Let $f(z)=P(x^{2}-y^{2})+iQ(x,y)$ be a holomorphic function where P and Q are of class $C^{2}$

I tried to use the fact that $P$ is harmonic , but couldn't continue


closed as off-topic by cmk, Lee David Chung Lin, Xander Henderson, nmasanta, Leucippus Jul 30 at 5:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – cmk, Lee David Chung Lin, Xander Henderson, nmasanta, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.


Let $u = \Re(h)$ where $h$ is analytic and non-constant. More generally, $P(u)$ is harmonic iff $P$ is linear.

Proof: The if direction is clear. Assuming $P(u)$ is harmonic, we have $(u_x^2+u_y^2)P''(u) = 0$ by direct calculation of the Laplacian. If $u_x(x_0, y_0) = u_y(x_0, y_0) = 0,$ we obtain $h'(z_0) = 0$ where $z_0 = x_0+iy_0.$

Let $K = \text{Ker}(h').$ If $K$ is dense in $\mathbb{C},$ we have $h' \equiv 0$ on all of $\mathbb{C}$ by any number of arguments, contradiction. Otherwise, there must be an open neighborhood $S \subseteq \mathbb{C}$ on which $h'$ is non-zero. This corresponds to an open neighborhood $S' \subseteq \mathbb{R}^2$ on which $R(u) \equiv 0,$ where $R = P''.$ By the open mapping theorem for harmonic functions, this corresponds to an open set $S''=u(S') \subseteq \mathbb{R}$ on which $R \equiv 0.$ We wish to show that $R \equiv 0$ on all of $\mathbb{R}.$

Unfortunately, we are thwarted at this point by the existence of bump functions. We must strengthen the hypothesis to $P$ being analytic (here, having a convergent Taylor series) to get rid of them. Then $R$ is analytic too, and hence $R \equiv 0.$

I do not claim to have come up with the general case immediately. Here is the original draft:

Note that $P(x^2-y^2) = P(\Re((x+iy)^2)) = P(\Re(z^2)).$ With that in mind, we should aim to prove that $P$ is constant or linear so that $f(z) = az^2+b$ or $f(z) \equiv c$ are the only solutions (by applying the Cauchy-Riemann equations).

It turns out that your abandoned line of reasoning works wonders: We have $0 = \Delta P(x^2-y^2) = 4(x^2+y^2)P''(x^2-y^2).$ Choosing $(x,y) = (\sqrt{t}, 0), (0, \sqrt{t}), (1,1)$ as $t$ traverses the positive reals shows that $P'' \equiv 0,$ which implies the result.


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