- Give another proof of Lemma 1.3 by showing directly that the Laplacian of \(u \circ F\) is zero.
[Hint: The real and imaginary parts of F satisfy the Cauchy-Riemann equations.]
Solution
Write \[ F(x,y)=(p(x,y),q(x,y)), \] where \(F\) is holomorphic, and let \(u\) be harmonic on the target of \(F\). Put \(v=u\circ F\). By the two-variable chain rule, \[ \begin{aligned} \Delta v ={}&u_{pp}(p_x^2+p_y^2) +2u_{pq}(p_xq_x+p_yq_y) +u_{qq}(q_x^2+q_y^2)\\ &+u_p\Delta p+u_q\Delta q. \end{aligned} \] Because \(F\) is holomorphic, the Cauchy–Riemann equations give \[ p_x=q_y,\qquad p_y=-q_x. \] Hence \[ p_xq_x+p_yq_y=0, \qquad p_x^2+p_y^2=q_x^2+q_y^2=|F'|^2. \] Moreover the real and imaginary parts \(p,q\) of a holomorphic function are harmonic, so \(\Delta p=\Delta q=0\). Therefore \[ \Delta(u\circ F) =|F'|^2(u_{pp}+u_{qq})\circ F =|F'|^2(\Delta u)\circ F=0. \] Thus \(u\circ F\) is harmonic.