Let \(f:\mathbb C\to\mathbb C\) be entire and suppose that for some nonnegative integers \(k,m\), \[ \frac{f^{(m)}(z)}{1+|z|^k} \] is bounded on \(\mathbb C\). Prove that \(f^{(n)}\) is identically zero for all sufficiently large \(n\). How large must \(n\) be in terms of \(k\) and \(m\)?
Solution
Put \[ g=f^{(m)}. \] By hypothesis there is a constant \(C>0\) such that \[ \abs{g(z)} \leq C(1+\abs{z}^k) \] for every \(z\in\CC\).
1. For every \(z_0\in\CC\) and every \(R>0\),\[ \abs{g^{(k+1)}(z_0)} \leq \frac{(k+1)!C}{R^{k+1}} \left(1+(R+\abs{z_0})^k\right). \]
On the circle \[ \abs{z-z_0}=R, \] the triangle inequality gives \[ \abs{z}\leq R+\abs{z_0}. \] Hence \[ \max_{\abs{z-z_0}=R}\abs{g(z)} \leq C\left(1+(R+\abs{z_0})^k\right). \] Cauchy’s derivative estimate of order \(k+1\) therefore gives \[ \abs{g^{(k+1)}(z_0)} \leq \frac{(k+1)!}{R^{k+1}} \max_{\abs{z-z_0}=R}\abs{g(z)}, \] which is the claimed inequality.
2. One has\[ g^{(k+1)}\equiv0. \]
Fix \(z_0\in\CC\). In step 1, let \(R\to\infty\). Since \[ \frac{1+(R+\abs{z_0})^k}{R^{k+1}} \longrightarrow0, \] we obtain \[ \abs{g^{(k+1)}(z_0)}=0. \] The point \(z_0\) was arbitrary, so \(g^{(k+1)}\) is identically zero.
3. One has\[ f^{(m+k+1)}\equiv0. \]
Since \(g=f^{(m)}\), \[ g^{(k+1)} = f^{(m+k+1)}. \] Now apply step 2.
4. For every integer\[ n\geq m+k+1, \]one has\[ \boxed{f^{(n)}\equiv0}. \]
Step 3 gives the assertion for \(n=m+k+1\). Every higher derivative is a derivative of the zero function and is therefore also identically zero.
5. The threshold \(m+k+1\) is the best possible uniform bound.
Take \[ f(z)=z^{m+k}. \] Then \(f^{(m)}\) is a polynomial of degree \(k\), so for some constant \(C>0\), \[ \abs{f^{(m)}(z)} \leq C(1+\abs{z}^k). \] Thus the hypothesis is satisfied. However, \[ f^{(m+k)} \] is a nonzero constant, while \[ f^{(m+k+1)}\equiv0. \] Hence no smaller derivative order works for all functions satisfying the hypothesis.
6. Q.E.D.
Step 4 gives the required vanishing range, and step 5 shows that the bound is sharp.