@masfenix For any $p$?

yes, $1 \leq p \leq \infty$

Well, if a function is in $L^p$ then its in $L^q$ with $q > p$ no?

I dont remember if thats true for the infinite dimensional case

@masfenix And $L^p(\Bbb R)$; yes?

$L^p(R^n)$ to be more precise, but $R$ is fine also.

Oh.

Maybe you can ask in main. It shouldn't be hard.

The last paragraph of page 5 gives an argument math.uchicago.edu/~vipul/sometransforms.pdf but I was wondering if you could clarify something for me?

@masfenix OK.

i love real and complex analysis for walter rudin

How to prove that $\frac d{dx}e^x=e^x$ using it's infinite series? I got $\sum^\infty_{n=0}\frac{x^{n-1}}{(n-1)!}$ so far but I am having trouble saying that it is equal to the series expansion of $e$

@Alizter how did you get a $\frac{1}{x}$ term?

Well just so I am understanding this correctly. There is a function $f \in S$ which is multiplied with some polynomial in the form $\frac{1}{x_1^2x_2^2...}$. Now my first question is that what do they mean "outside of a compact set, f is being dominated by..."

@KevinDriscoll Where?

@Kevin: NO. Try $1/\sqrt x$ on $[0,1]$.

@Alizter the n=0 term

sigh ... I have to write this out

@Ted AH yes. Integrable singularities then become non-integrable. Thanks

The reverse inequality on finite measure spaces is right, though.

@Alizter I think it should be $\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}$

@PedroTamaroff I posted by question a few posts above

@TedShifrin In the case @masfenix is interested in, the functions are all $C^{\infty}(\mathbb{R})$. Is my claim then true?

@KevinDriscoll $$\frac d{dx}\sum^\infty_{n=0}\frac{x^n}{n!}=\sum^\infty_{n=0}\frac1{n!}\frac d{dx}x^n=\sum^\infty_{n=0}\frac{nx^{n-1}}{n!}=\sum^\infty_{n=0}\frac{x^{n-1}}{(n-1)!}$$

and the $ 1 $ term

@Alizter The last step fails for n=0

??

@masfenix They mean that there is a compact set $K\subseteq \Bbb R^n$ such that over $\Bbb R^n\setminus K$ your function is dominated by $$\frac{1}{x_1^{2p}\cdots x_n^{2p}}$$ meaning it is less than that. Being positive, it is integrable by comparison.

Continuous would be the issue, I guess. I haven't been paying attention.

@KevinDriscoll Yeah this is where I need help I don't know what went wrong

Then inside $K$ we can integrate the function for it is continuous.

@ALizter if $n=0$ then $$\frac{n}{n!} \neq \frac{1}{(n-1)!}$$

@PedroTamaroff so how can we say that is it dominated? Is it because its a Schwartz function?

@KevinDriscoll Oops

errrr well I guess you could say it does.....but its ambiguous

@masfenix Indeed.

Just make a limit argument.

@PedroTamaroff and because its a schwarts function (lets say $f$) then $f$ times the expression above makes it go to zero fast?

@masfenix times the inverse of that function

Say $$(x_1^{2p}\cdots x_n^{2p}){f({\bf x})}\to 0$$

@masfenix which proves that $f$ decreases FASTER than @Pedro 's given function

Then we out of some big ball $K=\bar{B}({\bf 0},R)$ we have

$$(x_1^{2p}\cdots x_n^{2p}){f({\bf x})}<1$$

@PedroTamaroff Yes, I was trying to think of how to make the argument that SOMEWHERE the product is less than 1 formal

Thus $$\int_{\Bbb R^n\smallsetminus K}f({\bf x})d{\bf x}<\int_{\Bbb R^n\smallsetminus K}\frac{1}{x_1^{2p}\cdots x_n^{2p}}d{\bf x}<+\infty$$

@KevinDriscoll Take $\varepsilon =1$.

