Let \(f\) be entire and let \[ \Omega=\mathbb C\setminus(-\infty,0]. \] Suppose that for every \(z\in\Omega\), \[ |f(z)|\le |\log z|, \] where \(\log z\) is the principal branch. What can be concluded about \(f\)?
Solution
1. One has\[ f(1)=0. \]
The point \(1\) lies in \(\Omega\), and \[ \log1=0. \] The hypothesis therefore gives \[ \abs{f(1)} \leq0, \] so \(f(1)=0\).
2. Fix \(R>1\). For every point\[ z=Re^{i\theta} \]on the circle with\[ -\pi<\theta<\pi, \]one has\[ \abs{f(z)} \leq \sqrt{(\log R)^2+\pi^2}. \]
On the principal branch, \[ \log z = \log R+i\theta. \] Hence \[ \abs{\log z}^2 = (\log R)^2+\theta^2 \leq (\log R)^2+\pi^2. \] Apply the assumed inequality for \(f\).
3. The same bound holds at the remaining point\[ z=-R \]of the circle.
Approach \(-R\) along the circle from within \(\Omega\), for example with \[ z_\theta=Re^{i\theta}, \qquad \theta\uparrow\pi. \] Since \(f\) is entire, \[ f(z_\theta)\longrightarrow f(-R). \] Step 2 bounds every \(\abs{f(z_\theta)}\) by the same constant. Passing to the limit gives \[ \abs{f(-R)} \leq \sqrt{(\log R)^2+\pi^2}. \]
4. If\[ f(z)=\sum_{n=0}^{\infty}a_nz^n, \]then for every \(n\geq1\) and every \(R>1\),\[ \abs{a_n} \leq \frac{\sqrt{(\log R)^2+\pi^2}}{R^n}. \]
Steps 2 and 3 give \[ \max_{\abs z=R}\abs{f(z)} \leq \sqrt{(\log R)^2+\pi^2}. \] Cauchy’s coefficient estimate at the origin gives the displayed bound.
5. For every \(n\geq1\),\[ a_n=0. \]
4. Since\[ \frac{\sqrt{(\log R)^2+\pi^2}}{R^n} \longrightarrow0, \]one obtains \(\abs{a_n}=0\).
6. The entire function \(f\) is constant.
Step 5 shows that every positive-degree Taylor coefficient of \(f\) at the origin vanishes. Hence \[ f(z)=a_0 \] for all \(z\).
7. One has\[ \boxed{f\equiv0}. \]
Step 6 makes \(f\) constant, and step 1 shows that this constant is zero.
8. Q.E.D.
Step 7 is the conclusion forced by the hypotheses.