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Spring 2017 - Problem 12

Cauchy estimates

Let 0<α<10 < \alpha < 1 and let f(z)f\p{z} be an analytic function on the unit disc D\D. Prove that if

∣f(z)−f(w)∣≤C∣z−w∣α\abs{f\p{z} - f\p{w}} \leq C\abs{z - w}^\alpha

for all z,w∈Dz, w \in \D and some constant C∈RC \in \R, then there is a constant A=A(C)<∞A = A\p{C} < \infty such that

∣f′(z)∣≤A(1−∣z∣)α−1.\abs{f'\p{z}} \leq A\p{1 - \abs{z}}^{\alpha-1}.
Solution.

First, notice that because w↦1(z−w)2w \mapsto \frac{1}{\p{z - w}^2} has a primitive, it is conservative, so

12πi∫γ1(z−w)2 dw=0\frac{1}{2\pi i} \int_\gamma \frac{1}{\p{z - w}^2} \,\diff{w} = 0

for any closed γ\gamma. By the generalized Cauchy integral formula, if B(z,r)‾⊆D\cl{B\p{z, r}} \subseteq \D, then

∣f′(z)∣=∣12πi∫∂B(z,r)f(ζ)(ζ−z)2 dζ∣=∣12πi∫∂B(z,r)f(ζ)−f(z)(ζ−z)2 dζ∣≤12π∫∂B(z,r)∣f(ζ)−f(z)∣∣ζ−z∣2 d∣ζ∣≤12π∫∂B(z,r)C∣ζ−z∣α∣ζ−z∣2 d∣ζ∣=C2π⋅2πr⋅rα−2=Crα−1.\begin{aligned} \abs{f'\p{z}} &= \abs{\frac{1}{2\pi i} \int_{\partial B\p{z, r}} \frac{f\p{\zeta}}{\p{\zeta - z}^2} \,\diff\zeta} \\ &= \abs{\frac{1}{2\pi i} \int_{\partial B\p{z, r}} \frac{f\p{\zeta} - f\p{z}}{\p{\zeta - z}^2} \,\diff\zeta} \\ &\leq \frac{1}{2\pi} \int_{\partial B\p{z, r}} \frac{\abs{f\p{\zeta} - f\p{z}}}{\abs{\zeta - z}^2} \,\diff\abs{\zeta} \\ &\leq \frac{1}{2\pi} \int_{\partial B\p{z, r}} \frac{C\abs{\zeta - z}^\alpha}{\abs{\zeta - z}^2} \,\diff\abs{\zeta} \\ &= \frac{C}{2\pi} \cdot 2\pi r \cdot r^{\alpha-2} \\ &= Cr^{\alpha-1}. \end{aligned}

If we set r=1−∣z∣2r = \frac{1 - \abs{z}}{2}, then B(z,r)‾⊆D\cl{B\p{z, r}} \subseteq \D, and so

∣f′(z)∣≤C2α−1(1−∣z∣)α−1.\abs{f'\p{z}} \leq \frac{C}{2^{\alpha-1}}\p{1 - \abs{z}}^{\alpha-1}.