Solution.
First, notice that this set is non-empty: if f ( z ) = z − i z + i f\p{z} = \frac{z-i}{z+i} f ( z ) = z + i z − i f f f Ω → D \Omega \to \D Ω → D ∣ f ′ ( i 2 ) ∣ \abs{f'\p{\frac{i}{2}}} ∣ ∣ f ′ ( 2 i ) ∣ ∣ Re f ′ ( i 2 ) \Re{f'\p{\frac{i}{2}}} Re f ′ ( 2 i )
Let f : Ω → D \func{f}{\Omega}{\D} f : Ω → D a ∈ D a \in \D a ∈ D φ a : D → D \func{\phi_a}{\D}{\D} φ a : D → D
φ a ( z ) = z − a 1 − a ‾ z \phi_a\p{z} = \frac{z - a}{1 - \conj{a}z} φ a ( z ) = 1 − a z z − a which is an automorphism of D \D D a a a 0 0 0
φ a ′ ( z ) = 1 − a ‾ z + ( z − a ) a ‾ ( 1 − a ‾ z ) 2 = 1 − ∣ a ∣ 2 ( 1 − a ‾ z ) 2 ⟹ φ a ′ ( a ) = 1 − ∣ a ∣ 2 ( 1 − ∣ a ∣ 2 ) 2 = 1 1 − ∣ a ∣ 2 . \phi_a'\p{z}
= \frac{1 - \conj{a}z + \p{z - a}\conj{a}}{\p{1 - \conj{a}z}^2}
= \frac{1 - \abs{a}^2}{\p{1 - \conj{a}z}^2}
\implies
\phi_a'\p{a}
= \frac{1 - \abs{a}^2}{\p{1 - \abs{a}^2}^2}
= \frac{1}{1 - \abs{a}^2}. φ a ′ ( z ) = ( 1 − a z ) 2 1 − a z + ( z − a ) a = ( 1 − a z ) 2 1 − ∣ a ∣ 2 ⟹ φ a ′ ( a ) = ( 1 − ∣ a ∣ 2 ) 2 1 − ∣ a ∣ 2 = 1 − ∣ a ∣ 2 1 . Let ψ − 1 ( z ) = z − i 2 z + i 2 \psi^{-1}\p{z} = \frac{z - \frac{i}{2}}{z + \frac{i}{2}} ψ − 1 ( z ) = z + 2 i z − 2 i Ω → D \Omega \to \D Ω → D i 2 \frac{i}{2} 2 i 0 0 0
ψ ( z ) = i 2 ( 1 + w 1 − w ) . \psi\p{z} = \frac{i}{2} \p{\frac{1 + w}{1 - w}}. ψ ( z ) = 2 i ( 1 − w 1 + w ) . We also have
ψ ′ ( z ) = i 2 ( 1 − w + 1 + w ( 1 − w ) 2 ) = i ( 1 − w ) 2 . \psi'\p{z}
= \frac{i}{2} \p{\frac{1 - w + 1 + w}{\p{1 - w}^2}}
= \frac{i}{\p{1 - w}^2}. ψ ′ ( z ) = 2 i ( ( 1 − w ) 2 1 − w + 1 + w ) = ( 1 − w ) 2 i . Hence, g = φ f ( i / 2 ) ∘ f ∘ ψ : D → D g = \func{\phi_{f\p{i/2}} \circ f \circ \psi}{\D}{\D} g = φ f ( i /2 ) ∘ f ∘ ψ : D → D g ( 0 ) = 0 g\p{0} = 0 g ( 0 ) = 0
1 ≥ ∣ ( φ f ( i / 2 ) ∘ f ∘ ψ ) ′ ( 0 ) ∣ = ∣ φ f ( i / 2 ) ′ ( ( f ∘ ψ ) ( 0 ) ) f ′ ( ψ ( 0 ) ) ψ ′ ( 0 ) ∣ = ∣ φ f ( i / 2 ) ′ ( f ( i 2 ) ) f ′ ( i 2 ) i ∣ = ∣ f ′ ( i 2 ) ∣ 1 − ∣ f ( i 2 ) ∣ 2 ≥ ∣ f ′ ( i 2 ) ∣ . \begin{aligned}
1
&\geq \abs{\p{\phi_{f\p{i/2}} \circ f \circ \psi}'\p{0}} \\
&= \abs{\phi_{f\p{i/2}}'\p{\p{f \circ \psi}\p{0}} f'\p{\psi\p{0}} \psi'\p{0}} \\
&= \abs{\phi_{f\p{i/2}}'\p{f\p{\frac{i}{2}}} f'\p{\frac{i}{2}} i} \\
&= \frac{\abs{f'\p{\frac{i}{2}}}}{1 - \abs{f\p{\frac{i}{2}}}^2} \\
&\geq \abs{f'\p{\frac{i}{2}}}.
\end{aligned} 1 ≥ ∣ ∣ ( φ f ( i /2 ) ∘ f ∘ ψ ) ′ ( 0 ) ∣ ∣ = ∣ ∣ φ f ( i /2 ) ′ ( ( f ∘ ψ ) ( 0 ) ) f ′ ( ψ ( 0 ) ) ψ ′ ( 0 ) ∣ ∣ = ∣ ∣ φ f ( i /2 ) ′ ( f ( 2 i ) ) f ′ ( 2 i ) i ∣ ∣ = 1 − ∣ ∣ f ( 2 i ) ∣ ∣ 2 ∣ ∣ f ′ ( 2 i ) ∣ ∣ ≥ ∣ ∣ f ′ ( 2 i ) ∣ ∣ . So 1 1 1
( ψ − 1 ) ′ ( z ) = z + i 2 − z + i 2 ( z + i 2 ) 2 = i ( z + i 2 ) 2 ⟹ ∣ ( ψ − 1 ) ′ ( i 2 ) ∣ = 1 , \p{\psi^{-1}}'\p{z}
= \frac{z + \frac{i}{2} - z + \frac{i}{2}}{\p{z + \frac{i}{2}}^2}
= \frac{i}{\p{z + \frac{i}{2}}^2}
\implies
\abs{\p{\psi^{-1}}'\p{\frac{i}{2}}}
= 1, ( ψ − 1 ) ′ ( z ) = ( z + 2 i ) 2 z + 2 i − z + 2 i = ( z + 2 i ) 2 i ⟹ ∣ ∣ ( ψ − 1 ) ′ ( 2 i ) ∣ ∣ = 1 , so the upper bound is attained. Hence, the supremum is 1 1 1