0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $D(0,1).$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.