2

I am trying to prove that if a continuous function $F$ is holomorphic everywhere in a region $D$ except on line $L$ that passes through $D$, then $F$ is holomorphic on $D$.

However, I am having trouble in coming up with a rigorous prove.

We can assume that $D$ is an open ball with $L$ running through $D$

Possible proof: My method was to use Morera's to show that $\int_TFdz=0$ for all triangle $T\subseteq D$. I have shown that if T has a vertex on L then $\int_TFdz=0$. The next case I considered was when T had an edge lying on the line. I tried to use the method here: Function holomorphic except for real line and continuous everywhere is entire

enter image description here

So I'm trying to show that $|\int_TFdz|=|\sum \int_{T_i}Fdz|<\epsilon $,$\forall \epsilon>0 $ where $T_i$ are the smaller triangles with edges on the line L. The integrals over the other triangles would be 0 as they would either have a vertex on the line or are completely in a holomorphic region. I think we can bound the $\int_{T_i}Fdz$ to be arbitrarily small. However, I can't come up with something rigourous. i.e. I tried to cover the edge with finite number of open balls to bound the smaller triangles but got stuck. Can anyone come up with a rigorous way of proving that $\int_TFdz=0$ if T has an edge on L?

Jhon Doe
  • 2,902
  • The line is the real line (with an affine transformation we may wlog assume to be in that case). We want to show that the integral along the trapezoid is going to vanish if we let the hight go to zero. We can split the trapezoid into a rectangle and two small triangle. The integral along the rectangle is small as the integrals along the long side essentially cancel due to uniform convergence and the integral along the two small sides are essentially bounded by the height. The sum of the length of the sides of the small triangles is also comparable to the hight, thus the integral is small too. – Severin Schraven Dec 26 '19 at 10:54
  • I meant "the integrals along the long sides essentially cancel due to uniform continuity". Because this is of the form $$\int (F(x)-F(i\varepsilon+x))$$ – Severin Schraven Dec 26 '19 at 11:05
  • Thanks. But why does it suffice to consider the line to be the real line? – Jhon Doe Dec 26 '19 at 11:43
  • Because you can precompose with an affine transformation that transforms your line into the real line. – Severin Schraven Dec 26 '19 at 12:31

0 Answers0