just take $(1,0,...),(0,1,0,...),...$

Calvin Khor

must be the last bit, $L^2$ norm should blow up

or theres only finitely many polynomials and annals of math will be in contact shortly

porridgemathematics

this was the precise argument for the $L^2$ norm thing: Let $C \subset \Omega$ be compact, then $|f_{n_k}-f|^2 \mathbf{1}_{C} \rightarrow 0 $ pointwise, and $|f_{n_k} - f|^2 \leq 2^2 |f_{n_k}|^2 + 2^2 |f|^2 $, the RHS converges pointwise to $2^3 |f|^2$ on $C$, which has a finite integral because $C$ is compact, so by DCT $f_{n_k} \rightarrow_{L^2} f$ at least on compact subsets of $\Omega$

Calvin Khor

well locally uniformly already gives $L^2_{\text{loc}}$

but the uniform bound depends on distance to boundary

porridgemathematics

sure, but we should only need local boundedness to give that $B$ is normal by montels theorem

and at least locally, we can fix a small ball, and control $\sigma(x)$ for all $x$ in that small ball no?

Calvin Khor

08:55

i have to google montel :') one sec

right ok, i agree that you can get an analytic limit for some subsequence

porridgemathematics

and you also agree that that analytic limit should be in $L^2_{loc}$

Calvin Khor

yes. but unless $\sigma^{-1}\in L^2$ I don't believe this gets you all the way to $L^2$

porridgemathematics

it shouldn't, but my argument was that on compact subsets of $\Omega$, $||f||$ should be less than or equal to $1$

because $f$ is the $L^2$ limit on this compact subset of functions in $B$

and those functions in $B$ have $L^2$ norm on all of $\Omega$ equal to $1$, so on some compact subset their $L^2$ norm should be at most $1$

Calvin Khor

hmmm.................ok you know what I agree that the limit is also in $L^2$. So I'm confused

do you want to write jointly to Inventiones

porridgemathematics

maybe $H^2(\Omega)$ is just not hilbert

im 99% sure im wrong haha

Calvin Khor

09:01

oh im 200%....but I can't figure out the bug in the proof

porridgemathematics

im just revising for an oral on kahler geometry in two days, and this came up when looking at stuff related to the bergman metric on bounded domains

the claim maid there was that for such domains $H^2(\Omega)$ is a closed subspace of $L^2(\Omega)$, because of the mean value inequality I stated before

made*

Calvin Khor

I was happy enough to have proved that without asking you why its true

porridgemathematics

fair enough, so $H^2(\Omega)$ is definitely hilbert

hmmm, what about this then? This is a well established result in the literature: If $X$ is a compact complex manifold and $E$ is a holomorphic vector bundle on $X$, then the space of holomorphic sections of $E$ is finite dimensional

what if we take $X = \mathbb{C} \cup \{\infty \}$ and $E = X \times \mathbb{C}$

ah no, thats fine, these are just constant functions :)

Calvin Khor

hahaha

porridgemathematics

i thought for a second we can just use polynomials here too, but they blowup at $\infty$

Calvin Khor

09:06

its possible for $f_n$ to be super nice and for the limit $f$ to be in $L^2$ without the convergence happening in $L^2$

porridgemathematics

im going to boil this down to , like you said, something wrong with my $L^2$ convergence argument

thats true

Calvin Khor

purely from bounded in $L^2$ they already converge weakly in $L^2$

porridgemathematics

right

Calvin Khor

you should ask them in your exam

hey, i gotta question for you

porridgemathematics

oh, i figured out the error

Calvin Khor

09:10

oh, what is it?

porridgemathematics

to apply DCT on compact subsets $C$ as I wanted to, I need to show that $\int_{C} 2^2|f_{n_k}|^2 + 2^2|f|^2$ converges to $2^3 \int_{C} |f|^2$, but that means I need to show $f_{n_k} \rightarrow_{L^2} f$ on $C$, which is what I was applying DCT to try to show in the first place

Calvin Khor

tbh i didn't read that because $L^2_{\text{loc}}$ follows just from convergence locally uniformly

porridgemathematics

hmm, wait a sec, but if $f_{n_k}$ are locally uniformly bounded, they are uniformly bounded on compact subsets, so I could use DCT to show this, since $|f_{n_k}|^2 \leq M$, and $f_{n_k} \rightarrow f$ almost everywhere, this use of DCT should still go through

ugh, now im even more confused

yeah, $L^2_{loc}$ follows because the limit is continuous, we dont even need holomorphic

Calvin Khor

$\|f_n-f\|_{L^2(C)}\le |C|^{1/2} \|f_n - f\|_{L^\infty(C)} \to 0$ for fixed $C\Subset \Omega$

err....wrong power

but if you apply holder correctly

porridgemathematics

ah yeah, and holder also shows this, true

Calvin Khor

09:14

and some people don't call the $L^\infty-L^1$ case Holder but wtv

porridgemathematics

okay, it has to be with the last bit then, so I think I can't say $||f_{n_k} - f||_{2} \rightarrow 0$ on $\Omega$ using DCT, because I don't have boundedness of $f_{n_k}$ on all of $\Omega$

and $f_{n_k}$ at least being dominated by something known to be in $L^2$ on $\Omega$ is necessary for that final application

Calvin Khor

yeah i think so

porridgemathematics

ugh, i feel so stupid spending so long on this lol

thanks for working thorugh it with me!

