Talk:Aufgaben:Problem 14

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.

I don't think you can say that the whole thing is smooth can you? if you only know that f is double periodic on the boundary, your continuation would be continues, but how do you know it is smooth? I just added a third proof. But it still has the same problem. -Carl

Isn't what you said already the deal? Let's do everything with \( f ' \). We know that this is a holomorphic function as well. Let's have a look at the continuation: w.l.o.g let's take \( z \in [0, 1] \) and \( [i, i + 1] \): \( f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z + \Delta z ) - f(z)}{\Delta z} = \lim_{\Delta z \rightarrow 0} \frac{f(z + i + \Delta z ) - f(z + i)}{\Delta z} = f'(z+i) \). So \( f' \) is doubly periodic as well on the boundary. Since you agree with me that the continuation of a holomorphic function has to be continuous, \( f' \) has to be continuous on the elliptic continuation. Thus, the continuation of \( f \) is holomorphic.

Can I convince you with that? Best, A.