Solution.
" ⟹ \implies ⟹
Suppose u u u u u u u : D ‾ → R \func{u}{\cl{\D}}{\R} u : D → R
u ( z ) = 1 2 π ∫ 0 2 π Re ( e i θ + z e i θ − z ) u ( e i θ ) d θ , u\p{z}
= \frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} u\p{e^{i\theta}} \,\diff\theta, u ( z ) = 2 π 1 ∫ 0 2 π Re ( e i θ − z e i θ + z ) u ( e i θ ) d θ , and we may simply take f = u f = u f = u ∂ D \partial\D ∂ D
" ⟸ \impliedby ⟸
To show that u u u u u u D ‾ \cl{\D} D u ( e i θ ) = f ( e i θ ) u\p{e^{i\theta}} = f\p{e^{i\theta}} u ( e i θ ) = f ( e i θ ) u u u D ‾ \cl{\D} D D \D D
First, observe that applying the first direction to the harmonic function u ( z ) = 1 u\p{z} = 1 u ( z ) = 1
1 = 1 2 π ∫ 0 2 π Re ( e i θ + z e i θ − z ) d θ . 1 = \frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \,\diff\theta. 1 = 2 π 1 ∫ 0 2 π Re ( e i θ − z e i θ + z ) d θ . We also have
Re ( e i θ + z e i θ − z ) = 1 2 ( e i θ + z e i θ − z + e − i θ + z ‾ e − i θ − z ‾ ) = 1 2 1 + z e − i θ − z ‾ e i θ − ∣ z ∣ 2 + 1 + z ‾ e i θ − z e − i θ − ∣ z ∣ 2 ∣ e i θ − z ∣ 2 = 1 − ∣ z ∣ 2 ∣ e i θ − z ∣ 2 , \begin{aligned}
\Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}}
&= \frac{1}{2}\p{\frac{e^{i\theta} + z}{e^{i\theta} - z} + \frac{e^{-i\theta} + \conj{z}}{e^{-i\theta} - \conj{z}}} \\
&= \frac{1}{2}\frac{1 + ze^{-i\theta} - \conj{z}e^{i\theta} - \abs{z}^2 + 1 + \conj{z}e^{i\theta} - ze^{-i\theta} - \abs{z}^2}{\abs{e^{i\theta} - z}^2} \\
&= \frac{1 - \abs{z}^2}{\abs{e^{i\theta} - z}^2},
\end{aligned} Re ( e i θ − z e i θ + z ) = 2 1 ( e i θ − z e i θ + z + e − i θ − z e − i θ + z ) = 2 1 ∣ e i θ − z ∣ 2 1 + z e − i θ − z e i θ − ∣ z ∣ 2 + 1 + z e i θ − z e − i θ − ∣ z ∣ 2 = ∣ e i θ − z ∣ 2 1 − ∣ z ∣ 2 , which is non-negative for z ∈ D z \in \D z ∈ D 0 0 0 z = e i θ z = e^{i\theta} z = e i θ
Let ε > 0 \epsilon > 0 ε > 0 z 0 ∈ ∂ D z_0 \in \partial \D z 0 ∈ ∂ D f f f δ > 0 \delta > 0 δ > 0 ∣ z − z 0 ∣ < δ \abs{z - z_0} < \delta ∣ z − z 0 ∣ < δ ∣ f ( z ) − f ( z 0 ) ∣ < ε \abs{f\p{z} - f\p{z_0}} < \epsilon ∣ f ( z ) − f ( z 0 ) ∣ < ε C = { z ∈ ∂ D ∣ ∣ z − z 0 ∣ < δ } C = \set{z \in \partial\D \mid \abs{z - z_0} < \delta} C = { z ∈ ∂ D ∣ ∣ z − z 0 ∣ < δ } M M M f f f
∣ u ( z ) − f ( z 0 ) ∣ = ∣ 1 2 π ∫ 0 2 π Re ( e i θ + z e i θ − z ) ( f ( e i θ ) − f ( z 0 ) ) d θ ∣ ≤ 1 2 π ∫ 0 2 π Re ( e i θ + z e i θ − z ) ∣ f ( e i θ ) − f ( z 0 ) ∣ d θ = 1 2 π ∫ C Re ( e i θ + z e i θ − z ) ∣ f ( e i θ ) − f ( z 0 ) ∣ d θ + 1 2 π ∫ [ 0 , 2 π ] ∖ C Re ( e i θ + z e i θ − z ) ∣ f ( e i θ ) − f ( z 0 ) ∣ d θ ≕ I 1 + I 2 . \begin{aligned}
\abs{u\p{z} - f\p{z_0}}
&= \abs{\frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \p{f\p{e^{i\theta}} - f\p{z_0}} \,\diff\theta} \\
&\leq \frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \abs{f\p{e^{i\theta}} - f\p{z_0}} \,\diff\theta \\
&= \frac{1}{2\pi} \int_C \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \abs{f\p{e^{i\theta}} - f\p{z_0}} \,\diff\theta + \frac{1}{2\pi} \int_{\br{0,2\pi} \setminus C} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \abs{f\p{e^{i\theta}} - f\p{z_0}} \,\diff\theta \\
&\eqqcolon I_1 + I_2.
