Question 2.2. Let \(f ( z )\) be holomorphic on \(\mathbb { C } \backslash\) R and continuous on \(\mathbb { C } .\) Show that f extends to an entire analytic function.
Solution
1. The integral around any polygon in either closed half-plane is zero.
Let \(P\) be a closed polygonal region contained in \(\{\operatorname{Im}z\geq0\}\), with positively oriented boundary. For \(\varepsilon>0\), the translate \(P+i\varepsilon\) lies in the open upper half-plane, where \(f\) is holomorphic. Cauchy’s theorem therefore gives \[ 0=\int_{\partial(P+i\varepsilon)}f(w)\,dw =\int_{\partial P}f(z+i\varepsilon)\,dz \] [@SS03]. Continuity of \(f\) is uniform on the compact set swept out by these boundaries for \(0\leq\varepsilon\leq1\). Hence \[ \left|\int_{\partial P}(f(z+i\varepsilon)-f(z))\,dz\right| \leq\operatorname{length}(\partial P) \sup_{z\in\partial P}|f(z+i\varepsilon)-f(z)|\longrightarrow0. \] Thus \(\int_{\partial P}f=0\). Translating downward proves the corresponding assertion for the closed lower half-plane.
2. Every triangle has zero boundary integral, so \(f\) is entire.
If a triangle lies in one closed half-plane, step 1 applies directly. Otherwise its intersection with each closed half-plane is a polygonal region. Apply step 1 to the two pieces and add their boundary integrals. The common real segment occurs with opposite orientations and cancels, leaving exactly the original triangle boundary. Thus its integral is zero as well. A degenerate triangle has zero boundary integral by cancellation of its segments.
The function is continuous on \(\mathbb C\) by hypothesis. Morera’s theorem now makes it holomorphic on \(\mathbb C\) [@SS03]. Its given continuous values on the real line are therefore the values of the required entire extension.