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$U$ is the square with verices in $0, 1, i, i+1$. I though immediately using the identity principle. The thing is what I know about it is that, if U is an open and connected set, holomorph in U such that exists $z_n, f, g$ such that $f(z_n)=g(z_n)$ and $\lim_\limits{n\to\infty}z_n\in U$ then $f=g$.

The problem I have here, is that the simple $z_n=1/n$ is not un $U$, but in $\bar{U}$. Can I use it anyway?

Thanks

Lorenzo B.
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Silkking
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  • Do you know the Schwarz reflection principle? – Paul Frost May 18 '18 at 14:32
  • I don't. I just read what it says, and, if I'm not mistaken, I could extend $f$ adding the square with vertices in $0, -i, 1, 1-i$ using the Schwarz reflection principle and there I could use what I said, taking $z_n=0.1+{1 \over n+1}$. Am I right? – Silkking May 18 '18 at 15:29
  • Yes, you are right! – Paul Frost May 18 '18 at 15:51
  • Formally your question is still unanswered (if you look at https://math.stackexchange.com/ you see that is has $0$ answers). I therefore suggest that you write down as an official answer what you learnt by the above comments. Yes, you can answer you own questions! – Paul Frost May 18 '18 at 21:57
  • The purpose is that everybody can see at first glance that the queation is no longer open. – Paul Frost May 18 '18 at 22:05
  • Thanks. I will. I have also remembered other way to do it. I'll see now if I write both or just one. Thanks – Silkking May 19 '18 at 15:59

1 Answers1

1

After chatting with different people, I've reached two ansers. In both it can be easily seen that $f(z)=e^z+cos(z)$ is a solution. Let $\bar{f}$ be another function that satisfies the condition:

1) Using the Schwarz reflection principle, I can extend both functions $f$ and $\bar{f}$ to the square with vertices in $0, 1, -i, 1-i$ and consider the function $g(z)=f(z)-\bar{f}(z)$. I call the new domain as $V$. Then, I can define $z_n=0.1+\frac{1}{n+1}$. This sequence is contained in the interior of $V$, and converges in $0.1$ that is in the interior of $V$. Then, the Identity Principle says the new function is $0$ for every $z$ in the domain. Then $g \equiv 0$. Then $f \equiv \bar{f}$

2) I define the function $g(z)=i\cdot(z-\frac{1}{2}-\frac{i}{2}) + \frac{1}{2} + \frac{i}{2}$. It's not hard to prove that this rotates the square in $\frac{\pi}{2}$. I also define the functions $h_0(z)=f(z)-\bar{f}(z)$, $h_1(z)=h(g(z))$, $h_2(z)=h(g^2(z))$, $h_3(z)=h(g^3(z))$. I define $h(z)=h_0(z)\cdot h_1(z) \cdot h_2(z) \cdot h_3(z)$. Then, $z\in\partial U \Rightarrow h(z)=0$. Then, using the maximum module principle, $h\equiv 0$. Then, since $h$ is product of holomorph functions, it must be that one of them is the zero function. But if one of them is, the other ones are too, because I can obtain either of them from the others composing with $g$ enough times. Then $h_0 \equiv 0$. Then $f \equiv \bar{f}$

Silkking
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  • My understanding of your question is that one has to find all continuous functions $f : \overline{U} \to \mathbb{C}$ such that $f \mid_U$ is holomorphic and $f(t) = \cos(t) + e^t$ for $t \in [0,1]$. This cannot be $f = 0$. Clearly, $f(z) = \cos(z) + e^z$ will do and it remains to show that this is the only solution. Assume you have another solution $\tilde{f}$. Then it suffices to show $f - \tilde{f} = 0$ which can be done as in 1) above. Concerning 2): The product of non-zero continuous functions may be $0$. – Paul Frost May 20 '18 at 15:59
  • Thanks. I wrote the answer remembering the idea, but not reading the question. I'll edit it right now – Silkking May 21 '18 at 12:54
  • You have to replace $h$ by $h_0$ in the definitions of $h_1, h_2,h_3$. Then define $h = h_0 \cdot h_1 \cdot h_2 \cdot h_3$. Moreover, you have use the fact the the $h_i$ are holomorphic in $U$ to conclude that some $h_i = 0$ - the product of nonzero continuous functions may be zero. To see that if $a(z) \cdot b(z) \equiv 0$ implies $a(z) \equiv 0$ or $b(z) \equiv 0$ for holomorphic $a , b : U \to \mathbb{C}$ consider any sequence $(z_n)$ in $U$ converging to $z' \in U$. Then one af $a$ or $b$ must be zero at infinitely many $z_n$. W.l.o.g. assume that $a(z_n) = 0$ for all $n$. – Paul Frost May 22 '18 at 08:25
  • Thanks for the corrections. I've edited again. One question: what is W.l.o.g? Trying to guess... without loss of generality? – Silkking May 22 '18 at 13:25
  • Yes, your guess is correct! – Paul Frost May 22 '18 at 13:40