$$\iint_D u\Delta u=\int_{\partial D^+}u\nabla u\cdot \vec n-\iint_D \nabla u\cdot \nabla u$$

That's also Gauss.

@Kaj: I'm not your adviser, but I approve. :)

That shows that if you have an harmonic function that vanishes on the boundary of a open connected bounded subset of the plane, it must be zero there.

It was cool.

Yup @Pedro, plus product rule.

@TedShifrin Then I had to show that the solutions of $x''=Ax$ form a subspace of dimension $2n$.

Next prove mean value property of harmonic fns, @Pedro.

Where $A\in M_n(R)$.

@TedShifrin Just like holomorphic functions, yes?

But works for harmonic fns in any dimension, @Pedro

@BalarkaSen, I'm not so sure. You probably know more algebra than I do, at the very least.

@TedShifrin Cool beans.

@DanielFischer It seems like a case-by-case analysis does the trick. We want to prove that $d'(x, y) < d'(x, z) + d'(z, y)$. Assume $d(x, z) < d(x, y)$, then $d'(x, z) > d'(x, y)$ by monotonicity. Hence $d'(x, y) < d'(x, z) < d'(x, z) + d'(z, y)$, done. This also holds true similarly for $d(x, z) < d(z, y)$. Thus WLOG $d(x, z) > d(x, y)$ and $d(x, z) > d(z, y)$, in which case just normal triangle ineq for $d$ does the trick.

I believe I need another redo at my module theory before I can claim I know any algebra, @Kaj

I have already redoed groups.

@TedShifrin So I just have to use the Hodge star and to get the general Gauss trick, yes?

@BalarkaSen Or $$\frac{d(x,y)}{1+d(x,y)} \leqslant \frac{d(x,z) + d(z,y)}{1+d(x,z)+d(z,y)} \leqslant \frac{d(x,z)}{1+d(x,z)} + \frac{d(z,y)}{1+d(z,y)}$$

Oh that's beautiful @DanielFischer

@DanielFischer Daniel, I want to learn German.

I want to read dis archive.org/stream/handbuchderlehre01landuoft#page/366/mode/2up

PNT, yes

@Pedro Why don't you look at a bit modern and english books?

Like Titchmarsh?

I want to read Landau's work.

He has only 3 translated works, which I have already looked into.

Plus I want to visit Cologne.

Blah. As I can see it, he used Perron, so potato-pohtato.

@PedroTamaroff Warum nicht.

You'll just get some stronger/weaker estimates.

@DanielFischer I know that "nicht" means no.

So, I got half of what you said

Hm...

Sup people of the world

No, it means night. Or not?

OMG :O Hi Prof. @TedShifrin

@PedroTamaroff "Why not"

@Sabಠ_ಠ Inf people not of the word.

You need to use spherically symmetric harmonic functions in dimension $n$, @Pedro, and use Green's formula for two harmonic functions.

@DanielFischer Why not, what?

@TedShifrin Spherically symmetric?

You're losing me Ted!

@PedroTamaroff Why not learn German. It can be useful.

@DanielFischer I want to, yes!

Hi @Sab

Long time @Ted

Exams are 10 days away :D

In dim 2, it's $\log r$; in dim 3 it's $1/r$. @Pedro

You learning lots, @Sab?

Repost : Can anyone have a look and confirm that this answer is BS?

@Pedro: Look at exercises in section 8.6 of my text.

@TedShifrin OK.

looks

LOL

@TedShifrin Be careful when Balarka says topology is cool; he's talking about point-set, the poor fellow.

I know, @Mike. And I like point set.

YES

@TedShifrin I'm looking at it.

I have a "problem."

I taught it 4 or 5 times and had some great classes.

@BalarkaSen No, it's okay, $\sum 5^k = \frac{1}{1-5} = -\frac{1}{4}$ in the $5$-adic metric.

@TedShifrin What topics did you cover?

@MikeMiller Point set topology is fun.

DOUBLE YES

When I pulled back, I got $\cos s ds\wedge dt$. But maybe we used different spherical coordinates.

Through Tietze Extension in Munkres, plus extra analysis a few courses.

I used the one where $s\in [-\pi/2,\pi/2]$.

@DanielFischer Is it OK to manipulate your expressions that way to result in those stuffs?

I mean, one needs a rigorous approach to prove that $\sum 5^k$ indeed does converge in the 5-adic sense

@TedShifrin OK, that's reasonable... it's the stuff after that (and the stuff that Brian Scott used to answer, say) that I find absolutely unfathomably dull.

@BalarkaSen $5^n\to 0$.

