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.

Is this allowed: \((f(z + \Delta z ) = f(z + i + \Delta z )\) , since z+\(\Delta z \) is not necessarily on the line \( [0, 1] \) ? You might still be right, but I just thought of a different explanation: We know that f is analytic in a neighborhood of Q, if we take a point for example on the bottom line \( z \in [0, 1] \) and the point \(z + i\) we know that f has a converging power series at both points. We know that these power series have the same values on the horizontal line, which means by the identity principle they must be the same (by the same i mean they have the same coefficients and take the same values if you shift between z and z+i). this way we can continue f just like we initially wanted, does this make sense? best, carl

Okay, there seemed to be quite some confusion about this, so I put it clearly in the proof. I think the thing with the power series doesn't work exactly because by strict use of the identity principle it is just equivalent on a neighborhood (the identity principle for domains as we've seen in class would lead to a circulus in probando). But I've combined the two ideas we had to make it as clear as possible. Best, A.