Let \(f\) be an entire function and suppose that \(|f(z)| \leq A|z|^2\) for all \(z\) and some constant \(A\). Show that \(f\) is a polynomial of degree \(\leq 2\).
Solution
We prove the stronger statement \(f(z)=c_2z^2\) for a constant \(c_2\in\mathbb C\).
Step 1: Taylor Series and Cauchy’s Estimates
1. \(f\) is represented by its Taylor series centered at the origin for all \(z \in \mathbb{C}\).
1.1. Since \(f\) is entire, \(f(z) = \sum_{n=0}^\infty c_n z^n\) converges everywhere on \(\mathbb{C}\), with coefficients given by \(c_n = \frac{f^{(n)}(0)}{n!}\).Proof: The Taylor series of a holomorphic function at \(0\) converges to it on every open disk centered at \(0\) contained in its domain, here \(\mathbb C\).
1.2. For any \(R > 0\) and any \(n \geq 0\), Cauchy’s coefficient formula gives: \[c_n = \frac{1}{2\pi i} \oint_{|z|=R} \frac{f(z)}{z^{n+1}} \, dz.\] Proof: Cauchy integral formula for derivatives.
1.3. Q.E.D.
2. Apply Cauchy’s Estimates for each coefficient \(c_n\).
2.1. By the \(ML\)-inequality on the circle \(|z| = R\): \[|c_n| \leq \frac{1}{2\pi} \cdot \left( \sup_{|z|=R} \frac{|f(z)|}{|z|^{n+1}} \right) \cdot (2\pi R) = \frac{\sup_{|z|=R} |f(z)|}{R^n}.\] Proof: On \(|z|=R\) the integrand of step <1>1.<2>2 has modulus at most \(\sup_{|z|=R}|f(z)|/R^{n+1}\), and the circle has length \(2\pi R\).
2.2. By the growth hypothesis, on \(|z| = R\), \(|f(z)| \leq A R^2\).Proof: Given assumption \(|f(z)| \leq A|z|^2\).
2.3. Substituting 2 into 1 yields the bound: \[|c_n| \leq \frac{A R^2}{R^n} = A R^{2-n} \quad \text{for all } R > 0.\] Proof: Algebra.
2.4. Q.E.D.
Step 2: Vanishing of Higher-Order Coefficients
3. \(c_n = 0\) for all \(n \geq 3\).
3.1. For \(n \geq 3\), the exponent \(2 - n \leq -1 < 0\).Proof: \(n \geq 3 \implies 2 - n \leq -1\).
3.2. Since the bound \(|c_n| \leq A R^{2-n}\) holds for all \(R > 0\), taking the limit as \(R \to \infty\): \[|c_n| \leq \lim_{R \to \infty} A R^{2-n} = 0.\] Proof: \(\lim_{R\to\infty} R^{-k} = 0\) for \(k \geq 1\).
3.3. Therefore, \(|c_n| = 0 \implies c_n = 0\) for all \(n \geq 3\).Proof: Absolute value is non-negative.
3.4. Q.E.D.
Step 3: Behavior at the Origin and Lower-Order Coefficients
4. \(c_0 = 0\) and \(c_1 = 0\).
4.1. Evaluating the growth condition at \(z = 0\): \(|f(0)| \leq A |0|^2 = 0 \implies f(0) = 0\), so \(c_0 = 0\).Proof: Evaluation at \(z=0\).
4.2. For \(c_1\), apply the bound from <1>2.<2>3 with \(n = 1\): \(|c_1| \leq A R^{2-1} = A R\) for all \(R > 0\).Proof: Setting \(n=1\) in <1>2.<2>3.
4.3. Taking the limit as \(R \to 0^+\): \[|c_1| \leq \lim_{R \to 0^+} A R = 0 \implies c_1 = 0.\] Proof: \(R\) can be taken arbitrarily small.
4.4. Q.E.D.
Step 4: Conclusion
5. \(f(z)\) is a polynomial of degree at most 2 (in fact \(f(z) = c_2 z^2\)).
5.1. From 3 and 4, \(c_n = 0\) for all \(n \neq 2\).Proof: Combining \(c_0 = c_1 = 0\) and \(c_n = 0\) for \(n \geq 3\).
5.2. Thus \(f(z) = c_2 z^2\), which is a polynomial of degree \(\leq 2\) (with \(|c_2| \leq A\)). Proof: Direct substitution into Taylor series.
5.3. Q.E.D.