Calvin Khor

on the other hand, i have only ever seen finitely many polynomials....

porridgemathematics

xD

Calvin Khor

09:18

np :) it was fun

robjohn

10:52

@CalvinKhor a Generalization of Hölder perhaps.

$\|f\cdot1\|_2\le\|f\|_\infty\|1\|_2$

Calvin Khor

right, in my head I did do $\frac12 = \frac12 + \frac1\infty$, so even this was a stretch :)

but you could just use holder and get there

$ \|f\|^2_{2} = \|f^2\|_1 \le \| f\|^2_\infty \|1\|_1$ then root

robjohn

well, yes. But that would include the work that was done in proving the generalization. I guess we could say that Fermat's Last Theorem can be proven with basic arithmetic, if we don't mind doing all the intervening work ;-)

@CalvinKhor that is what was done in the linked answer.

Calvin Khor

:D the issue is not having a name for it...

also I can't quantify this but surely FLT should be further from basic arithmetic than the two versions of holder :P

robjohn

sure, I was just extrapolating for emphasis.

porridgemathematics

this is only true for finite measure spaces though, no?

Calvin Khor

11:00

yes

robjohn

@porridgemathematics $\|1\|_2\lt\infty$

porridgemathematics

whereas holder works with whatever measure space you want

Calvin Khor

well ok. the application of holder is always true. whether or not the result is useful depends on if 1 has finite integral

porridgemathematics

sure, I like to see this result as essentially the statement : in a finite measure space, the $L^p$ 'averages' of a function converge upwards to $||f||_{\infty}$

i think this can be proven without holder

errr, well, the upwards part requires holder, so this is untrue

just convergence doesn't need holder though

Calvin Khor

on finite measure spaces, $L^\infty$ embeds into $L^1$, and on....atomic? spaces? or whatever collection $\ell^p$ falls in the reverse inequality holds

porridgemathematics

11:06

yeah, this comes down to this problem in folland i think, basically for $p<q$ finite, you get $L^p \subset L^q$ if and only if the measure space has measurable sets of arbitrarily small positive measure

and I think the reverse inclusion holds with a similar sort of condition at least for finite $p,q$

oh whoops, *if the measure spaces does NOT have measurable sets of arbitrarily small positive measure

Calvin Khor

to match with $L^\infty\subset L^1$ you'll need $p>q$

right ok

porridgemathematics

so yeah, atomic seems intuitive

Calvin Khor

yes, that was what I meant by atomic. there is some constant $c$ so that $\mu(A)>0$ implies in fact $\mu(A)>c$

porridgemathematics

yeah thats precisely it

its an iff too

Calvin Khor

is this how this goes for the lower bound - say $0<C<1$, then $\|f\|_p^p \ge \int{|f|\ge
C \|f\|_\infty} |f|^p\ge C^p \|f\|_{\infty}^p \mu(|f|\ge
C \|f\|_\infty)$

take $p$th roots and $p\to\infty$ the measure goes to 1, then $C$ arbitrary less than 1 so result?

ok sounds good

....converges upwards though....hmm

I only get convergence with squeeze rule (above mentioned generalization of Holder for the upper bound)

the inequality chain should read
$$\|f\|_p^p \ge \int_{|f|\ge C \|f\|_\infty} |f|^p\ge C^p \|f\|_{\infty}^p \mu(|f|\ge C \|f\|_\infty)$$

William John

11:57

I'd like to sniff Heron.

robjohn

12:47

How do you feel about L'Huilier?

Oliver Díaz

14:27

@porridgemathematics: Here is a solution to $\lim_{p\rightarrow\infty}\|f\|_p=\|f\|_\infty$ that based on Chabyshev's Markov inequality.

porridgemathematics

16:56

@CalvinKhor for convergence upwards i think its really just a matter of applying Holder to show that $(\frac{1}{\mu(X)} \int_{X} |f|^p)^{\frac{1}{p}}$ is increasing in $p$

@OliverDíaz nice, it makes sense to use markov there, i think most proofs essentially come down to using it in some form

Derivative

17:10

Why does Terence Tao call the Elliott-Halberstam conjecture a "sort of super-generalized Riemann Hypothesis" in a blog post? Does it imply the GRH? terrytao.wordpress.com/2007/06/05/…

leslie townes

good question. i dunno if anyone in the chat is a number theorist.

robjohn

@porridgemathematics In this answer, I use Jensen's Inequality, but that is pretty close to Hölder. They are all convexity arguments, and so are closely related.

leslie townes

mathoverflow.net/questions/153078/… is the closest i could find from google.

Oliver Díaz

@robjohn: For finite measure, it does make sense to use either Jensen's (or Holder's, which is in a sense an application of Jensen's inequality)

robjohn

17:35

Another approach is to use $$\|f\|_p=\left(p\int_0^\infty\mu\left\{x:|f(x)|\ge\alpha\right\}\alpha^{p-1}\,\mathrm{d}\alpha\right)^{1/p}$$

geocalc33

18:46

does $\sum_{n=1}^\infty e^{zn^{-s}}$ converge in a distributional sense?

robjohn

20:00

@geocalc33 On what domain?

I am not sure any domain works, but it would be nice to know what domain you are looking at.

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