# Proving $f$ is holomorphic in $\Omega$ using symmetry and Morera

complex-analysis

Suppose $$\Omega$$ be an open subset of $$\mathbb{C}$$, and let $$S$$ be a finite union of line segments in $$\Omega$$. Also, suppose we have a function $$f$$ which is continuous in $$\Omega$$ and holomorphic in $$\Omega \setminus S$$. We want to show that $$f$$ is holomorphic in $$\Omega$$.

I'm thinking of using Morera and the symmetry principle, but I'm confused on what would happen to the possible intersections and in the 'corners' where the line segments meet.

First a preliminary: the theorem is still true if the line segments are not contained in $$\Omega$$, and it's also easier to prove this way. So that's what we're doing here.
We can do this via induction over the number $$n$$ of line segments. For $$n=0$$ the function is holomorphic by assumption. So we can go right to the induction step, that is, assume the claim is true for $$n$$ line segments, and consider the case of $$n+1$$ segments. Let $$L_i,i=1,\dots,n+1$$ be the line segments whose union is $$S$$. Consider just the line segment $$L_{n+1}$$. Extend this segment to a whole line, which cuts $$\Omega$$ in two halves, $$\Omega_1$$ and $$\Omega_2$$. On both halves we can apply the induction hypothesis, because $$f$$ is continuous on $$\Omega_{1/2}$$ and holomorphic on $$\Omega_{1/2}\backslash\bigcup_{i=1}^n L_i$$, which is just an open set with $$n$$ line segments removed. So $$f$$ is holomorphic on both $$\Omega_1$$ and $$\Omega_2$$. The question remains wether it is on the line dividing the two. We use Morera for this. Morera says that if a function on an open set is continuous and all integrals along triangle contours vanish, then the function is holomorphic. The function is continuous by assumption. Now take any triangle contour $$\Delta$$ and consider
$$\int_\Delta f(w)\mathrm dw.$$
If the contour is contained in either $$\Omega_1$$ or $$\Omega_2$$, then this integral vanishes by the Cauchy integral theorem. If it intersects the line, make two cuts, both parallel to $$L_{n+1}$$ at a distance of $$\varepsilon>0$$ to $$L_{n+1}$$. One cut on each side. You can recover up to two contours from this, a triangle and/or a quadrilateral. The integrals along these contours vanish because they're completely contained in one of the halves of $$\Omega$$, on which $$f$$ is holomorphic.
Now use the continuity of $$f$$ to show that as $$\varepsilon\to0$$, the sum of the integrals along the two contours converges to the integral along the original triangle, making that $$0$$. I leave this to you. Now Morera applies and the function is holomorphic.