@TedShifrin p $\ge$ 1

@TedShifrin $p\geqslant 1$.

@PedroTamaroff eeeeeeeewwwwwww.......... slanty line

Hmm ... need to be a bit careful here.

@KevinDriscoll Damn you. The slanty line is awesome.

@TedShifrin Oops?

I don't use slanty lines.

@TedShifrin You too? Oh, this is a tragedy.

LOL, good, Lady Macbeth can get bloody again :P

@TedShifrin What needs to be taken care of?

OUT DAMNED SPOT!!!!!!

@PedroTamaroff thank you. okay so let me recap. We want to show that every Schwartz function (a function that decreases rapidly when multiplied by inverse power). Now, consider some function $f \in S$ where S is the Schwartz space. Now clearly, inside some compact set, f is continuous so that argument is easy. But outside the compact support K (ie, consider some ball with radius R), we have that $f$ times $x_1^{2p}...x_n^{2p} < 1$ is so its bounded (correct?)

I admit I'm being stupid, but I'm switching to spherical coordinates. With trig stuff in the denominator, is it obvious there's no problem? @Kevin... You missed an OUT.

@Ted :-(

@TedShifrin Of course. Fubini ad nauseam?

It's been........ 6 or 7 years?

OK, I'm stupid. Do a big cube instead of a ball.

Time to go cook dinner and vanish.

@TedShifrin LAWLZ.

It's been 46 for me, @Kevin.

Poof

@PedroTamaroff so what does the integral say?

@TedShifrin I bow to your superior theatrical ability then

I mean, what is the intuition and interpretation?

@masfenix ?

@KevinDriscoll Eh?

@PedroTamaroff we were just musing about Macbeth

@KevinDriscoll Was I? Never read that one.

@PedroTamaroff I wrote a recap of everything which I think is correct, but I am a bit confused on what the interpretation of the integral you wrote down is

@PedroTamaroff You called it a tragedy!

@masfenix What do you mean "interpretation of the integral"?

@masfenix The interpretation is that all $f \in S$ decrease so fast at infinity that the area underneath them is bounded, even though the domain is unbounded

and we proved that by finding a function that also has bounded area from a large cube out to infinity and then argued that outside that cube all $f \in S$ are less than the function we found

thus their area is also bounded

ahh I see. Also last question and I think I have it then. When you wrote down $x_1^{2p}...x_n^{2p} \cdot f(x) < 1 $ what does that mean? is that what @KevinDriscoll just wrote?

@masfenix What does it mean?

I means the function $g({\bf x})={\bf x}^{2p}f({\bf x})$ is less than one outside the ball.

You're not a math student?

nevermind, $x^{2p}f(x) \rightarrow 0$ so clearly $x^{2p} < \epsilon = 1$.

@masfenix if $x_1^{2p}.....x_n^{2p} f(x) < 1$ then $f$ is less than $\frac{1}{x_1^{2p}...x_n^{2p}}$

@PedroTamaroff, @KevinDriscoll thank you very much.

and $\int_a^b \left| f(x)\right| dx < \int_a^b \left|g(x)\right| dx$ if $\left|f(x)\right| < \left|g(x)\right|$

and the two function are non-negative

there we go

@masfenix No problme.

@Pedro OOOOOOOh!

@PedroTamaroff I have a confession....... I am envious of all your badges

@KevinDriscoll LOL, see other top user's badges.

Phew!

@KevinDriscoll Look at Arturo, say.

@PedroTamaroff I know si0.twimg.com/profile_images/3088683507/…

@PedroTamaroff Now that just makes me feel bad...... I have not even 1 silver badge

@KevinDriscoll I reseted, got 4 heavenly cookies.

You know in the beginning I thought this place was just another way for me to waste time, but I find I al slowly learning

@Pedro congratulations???

@KevinDriscoll Oh, LOL, you don't play the game.

@pedro I did once for a few days, but I never got any heavenly cookies I dont think

It just didn't hook me

@KevinDriscoll It does get kinda boring.

