Suppose \(f: \DD \to \CC\) is holomorphic and let \(d \definedas \sup_{z, w\in \DD}\abs{f(z) - f(w)}\) be the diameter of the image of \(f\). Show that \(2 \abs{f'(0)} \leq d\), and that equality holds iff \(f\) is linear, so \(f(z) = a_1 z + a_2\).
Hint: \[ 2f'(0) = \frac{1}{2\pi i} \int_{\abs \xi = r} \frac{ f(\xi) - f(-\xi) }{\xi^2} ~d\xi \] whenever \(0<r<1\).
Solution
Let \[ d=\operatorname{diam}f(\mathbb D). \] If \(d=0\), then \(f\) is constant and the result is immediate. Assume \(d>0\) and define the odd part \[ h(z)=\frac{f(z)-f(-z)}{d}. \] Then \(h(0)=0\) and \(|h(z)|\le1\) on \(\mathbb D\). Schwarz’s lemma gives \[ |h'(0)|\le1. \] Since \(h'(0)=2f'(0)/d\), this is exactly \[ \boxed{2|f'(0)|\le d}. \]
Every affine function \(f(z)=a_1z+a_2\) satisfies \[ d=2|a_1|=2|f'(0)|, \] so it remains to prove rigidity in the equality case.
Assume \(2|f'(0)|=d\). After replacing \(f\) by \[ F(z)=\frac{f(z)-f(0)}{f'(0)}, \] we may assume \[ F(0)=0,\qquad F'(0)=1,\qquad \operatorname{diam}F(\mathbb D)=2. \] We prove \(F(z)=z\).
The odd part \[ F_o(z)=\frac{F(z)-F(-z)}2 \] maps \(\mathbb D\) into \(\overline{\mathbb D}\), satisfies \(F_o(0)=0\), and has \(F_o'(0)=1\). The equality case of Schwarz’s lemma therefore gives \[ \boxed{F(z)-F(-z)=2z.}\tag{1} \]
For \(0<r<1\), let \[ D_r=\operatorname{diam}F(r\mathbb D). \] The diameter is attained on the boundary circle \(|z|=r\): if one point of a maximizing pair were interior, the open mapping theorem would let its image be moved slightly farther from the other image point, contradicting maximality. Hence \[ D_r=\max_{|z|=|w|=r}|F(z)-F(w)|. \] For \(|u|=1\), set \[ H_u(z)=\frac{F(z)-F(-uz)}{z}, \] with the removable value at \(z=0\). By the maximum modulus principle, \[ \frac{D_r}{r} =\max_{|u|=1}\max_{|z|\le r}|H_u(z)|, \] so \(D_r/r\) is nondecreasing in \(r\). On the other hand, (1) gives \(D_r\ge2r\), while \(D_r\le2\). Therefore \[ 2\le\frac{D_r}{r}\le\frac2r. \] Since the left-hand side is nondecreasing and its limsup as \(r\uparrow1\) is at most \(2\), we obtain \[ \boxed{D_r=2r\qquad(0<r<1).}\tag{2} \]
Fix \(w\in\mathbb D\setminus\{0\}\) and put \(r=|w|\). Define \[ G_w(z)=\frac{F(z)-F(-w)}2. \] By (2), \(|G_w(z)|\le r\) for \(|z|\le r\), and by (1), \[ G_w(w)=w. \] Thus the function \[ \phi(\theta)=|G_w(we^{i\theta})|^2 \] has a maximum at \(\theta=0\). Differentiating there gives \[ 0=\phi'(0)=-2|w|^2\operatorname{Im}G_w'(w), \] so \[ \operatorname{Im}F'(w)=0. \] Hence the holomorphic function \(F'\) takes values in \(\mathbb R\). By the open mapping theorem it is constant; since \(F'(0)=1\), we get \(F'\equiv1\). Together with \(F(0)=0\), this yields \(F(z)=z\).
Undoing the normalization shows \[ \boxed{f(z)=a_1z+a_2.} \] Thus equality holds exactly for affine-linear \(f\).