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I want to show that the functions $|x|,|z|^\alpha(\alpha \geq0),\log(1+|z|^2):\mathbb C \to \mathbb R$ are all subharmonic. (This is an exercise from Ahlfors' text)


  • $|x|$

$x$ and $-x$ are harmonic functions, thus subharmonic as well. $|x|=\max(x,-x)$ is the maximum of two subharmonic functions, and is consequently subharmonic as well.


  • $|z|^\alpha (\alpha \geq 0) $

The case $\alpha=0$ is obvious, so we may assume $\alpha>0$.

Since subharmonicity is a local property, we will prove subharmonicity around each point: Given any $z_0 \neq 0$ we may construct a branch of $z^\alpha$ which is analytic around $z_0$, and for a suitable choice of an angle $\theta$, $\text{Log}(e^{i \theta} z^\alpha)$ is analytic around $z_0$ as well. (Log is the principal branch of the logarithm, with branch cut $(-\infty,0]$).

Now, $u=\Re \text{Log}(e^{i \theta} z^\alpha)= \log |e^{i \theta} z^\alpha|=\log |z^\alpha|$ is harmonic in some neighborhood of $z_0$, and $U=|z|^\alpha=\exp(u(x,y))$ is subharmonic around $z_0$ since its Laplacian is $$\Delta U=e^u( \Delta u+u_x^2+u_y^2 )=e^u \| \nabla u \|^2 \geq 0$$

If $U$ isn't subharmonic at $z_0=0$, there exists a neighborhood $\Omega_0$ of $0$ and a harmonic function $u_0:\Omega_0 \to \mathbb R$ such that $U-u_0$ has its maximal value in $\overline{\Omega_0}$ at $z=0$, which is $U(0)-u_0(0)=-u_0(0)$.

But since $U \geq 0$ for all $z \in \overline{\Omega_0}$ $$-u_0(z) \leq U(z)-u_0(z) \leq -u_0(0)$$
and this forces $u$ to be constant. Using the same inequality shows that $U \equiv 0$ in $\Omega_0$. This is a contradiction, thus $U=|z|^\alpha$ is subharmonic throughout $\mathbb C$.


  • $\log(1+|z|^2)$

$$\Delta \log(1+x^2+y^2)=\frac{4}{(1+x^2+y^2)^2}>0 $$


I hope all cases are correct. I've used some standard results from the text. If there are any errors, I'd like to know.

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The proofs are correct. I think the second one can be shorter, though. Observe that to verify the sub-mean-value property of any function $u$, it suffices to find, for every point $a\in\mathbb C$, a function $h$ that is harmonic in some neighborhood of $u$, satisfies $h\le u$ in that neighborhood, and has $h(a)=u(a)$. Indeed, a circular mean of $u$ is at least the corresponding mean of $h$, which is equal to $u(a)$.

For $u(z)=|z|^\alpha$, use

  • $h(z)=0$ if $a=0$. This obviously satisfies the above requirements.
  • $h(z)=|a|^{\alpha}\operatorname{Re} (z/a)^\alpha$ if $a\ne 0$. Here the branch of $(z/a)^\alpha$ is chosen so that that $1^\alpha=1$. We have $h(a)=u(a)$ and $h(z)\le |a|^\alpha |z/a|^\alpha = u(z)$ for other $z$.

Remark: the inequality $h\le u$ kind of goes against the word subharmonic, but that's how it works. A subharmonic function is under a harmonic one if they share boundary values; but it can be above a harmonic function with which it shares an interior value. Same with convexity: the graph of a convex function lies below its secants, and above its tangents.

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