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I have trouble proving that $h(z)=\overline{h(\overline{z})}$ for all $z \in \mathbb{C}$ under the assumption that $h$ is holomorphic and the real line maps itself or in other words: $h(z) \in \mathbb{R}$ if $z \in \mathbb{R}$.

I showed that $\overline{h(\overline{z})}$ is holomorphic and thus $\overline{h'(\overline{z})}=h'(z)$ for $z \in \mathbb{R}$ but don't know how to continue from here.

I was also wondering if this is true, does that mean $h(-z)=-h(z)$ for all $z \in \mathbb{C}$ if we also assume that the imaginary axis maps itself?

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    $\begingroup$ What does the identity theorem tell you about $h(z)$ and $g(z) = \overline{h(\overline{z})}$? $\endgroup$ – Daniel Fischer Feb 5 '16 at 16:02
  • $\begingroup$ I know it would hold if $\overline{h'(\overline{z})}=h'(z)$ holds for all $z \in \mathbb{C}$. However I only know how to show this in $\mathbb{R}$. $\endgroup$ – Jan Feb 5 '16 at 16:04
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    $\begingroup$ Forget the derivatives. Just look at the function values. $\endgroup$ – Daniel Fischer Feb 5 '16 at 16:07
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What you need is the identity theorem:

Let $U \subset \mathbb{C}$ be a connected open set, and $f, g \colon U \to \mathbb{C}$ two holomorphic functions. If the set $\{ z \in U : f(z) = g(z)\}$ has an accumulation point in $U$, then $f \equiv g$.

This strong identity theorem is particular to holomorphic [and antiholomorphic(1)] functions, and it is particular to (complex) one-dimensional domains. A weaker version of the identity theorem holds for real-analytic functions or for holomorphic functions of several complex variables.

Armed with the strong identity theorem, it only remains to note that the two holomorphic functions $h$ and $g \colon z \mapsto \overline{h(\overline{z})}$ coincide on $\mathbb{R}$ since by the premise we have $h(\mathbb{R}) \subset \mathbb{R}$, and as every point of $\mathbb{R}$ is an accumulation point of $\mathbb{R}$ it follows that indeed $g \equiv h$, i.e. $h(z) = \overline{h(\overline{z})}$ for all $z \in \mathbb{C}$.

I was also wondering if this is true, does that mean $h(−z)=−h(z)$ for all $z \in \mathbb{C}$ if we also assume that the imaginary axis maps itself?

Indeed, if we additionally assume that $h$ maps the imaginary axis to itself, then $h$ must be an odd function. Applying the identity $h(z) = \overline{h(\overline{z})}$ to purely imaginary $z$, we then find that the holomorphic functions $h$ and $k \colon z \mapsto -h(-z)$ coincide on the imaginary axis - since $\overline{w} = -w$ for $w \in i\mathbb{R}$ - and hence by the identity theorem $h \equiv k$, i.e. $h$ is odd.

If we only assume that $h$ maps the imaginary axis to itself, we find $h(z) = - \overline{h(-\overline{z})}$ for all $z\in \mathbb{C}$, since $w = -\overline{w}$ for purely imaginary $w$.


(1) An antiholomorphic function is a function whose conjugate is holomorphic. Since conjugation is a simple operation, the properties of antiholomorphic functions are easily derived from the properties of holomorphic functions, and antiholomorphic functions share many of the nice properties of holomorphic functions. The main point that makes holomorphic functions much nicer than antiholomorphic functions is that a composition of two holomorphic functions is again holomorphic, while the composition of two antiholomorphic functions is - unless it's (locally) constant - not antiholomorphic but holomorphic.


My search-fu is weak today. I'm sure essentially this question must have been answered here previously, but I could only find https://math.stackexchange.com/questions/1540431/f-overlinez-overlinefz-given-certain-conditions, which doesn't have an answer.

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    $\begingroup$ I looked extensively but could not find a clear explanation. Thank you for your elaborate answer. $\endgroup$ – Jan Feb 6 '16 at 17:42

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