Suppose \(f\) is holomorphic on \(U \coloneqq D(0,1) \setminus \{0\}\). Assume that the real part \(\operatorname{Re}(f)\) is bounded from below on \(U\). Prove that \(z = 0\) is a removable singularity.
Solution
Choose \(m\in\RR\) with \(\operatorname{Re}f(z)\geq m\) on the punctured disk \(U\).
1. A fractional transform of \(f\) is bounded and holomorphic on \(U\).
Let \(F=f-m+1\), so \(\operatorname{Re}F\geq1\), and put \[ q(z)=\frac{F(z)-1}{F(z)+1}. \] Since \(\operatorname{Re}(F+1)\geq2\), its denominator never vanishes. Moreover, \[ \abs{F+1}^2-\abs{F-1}^2=4\operatorname{Re}F\geq4>0, \] so \(\abs{q}<1\) throughout \(U\). Riemann's removable singularity theorem extends \(q\) holomorphically to the full unit disk; write \(Q\) for this extension. Continuity gives \(\abs{Q(0)}\leq1\).
2. The inverse transform extends \(f\) across zero.
In fact \(\abs{Q(0)}<1\). Otherwise \(Q\) would attain its maximum modulus one at an interior point of the disk. The maximum modulus principle would make \(Q\) constant of modulus one, contradicting \(\abs{q(z)}<1\) on \(U\). Together with the original strict bound, this gives \(\abs{Q}<1\) everywhere on the full disk. Hence \(1-Q\) is nowhere zero there, and \[ \widetilde f(z)=m+\frac{2Q(z)}{1-Q(z)} \] is holomorphic on that disk. On \(U\), solving \(q=(F-1)/(F+1)\) gives \(F=(1+q)/(1-q)\) and consequently \(f=m+2q/(1-q)\). Thus \(\widetilde f=f\) on \(U\). The displayed function is a holomorphic extension of \(f\) through zero, proving removability.
3. Q.E.D.
Step 2 constructs the required holomorphic extension across zero.