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Let $u(x,y)$ be a harmonic function on $\mathbb{R^2}$ and $v(x,y)$ be a harmonic conjugate of $u(x,y)$ on $\mathbb{R^2}$. Suppose that the partial derivative $v_x(x,y)<C$ for a real constant $C$. Show that $u(x,y)$ is a linear function.

I have derived the following:

Since $v$ is the harmonic conjugate of $u$, we have $u_x=v_y$ & $u_y=-v_x$

Since both $u$ and $v$ are harmonic, we also have $u_{xx}+u_{yy}=0$ & $v_{xx}+v_{yy}=0$

However, I cant establish a prove given the following identities.

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$v_x$ is harmonic as well: $$ (v_x)_{xx} + (v_x)_{yy} = (v_{xx})_x + (v_{yy})_x = (v_{xx} + v_{yy})_x = 0 $$ behause a harmonic function is infinitely often differentiable and the order of partial derivatives can be changed arbitrarily.

Then $C - v_x$ is a non-negative harmonic function in the plane, and therefore constant. It follows that $v$ is a linear function. Now use the Cauchy Riemann equations to conclude that $u$ is a linear function.

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  • $\begingroup$ May I know why is $v_x$ harmonic? $\endgroup$ – Jayne Leblanc Mar 17 '19 at 10:00
  • $\begingroup$ @JayneLeblanc: See update. $\endgroup$ – Martin R Mar 17 '19 at 10:04
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Let $f=u+iv $, then $f $ is entire and $f'=u_x+iv_x $. Since $v_x <C $, $f'$ is constant, by Picard.

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