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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1You 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. – 2012-06-14
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1You can use Morera's theorem. – 2012-06-14
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0@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) – 2012-06-14
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0The 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). – 2012-06-14
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0@benmachine: No I don't have this information, $f$ is not necessary real-on-reals. – 2012-06-14
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0Unless $f$ is real-valued on $\mathbb R$, $\overline{f}(\overline{z})$ won't be the same as $f$ on $\mathbb R$. As for $f(\overline{z})$, that's not analytic if $f$ is analytic and non-constant. – 2012-06-14
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0@Mate: if $f$ is not real on the reals, then $F$ as in the question won't necessarily be continuous, let alone entire. Whether or not it's real on the reals, $F$ as in your comment probably won't be entire. – 2012-06-14
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0Without the hypothesis that $f(\mathbb{R}) \subseteq \mathbb{R}$, the desired result need not hold. For example if $f(z) = i$ for all $z$ with $\operatorname{Im}(z) \geq 0$, then $f$ is certainly analytic on its domain and continuous on the real axis, but the function $f^{\#}$ is the constant $-i$ and the resulting $F$ is not continuous on $\mathbb{C}$, let alone entire. When $f(\mathbb{R}) \subseteq \mathbb{R}$ this result is often called the Schwarz reflection principle. http://en.wikipedia.org/wiki/Schwarz_reflection_principle – 2012-06-14
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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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0If 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!! – 2012-06-14
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0@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. – 2012-06-14
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0I 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? – 2012-06-14
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0@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 $g$. – 2012-06-14
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0Why there should be only one such $g$? – 2012-06-14
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0Essentially because of the [identity theorem](http://en.wikipedia.org/wiki/Identity_theorem). – 2012-06-19