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If we have an analytic function $f(z)$ in the upper half-plane, which is also continuous on the real line. If we define a new function as $$ F(z) = \left\{ \begin{array}{lr} f(z) &: z\in \mathbb U \cup \mathbb R\\ f^{\#}(z) & :z\in \mathbb L \;\;\;\;\;\;\; \end{array} \right. $$ where $\mathbb U $ is the open half-plane, and $\mathbb L $is the open lower half plane, $f^{\#}(z)=\bar{f}(\bar{z})$. Is the function $F(z)$ an entire function? Why?

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You may want the additional hythesis that the restriction of $f$ to the real line is real-valued. Also, if this is homework, please tell it, and explain what you have done; otherwise, I'd like to know the motivation. –  D. Thomine Jun 14 '12 at 0:47
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You can use Morera's theorem. –  Jonas Meyer Jun 14 '12 at 0:48
    
@D.Thomine: You mean the problem could be in the $\#$ sign? If this is the case I think we can use the definition $$ F(z) = \left\{ \begin{array}{lr} f(z) &: z\in \mathbb U \cup \mathbb R\\ f(\bar{z}) & :z\in \mathbb L \;\;\;\;\;\;\; \end{array} \right. $$ instead. (BTW, this is not a homework) –  Mate Jun 14 '12 at 0:51
    
The new definition is very different, and I'm pretty sure isn't in general entire (whereas your first one was, if you add the real-on-reals condition). –  Ben Millwood Jun 14 '12 at 0:55
    
@benmachine: No I don't have this information, $f$ is not necessary real-on-reals. –  Mate Jun 14 '12 at 1:00
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1 Answer

To summarise the conclusions of the comments:

  • If $f$ isn't real-valued on $\mathbb R$, say $f(a) = b + ci$ for some $a,b,c\in \mathbb R$, $c \not= 0$, then we have $$\begin{align} \lim_{n\to\infty} F(a+i/n)&=\lim_{n\to\infty}f(a+i/n)\\ &= f(a) = b + ci\\ \lim_{n\to\infty} F(a-i/n)&=\;\lim_{n\to\infty}\overline{f(\overline{a-i/n})} \\ &=\lim_{n\to\infty} \overline{f(a+i/n)} \\ &= \overline{f(a)} = b - ci \end{align}$$. Hence the limits from above the real axis and below disagree at $a$, so $F$ is not continuous, hence cannot possibly be analytic.
  • If $f$ is real-valued on $\mathbb R$ then Morera's theorem will give that $F$ is analytic (hence entire): this result is called the Schwarz reflection principle.
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If I understand the argument well, then the function $F$ defined by $$ F(z) = \left\{ \begin{array}{lr} f(z) &: z\in \mathbb U \cup \mathbb R\\ g(z) & :z\in \mathbb L \;\;\;\;\;\;\; \end{array} \right. $$ is entire if $f$ is analytic in $\mathbb U\cup\mathbb R$, $g$ is analytic in $\mathbb L\cup\mathbb R$, and $f(x)=g(x)$ for all $x\in \mathbb R$, I'm right!! –  Mate Jun 14 '12 at 1:54
    
@Mate: I think that's correct, but note that for each $f$ there will (unless I'm mistaken) only be at most one $g$ that satisfies those conditions. –  Ben Millwood Jun 14 '12 at 11:40
    
I didn't get it? What I have is that, given an analytic functions $f$ in the closed upper half plane and another analytic function $g$ in the closed lower half plane which agrees with $f$ on the real axis, then I ddefine a function $F$ asin my comment above, and I think it must be entire. Is this true? –  Mate Jun 14 '12 at 16:06
    
@Mate: yes. My comment only meant that if you're given $f$, there might not be an analytic $g$ that is equal to it on the real axis, and if there is such a $g$, then there's only one such $g$. –  Ben Millwood Jun 14 '12 at 18:18
    
Why there should be only one such $g$? –  Mate Jun 14 '12 at 20:23
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