\end{aligned} ∣ u ( z ) − f ( z 0 ) ∣ = ∣ ∣ 2 π 1 ∫ 0 2 π Re ( e i θ − z e i θ + z ) ( f ( e i θ ) − f ( z 0 ) ) d θ ∣ ∣ ≤ 2 π 1 ∫ 0 2 π Re ( e i θ − z e i θ + z ) ∣ ∣ f ( e i θ ) − f ( z 0 ) ∣ ∣ d θ = 2 π 1 ∫ C Re ( e i θ − z e i θ + z ) ∣ ∣ f ( e i θ ) − f ( z 0 ) ∣ ∣ d θ + 2 π 1 ∫ [ 0 , 2 π ] ∖ C Re ( e i θ − z e i θ + z ) ∣ ∣ f ( e i θ ) − f ( z 0 ) ∣ ∣ d θ = : I 1 + I 2 . On I 1 I_1 I 1 ∣ e i θ − z 0 ∣ < δ \abs{e^{i\theta} - z_0} < \delta ∣ ∣ e i θ − z 0 ∣ ∣ < δ
I 1 ≤ ε 2 π ∫ 0 2 π Re ( e i θ + z e i θ − z ) d θ = ε . I_1
\leq \frac{\epsilon}{2\pi} \int_0^{2\pi} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \,\diff\theta
= \epsilon. I 1 ≤ 2 π ε ∫ 0 2 π Re ( e i θ − z e i θ + z ) d θ = ε . For I 2 I_2 I 2 ∣ z − z 0 ∣ < δ 2 \abs{z - z_0} < \frac{\delta}{2} ∣ z − z 0 ∣ < 2 δ θ ∈ [ 0 , 2 π ] ∖ C \theta \in \br{0, 2\pi} \setminus C θ ∈ [ 0 , 2 π ] ∖ C
δ ≤ ∣ e i θ − z 0 ∣ ≤ ∣ e i θ − z ∣ + ∣ z − z 0 ∣ ≤ ∣ e i θ − z ∣ + δ 2 ⟹ ∣ e i θ − z ∣ ≥ δ 2 . \delta
\leq \abs{e^{i\theta} - z_0}
\leq \abs{e^{i\theta} - z} + \abs{z - z_0}
\leq \abs{e^{i\theta} - z} + \frac{\delta}{2}
\implies \abs{e^{i\theta} - z} \geq \frac{\delta}{2}. δ ≤ ∣ ∣ e i θ − z 0 ∣ ∣ ≤ ∣ ∣ e i θ − z ∣ ∣ + ∣ z − z 0 ∣ ≤ ∣ ∣ e i θ − z ∣ ∣ + 2 δ ⟹ ∣ ∣ e i θ − z ∣ ∣ ≥ 2 δ . Moreover,
Re ( e i θ + z e i θ − z ) = 1 − ∣ z ∣ 2 ∣ e i θ − z ∣ 2 ≤ 1 − ∣ z ∣ 2 ( δ / 2 ) 2 → ∣ z ∣ → 1 0 \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}}
= \frac{1 - \abs{z}^2}{\abs{e^{i\theta} - z}^2}
\leq \frac{1 - \abs{z}^2}{\p{\delta/2}^2} \xrightarrow{\abs{z}\to1} 0 Re ( e i θ − z e i θ + z ) = ∣ e i θ − z ∣ 2 1 − ∣ z ∣ 2 ≤ ( δ /2 ) 2 1 − ∣ z ∣ 2 ∣ z ∣ → 1 0 uniformly in θ \theta θ
I 2 ≤ 2 M 2 π ∫ [ 0 , 2 π ] ∖ C Re ( e i θ + z e i θ − z ) d θ → z → z 0 0. I_2
\leq \frac{2M}{2\pi} \int_{\br{0,2\pi} \setminus C} \Re\p{\frac{e^{i\theta} + z}{e^{i\theta} - z}} \,\diff\theta \xrightarrow{z \to z_0} 0. I 2 ≤ 2 π 2 M ∫ [ 0 , 2 π ] ∖ C Re ( e i θ − z e i θ + z ) d θ z → z 0 0. In summary, lim sup z → z 0 ( I 1 + I 2 ) ≤ ε \limsup_{z \to z_0} \p{I_1 + I_2} \leq \epsilon lim sup z → z 0 ( I 1 + I 2 ) ≤ ε
lim z → z 0 u ( z ) = f ( z 0 ) . \lim_{z \to z_0} u\p{z} = f\p{z_0}. z → z 0 lim u ( z ) = f ( z 0 ) . Thus, u u u D ‾ \cl{\D} D