Solution.
For 0 < R < 1 0 < R < 1 0 < R < 1 u R ( z ) = u ( R z ) u_R\p{z} = u\p{Rz} u R ( z ) = u ( R z ) D ‾ \cl{\D} D
u R ( r e i θ ) = 1 2 π ∫ 0 2 π 1 − r 2 1 − 2 r cos ( θ − t ) + r 2 u R ( e i t ) d t ⟹ u R ( x , 0 ) = 1 2 π ∫ 0 2 π 1 − x 2 1 − 2 x cos ( t ) + x 2 u R ( e i t ) d t . \begin{gathered}
u_R\p{re^{i\theta}}
= \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos\p{\theta - t} + r^2} u_R\p{e^{it}} \,\diff{t} \\
\implies
u_R\p{x, 0}
= \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - x^2}{1 - 2x\cos\p{t} + x^2} u_R\p{e^{it}} \,\diff{t}.
\end{gathered} u R ( r e i θ ) = 2 π 1 ∫ 0 2 π 1 − 2 r cos ( θ − t ) + r 2 1 − r 2 u R ( e i t ) d t ⟹ u R ( x , 0 ) = 2 π 1 ∫ 0 2 π 1 − 2 x cos ( t ) + x 2 1 − x 2 u R ( e i t ) d t . Since u R u_R u R
∂ u R ∂ x ( x , 0 ) = 1 2 π ∫ 0 2 π − 2 x ( 1 − 2 x cos ( t ) + x 2 ) + ( 1 − x 2 ) ( 2 cos ( t ) + 2 x ) ( 1 − 2 x cos ( t ) + x 2 ) 2 u R ( e i t ) d t . \pder{u_R}{x}\p{x, 0}
= \frac{1}{2\pi} \int_0^{2\pi} \frac{-2x\p{1 - 2x\cos\p{t} + x^2} + \p{1 - x^2}\p{2\cos\p{t} + 2x}}{\p{1 - 2x\cos\p{t} + x^2}^2} u_R\p{e^{it}} \,\diff{t}. ∂ x ∂ u R ( x , 0 ) = 2 π 1 ∫ 0 2 π ( 1 − 2 x cos ( t ) + x 2 ) 2 − 2 x ( 1 − 2 x cos ( t ) + x 2 ) + ( 1 − x 2 ) ( 2 cos ( t ) + 2 x ) u R ( e i t ) d t . Hence, because u ( z ) ∈ [ 0 , 1 ] u\p{z} \in \br{0, 1} u ( z ) ∈ [ 0 , 1 ] cos t \cos{t} cos t
∣ ∂ u R ∂ x ( 0 , 0 ) ∣ = ∣ 1 2 π ∫ 0 2 π 2 cos ( t ) u R ( e i t ) d t ∣ ≤ 1 2 π ∫ − π / 2 π / 2 2 cos ( t ) d t ≤ 2 π . \begin{aligned}
\abs{\pder{u_R}{x}\p{0, 0}}
&= \abs{\frac{1}{2\pi} \int_0^{2\pi} 2\cos\p{t} u_R\p{e^{it}} \,\diff{t}} \\
&\leq \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} 2\cos\p{t} \,\diff{t} \\
&\leq \frac{2}{\pi}.
\end{aligned} ∣ ∣ ∂ x ∂ u R ( 0 , 0 ) ∣ ∣ = ∣ ∣ 2 π 1 ∫ 0 2 π 2 cos ( t ) u R ( e i t ) d t ∣ ∣ ≤ 2 π 1 ∫ − π /2 π /2 2 cos ( t ) d t ≤ π 2 . By the chain rule, ∂ u R ∂ x ( x , y ) = R ∂ u ∂ x ( x , y ) \pder{u_R}{x}\p{x, y} = R\pder{u}{x}\p{x, y} ∂ x ∂ u R ( x , y ) = R ∂ x ∂ u ( x , y ) R → 1 R \to 1 R → 1 ∣ ∂ u ∂ x ( 0 , 0 ) ∣ ≤ 2 π \abs{\pder{u}{x}\p{0, 0}} \leq \frac{2}{\pi} ∣ ∣ ∂ x ∂ u ( 0 , 0 ) ∣ ∣ ≤ π 2 f : S 1 → R \func{f}{S^1}{\R} f : S 1 → R f ( e i θ ) = 1 f\p{e^{i\theta}} = 1 f ( e i θ ) = 1 cos ( θ ) ≥ 0 \cos\p{\theta} \geq 0 cos ( θ ) ≥ 0 0 0 0 f ∈ L 1 ( S 1 ) f \in L^1\p{S^1} f ∈ L 1 ( S 1 )
u ( z ) = 1 2 π ∫ 0 2 π 1 − r 2 1 − 2 r cos ( θ − t ) + r 2 f ( e i t ) d t u\p{z}
= \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos\p{\theta - t} + r^2} f\p{e^{it}} \,\diff{t} u ( z ) = 2 π 1 ∫ 0 2 π 1 − 2 r cos ( θ − t ) + r 2 1 − r 2 f ( e i t ) d t defines a harmonic function with ∂ u ∂ x ( 0 , 0 ) = 2 π \pder{u}{x}\p{0, 0} = \frac{2}{\pi} ∂ x ∂ u ( 0 , 0 ) = π 2 u ( z ) ≥ 0 u\p{z} \geq 0 u ( z ) ≥ 0
u ( z ) ≤ 1 2 π ∫ 0 2 π 1 − r 2 1 − 2 r cos ( θ − t ) + r 2 d t = 1 , u\p{z}
\leq \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos\p{\theta - t} + r^2} \,\diff{t}
= 1, u ( z ) ≤ 2 π 1 ∫ 0 2 π 1 − 2 r cos ( θ − t ) + r 2 1 − r 2 d t = 1 , so u ( z ) ∈ [ 0 , 1 ] u\p{z} \in \br{0, 1} u ( z ) ∈ [ 0 , 1 ]