Find, with proof, all entire functions satisfying \(| f ( z ) | \leq \sqrt { | z | } \mathrm { ~ f o r ~ } | z | > 1 0\)
Solution
Write \[ f(z)=\sum_{n=0}^{\infty}a_nz^n. \]
1. For every integer \(n\geq1\) and every \(R>10\),\[ \abs{a_n}\leq R^{1/2-n}. \]
Cauchy’s coefficient formula on the circle \(\abs{\zeta}=R\) gives \[ a_n = \frac{1}{2\pi i} \int_{\abs{\zeta}=R} \frac{f(\zeta)}{\zeta^{n+1}} \,d\zeta. \] Since \(R>10\), the hypothesis gives \[ \abs{f(\zeta)}\leq\sqrt R \] on this circle. Therefore \[ \begin{aligned} \abs{a_n} &\leq \frac{1}{2\pi}(2\pi R) \frac{R^{1/2}}{R^{n+1}}\\ &= R^{1/2-n}. \end{aligned} \]
2. Every Taylor coefficient \(a_n\) with \(n\geq1\) is zero.
Fix \(n\geq1\). The estimate in step 1 holds for every \(R>10\), and \[ R^{1/2-n}\longrightarrow0 \] as \(R\to\infty\). Hence \(\abs{a_n}=0\).
3. Every entire function satisfying the hypothesis is constant.
By step 2, \[ f(z)=a_0 \] for every \(z\in\CC\).
4. If \(f\equiv c\) satisfies the hypothesis, then\[ \abs{c}\leq\sqrt{10}. \]
For every real \(R>10\), choose \(z\) with \(\abs{z}=R\). The hypothesis gives \[ \abs{c}\leq\sqrt R. \] Letting \(R\downarrow10\) yields \(\abs{c}\leq\sqrt{10}\).
5. Conversely, every constant \(c\in\CC\) with\(\abs{c}\leq\sqrt{10}\) satisfies the hypothesis.
If \(\abs{z}>10\), then \[ \abs{c} \leq \sqrt{10} < \sqrt{\abs{z}}. \] Thus \(f\equiv c\) satisfies the required inequality.
6. The complete list is\[ \boxed{ f(z)=c, \qquad c\in\CC, \qquad \abs{c}\leq\sqrt{10}. } \]
Steps 3 and 4 show that every solution is on the displayed list, and step 5 shows that every function on the list is a solution.
7. Q.E.D.
Step 6 is the requested classification.