Difference between revisions of "Talk:Aufgaben:Problem 14"

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But you have only proved that the function f is doubly periodic on the boundary of [0,1]x[0,1], not on the interior... So you cannot say that is elliptic on \( \mathbb{C} \).
 
But you have only proved that the function f is doubly periodic on the boundary of [0,1]x[0,1], not on the interior... So you cannot say that is elliptic on \( \mathbb{C} \).
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I think I can anyway: For the holomorphic continuation of \( f \) I'll just define \( f(z + i) = f(z) \) and \( f(z + 1) = f(z) \). If we have an interior point, we just sort of "copy" that point to the area outside. The whole thing is smooth since \( f \) is smooth on the closure of Q. From the identity principle it follows that this continuation is unique, so \( f \) after all has to be constant. Does that make sense? Best, A.

Revision as of 16:23, 8 January 2015

What should this "Pierre's lemma" be? It's clearly false: let \( f = u + iv \) (with \( u \) and \( v \) real-valued) be any non-constant analytic function; then clearly \( u \) or \( v \) is not constant. Following "Pierre's lemma" they should both be analytic, but we've showed in Ch. II.3 that any analytic real-valued function is constant, which is a contradiction. (And in any case, in the "proof" \( a \) and \( b \) should be different from \( 0 \)...)

Cheers, the tables

Let's say u and v are meromorphic and everything is okay again, I guess. f has a Taylorexpansion, can be divided into an a and ib part, yadiyadiya, v is elliptic and so on.

Cheers, the guy hating you for pointing out his mistakes

I'm sorry, but I guess somebody has to point out mistakes when there are any, even if it's a tough job... Anyway, I fear it still doesn't work: if they're meromorphic, then they're holomorphic at "almost every point", which for the previous argument means that they're constant almost everywhere, which of course is not the case for all holomorphic functions f. I didn't want to waste your work, but we can't write it in the MMP proof like that, don't you agree?

I was just joking. Due to my strange academic sense of humor, people usually think I can't handle criticism but I actually love it. That's why I like cooking with my girlfriend so much.

Now back to the serious stuff: It's a sketch so far I did for the people solving the problem. They'll work on it, don't worry. If f has a Taylorexpansion, then u and v do from the argument already there, so v has no poles. Then use part a). That should make you happy, I guess.

Otherwise please correct me again. Happy new year by the way. A.


I'm not sure that we can apply part (a) at the end of the proof 1 (in (b)). The function must be elliptic on C, in the sense that must be doubly periodic on C, not on Z+iZ. Otherwise the proof of part (a) doesn't work, because we can't apply Liouville's theorem.


Maybe the notation is not the best. The holomorphic continuation of \( f \) is of course elliptic on \( \mathbb{C} \). What I wanted to say with that was that define the elliptic continuation on the lattice Z + iZ. I'll change that. Best, A.


But you have only proved that the function f is doubly periodic on the boundary of [0,1]x[0,1], not on the interior... So you cannot say that is elliptic on \( \mathbb{C} \).


I think I can anyway: For the holomorphic continuation of \( f \) I'll just define \( f(z + i) = f(z) \) and \( f(z + 1) = f(z) \). If we have an interior point, we just sort of "copy" that point to the area outside. The whole thing is smooth since \( f \) is smooth on the closure of Q. From the identity principle it follows that this continuation is unique, so \( f \) after all has to be constant. Does that make sense? Best, A.