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I'm reading a proof for the statement: let $G\subset\mathbb{C}$ be an open set containing all inner regions of piecewise horizontal or vertical closed linear curves in $G$, and let $u$ be harmonic on $G$. Then there exists a harmonic conjugate $v$ of $u$ on $G$.

The proof goes on as follows.

Since $u$ is harmonic, it satisfies $\frac{\partial^2u}{\partial\overline{z}\partial z}=0$ on $G$, thus $g=\frac{\partial u}{\partial z}$ satisfies C-R. Define the holomorphic function $F(z)=\int_a^z g(w)\,\mathrm{d}w$, where we integrate over a piecewise horizontal or vertical linear curve (it follows from Cauchy that it doesn't matter which one). Then $\frac{\partial F}{\partial z}=\frac{\partial u}{\partial z}=g(z)$. Then \begin{align*}\frac{\partial }{\partial z}(F+\overline{F})&=\frac{\partial u}{\partial z};\\\frac{\partial }{\partial \overline{z}}(F+\overline{F})&=\frac{\partial u}{\partial \overline{z}}.\end{align*} The last equation follows because $u$ is a real valued function. Thus $F+\overline{F}-u=C\in\mathbb{R}$, thus $u=2\mathrm{Re}(F)-C$, and we could take $v=2\mathrm{Im}(F)$.

I can follow this proof, except for the step $\frac{\partial }{\partial z}(F+\overline{F})=\frac{\partial u}{\partial z}$. They mention that this is true because $\frac{\partial\overline{F}}{\partial z}=0$. I don't understand how we could just take the derivative of a complex conjugate of a function; also in general this step does not seem to make sense. Similar problems arise at the second line of the aligned part. I'm looking for some explanation for this step. Any help or reference is much appreciated!

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  • $\begingroup$ What does "containing all piecewise horizontal or vertical linear curves" mean??? $\endgroup$ – David C. Ullrich Apr 24 '19 at 18:37
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Write \begin{align*} F(z) & = F(x,y) = u(x,y) + iv(x,y) \\ \overline{F}(z) & = \overline{F}(x,y) = u(x,y) - iv(x,y). \end{align*} Then: \begin{align*} \frac{\partial F}{\partial z} & = \frac{1}{2}\left( \frac{\partial F}{\partial x} - i \frac{\partial F}{\partial y} \right) \\ & = \frac{1}{2}\left( u_x + iv_x - i(u_y + iv_y) \right) \\ & = \frac{1}{2}\left( (u_x + v_y) + i(v_x - u_y) \right) \tag{1} \end{align*} while on the other hand \begin{align*} \frac{\partial \overline{F}}{\partial \overline{z}} & = \frac{1}{2}\left( \frac{\partial \overline{F}}{\partial x} + i \frac{\partial \overline{F}}{\partial y} \right) \\ & = \frac{1}{2}\left( u_x - iv_x + i(u_y - iv_y) \right) \\ & = \frac{1}{2}\left( (u_x + v_y) + i(- v_x + u_y) \right) \\ & = \frac{1}{2}\left( (u_x + v_y) - i(v_x - u_y) \right) \tag{2} \end{align*} Expressions (1) and (2) are complex conjugates of each other. We deduce that: $$\overline{\frac{\partial F}{\partial z}} = \frac{\partial \overline{F}}{\partial \overline{z}}$$

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  • $\begingroup$ Thank you, this makes it clear to me! $\endgroup$ – Václav Mordvinov Apr 24 '19 at 19:14
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They are using the Wirtinger derivative: $$\dfrac{\partial}{\partial z}=\dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\bigg)\\ \dfrac{\partial}{\partial \bar z}=\dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\bigg) $$

It is then easily calculated $$\dfrac{\partial}{\partial z}\bar F(z)=\dfrac{\partial}{\partial \bar z}F(z) =0.$$

EDIT: The calculation. Write $F(x,y)=u(x,y)+iv(x,y)$ so $\bar F(x,y)=u(x,y)-iv(x,y)$. Then $$\begin{align} \dfrac{\partial}{\partial z}\bar F & = \dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\bigg)(u-iv)\\ & = \dfrac{1}{2}(u_x-v_y)-\dfrac{i}{2}(u_y+v_x) \\ & = \overline{\dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\bigg)(u+iv)}\\ & = \overline{\dfrac{\partial}{\partial\bar z}F}\\ & = \bar 0 =0. \end{align} $$So it would be better to say $$\dfrac{\partial}{\partial \bar z}F(z)=0=\bar 0 =\dfrac{\partial}{\partial z}\bar F(z).$$

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  • $\begingroup$ I understand that they are using the Wirtinger derivative, but the part you state in your last line, $\dfrac{\partial}{\partial z}\bar F(z)=\dfrac{\partial}{\partial \bar z}F(z)$ is exactly the part I do not understand. In general also $\overline{f(z)}\neq f(\overline{z})$, right? $\endgroup$ – Václav Mordvinov Apr 24 '19 at 17:26
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"where we integrate over a piecewise horizontal or vertical linear curve (it follows from Cauchy that it doesn't matter which one)": It certainly does matter which one, unless you know something you're not telling us. If $g(z)=1/z$ then the integral of $g$ over the piecewise linear curve joining $-1-i$, $-1+i$ and then $1+i$ is not the same as the integral over the curve from $-1-i$ to $1-i$ and then to $1+i$.

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  • $\begingroup$ Okay, I meant that $G$ contains all inner regions of piecewise linear curves in $G$. My apologies since I did not state it in the original post. I have some difficulties translating since my syllabus is not English. Thanks, I will edit $\endgroup$ – Václav Mordvinov Apr 24 '19 at 18:47

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