# if $u$ is a harmonic function then $u$ has a harmonic conjugate.

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)

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