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.

## Best Answer

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.