- Let \(I = \{ t \in \mathbb { R } : 0 \leq t \leq 1 \} \subset \mathbb { C }\) . Suppose that \(f : \mathbb { C } \to \mathbb { C }\) is a continuous function such that f is holomorphic on \(\mathbb { C } \setminus I\) . Prove that f is an entire function \(( \mathrm { i . e . , ~ } f\) is holomorphic on all of C).
Solution
Since \(I\subset\mathbb R\), the function \(f\) is holomorphic in each open half-plane. We prove that its continuity across the whole real line is enough to make it entire.
1. The boundary integral of \(f\) is zero for every polygonal region in a closed half-plane.
Let \(P\) be a closed polygonal region in \(\{\operatorname{Im}z\geq0\}\), oriented positively. For \(\varepsilon>0\), the region \(P+i\varepsilon\) lies in the open upper half-plane. Cauchy’s theorem gives \[ 0=\int_{\partial(P+i\varepsilon)}f(w)\,dw =\int_{\partial P}f(z+i\varepsilon)\,dz \] [@SS03]. The set of all \(z+it\) with \(z\in\partial P\) and \(0\leq t\leq1\) is compact. Uniform continuity of \(f\) on that set therefore gives \[ \left|\int_{\partial P}\bigl(f(z+i\varepsilon)-f(z)\bigr)\,dz\right| \leq \operatorname{length}(\partial P) \sup_{z\in\partial P}|f(z+i\varepsilon)-f(z)|\longrightarrow0. \] Thus \(\int_{\partial P}f(z)\,dz=0\). For a region in the closed lower half-plane, translate by \(-i\varepsilon\) and use exactly the same estimate with the downward translates, obtaining the same zero integral.
2. Every triangle has zero boundary integral, and Morera’s theorem applies.
If a triangle lies in a closed half-plane, step 1 already gives its zero boundary integral. Otherwise, cut its closed triangular region along the real line. The two pieces are polygonal regions, one in each closed half-plane. Adding their zero boundary integrals cancels the shared real segment, whose two orientations are opposite. The remaining segments form the original triangle boundary, so its integral is also zero. Degenerate pieces have zero boundary integral by cancellation and cause no exception.
The given function is continuous on \(\mathbb C\) and has zero integral around every triangle. Morera’s theorem makes it holomorphic throughout \(\mathbb C\) [@SS03]. The argument includes triangles meeting either endpoint of \(I\); no endpoint extension remains to be proved.