MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f(z)$ is an analytic function on a domain $D$ which maps negative axis to negative axis. For $s>1$ consider the function $$u(z)=\Re \sqrt[s]{f(z)}$$ with the branch cut along the negative axis. Is $u(z)$ subharmonic on $D\setminus\lbrace z\in D: f(z)=0\rbrace?$ When applying the maximum principle to $u(z)$ must one check values in $D$ along the negative axis?

share|cite|improve this question
Why is $u(z)$ a well-defined function? You can define it in regions away from the zeros of $f(z)$, but in any neighborhood of $a zero of $f(z)$ there is no way to define it in a continuous way in general, so I wouldn't expect it to be subharmonic, however it is defined. – John Jiang Apr 22 '12 at 5:03
If using the principal branch of the $1/s$ power, you do get a continuous function: note that $$\lim_{z \to -x} \Re z^{1/s} = x^{1/s} \cos(\pi/s)$$ – Robert Israel Apr 22 '12 at 5:20
John: I think you are correct. The function cannot be subharmonic near $f(z)=0$ so I revised the question. – Daniel Parry Apr 22 '12 at 20:10
It should also be subharmonic where $f(z)=0$, again if you are using the principal branch. For example with $f(z)=z$, $\frac{1}{2\pi} \int_{-\pi}^\pi \Re(r^{1/s} e^{i\theta/s})\ d\theta \ge 0$. – Robert Israel Apr 22 '12 at 22:02
up vote 2 down vote accepted

Assuming you are using the principal branch of the $s$'th root, $u(z)$ is indeed a subharmonic function. You do need to "check the branch cut", but it works out ok.

Consider a point $p$ where $f(p)$ is on the negative real axis (so $p$ is on a branch cut of $f(z)^{1/s}$), and let $v(z)$ be the version of $\Re f(z)^{1/s}$ obtained by moving the branch cut slightly away from $p$ in some direction. Then $v(z)$, being the real part of an analytic function, is harmonic near $p$, and $u(z) \ge v(z)$. So for small $r > 0$, $$u(p) = v(p) = \frac{1}{2\pi} \int_0^{2\pi} v(p + r e^{i\theta})\ d\theta \le \frac{1}{2\pi} \int_0^{2\pi} u(p + r e^{i\theta})\ d\theta $$

share|cite|improve this answer
To be clear, $u(z)=v(z)$ identically correct? – Daniel Parry Apr 22 '12 at 20:07
No, $u(z) = v(z)$ on one side of the original branch cut, not on the other. For example if $f(z) = z$ and you move the branch cut down, $v(r e^{i\theta}) = r^{1/s} \cos(\theta/s)$ for $r > 0$ and $\theta$ near $\pi$, but $u(r e^{i \theta}) = r^{1/s} \cos(\theta/s)$ for $\theta \le \pi$, $r^{1/s} \cos((\theta-2\pi)/s)$ for $\theta > \pi$. – Robert Israel Apr 22 '12 at 21:55
One last question. You said that I still need to "check the branch cut" even though it may not be part of the boundary of $D.$ I am a little confused because it contradicts the maximum principle I know of for subharmonic functions. – Daniel Parry Apr 23 '12 at 3:28
When I said "check the branch cut", I meant that you have to check whether the function is subharmonic there. It wasn't obvious, and it might not be true if you didn't use the principal branch. Once you know the function is subharmonic on a domain, you can use the maximum principle on that domain. – Robert Israel Apr 23 '12 at 5:00
Thank you Robert you have been very helpful. – Daniel Parry Apr 24 '12 at 2:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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