Actually, 1 follows from 2. Anyway, 1 and 3 are obvious.

(missing quantifiers, sorry :P)

@SohamChowdhury Is this for any metric coming from a norm? Or metric from the sup-norm?

sup-norm.

(I do not see definition of $d(f,g)$ but I might have missed something.)

Then I'd say 2 is obvious two.

Hi @SohamChowdhury

If you now $\sup\limits_{x\in X} |f(x)-g(x)| = 0$, then ...

@evinda hallo, wie geht's?

@SohamChowdhury Gut :) Und dir?

@MartinSleziak f(x)-g(x) must be identically zero, so it is the constant zero function, implying f = g.

And 4. is basically a combination of triangle inequality for absolute velue and properties of supremum.

@SohamChowdhury Exactly.

@evinda Im Moment, toll. Es regnet nicht, so . . . :)

@MartinSleziak I just realised that a while back when you posted your first message.

Fine, so that's settled.

Yeah :)

Is this some kind of introductory functional analysis course?

I'd been reading it as $\sup f(x) - \sup g(x)$ all along, so I was like "same sup implies same function?!" the first time Balarka gave me this.

@MartinSleziak no, no.

I'm studying topology. Balarka said if I could do this problem, then he'd tell me something cool about the p-adic metric.

Many people mention various p-adic things a lot. Probably I should try to learn a bit about this topic...

@SohamChowdhury Hier ist es um die 30 Grad...

Bei uns ist das Wetter auch ziemlich heiss.

@MartinSleziak Gut... Geht es dir besser?

@evinda Zu viel arbeit. (Und ich bin zu faul.)

@MartinSleziak Hast du bald Urlaub?

Ja. Aber auch dort werde ich wahrscheinlich ziemlich viel zu tun haben.

@evinda Du wusstest, ich in Indien leben, ja? :P

33 Grad.

@MartinSleziak Für die Arbeit?

@lopata Judging [by your username], you might speak Czech or Slovak (or some other Slavic language). So maybe you could simply look for some lecture notes for an introductory course in number theory. I have no doubt, that you can find several good text in Slovak or in Czech.

Or you can restrict the search by language rather by domain, like this.

@Balarka, done.

ok?

@SohamChowdhury Ja, das hattest du mir erzählt. Also habt ihr Schwimmwetter. Oder nicht?

@evinda Vor allem wahrscheinliche Dinge ums Haus.

@MartinSleziak Bist du neu umgezogen?

@SohamChowdhury 1. is superflous, given 2.

@BalarkaSen I said as much.

2. and 3. are obvious. how did you prove 4?

Hi @BalarkaSen What's up?

@evinda Nein, aber im Sommer werde ich meistens im Elternhaus sein. Dort gibt es ziemlich viel zu tun.

@MartinSleziak Achso...

@MartinSleziak Wo warst du schon mal im Urlaub?

Ich fahre sehr selten nach Ausland. Also verbringe ich den Urlaub meistens hier in der Slowakei. Wenn ich Zeit habe, mache ich auch ein bisschen Turistik.

Gehst du oft schwimmen? @MartinSleziak

@evinda Nur selten. Ab und zu wenn meine Schwester geht mit ihre Familie, dann gehe ich mit.

Es ist ein bischen ungewoehnlinch dass wir heir auf Deutsch sprechen.

Mann sieht hier meistens nur Englisch.

Kein Slowakisch, @MartinSleziak?

@SohamChowdhury Falls du willst...

@Soham did you see my message above?

"how did you prove 4?"

Ich sollte spellcheck fuer Deutsch installieren.

:D

@Balarka: $|(f-g)(x)| \leq |(f-h)(x)| + |(g-h)(x)|$ for all $x\in[0,1]$. Take sup of both sides?

Or do I have to work from the def of sup?

skillpatrol is right, we should move there if we want make smalltalk in German. In any case, I will probably leave now, I have some stuff to do.

yes, but you still have to prove something about sup

See you later!

bye @MartinSleziak

@BalarkaSen what, exactly?

Later pal

Also, 3 was not obvious to me when you asked me the first time. I'm getting less dumb, I guess. :P

sup of sum is smaller than sum of sups

yeah, well, $|f - g| = |g - f|$

Oh, no. I meant 2.

