Does there exist a function \(f\), holomorphic on \(\mathbb C\setminus\{0\}\), such that \[ |f(z)|\ge\frac1{\sqrt{|z|}} \] for every \(z\ne0\)?
Solution
No such function exists.
1. If such an \(f\) existed, then \(f\) would have no zeros on\[ \CC\setminus\{0\}, \]and\[ g(z)\coloneqq\frac1{f(z)} \]would be holomorphic there with\[ \abs{g(z)}\leq\sqrt{\abs z}. \]
The assumed lower bound is strictly positive for every \(z\neq0\), so \(f(z)\neq0\). Taking reciprocals of \[ \abs{f(z)} \geq \frac1{\sqrt{\abs z}} \] gives \[ \abs{g(z)} \leq \sqrt{\abs z}. \]
2. The singularity of \(g\) at \(0\) is removable, and the holomorphicextension satisfies\[ g(0)=0. \]
Step 1 gives \[ \abs{g(z)} \leq \sqrt{\abs z} \longrightarrow0 \] as \(z\to0\). Thus \(g\) is bounded near \(0\), so the removable singularity theorem extends it holomorphically across \(0\). The displayed limit forces the extension to have value \(0\) there.
3. Write the resulting entire function as\[ g(z)=\sum_{n=0}^{\infty}a_nz^n. \]Then\[ a_n=0 \]for every \(n\geq1\).
For every \(R>0\), step 1 gives on the circle \(\abs z=R\) \[ \max_{\abs z=R}\abs{g(z)} \leq \sqrt R. \] Cauchy’s coefficient estimate therefore gives \[ \abs{a_n} \leq \frac{\sqrt R}{R^n} = R^{1/2-n}. \] If \(n\geq1\), the right side tends to \(0\) as \(R\to\infty\). Hence \(a_n=0\).
4. One has\[ g\equiv0. \]
Step 3 shows that \(g\) is constant, while step 2 gives \[ g(0)=0. \] Therefore the constant is zero.
5. The assumption that such an \(f\) exists is impossible.
On \(\CC\setminus\{0\}\), the definition of \(g\) gives \[ g(z)=\frac1{f(z)}\neq0. \] This contradicts step 4.
6. Q.E.D.
Step 5 proves that no holomorphic function with the stated lower bound exists.