Let \(f\) be entire and suppose \[ |f(z)|\le |\operatorname{Re}z|^{-1/2} \] whenever \(\operatorname{Re}z\ne0\). Prove that \(f\) is constant.
Solution
1. The constant\[ C\coloneqq\int_0^{2\pi}\abs{\cos\theta}^{-1/2}\,d\theta \]is finite.
The only singularities occur at \[ \theta=\frac\pi2, \qquad \theta=\frac{3\pi}{2}. \] Near either point, \(\abs{\cos\theta}\) is comparable to the distance from that point, so the integrand is comparable to \(\abs t^{-1/2}\), which is locally integrable.
2. For every \(y\in\RR\) and every \(R>0\),\[ \abs{f(iy)}\le\frac{C}{2\pi\sqrt R}. \]
Cauchy’s formula on the circle \[ z=iy+Re^{i\theta} \] gives \[ f(iy) = \frac1{2\pi} \int_0^{2\pi}f(iy+Re^{i\theta})\,d\theta. \] For all \(\theta\) except the two points where \(\cos\theta=0\), \[ \operatorname{Re}(iy+Re^{i\theta})=R\cos\theta\ne0, \] so the hypothesis gives \[ \abs{f(iy+Re^{i\theta})} \le R^{-1/2}\abs{\cos\theta}^{-1/2}. \] The two exceptional angles have measure zero. Therefore, by step 1, \[ \abs{f(iy)} \le \frac1{2\pi} \int_0^{2\pi}\abs{f(iy+Re^{i\theta})}\,d\theta \le \frac{C}{2\pi\sqrt R}. \]
3. The function \(f\) vanishes on the imaginary axis.
Fix \(y\in\RR\). The estimate in step 2 holds for every \(R>0\). Letting \(R\to\infty\) gives \[ \abs{f(iy)}=0. \] Thus \(f(iy)=0\) for every real \(y\).
4. Hence\[ \boxed{f\equiv0}. \]
By step 3, the zero set of the entire function \(f\) contains the imaginary axis, which has accumulation points in \(\CC\). The identity theorem therefore gives \(f\equiv0\).
5. In particular, \(f\) is constant.
Step 4 shows that the only possible function is the constant zero function.
6. Q.E.D.
Step 5 is the required conclusion.