- Suppose \(f : \mathbb { D } \to \mathbb { C }\) is holomorphic. Show that the diameter \(d =\) \(\begin{array} { r } { \operatorname* { s u p } _ { z , w \in \mathbb { D } } | f ( z ) - f ( w ) | } \end{array}\) of the image of f satisfies
\[ 2 | f ^ {\prime} (0) | \leq d. \]
Moreover, it can be shown that equality holds precisely when f is linear, \(f ( z ) =\) \(a _ { 0 } + a _ { 1 } z\)
Note. In connection with this result, see the relationship between the diameter of a curve and Fourier series described in Problem 1, Chapter 4, Book I.
[Hint: \(\begin{array} { r } { 2 f ^ { \prime } ( 0 ) = \frac { 1 } { 2 \pi i } \int _ { | \zeta | = r } \frac { f ( \zeta ) - f ( - \zeta ) } { \zeta ^ { 2 } } d \zeta } \end{array}\) whenever \(0 < r < 1 . ]\)
Solution
Fix \(0<r<1\). By the Cauchy integral formula for first derivatives, \(\frac1{2\pi i}\int_{|\zeta|=r}\frac{f(\zeta)}{\zeta^2}\,d\zeta=f'(0)\), and applied to \(\zeta\mapsto f(-\zeta)\), whose derivative at \(0\) is \(-f'(0)\), it gives \(\frac1{2\pi i}\int_{|\zeta|=r}\frac{f(-\zeta)}{\zeta^2}\,d\zeta=-f'(0)\). Subtracting, \[ 2f'(0)=\frac1{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-f(-\zeta)}{\zeta^2}\,d\zeta. \] Since both \(\zeta\) and \(-\zeta\) lie in \(\mathbb D\), \[ |f(\zeta)-f(-\zeta)|\le d. \] Therefore \[ \begin{aligned} 2|f'(0)| &\le \frac1{2\pi}\int_{|\zeta|=r} \frac{d}{r^2}|d\zeta|\\ &=\frac1{2\pi}\frac d{r^2}(2\pi r) =\frac dr. \end{aligned} \] Letting \(r\uparrow1\) yields \[ \boxed{2|f'(0)|\le d}. \]