Let \(u\) be a subharmonic function on \(\mathbb C\), and put \(M(r)=\max_{|z|=r}u(z)\). For \(0<r_1<r_2\), prove that
\[ u(z)\leq \frac{\log r_2-\log|z|}{\log r_2-\log r_1}M(r_1) +\frac{\log|z|-\log r_1}{\log r_2-\log r_1}M(r_2) \]
for \(r_1\leq|z|\leq r_2\).
The displayed fractions require \(r_1<r_2\). If the radii coincide, the annulus is a circle and the corresponding bound is simply \(u(z)\leq M(r_1)\). For extended-valued subharmonic functions, interpret the boundary bounds directly as \(u(z)\leq M(r_j)\), so no product \(0\cdot(-\infty)\) is needed.
Solution
We use upper semicontinuity and the submean inequality that define subharmonicity. First suppose \(u\) is finite-valued.
1. The right-hand side is a harmonic function on the open annulus.
Upper semicontinuity implies that \(u\) attains a finite maximum on each compact circle. Let \(h(z)\) be the displayed linear interpolation of \(M(r_1)\) and \(M(r_2)\). It has the form \(a\log|z|+b\) for real constants \(a,b\). The radial Laplacian of \(\log|z|\) for \(z\ne0\) is \(-1/|z|^2+(1/|z|)(1/|z|)=0\), so \(h\) is harmonic. It is continuous on the closed annulus and equals \(M(r_j)\) on its \(j\)th boundary circle. Consequently \(v=u-h\) is upper semicontinuous, satisfies the submean inequality, and is nonpositive on the boundary. The submean inequality follows by subtracting the mean-value equality for the harmonic function \(h\) [@SS03].
2. The comparison \(u\leq h\) holds throughout the closed annulus.
For \(\varepsilon>0\), put \[ v_\varepsilon(z)=v(z)+\varepsilon(|z|^2-r_2^2). \] It is nonpositive on both boundary circles. If it were positive somewhere, upper semicontinuity and compactness would give a positive global maximum at an interior point \(a\). Choose a small circle \(a+\rho e^{it}\) whose disk lies in the open annulus. The submean inequality for \(v\) and the identity \[ \frac1{2\pi}\int_0^{2\pi}|a+\rho e^{it}|^2\,dt =|a|^2+\rho^2 \] give \[ v_\varepsilon(a) \leq\frac1{2\pi}\int_0^{2\pi}v_\varepsilon(a+\rho e^{it})\,dt -\varepsilon\rho^2 \leq v_\varepsilon(a)-\varepsilon\rho^2, \] a contradiction. Thus \(v_\varepsilon\leq0\). Letting \(\varepsilon\downarrow0\) proves \(v\leq0\), or \(u\leq h\).
3. The argument also covers the value \(-\infty\).
For an extended-valued subharmonic \(u\), set \(u_N=\max(u,-N)\) for positive integers \(N\). This is finite-valued and upper semicontinuous. It satisfies the submean inequality: at a center where \(u_N=u\), use the inequality for \(u\) and \(u\leq u_N\) on the circle; at a center where \(u_N=-N\), use \(u_N\geq-N\) on the circle. Its circular maximum is \(M_N(r)=\max(M(r),-N)\).
Apply steps <1>1–<1>2 to \(u_N\). At an interior point of the annulus both interpolation coefficients are positive. Letting \(N\to\infty\) gives the required inequality in the extended real numbers, since \(u_N\downarrow u\) and \(M_N(r_j)\downarrow M(r_j)\). On a boundary circle the bound is the defining maximum inequality. Thus all the stated cases are covered.