- Let \(f\) be an entire function \(| R e ( f ( z ) ) | \leq C ( 1 + | z | ) ^ { p }\) for some \(p > 0 , C > 0\) . Show that \(f\) is a polynomial.
Solution
Let \(u=\Re f\). Then \(u\) is harmonic on \(\mathbb C\) and \[ |u(z)| \le C(1+|z|)^p. \]
1. Every partial derivative of \(u\) of order \(m>p\) vanishes identically.
Fix \(z_0\in\mathbb C\) and \(r>0\). The interior derivative estimate for harmonic functions gives a constant \(C_m\), depending only on \(m\), such that for every multi-index \(\alpha\) with \(|\alpha|=m\), \[ |\partial^\alpha u(z_0)| \le\frac{C_m}{r^m}\sup_{|z-z_0|\le r}|u(z)| \le C_m\frac{C(1+|z_0|+r)^p}{r^m}. \] If \(m>p\), the right-hand side tends to zero as \(r\to\infty\). Hence \(\partial^\alpha u(z_0)=0\) for every \(z_0\).
2. Every complex derivative of \(f\) of order \(k>p\) vanishes identically.
Write \(f=u+iv\). The Cauchy–Riemann equations give \(f'=u_x-iu_y\). Since \(f^{(k)}=\partial_x^{k-1}f'\), the real and imaginary parts of \(f^{(k)}\) are \(\partial_x^{k}u\) and \(-\partial_x^{k-1}\partial_yu\), which are partial derivatives of \(u\) of order \(k\). Step 1 makes both vanish when \(k>p\).
3. \(f\) is a polynomial of degree at most \(\lfloor p\rfloor\).
The Taylor series of \(f\) at \(0\) represents \(f\) on all of \(\mathbb C\) [@SS03]. By step 2 its coefficients \(f^{(k)}(0)/k!\) vanish for \(k>p\), so \[ f(z)=\sum_{k=0}^{\lfloor p\rfloor}\frac{f^{(k)}(0)}{k!}z^k. \]