- Find all entire functions f such that \(| f ( z ) | \le | z | ^ { 3 / 2 }\) whenever \(| z | \geq 1\) . Give explicit formulas for the functions and give a proof for your answer. (An entire function is a holomorphic function on C.)
Solution
The functions are precisely \[ \boxed{f(z)=az+b,\qquad a,b\in\mathbb C,\quad |a|+|b|\leq1.} \]
1. Every solution is affine.
Write the entire Taylor expansion as \(f(z)=\sum_{k\geq0}c_kz^k\). For each \(R\geq1\), the hypothesis gives \(\max_{|z|=R}|f(z)|\leq R^{3/2}\). Cauchy’s coefficient estimate therefore yields \[ |c_k|\leq R^{3/2-k} \] [@SS03]. For every \(k\geq2\), the right-hand side tends to zero as \(R\to\infty\), so \(c_k=0\). Consequently \(f(z)=az+b\) for \(a=c_1\) and \(b=c_0\).
2. The unit-circle bound is equivalent to \(|a|+|b|\leq1\).
The triangle inequality gives \(|az+b|\leq|a|+|b|\) on \(|z|=1\). Equality is attained: if \(a\) and \(b\) are nonzero, take \(z=(b/|b|)(\overline a/|a|)\), so \(az\) and \(b\) have the same argument. If either coefficient is zero, every point of the circle attains the sum. Hence \[ \max_{|z|=1}|az+b|=|a|+|b|. \] The given bound at \(|z|=1\) implies that this sum is at most one.
3. Every stated affine function obeys the original exterior bound.
If \(r=|z|\geq1\) and \(|a|+|b|\leq1\), then \[ |az+b|\leq |a|r+|b|\leq r(|a|+|b|)\leq r\leq r^{3/2}. \] Thus every displayed function satisfies the inequality for every required \(z\). Together with steps 1 and 2, this proves the exhaustive classification.