Solution.
Given B ( z , r ) ⊆ D B\p{z, r} \subseteq \D B ( z , r ) ⊆ D
∣ f n ( z ) ∣ 2 = ∣ 1 2 π ∫ 0 2 π f n ( z + r e i θ ) d θ ∣ 2 ≤ 1 4 π 2 ( ∫ 0 2 π ∣ f n ( z + r e i θ ) ∣ 2 d θ ) ( ∫ 0 2 π d θ ) = 1 2 π ∫ 0 2 π ∣ f n ( z + r e i θ ) ∣ 2 d θ , \begin{aligned}
\abs{f_n\p{z}}^2
&= \abs{\frac{1}{2\pi} \int_0^{2\pi} f_n\p{z + re^{i\theta}} \,\diff\theta}^2 \\
&\leq \frac{1}{4\pi^2} \p{\int_0^{2\pi} \abs{f_n\p{z + re^{i\theta}}}^2 \,\diff\theta} \p{\int_0^{2\pi} \,\diff\theta} \\
&= \frac{1}{2\pi} \int_0^{2\pi} \abs{f_n\p{z + re^{i\theta}}}^2 \,\diff\theta,
\end{aligned} ∣ f n ( z ) ∣ 2 = ∣ ∣ 2 π 1 ∫ 0 2 π f n ( z + r e i θ ) d θ ∣ ∣ 2 ≤ 4 π 2 1 ( ∫ 0 2 π ∣ ∣ f n ( z + r e i θ ) ∣ ∣ 2 d θ ) ( ∫ 0 2 π d θ ) = 2 π 1 ∫ 0 2 π ∣ ∣ f n ( z + r e i θ ) ∣ ∣ 2 d θ , so ∣ f n ∣ 2 \abs{f_n}^2 ∣ f n ∣ 2 ∣ f n ∣ 2 \abs{f_n}^2 ∣ f n ∣ 2 f n f_n f n ∣ f n ∣ 2 \abs{f_n}^2 ∣ f n ∣ 2 F F F
It remains to show locally uniformly convergence. First, we will show that if u u u B ( z 0 , r ) ‾ ⊆ D \cl{B\p{z_0, r}} \subseteq \D B ( z 0 , r ) ⊆ D v v v B ( z 0 , r ) B\p{z_0, r} B ( z 0 , r ) u u u
v ( z ) ≤ u ( z ) = 1 2 π ∫ 0 2 π Re ( r e i θ + z − z 0 r e i θ − ( z − z 0 ) ) v ( z 0 + r e i θ ) d θ v\p{z} \leq u\p{z} = \frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{re^{i\theta} + z - z_0}{re^{i\theta} - \p{z - z_0}}} v\p{z_0 + re^{i\theta}} \,\diff\theta v ( z ) ≤ u ( z ) = 2 π 1 ∫ 0 2 π Re ( r e i θ − ( z − z 0 ) r e i θ + z − z 0 ) v ( z 0 + r e i θ ) d θ on ∂ B ( z 0 , r ) \partial B\p{z_0, r} ∂ B ( z 0 , r ) v ( z ) ≤ u ( z ) v\p{z} \leq u\p{z} v ( z ) ≤ u ( z ) B ( z 0 , r ) B\p{z_0, r} B ( z 0 , r )
Let K = B ( 0 , R ) ‾ K = \cl{B\p{0, R}} K = B ( 0 , R ) 0 < R < 1 0 < R < 1 0 < R < 1 B ( 0 , 1 + R 2 ) ‾ ⊆ D \cl{B\p{0, \frac{1 + R}{2}}} \subseteq \D B ( 0 , 2 1 + R ) ⊆ D R ′ = 1 + R 2 R' = \frac{1 + R}{2} R ′ = 2 1 + R z ∈ K z \in K z ∈ K
v ( z ) ≤ 1 2 π ∫ 0 2 π Re ( R ′ e i θ + z R ′ e i θ − z ) v ( R ′ e i θ ) d θ = 1 2 π ∫ 0 2 π R ′ − ∣ z ∣ 2 ∣ R ′ e i θ − z ∣ 2 v ( R ′ e i θ ) d θ ≤ 1 2 π ∫ 0 2 π ( R ′ ) 2 − R ( R ′ − R ) 2 v ( R ′ e i θ ) d θ ≤ R ′ + R R ′ − R v ( 0 ) ≕ M K v ( 0 ) . \begin{aligned}
v\p{z}
&\leq \frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{R'e^{i\theta} + z}{R'e^{i\theta} - z}} v\p{R'e^{i\theta}} \,\diff\theta \\
&= \frac{1}{2\pi} \int_0^{2\pi} \frac{R' - \abs{z}^2}{\abs{R'e^{i\theta} - z}^2} v\p{R'e^{i\theta}} \,\diff\theta \\
&\leq \frac{1}{2\pi} \int_0^{2\pi} \frac{\p{R'}^2 - R}{\p{R' - R}^2} v\p{R'e^{i\theta}} \,\diff\theta \\
&\leq \frac{R' + R}{R' - R} v\p{0} \\
&\eqqcolon M_K v\p{0}.
\end{aligned} v ( z ) ≤ 2 π 1 ∫ 0 2 π Re ( R ′ e i θ − z R ′ e i θ + z ) v ( R ′ e i θ ) d θ = 2 π 1 ∫ 0 2 π ∣ R ′ e i θ − z ∣ 2 R ′ − ∣ z ∣ 2 v ( R ′ e i θ ) d θ ≤ 2 π 1 ∫ 0 2 π ( R ′ − R ) 2 ( R ′ ) 2 − R v ( R ′ e i θ ) d θ ≤ R ′ − R R ′ + R v ( 0 ) = : M K v ( 0 ) . for any non-negative subharmonic v v v F N = ∑ n = 1 N ∣ f n ∣ 2 F_N = \sum_{n=1}^N \abs{f_n}^2 F N = ∑ n = 1 N ∣ f n ∣ 2 F N F_N F N N ≤ M N \leq M N ≤ M
∣ F M ( z ) − F N ( z ) ∣ ≤ M K ∣ F M ( 0 ) − F N ( 0 ) ∣ , \abs{F_M\p{z} - F_N\p{z}}
\leq M_K \abs{F_M\p{0} - F_N\p{0}}, ∣ F M ( z ) − F N ( z ) ∣ ≤ M K ∣ F M ( 0 ) − F N ( 0 ) ∣ , so { F N } n \set{F_N}_n { F N } n K K K 0 0 0 F F F