Let \(f(z)\) be entire and assume that \(f(z) \leq M |z|^2\) outside some disk for some constant \(M\). Show that \(f(z)\) is a polynomial in \(z\) of degree \(\leq 2\).
Solution
Goal: Prove that if \(f\) is entire and \(\abs{f(z)} \leq M \abs z^2\) for all \(z\) outside some disk (i.e. for \(\abs z \geq R_0\)), then \(f\) is a polynomial of degree \(\leq 2\).
1. Write \(f(z) = \sum_{n=0}^{\infty} a_n z^n\).
\(f\) is entire.
2. For \(R \geq R_0\), \(M(R) := \max_{\abs z = R} \abs{f(z)} \leq M R^2\).
By hypothesis applied on the circle \(\abs z = R\).
3. For \(n \geq 3\) and \(R \geq R_0\), \(\abs{a_n} \leq \frac{M(R)}{R^n} \leq \frac{M R^2}{R^n} = M R^{2-n}\).
Cauchy’s estimate 2.
4. \(a_n = 0\) for all \(n \geq 3\).
3 holds for arbitrarily large \(R\), and \(R^{2-n} \to 0\) as \(R \to \infty\) for \(n \geq 3\).
5. \(f(z) = a_0 + a_1 z + a_2 z^2\), a polynomial of degree at most \(2\).
1 and 4.
6. Q.E.D.
5 is the claim.
Solution
Take a Laurent expansion at zero: \[ f(z) = \sum_{k\geq 0} c_k z^k,\qquad c_k = {1\over k!} f^{(k)}(0) = {1\over 2\pi i}\oint_{\abs{\xi} = R} {f(\xi) \over \xi^{k+1}}\dxi .\] The usual estimate: \[ 2\pi i\abs{c_k} \leq \oint_{\abs{\xi} = R} R^{-(k+1)}\abs{f(\xi)} \dxi &\leq \oint_{\abs{\xi} = R}R^{-(k+1)} M R^2 \dxi \\ &= M R^{-(k-1)} \cdot 2\pi R \\ &= 2\pi M R^{-k+2} \\ &\convergesto{R\to\infty}0 ,\] provided \(-k+2<0 \iff k>2\).