[April 1999 Problem \(\# 7 ]\) Let \(S : = \{ z \in \mathbb { D } : \operatorname { I m } ( z ) \geq 0 \}\) . Suppose \(f : S \to \mathbb { C }\) is continuous on \(S\), real on \(S \cap \RR\), and holomorphic on the interior of \(S\). Prove that \(f\) is the restriction of a holomorphic function on the open unit disk.
Solution
Define \(F:\DD\to\CC\) by \[ F(z) = \begin{cases} f(z),&\operatorname{Im}z\geq0,\\ \overline{f(\bar z)},&\operatorname{Im}z<0. \end{cases} \]
1. The two formulas defining \(F\) agree on the real diameter\((-1,1)\).
If \(x\in(-1,1)\) is real, then \(\bar x=x\). Since \(f\) is real-valued on \(S\cap\RR\), \[ \overline{f(\bar x)} = \overline{f(x)} = f(x). \] Thus the reflected lower-half-disk formula has the same boundary value as the upper-half-disk formula.
2. The function \(F\) is continuous on \(\DD\).
On the open upper half-disk and the open lower half-disk, continuity follows from continuity of \(f\) and of conjugation. Along the real diameter, step 1 shows that the two formulas have the same value, and the continuity of \(f\) on \(S\) makes both one-sided limits equal to that value. Hence \(F\) is continuous everywhere in \(\DD\).
3. The function \(F\) is holomorphic on the open upper and lowerhalf-disks.
It equals \(f\) on the open upper half-disk, where \(f\) is holomorphic. If \(z_0\) lies in the open lower half-disk, then \(\bar z_0\) lies in the open upper half-disk. Writing the local power series \[ f(w)=\sum_{n=0}^{\infty}a_n(w-\bar z_0)^n \] near \(\bar z_0\) gives \[ \overline{f(\bar z)} = \sum_{n=0}^{\infty}\bar a_n(z-z_0)^n \] near \(z_0\). Hence the reflected formula is holomorphic there.
4. The function \(F\) is holomorphic across the real diameter.
Steps <1>1–<1>3 give exactly the hypotheses of the Schwarz reflection principle across the real axis: the upper-half-disk function is holomorphic, extends continuously to the real diameter, and is real-valued there. Therefore its reflected extension \(F\) is holomorphic through every point of \((-1,1)\).
5. The function \(F\) is holomorphic on the entire open unit disk andrestricts to \(f\) on \(S\).
Step 3 gives holomorphy away from the real diameter, and step 4 gives holomorphy on the diameter. Hence \(F\) is holomorphic on \(\DD\). By its definition, \(F=f\) wherever \(\operatorname{Im}z\geq0\), which is precisely \(S\).
6. Q.E.D.
Step 5 is the required holomorphic extension.