Subharmonicity is equivalent to harmonic comparison on compactly contained discs

Assume condition 1. Let $h$ be continuous on $\overline{D(a,r)}$, harmonic on $D(a,r)$, and satisfy $h\ge u$ on $\mathrm{\partial }D(a,r)$. On $D(a,r)$ define $v:=u-h$. For $z\in D(a,r)$ and every $0<\rho <\mathrm{dist}(z,\mathrm{\partial }D(a,r))$, the submean inequality for $u$ and the circle mean-value property for $h$ give [L1, L4, given] $v(z)=u(z)-h(z)\le \frac{1}{2\pi }{\int }_0^{2\pi }(u(z+\rho e^{it})-h(z+\rho e^{it}))dt\mathrm{.}$ Thus $v$ is subharmonic on $D(a,r)$. If some point of $D(a,r)$ satisfied $v>0$, then upper semicontinuity on the compact disc would make $v$ attain a positive interior maximum there, contradicting [L6] because $v\le 0$ on $\mathrm{\partial }D(a,r)$. Hence $v\le 0$ on $D(a,r)$, so $h\ge u$ throughout the disc. This is condition 2.

Assume condition 2. Fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ and write $g(\zeta )=u(\zeta )$ on $\mathrm{\partial }D(a,r)$. By [L2], $g$ is Borel measurable and bounded above. On the compact circle, define [given, L2, construct] ${\varphi }_n(\zeta ):={\sup }_{\eta \in \mathrm{\partial }D(a,r)}(g(\eta )-n\mathrm{\mid }\zeta -\eta \mathrm{\mid })\mathrm{.}$ Each ${\varphi }_n$ is finite and continuous, satisfies ${\varphi }_n\ge g$, and decreases pointwise to $g$ because $g$ is upper semicontinuous.

Transporting the Poisson solution from the unit disc by [L3] and [L4], let $h_n$ be the harmonic function on $D(a,r)$, continuous on $\overline{D(a,r)}$, whose boundary values are ${\varphi }_n$. Since ${\varphi }_n\ge g=u$ on $\mathrm{\partial }D(a,r)$, condition 2 gives $u\le h_n$ on $D(a,r)$. Evaluating at the center and using the Poisson formula at the center of a disc, [step 1.2, L3, L4] $u(a)\le h_n(a)=\frac{1}{2\pi }{\int }_0^{2\pi }{\varphi }_n(a+r{e}^{it})dt\mathrm{.}$

Let $M$ be an upper bound for ${\varphi }_1$ on the circle. Then $M-{\varphi }_n$ is an increasing sequence of nonnegative boundary functions, so [L5] gives [step 1.2, step 2.1, L1, L5] ${\lim }_{n\to \mathrm{\infty }}\frac{1}{2\pi }{\int }_0^{2\pi }{\varphi }_n(a+r{e}^{it})dt=\frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$ Passing to the limit in step 2.1 yields the circle submean inequality at $a$. Since the disc was arbitrary and upper semicontinuity is already part of condition 2, condition 1 follows.