Was supposed to learn from users interacting with it over Twitter. Unfortunately it was quickly turned into a genocidal ultra fascist once a certain crowd caught wind of it.
I want to show that a function of several complex variables, presented as an integral over $\partial B_\epsilon(0)\subset \Bbb C$ of a holomorphic function, is holomorphic.
So say we have this integrand, call it $f$ (the integral will be called $g$). To show that $g$ is holomorphic, we do the following: Since $f$ is holom., it has a power series expansion on $B_{\delta(\xi)}\times B_{\delta'(\xi)}(z)\subset V\times U$, where $U,V$ are open in $\Bbb C^n,\Bbb C$.
Now, I just need to subdivide the integral into parts that lie inside this "domain of power series"
So what the proof does is consider $\partial B_\epsilon(0)$, and cover it in half-radius disks (also in the plane) $B_{\delta(\xi)/2}(\xi)$. Then, Huybrechts states the following:
There's the proof of Lagrange four-squares theorem out there. But that's not a geometric application of quaternions so not sure if you'd want to teach him that.
This is a problem for me: If I just use compactness in the obvious way, I get an open subcovering which is clearly not going to be disjoint (no covering can be disjoint because the circle isn't)
But then I don't see if/how I can make it disjoint, while ensuring condition 2.
The book doesn't say anything about it, and I'm pretty lost.
The closed sets cannot be used for a covering because then I can't guarantee a finite subcovering.
First, @Danu, I will reiterate that you shouldn't bog down in every picky detail, or you'll never learn the amount you need to. I have no idea why he's making this so obscure. But I think it's easy. I just need to think.
@TedShifrin: off topic: how do mathematicians store their books & folders? in a (home) office on shelves? I don't have a lot of space but am worried to put a shelf in the basement due to moisture.
But he's making this so obscure when it's not. Just cover the circle with finitely many closed "intervals" and choose the balls appropriately to get those intervals.
@TedShifrin So that'd be slightly different from the proof he does, right? You'd no longer be working with $\delta(\xi)/2$ in every case (but it doesn't matter because shrinking is okay)
The $\delta/2$ is done (as happens in complex analysis frequently) to insure uniform convergence on the closed ball. But anything less than $\delta$ will do there, anyhow.
If you have a continuous function and you restrict it to a compact set, it becomes uniformly continuous on that set. But what if you restrict it to a set that simply bounded?
Pedro: They're complaining that I keep morphing back and forth between orange and green on here.
What do you mean, @0celo? He doesn't assign homework? Or he assigns homework at the beginning of the week due in a week and doesn't cover everything until the end of the week? (I did that all the time.)
Yeah, I don't recommend one day per course. I recommend cycling through the courses. But my students usually could tell from which section I assigned the problems from.
I generally tried to keep the homework slightly behind the lectures, by a day or so, but not more.
@TedShifrin Yeah, but that's much harder to plan. Especially when I'm taking 5 courses at a time, I need to be able to set hard cut-offs for when to stop working on something...
That gets tricky if you have 2-3 weekly "moments" for each course, instead of 1 block of a few hours time
I've never thought about your geometry question, @0celo. I guess I hadn't thought about it: Given a metric space $X$ which is not complete, is there a metric inducing the same topology which is complete?
(That is to say, a research problem wouldn't contain much information on what you should use and what you shouldn't. It need not even make much sense for a person who's just started doing research, from what I gather from new grad students. But all of this is from a person who's nowhere near a researcher, so feel free to not pay attention)
Locally Lipschitz on a compact set implies globally Lipschitz. Just take a finite subcover of Lipschitz nbds, then take the minimum of Lipschitz constants?
re: this question, am I losing my mind or isn't it clear that permutations in the symmetric group $S_g$ acting on $\{a_i\}$ and $\{b_i\}$ simultaneously induce automorphisms of $$\Gamma_g=\langle a_1,b_1,\cdots,a_g,b_g:[a_1,b_1],\cdots,[a_g,b_g]\rangle?$$
it's actually a great movie shot in 2006 by Alfonso Cuarón (director of Gravity)
I'll copy/paste the summary from imdb: In 2027, in a chaotic world in which women have become somehow infertile, a former activist agrees to help transport a miraculously pregnant woman to a sanctuary at sea.
the plot is simple but solid, the interpretations are great and I really like Cuaron's style, there are a couple of scenes in which I really loved some stylistic choices, a tribute to pink floyd and a fantastic long shot comes to mind
@Jasper Trying to email you via Windows 10 (I can answer any of my email accounts from there...but it keeps freezing up on me...email me and let me know if you're okay when you have the chance...I'll reply!
Then they just switch the conjugates, yeah? So product is the same.
Seems right to me.
In fact the corresponding homeomorphisms of the surface $\Sigma_g$ are not hard to see either, I don't think. You're switching the torii around before connected summing.
Need to demonstrate:
Let $A$ be open in $\mathbb{R}^m$; let $g:A \longrightarrow \mathbb{R}^n$ a function locally Lipschitz. Show that if $C$ is a compact subset of $A$, then $g$ satisfies the Lipschitz condition on $C$.
Someone can help me?
@0celo7 Pick the Lipschitz nbhds so that $\|x - x_0\| \leq \|x - y\|$: argue by saying that otherwise $y$ would be closer to $x$ than $x_0$ hence they'd fit in the same Lipschitz nbhd, in which case no issue.
Same for $\|x_0 - y\| \leq \|x - y\|$.
Cool, my argument seems like the one Willie Wong is proposing.
@0celo7 Sorry, I might have sounded a bit rude there; didn't mean to. I don't want to think about it right now, especially after a week of rather draining, not-so-interesting high-school physics.
I went through it quickly: all I can say is that it seems believable. I am sure you can figure it out.
It's not an argument I'd come up with for sure (the one I'd come up with is sketched in what I wrote).
@Alessandro The filming is indeed great, some shots are just magnificent. But I did not like all of the acting and although the plot is profound, biblical and deep it also has a lot of holes and uses deus ex machinae.
Hi guys! Let us say we have a differentiable function $f:U\subset \mathbb{R}^m\longrightarrow \mathbb{R}^n$ where $U$ is an OPEN set. Then, is $df:U\longrightarrow \mathscr{L}(\mathbb{R}^m;\mathbb{R}^n)$ NECESSARILY continuous? At a first sight I would say it is not, but I couldn't find any example...
Need to demonstrate:
Let $A$ be open in $\mathbb{R}^m$; let $g:A \longrightarrow \mathbb{R}^n$ a function locally Lipschitz. Show that if $C$ is a compact subset of $A$, then $g$ satisfies the Lipschitz condition on $C$.
Someone can help me?