I am playing Dark Souls now.

Holy shit that game is amazing!

Where are you?

@KevinDriscoll Currently, tearing over Blighttown.

=(

@Pedro have you rung the 1st bell yet?

@KevinDriscoll Yeah.

OKay cool. You're following the kind of 'recommended beginner path' then

Blighttown is NO FUN

Well...... its actually a lot of fun. But its quite difficult.

Have you gotten to the bottom yet?

Hi, if I have a morphism of vector bundles wich is a projection on fibers, is it true that it has constant rank?

woooooooooooooooosh /me ducks as that question nearly decapitates me

@KevinDriscoll Nah, I am actually up looking for the Eagle Shield. I keep dying in silly ways. I lvl 41 so the guys are no match.

@KevinDriscoll Heh.

@ADR Ask @TedShifrin

@PedroTamaroff Oh okay. Eagle shield is quite nice if you're strong enough for it. Eventually that it lets through 5% dmg gets really annoying though

@KevinDriscoll Oh. I am using the Spider Shield ATM.

totally reasonable considering where you are

I think the only 100% physical reduction shield you couldve gotten at this point is the Black Knight Shield

@PedroTamaroff ehmm, How can I do that?

@ADR Ted will probably be around later

You could always ask on Main as well

@PedroTamaroff I beat NG++ a while ago. Gonna go back and get all the boss weapons at some point

@KevinDriscoll LAWL.

@KevinDriscoll Ah, I fought some of those, even being a thief, no loot whatsoever.

And being a theif gave me awesome loot.

I have a full Balder's set.

Except I am using the Gargoyle's Helmet.

@PedroTamaroff Ya drop rates for Black Knight stuff is pretty low. THey drop Titanite somewhat regularly but the weapons and shield are all <10% I think

well, base <10%

@PedroTamaroff

is the dimension of $\Bbb C^3$ over $\Bbb R$ 6?

@Twink Yes.

uff :D

thanks

Think about how $\Bbb C^3$ is really $\Bbb R^6$.

Is its canonical basis $(1,0,0),...,(i,0,0),...(0,0,i)$

Asker tagged this question String Theory

hee hee

@KevinDriscoll He did it for the LULZ.

@PedroTamaroff Think so?

@KevinDriscoll How do you fall quickly off ladders?

@PedroTamaroff I played with mouse and keyboard, so I think it was hold space

With a controller..... hold A???

@KevinDriscoll Oh, I have an XBOX controller.

How do I get through the poisonous swampy area?

Fuck, I am dead again.

-_-

@PedroTamaroff You just have to explore a bit. You'll find a bonfire down there eventually

@KevinDriscoll Yeah, I did. But to reduce tox. I use the Shadow suit.

@PedroTamaroff Yup. It also helps to have some purple moss clumps

@KevinDriscoll and @PedroTamaroff what kind of music do you like?

@PedroTamaroff if you're careful you can plot your route through the much so you dont get poisoned

@Twink I listen to just about everything except country. Usually clumps of Top 40, Classic Rock, Jazz, and some electronic stuff

cool :D

I have stopped playing the cookie game. Mark this moment.

@Twink This.

@Pedro: You've kicked it? :)

@TedShifrin Yeah.

I was thinking we'd need an intervention!

@TedShifrin We do, but for coffee and real analysis instead

Hi, could it be that the approximations for the Mod. Bessel function of the second kind is wrong here? en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms I get K_0(x)/K_1(x) -> -inf here, whereas I would expect 0 (for x->0)

@Pedro has man vices

@KevinDriscoll ?

@KevinDriscoll Grrrrrrrrrr

OH wait sorry at 0, not $\infty$

yes :-)

So, $K_0$ goes like $\frac{1}{\sqrt{x}}$ if I remember corrently

and $K_1$ like $\frac{1}{x}$

well, Wikipedia says $K_0 \approx -\ln(x)$

@quazgar hmmm lemme check