Find all entire functions \(f\) which satisfy \(\abs{f(z)} \geq \abs{z}\) for all \(z \in \CC\). Be sure to prove your list is complete.
Solution
1. One has\[ f(0)=0. \]
Suppose instead that \(f(0)\neq0\). For every \(z\neq0\), the hypothesis gives \[ \abs{f(z)}\geq\abs{z}>0, \] so together with \(f(0)\neq0\), the function \(f\) has no zeros on \(\CC\). Hence \[ g(z)=\frac{z}{f(z)} \] is entire. For \(z\neq0\), \[ \abs{g(z)} = \frac{\abs{z}}{\abs{f(z)}} \leq1, \] and \(g(0)=0\). Thus \(g\) is bounded and entire. By Liouville’s theorem, \(g\) is constant, and since \(g(0)=0\), one has \(g\equiv0\). This is impossible for any \(z\neq0\), because \(z/f(z)\neq0\). Therefore \(f(0)=0\).
2. Define\[ h(z) = \begin{cases} \dfrac{f(z)}{z},&z\neq0,\\ f'(0),&z=0. \end{cases} \]Then \(h\) is entire and\[ \abs{h(z)}\geq1 \]for every \(z\in\CC\).
By step 1, \(f(0)=0\). The quotient \(f(z)/z\) therefore has a removable singularity at \(0\), and its limiting value there is \[ \lim_{z\to0}\frac{f(z)}z=f'(0). \] Thus the displayed definition makes \(h\) entire.
For \(z\neq0\), the given inequality yields \[ \abs{h(z)} = \frac{\abs{f(z)}}{\abs{z}} \geq1. \] Taking \(z\to0\) and using continuity of \(h\) gives \(\abs{h(0)}\geq1\) as well.
3. The function \(h\) is constant.
Step 2 shows that \(h\) has no zeros. Hence \(1/h\) is entire, and \[ \abs{\frac1{h(z)}}\leq1 \] for every \(z\in\CC\). Liouville’s theorem implies that \(1/h\) is constant. Therefore \(h\) is constant.
4. Every function satisfying the hypothesis has the form\[ f(z)=cz \]for some \(c\in\CC\) with \(\abs{c}\geq1\).
By step 3, write \(h\equiv c\). Step 2 gives \(\abs{c}\geq1\), and the definition of \(h\) gives \(f(z)=zh(z)=cz\) for \(z\neq0\). The same formula also holds at \(z=0\) by step 1.
5. Conversely, every function \(f(z)=cz\) with \(\abs{c}\geq1\) satisfiesthe required inequality.
Such a function is entire, and \[ \abs{f(z)} = \abs{c}\abs{z} \geq \abs{z} \] for every \(z\in\CC\).
6. The complete list is\[ \boxed{ f(z)=cz, \qquad c\in\CC, \qquad \abs{c}\geq1. } \]
Step 4 proves that every solution is on this list, and step 5 proves that every function on the list is a solution.
7. Q.E.D.
Step 6 is the requested complete classification.