Let \[ M(r)=\sup_{|z|=r}|f(z)|, \] where \(f:\mathbb C\to\mathbb C\) is entire and not identically zero. Suppose that, for some \(r>0\) with \(r\ne1\), \[ M(r^2)^2=M(r)M(r^3). \] Prove that \[ f(z)=az^n \] for some \(a\ne0\) and integer \(n\ge0\).
Solution
Set \[ R_1=\min(r,r^3),\qquad R_2=r^2,\qquad R_3=\max(r,r^3). \] Since \(r\neq1\), \[ R_1<R_2<R_3, \qquad \log R_2=\frac{\log R_1+\log R_3}{2}. \]
1. Let \(h(z)=\alpha\log|z|+\beta\) be the harmonic radial function on\(R_1<|z|<R_3\) satisfying\[ h=\log M(R_1)\quad\text{on }|z|=R_1, \qquad h=\log M(R_3)\quad\text{on }|z|=R_3. \]Then\[ \log|f(z)|\le h(z) \]throughout the annulus.
The function \(\log|f|\) is subharmonic. On each boundary circle it is at most the corresponding value of \(\log M\), hence at most \(h\). The maximum principle for subharmonic functions gives the inequality in the annulus. This is the usual proof of Hadamard’s three-circles theorem.
2. The hypothesis implies\[ h(z)=\log M(R_2) \]on \(|z|=R_2\).
Because \(\log R_2\) is the midpoint of \(\log R_1\) and \(\log R_3\), \[ h|_{|z|=R_2} =\frac12\bigl(\log M(R_1)+\log M(R_3)\bigr). \] The given equality \[ M(r^2)^2=M(r)M(r^3) \] is exactly \[ 2\log M(R_2)=\log M(R_1)+\log M(R_3), \] so the two quantities agree.
3. One has\[ \log|f(z)|=h(z) \]throughout \(R_1<|z|<R_3\).
Choose \(z_0\) on \(|z|=R_2\) with \[ |f(z_0)|=M(R_2). \] By 2, \[ \log|f(z_0)|-h(z_0)=0. \] By 1, the subharmonic function \[ u(z)=\log|f(z)|-h(z) \] is at most zero on the annulus, so it attains its maximum at the interior point \(z_0\). The strong maximum principle for subharmonic functions forces \(u\equiv0\). In particular, \(f\) has no zeros in this annulus.
4. On the annulus,\[ z\frac{f'(z)}{f(z)}=\alpha. \]
Since \(f\) is nowhere zero there, \(f'/f\) is holomorphic. From 3, \[ \log|f(z)|=\alpha\log|z|+\beta. \] Writing \(z=re^{i\theta}\), differentiation in the radial direction gives \[ \Re\!\left(z\frac{f'}f\right)=\alpha, \] while differentiation in \(\theta\) gives \[ \Im\!\left(z\frac{f'}f\right)=0. \] Hence the holomorphic function \(zf'/f\) is identically the real constant \(\alpha\).
5. The number \(\alpha\) is an integer.
For any circle \(|z|=\rho\) contained in the annulus, 4 gives \[ \frac1{2\pi i}\int_{|z|=\rho}\frac{f'(z)}{f(z)}\,dz =\frac1{2\pi i}\int_{|z|=\rho}\frac{\alpha}{z}\,dz =\alpha. \] The left side is the winding number of \(f(|z|=\rho)\) about \(0\), hence is an integer. Thus \(\alpha=n\in\mathbb Z\).
6. There is a constant \(a\neq0\) such that\[ f(z)=az^n \]on the annulus.
By 4 and 5, \[ \frac{f'}f=\frac nz. \] Thus \[ \left(\frac{f(z)}{z^n}\right)'=0 \] on the connected annulus, so \(f(z)/z^n=a\) there for some nonzero constant \(a\).
7. In fact \(n\ge0\) and the same identity holds on all of \(\mathbb C\).
The Laurent expansion of the entire function \(f\) about \(0\) has no negative powers. Since it agrees on an annulus with the single Laurent monomial \(az^n\), one must have \(n\ge0\). Then both \(f\) and \(az^n\) are entire, and they agree on a nonempty open annulus. The identity theorem gives \[ f(z)=az^n \] for all \(z\in\mathbb C\).