Suppose \(f\) is entire and satisfies \[ |f(z)|\le |z|^2 \] for every \(z\in\mathbb C\). Find all possibilities for \(f\), and justify your answer.
Solution
1. One has\[ f(0)=0 \]and\[ f'(0)=0. \]
Evaluating the hypothesis at \(z=0\) gives \[ \abs{f(0)}\leq0, \] so \(f(0)=0\). Hence \[ f'(0) = \lim_{z\to0}\frac{f(z)-f(0)}{z} = \lim_{z\to0}\frac{f(z)}{z}. \] For \(z\neq0\), \[ \abs{\frac{f(z)}{z}} \leq \abs{z}, \] so the last limit is \(0\).
2. The function\[ h(z) \coloneqq \begin{cases} f(z)/z^2,&z\neq0,\\ f''(0)/2,&z=0 \end{cases} \]is entire.
Step 1 shows that the Taylor series of \(f\) at \(0\) has no constant or linear term. Thus \[ f(z) = z^2 \left( \frac{f''(0)}{2} + \frac{f^{(3)}(0)}{3!}z + \cdots \right), \] which proves that the displayed definition removes the apparent singularity at \(0\) and gives an entire function.
3. The entire function \(h\) satisfies\[ \abs{h(z)}\leq1 \]for every \(z\in\CC\).
For \(z\neq0\), the hypothesis gives \[ \abs{h(z)} = \frac{\abs{f(z)}}{\abs{z}^2} \leq 1. \] By continuity of \(h\), the same inequality holds at \(z=0\).
4. There is a constant \(a\in\CC\) with \(\abs{a}\leq1\) such that\[ f(z)=az^2 \]for every \(z\in\CC\).
By step 3, \(h\) is a bounded entire function. Liouville’s theorem implies that \[ h(z)=a \] for some constant \(a\). The same step gives \(\abs a\leq1\). Since \(f(z)=z^2h(z)\), the stated formula follows.
5. Conversely, every function\[ f(z)=az^2 \]with \(\abs{a}\leq1\) satisfies the hypothesis.
For every \(z\in\CC\), \[ \abs{f(z)} = \abs{a}\,\abs{z}^2 \leq \abs z^2. \]
6. The complete list is\[ \boxed{ f(z)=az^2 \quad\text{with}\quad a\in\CC, \abs{a}\leq1 }. \]
Step 4 shows that every admissible entire function has this form, and step 5 shows that every function of this form is admissible.
7. Q.E.D.
Step 6 is the required classification.