Let \(\Delta=\{z\in\mathbb C:|z|<1\}\), and let \(f:\Delta\to\Delta\) be holomorphic and satisfy \(f(0)=0\). Prove that \[ |f(z)+f(-z)|\le 2|z|^2 \] for every \(z\in\Delta\).
If equality holds for some nonzero \(z\in\Delta\), prove that \[ f(z)=e^{i\theta}z^2 \] for some real \(\theta\).
As printed, the source omits \(f(0)=0\), so the inequality is false at \(z=0\); \(f\equiv1/2\) is a counterexample. It also says only “equality holds for some \(z\)”. After the necessary normalization, equality at \(z=0\) is automatic, so the equality statement must require a nonzero point; \(f(z)=z\) otherwise gives a counterexample to the claimed conclusion.
Solution
1. The even part factors holomorphically through \(z^2\) as a disk map fixing zero.
Set \[ E(z)=\frac{f(z)+f(-z)}2. \] The unit disk is convex, and \(f(z),f(-z)\in\Delta\), so \[ |E(z)|\le\frac{|f(z)|+|f(-z)|}{2}<1. \] Thus \(E:\Delta\to\Delta\) is holomorphic. It is even and, by the normalization, \(E(0)=f(0)=0\).
Write the Taylor series of \(E\) at zero as \[ E(z)=\sum_{n=0}^\infty b_nz^n. \] Evenness gives \(b_{2m+1}=0\), and \(b_0=0\). Hence \[ E(z)=G(z^2),\qquad G(w)=\sum_{m=1}^\infty b_{2m}w^m. \] The series for \(G\) converges for \(|w|<1\): if \(|w|<1\), choose \(z\) with \(z^2=w\), so the corresponding even subseries is absolutely convergent. Therefore \(G\) is holomorphic on \(\Delta\), satisfies \(G(0)=0\), and obeys \(|G(w)|<1\) because \(G(w)=E(z)\) for either square root \(z\) of \(w\). Thus \(G:\Delta\to\Delta\) is a holomorphic self-map fixing zero.
2. Schwarz’s lemma gives the required inequality.
Schwarz’s lemma [@SS03] applied to \(G\) gives \[ |G(w)|\le|w|\qquad(w\in\Delta). \] Putting \(w=z^2\) yields \[ \frac12|f(z)+f(-z)|=|E(z)|=|G(z^2)|\le|z|^2. \] Multiplying by two gives the asserted bound.
3. Equality at a nonzero point fixes the entire even part.
Suppose equality holds at \(z_0\ne0\). Then \[ |G(z_0^2)|=|z_0^2|, \] and \(z_0^2\ne0\). The equality case of Schwarz’s lemma [@SS03] gives \[ G(w)=\lambda w \] for some \(|\lambda|=1\). Therefore \[ E(z)=\lambda z^2 \] throughout the disk. If \[ f(z)=\sum_{n=1}^\infty a_nz^n, \] then \(E\) is precisely the even part of this Taylor series, so \[ a_2=\lambda,\qquad a_{2m}=0\quad(m\ne1). \] In particular \(|a_2|=1\).
4. A disk map with a Taylor coefficient of modulus one at degree two has no other coefficients.
For \(0<r<1\), the Taylor partial sums converge uniformly to \(f\) on \(|z|=r\) [@SS03]. Their squared moduli also converge uniformly there. Integrating the finite sums, using orthogonality of the exponentials and then passing to the limit therefore gives \[ \frac1{2\pi}\int_0^{2\pi}|f(re^{it})|^2\,dt =\sum_{n=1}^\infty |a_n|^2r^{2n}. \] The left side is at most one because \(f(\Delta)\subset\Delta\). Letting \(r\uparrow1\) and using monotone convergence for the nonnegative series gives \[ \sum_{n=1}^\infty|a_n|^2\le1. \] Step 3 gives \(|a_2|=1\), so every other coefficient must vanish. Thus \[ f(z)=a_2z^2=\lambda z^2=e^{i\theta}z^2 \] for some real \(\theta\), which is exactly the corrected equality conclusion.