We use spherical coords with $s\in [0,\pi]$.

Oh, OK. Then I guess I'm safe.

Oh, Brian is back.

@BalarkaSen That follows from $5^k \to 0$, since the $p$-adic metrics are ultrametrics.

@PedroTamaroff Is that enough in $5$-adic metric for convergence?

@BalarkaSen what's the 5-adic metric?

@MikeMiller $d(x,y) = 5^{-v_5(x-y)}$

@MikeMiller metric induced by $|x| = p^{-\nu_p(x)}$, $\nu_p(x)$ largest integer such that $p^{\nu_p(x)} | x$

@MikeMiller MIKE!

@DanielFischer I was not asking you... :P

@BalarkaSen So what does convergence mean?

@TedShifrin Yes.

@MikeMiller We need interpunction for Socratic questions ;)

It's good, Brian takes care of topology.

Well, mostly the point set. I agree with Mike that what comes after is more interesting ....

@MikeMiller Do I have to repeat the standard metric definition of convergence? :P

Yes, @DanielF ... We've had that problem before :)

@BalarkaSen If you can't immediately say why $\sum 5^k$ converges, then yes!

I suppose I should clarify that I don't hate the introductory point set... it's the stuff that you start doing when you don't stop doing point-set.

@MikeMiller $\{x_n\}$ coverges in $(X, d)$ $\iff$ For every $\epsilon > 0$, there exists a positive int $n_0$ such that for all $n > n_0$, $d(x_n, x) < \epsilon$

OK. Now what are the elements of $\Bbb Z_5$?

Gah. $\mathbf{Z}_5$, man.

@MikeMiller $X = \mathbb{Z}$ is the underlying set.

Is the underlying set of $\Bbb Z_5$?!

@MikeMiller OK, no I was thinking of the metric again.

$\prod \Bbb Z/5^k \Bbb Z$

That's closer to true, but still not quite so (it's a subgroup of that).

The answer I was hoping for was "$\Bbb Z_5$ is the completion of $\Bbb Z$ under the $5$-adic metric".

I haven't studied metric completion.

I am only familiar with the inverse limit definition.

@DanielFischer Surely not!

@DanielFischer thanks for the correction

@MikeMiller Bah, so the underlying set was $\Bbb Z$

Grrr

what do you mean by underlying set?

Because the answer is probably "no, it's not"

The set on which the metric acts?

the underlying set is $\Bbb Z_5 \neq \Bbb Z$...

@BalarkaSen Ok, by your inverse limit definition, we have an element $(1, 1+5, 1+5+25, \dots) \in \Bbb Z_5$. This is the $x$ to which your sequence should converge. Verify it.

OK, I am going to just stick to the inverse limit definition. Can you show me that $\mathbf{Z}_p$ (the algebraic object) and completion of $\Bbb Z$ under the $p$-adic metric coincides?

pops popcorn

@BalarkaSen Yes, I can (where the "algebraic object" means "the algebraic object with the $p$-adic metric on it"), but I won't, because I think you should do it. All I'll say on the matter is that elements of the inverse limit look a lot like Cauchy sequences...

@MikeMiller Aaaand... what is my sequence supposed to be?

... the sequence you wanted to show converges ...

$\sum_{k=0}^n 5^k$

Under the 5-adic metric?

Yes.

I am feeling all my ideas on p-adics are intuitive. Would it be a good idea to think on it and rigorify it or would it be better to just read up some text?

I don't see what the problem with the former is; you have all the appropriate definitions, and you've said them here.

I am now 70 kg. Will try to cut down to 65 kg by the end of this year, lol.

OH WAIT

We want to prove that $\lim|| \sum_{k = 0}^n 5^k + 1/4 ||_5 = 0$

Now $|| \sum_{k = 0}^n 5^k - (-1/4) ||_5 = || 1/4(5^{n+1} - 1) + 1/4 ||_5 = || 5^{n+1}/4 ||_5 = 1/4 \cdot 5^{-n-1} \to 0$ as $n \to \infty$ @Mike

Is that right?

yes

That was a great mental block. Thinking about $\Bbb Q$ with a different metric imposed.

It's almost as hard as thinking of it as coming up with a discontinuous unbounded function.

@BalarkaSen Now prove that $\sum a_i 5^k$ converges for any sequence of $a_i \in \{0,1,2,3,4\}$.

@MikeMiller $||x + y||_p \leq \min(||x||_p, ||y||_p)$

@MikeMiller HELLO

go on...

h

$||\sum^n a_i 5^k ||_5 \leq \text{min}(||a_i 5^k||_5) \leq \text{min}(k \cdot a_i)$