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?
Analytic and entire functions
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complex-analysis
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0@leslie townes: So could we make changes to the definition so that $F$ be entire, like deviding or multiplying $f$ and $f^{\#}$ by a factor? – 2012-06-14
1 Answers
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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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0Essentially because of the [identity theorem](http://en.wikipedia.org/wiki/Identity_theorem). – 2012-06-19