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I have a question about a theorem and its proof from the book, Functions of one complex variable(John B. Conway) 3.2 section.

Theorem:

Let $\pmb{G}$ be either the whole plane $\mathbb{C}$ or some open disk. if $ u: \pmb{G} \rightarrow \mathbb{R}$ is a harmonic function then $u$ has a harmonic conjugate.

Proof:

Let $\pmb{G} = \pmb{B}(0; R)$, $0\lt R \leq \infty$ and let $ u: \pmb{G} \rightarrow \mathbb{R}$ be a harmonic function. The proof will be accomplished by finding a harmonic function $v$ such that $u$ and $v$ satisfy the Cauchy-Riemann equation. so define

$\mathit{v}(x,y) = \int^y_0u_x(x,t)dt+\phi(x)$

and determine $\phi$ so that $v_x=-u_x$. differentiating both sides of this equation with respect to $x$ gives

$v_x(x,y)= \int^y_0u_{xx}(x,t)dt+ \phi'(x)$

$=-\int_0^yu_{yy}(x,t)dt+\phi'$

$=-u_y(x,y)+u_y(x,0)+\phi'(x)$

so it must be that $\phi'(x)=-u_y(x,0)$. It is easily checked that $u$ and

$v(x,y)=\int_0^yu_x(x,t)dt - \int_0^xu_y(s,0)ds$

do satisfy the Cauchy-Riemann equations.


So, my question is that where we used an open disk or whole $\mathbb{C}$ to prove this theorem? (according to the statement)

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  • 2
    $\begingroup$ simply connectedness of domain is necessary for the well-definedness of the $v$ constructed. $\endgroup$ – r9m Oct 21 at 18:45
  • $\begingroup$ So why an open ball and why not close? $\endgroup$ – Paul Oct 22 at 4:11
  • $\begingroup$ @Paul derivatives play nice on open sets since we can approach from all directions and thus use the definition of the derivative. $\endgroup$ – Brevan Ellefsen Oct 22 at 8:15

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