I have no clue why I did not get that back then.

@BalarkaSen wait, working on that.

oh, I see. yes you just proved it above: "$|f - g|$ is identically zero"

that works for 2?

oh, I think I have it.

$|f - g|$ identically zero => $f - g = 0$ => $f = g$.

no, I think I have a proof of 4.

oh. let me see.

I skipped a bit actually.

Let $f,g$ attain their maxes at $x, x'$ resp.

Then $|f(x)-h(x')| = |a-c|$. sup over $[0,1]$.

It's not clear to me why $||f - h|| = |a - c|$.

the last line gives $||f-h|| = |a-c|$, I think.

obvious typo in "Let $f, g$ . . . ", sorry.

$f : [0, 1] \to \Bbb R$ given by $f(x) = \sqrt{x}$ and $g : [0, 1] \to \Bbb R$ given by $g(x) = x$ both attains maximum at $1$. But $|f(1) - g(1)| = 0$, whereas $|f(1/2) - g(1/2)| > 0$

So clearly $|f(1) - g(1)|$ isn't the maximum.

Visualization : $f$ and $g$ both attains maximum at the same thing, but does crazy things below the maximum so that the graph of $f$ lies on top of $g$ all the way in $[0, 1)$

I drew a picture already.

I'm thinking about how to prove it.

So now do you see what goes wrong with your argument?

yes.

ok, good.

4 is the only thing that separates me from p-adics. I've never wanted to solve an analysis problem so much before :P

haha

I think I have to go back to the def of supremum.

recall that you also want to prove that $||f||$ is always finite for a given $f$ (<-- this is a bit nontrivial)

@SohamChowdhury I encourage that.

oh, I know a proof of that theorem.

ah?

closed bounded interval, continuous functions achieve maxes?

I know a proof from before.

well, not maxes, but sups. max is stronger.

how do the proof you know goes?

considering sequences, using BW, etc.

precisely

@BalarkaSen Hmm, does it need differentiable to be guaranteed a max?

don't remember the achieves bit too well. It goes like a typical $\epsilon-\delta$ thing iirc.

@Tobias no.

@SohamChowdhury what, the proof of $||f|| < \infty$?

nah, you just need sequences for that

@BalarkaSen oitto, told you. considering sequences.

$f(x_1) > 1, f(x_2) > 2, . . .$

yup

now "every sequence has a convergent subsequence" gets contradicted.

Using BW, some subseq is convergent.

Now $f(\lim x_n) = f(x) = \lim f(x_n)$.

@BalarkaSen etc. yes.

yes, don't bother writing it up. i believe that you know it.

anyway.

so am I done with everything except 4?

mhm

side-note : $\mathcal{C}([0, 1], \Bbb R)$ is an abelian group under the operation $(f, g) \mapsto f + g$. $\Bbb R$ acts on $\mathcal{C}([0, 1], \Bbb R)$ by $(r, f) \mapsto rf$ (scalar multiplication). This makes $\mathcal{C}([0, 1], \Bbb R)$ into an $\Bbb R$-vector space.

Since you also have a natural norm on this space, this is actually an example of a normed vector space.

yeah, MathWorld told me it's a "commutative Banach algebra with identity", which means the same thing, possibly.

just sayin', since topological vector spaces are central objects studied in functional analysis.

@Soham yeah. it's an example of an infinite-dimensional vector space.

you'll come to appreciate those when you study linear algebra from Artin

@SohamChowdhury Banach is stronger than just normed vector space, as it requires the space to be complete with respect to the norm (Hilbert space is then when the norm comes from an inner product)

hi @BalarkaSen, hi @SohamChowdhury

hey

can i introduce the limite Inder the integral directly ?

thank you

hi people, Im studying a bit of theory of numbers and I had a though that I want to share with you, if you please, and see what you think about

hi @iwriteonbananas

I was seeing a bit of dirichlet series and riemann zeta function... and I asked to myself where the analytic power of zeta function lie... what make it "different", and I noticed that maybe because is a relation between 2 kind of numbers that "cannot be mixed in the context": zeta function is a fraction of entire numbers... at least entire in their base

@Masacroso What do you mean by entire?

