Let $u:D(0,1)\to \mathbf{R}$ be harmonic on the unit disc, and suppose there exists a $z_0\in D(0,1)$ such that all partial derivatives of $u$ vanish. Show that $u$ is constant.

I found this problem on a complex analysis qual, so it shouldn't require much more than the harmonic mean value property, but I have no idea how to proceed.


Let's use the fact that $u = \text {Re} \,f$ for some $f$ holomorphic in $D(0,1).$ Write $f=u+iv.$ Then for any $z\in D(0,1),$

$$f'(z) = u_x(z) + iv_x(z) = u_x(z) - iu_y(z),$$

where have used the Cauchy-Riemann equations to get the second equality. This implies $f''(z) = u_{xx}(z) - iu_{yx}(z),$ $f'''(z) = u_{xxx}(z) - iu_{yxx}(z),$ etc. Hence all of $f$'s derivatives can be written in terms of the partial derivatives of $u.$ This tells us that at $z_0,$ the Taylor series of $f$ is just the one term $f(z_0).$ Thus $f=f(z_0)$ in some $D(z_0,r).$ By the identity principle, $f = f(z_0)$ in $D(0,1),$ which implies $u=u(z_0)$ in $D(0,1).$

  • $\begingroup$ how do we know that there exists a function $v$ such that $f=u+iv$ is holomorphic? $\endgroup$ – alpastor Apr 20 '19 at 2:57
  • 1
    $\begingroup$ @alpastor This is a standard result in complex analysis: Any real harmonic function in a simply connected domain is the real part of a holomorphic function there. $\endgroup$ – zhw. Apr 20 '19 at 3:27

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