A harmonic h is continuous and satisfies the mean value property, so h(z) = \frac{1}{2\pi}\int h(z + re^{i\phi})\,d\phi \le \frac{1}{2\pi}\int\max\{h, a\}(z + re^{i\phi})\,d\phi. For u + h: the sum of an upper semicontinuous and a continuous function is upper semicontinuous, and if |h| \le H on a small circle then \max\{u, a - H\} + h \le \max\{u + h, a\} there, so the mean value property of h gives u(z) + h(z) \le \frac{1}{2\pi}\int\max\{u, a-H\} + \frac{1}{2\pi}\int h \le \frac{1}{2\pi}\int\max\{u + h, a\}.

Upper semicontinuity follows from the continuity of f and of \log : [0, \infty) \to [-\infty, \infty). At a zero of f there is nothing to prove. At a point z with f(z) \ne 0 take r > 0 with \{|w - z| \le r\} \subseteq U; Jensen's formula gives \log|f(z)| = \dfrac{1}{2\pi}\int_0^{2\pi}\log|f(z + re^{i\phi})|\,d\phi - \sum_{|w - z| < r}\operatorname{ord}_w(f)\log\dfrac{r}{|w - z|}, the sum running over the zeros w of f in the open disc, and the sum is nonnegative. As \log|f| is integrable on the circle, its circle average is the infimum of the averages of its truncations, which differ from \log|f| only on the finite set of zeros of f on the circle.

Strong maximum principle: if u attains its supremum M over \Omega at z_0 \in \Omega, then u = M on \Omega. Indeed, for small r the sub-mean-value inequality gives M = u(z_0) \le \frac{1}{2\pi}\int_0^{2\pi}\max\{u(z_0 + re^{i\phi}), a\}\,d\phi \le M for every a \le M, so u = M almost everywhere on every small circle around z_0; by upper semicontinuity the set \{u \ge M\} is closed and contains, with almost every point of every small circle, a neighbourhood of z_0 (every point near z_0 lies on such a circle, and the set \{u < M\} is open, so it cannot meet the circles only in null sets unless it is empty near z_0); hence \{u = M\} is open and closed in the connected set \Omega (SubharmonicOn.eqOn_const_of_isMaxOn). Now let g(\zeta) = \limsup_{z \to \zeta,\ z \in \Omega} u(z) for \zeta \in \overline{\Omega}: g is upper semicontinuous, equals u on \Omega and is at most 0 on \partial\Omega, so it attains its maximum on the compact set \overline{\Omega}. If u were positive somewhere, this maximum would be positive and attained at a point of \Omega, so u would be a positive constant on \Omega, contradicting the boundary condition at a point of \partial\Omega \ne \emptyset.

\int_{\mathbb{R}}\dfrac{h\,dx}{(x-a)^2 + h^2} = \bigl[\arctan\dfrac{x - a}{h}\bigr]_{-\infty}^{\infty} = \pi.

The kernel is positive with total mass 1 (Lemma 3.4.6).

P[b] is the imaginary part of the Nevanlinna integral N[b](z) = \dfrac{1}{\pi}\int_{\mathbb{R}}\Bigl(\dfrac{1}{x - z} - \dfrac{x}{1 + x^2}\Bigr)b(x)\,dx, whose kernel is O((1 + x^2)^{-1}) locally uniformly in z \in \mathbb{H}, together with its z-derivative; differentiation under the integral sign shows that N[b] is holomorphic on \mathbb{H}, so P[b] = \operatorname{Im} N[b] is harmonic.

Given \varepsilon > 0 choose \delta > 0 with |b(x) - b(x_0)| \le \varepsilon for |x - x_0| < \delta; since the kernel has total mass 1 (Lemma 3.4.6), |P[b](z) - b(x_0)| \le \varepsilon + \int_{|x - x_0| \ge \delta} P(z, x)\,|b(x) - b(x_0)|\,dx, and on |x - x_0| \ge \delta one has P(z, x) \le C\,\operatorname{Im} z\,(1 + x^2)^{-1} for z near x_0, so the last integral tends to 0 as z \to x_0.

Let u \le M on \mathbb{H} and \varepsilon > 0. The function h(z) = \sum_{x_0 \in E}\log\Bigl|\dfrac{z - x_0}{z - x_0 + 2i}\Bigr| - \log|z + i| is harmonic on \mathbb{H}, nonpositive on the closed upper half-plane, tends to -\infty at the points of E, and satisfies h(z) \le -\log(|z| - 1) for |z| > 1. Consider u + \varepsilon h, subharmonic (Lemma 3.4.2) on the half-disc \Omega_R = \{|z| < R\} \cap \mathbb{H}, with R so large that M - \varepsilon\log(R - 2) \le 0: at real boundary points outside E its \limsup is at most 0 because \varepsilon h \le 0, at the points of E because u \le M and \varepsilon h \to -\infty, and at boundary points of modulus R because u + \varepsilon h \le M - \varepsilon\log(R - 2) near them. The maximum principle (Lemma 3.4.4) gives u \le -\varepsilon h on \Omega_R, hence on \mathbb{H}, for every \varepsilon > 0; let \varepsilon \to 0.

For n \in \mathbb{N} the truncation b_n = \max\{b, -n\} is bounded below, so P[b_n] \ge -n and u - P[b_n] is subharmonic on \mathbb{H} (Lemma 3.4.9, Lemma 3.4.2) and bounded above by M + n (Lemma 3.4.8). At a real point x \notin E, \limsup u \le b(x) \le b_n(x) = \lim P[b_n] (Lemma 3.4.10), so u \le P[b_n] on \mathbb{H} by the extended maximum principle (Lemma 3.4.11). Finally P[b_n] \downarrow P[b] as n \to \infty by monotone convergence.

Apply Theorem 3.4.12 to u = \log|f|, which is subharmonic by Lemma 3.4.3 and bounded above by \log\sup|f|.