so I though that it power come from this relation. If you reverse the zeta function it lost all it "power"... in some sense the same happen when you mix real numbers and infinitesimal in calculus... the "power" comes from the relation of these 2 kind of numbers that cannot be mixed easily, in the same way you cant mix easily entire numbers and fractional numbers

I have no idea what you're talking about.

the definition of zeta is a relation of entire numbers and fractionals... is just the definition : $\frac{1}{n}$, and $s$ can be used just as an index depending of the context

@Masacroso by entire do you mean integers?

yes @TobiasKildetoft

or naturals

@Masacroso that's just for $\Re[s] > 1$

Then what do you mean by "at least in their base"?

in the index, that is the base of the fraction and the base of the reversed power

@Masacroso My recommendation is that you forget completely about the Riemann Zeta until you have learned a lot more of the basics

Someone help me please

no @TobiasKildetoft, lol

@Masacroso What do you mean by "no"? That is really my recommendation

I second that.

lol, I dont second you guys ;)

What you have said above seems much like vague philosophy rather than something concretely mathematical. You really should learn more basic elementary number theory before diving into zeta function and whatnots, otherwise you'll turn into a crank, which is not good.

lol, you are mad @BalarkaSen or you doesnt understand me, or both

@Masacroso We don't understand you because you are not communicating in the language of mathematics.

shrug none of us understands what you have said above, @Masacroso

@Balarka, okay?

why are we talking about maxes, when we really want something weaker?

i.e., sups.

I find maxes easier to reason about, obviously.

And a max is a sup of course in $[0,1]$, by that theorem whose proof I know.

@masacroso out of curiosity, did that thing i linked to re: painleve equations prove interesting?

@Semiclassical what evil concoction have you most unholily linked that person to and caused him to lose his mind? :P

@Balarka, does that do it?

@SohamChowdhury Painleve equations

@SohamChowdhury so we want to prove $\sup |f(x) + g(x)| \leq \sup |f(x)| + \sup |g(x)|$, right?

and you want to reason with max.

it came up in this context

okay. hmm.

it works for this interval.

I'd imagine the sup argument is very similar.

yes, what you say works.

you only have to add the extra step of showing that any other bound is greater. so, now.

$$\huge{\text{p-adics pls Bulerka ;-;}}$$

(I'm still a kid lol)

OK, sure.

What were you going to tell me about the p-adic metric?

There're two aspects of p-adic numbers. One is topological, one is algebraic.

Okay.

Let's begin with the topological aspect : $\Bbb Z$ be your base set. Equip it with the metric $d(x, y) = p^{-n}$ where $n$ is the greatest integer s.t. $p^n$ divides $|x - y|$

Ah, remove the abs.

okay, understood.

So essentially what we write $v_p(x-y)$?

Notational convention : $d(\bullet, \bullet)$ is usually called $|\bullet, \bullet|_p$

Exercise : Prove that this defines a metric.

okay, got it.

wait, this should be quick.

It's not hard to do it. Get it done, and then we'll talk more.

nevermind.

1/2. $|x,y|_p = 0 \iff p^{\text{anything}} | x - y \iff x = y$.

3. obvious

Go on.

not sure i see the first equivalence there

for instance, $|3,2|_p=0$ for all $p$ since only 1 divides $3-2=1$. but $3\neq 2$.

$|3, 2|_p = p^{-0} = 1$

woops, misunderstood that

You're mixing up the valuation with the norm, yes.

read it as the exponent rather than the exponentiation, yeah

hmm. so would the terminology be that, for example, $|x,x|_p$ would have infinite valuation and zero norm?

yep.

Hello@Balarka , What are you talking with Soham about

valuations, by definition, have $\infty$ in their codomain.

neat. and sensible, given what i recollect of p-adic metrics re: symbolic dynamics

i.e. given the p-adic expression of a number, if i started comparing it to itself and looking for discrepancies i'd of course go on forever

i see.

whereas for any distinct pair i'd eventually see a disagreement. it comes up when you talk about the cantor set and ask what it means for two points in the set to be "far apart"

didn't mean to sidetrack you guys, though

nah, what you say sounds interesting

think i linked you something on it at one point, but i forget when

i don't recall that you did

searching chat, this is the conversation i had in mind: chat.stackexchange.com/transcript/36?m=18279723#18279723

isn't apparent that i linked you anything on it, though