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Let $f$ be an analytic function in an open set $U \subseteq \mathbb{C}$. Let $V=\{z\in\mathbb C:\overline z\in U\}$. Define $g$ on $V$ by $g(z)=\overline{f(\overline{z})}$. Show that $g$ is analytic on $V$. Note: It is not sufficient to show that $g$ is holomorphic because we haven't yet proved that holomorphic implies analytic.

Here is my attempted proof:

First, we will simply state what it means that $f$ is analytic on $U$. Applying the definition, we know that since $f$ is analytic on $U$, we know that for every $z_0 \in U$, there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(z)=\sum_{n=0}^\infty a_n\left(z-z_0\right)^n$ on the disc $D(z_0,r)$.

Next, note by definition of $V$ that for all $v \in V$, we have $\overline{v} \in U$. Therefore since $f$ is analytic on $U$, then for all $v \in V$ there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(u)=\sum_{n=0}^\infty a_n\left(u-\overline{v}\right)^n$ for all $u \in D(\overline{v},r)$.

Edit (Following Daniel Fischer's suggestion) Now if we realize that if $u \in D(\overline{v},r)$, then we see $u=\overline{w}$ for some $w \in D(v,r) \subseteq V$.

Therefore, putting everything together, we have that for all $v \in V$, there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(w)=\sum_{n=0}^\infty \overline{a_n}\left(w-v\right)^n$ for all $w \in D(v,r)\subseteq V$ (is this last part correct?).

I feel like I am getting close; I just need to obtain $g$ from this somehow. But I am losing direction here... Am I on the right track? Thanks!

EthanAlvaree
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  • Forget the last paragraphs. Recall that if $u \in D(\overline{v},r)$, then $u = \overline{w}$ for some $w\in D(v,r)$. – Daniel Fischer Feb 11 '15 at 10:59
  • @DanielFischer "Since $f$ is analytic on $U$, then for all $v \in V$ there exists $r>0$ and a sequence of complex numbers $\left(a_n\right){n=0}^\infty$ such that $f(u)=\sum{n=0}^\infty a_n\left(u-\overline{v}\right)^n$ for all $u \in D(\overline{v},r)\subseteq U$. But now we realize that if $u \in D(\overline{v},r)$, then $u=\overline{w}$ for some $w \in D(v,r) \subseteq V$. Thus for all $v \in V$, there exists $r>0$ and a sequence of complex numbers $\left(a_n\right){n=0}^\infty$ such that $f(w)=\sum{n=0}^\infty \overline{a_n}\left(w-v\right)^n$ for all $w \in D(v,r)\subseteq V$." – EthanAlvaree Feb 11 '15 at 11:17
  • Yes, that is correct now - except that you wrote $f$ in a couple of places where it ought to be $g$. It might be that you need to mumble "continuity" to justify that a convergent series can be conjugated termwise, but you may also be past that stage and can do that without comment. And thus the analyticity of $g$ is established. – Daniel Fischer Feb 11 '15 at 12:17
  • Just to add a point : If the sets $U$ and $V$ share a common boundary as in the schwartz reflection theorem and we need to prove that g is the analytic continuation of $f$ in $U \union V$, we need to further assumptions that $f$ takes real values on the real line and is continuous upto the boundary. – Srinivas K Feb 11 '15 at 12:37
  • Just trying to figure out where in my argument I should be introducing $g$... – EthanAlvaree Feb 11 '15 at 12:58
  • @EthanAlvaree Generally, $w \notin U$, so $f(w)$ is not defined. It's $g(w) = \overline{f(\overline{w})} = \overline{f(u)}$. – Daniel Fischer Feb 11 '15 at 13:23
  • In the last line "$..f(w)=...$" , should've been $g(w)$ – Srinivas K Feb 11 '15 at 14:37
  • Related: http://math.stackexchange.com/q/102885/ – Jonas Meyer Feb 11 '15 at 15:22

1 Answers1

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Splitting into real and imaginary parts probably isn't all that helpful - use the fact that $\overline{p(z)}=p(\bar{z})$ for every polynomial $p$ and the continuity of $z\mapsto\bar{z}$. Let $z_0\in U$. There exists $r>0$ and a sequence $(a_n)$ in $\mathbb{C}$ such that $$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$$ for all $z\in\overline{D(z_0,r)}$. Define $$f_n(z):=\sum_{k=0}^na_k(z-z_0)^k$$ for each $n\in\mathbb{N}$. Then $f_n\to f$ uniformly on $\overline{D(z_0,r)}$. We also define $$g_n(z):=\overline{f_n(\bar{z})}=\sum_{k=0}^n\bar{a}_k(z-\bar{z}_0)^k$$ We have that $\liminf_{n\to\infty}|\bar{a}_n|^{-1/n}=\liminf_{n\to\infty}|a_n|^{-1/n}>r>0$ so the series $$\sum_{k=0}^\infty\bar{a}_k(z-\bar{z}_0)^k$$ has radius of converge $\rho>r>0$. In particular, this implies that the above series defines an analytic function $g(z)$ on $D(\bar{z}_0,\rho)$ and $g_n\to g$ uniformly on $\overline{D(\bar{z}_0,r)}$. It follows from the continuity of $z\mapsto\bar{z}$ that $g(z)=\overline{f(\bar{z})}$ on $\overline{D(\bar{z_0},r)}$. This completes the proof since $z_0\mapsto\bar{z}_0$ is a bijection $U\rightarrow V$.

Jason
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