Analysis Problem of the Day 24

Today’s problem is Problem 1 on the Johns Hopkins Spring 2021 Complex Analysis Qual:

Problem 1: Suppose $f, g$ are holomorphic on an open set $U \subset \mathbb{C}.$ Show that if $|f(z)|^4+|g(z)|^4$ attains a maximum inside $U,$ then $f$ and $g$ are constant on $U.$

Solution 1: Note that at the maximum $z_0,$ $f(z_0) = e^{i\theta_1} |f(z_0)|, g(z_0) = e^{i \theta_2} |g(z_0)|.$ Thus, by the triangle inequality $h(z)=e^{-4i \theta_1} f^4(z)+e^{-4i\theta_2} g^4(z)$ achieves a maximum at $z_0$ and is therefore constant by the maximum modulus principle. Then,

$h(z)=C \leq |f(z)|^4+|g(z)|^4 \leq M$

implies that equality in triangle inequality holds, i.e. $f^4(z) = c g^4(z)$ for some $c \in \mathbb{C}.$ This implies $\frac{f}{g}$ is a meromorphic function with image contained in the set of fourth roots of $c.$ By the open mapping theorem, this implies that $\frac{f}{g}$ is locally constant, i.e. $f=c'g.$ Then, it follows that $|f|^4(1+|c|^4)$ achieves a maximum inside $U,$ so $|f|$ achieves a maximum inside $U.$ By the classical maximum modulus principle, this then implies that $f$ (and therefore also $g$ ) are constant.

Solution 2: Recall that $\log |f|$ is subharmonic for holomorphic $f.$ Moreover, if $u$ is subharmonic and $\phi$ is convex, then $\phi \circ u$ is subharmonic. Thus, $|f|^4 + |g|^4 = e^{4\log |f|}+e^{4\log |g|}$ is subharmonic. A subharmonic function that achieves a local maximum is constant, so $|f|^4+|g|^4$ is constant. Since we have two subharmonic functions whose sum is harmonic, it follows that both $|f^4|,|g^4|$ are harmonic. By the mean value theorem for holomorphic functions, the modulus is harmonic if and only if equality in the triangle inequality for integrals holds. This is true if and only if $f,g$ is constant.

Remark: Both methods show that for any number of holomorphic functions $f_i,$ if $\sum_i |f_i|^{p_i}$ achieves a maximum inside $U,$ then each $f_i$ is constant.

Leave a Reply Cancel reply