This is the natural continuation of this other question. I have this exercise
Let $L:=a+\Bbb R b$, for some $a,b\in\Bbb C$, a straight line in $\Bbb C$. Also let $f\in C(U,\Bbb C)$ is holomorphic on $U\setminus L$. Show that $f$ is holomorphic on all of $U$. (Hint: Morera's theorem.)
I want to know is the proof below is correct or if there is some mistake or it is incomplete.
The version of Morera's theorem that I used here use only triangular paths $\Delta$, where a triangular path is a closed path defined by the straight segments between three different points whenever the region bounded by the triangle belongs to the domain of the function.
It says
Morera's theorem: if $\int_\Delta f dz=0$ for every triangular path in $U$ then $f$ is analytic in $U$.
The proof:
Observe that we need to consider just the analyticity of $f$ on $U\cap L$.
Then we only need to see what happen to the paths that contain some point of $U\cap L$. There are three kinds of triangular paths $\Delta$ that contain a point of $U\cap L$:
- $|\Delta\cap L|=1$, then necessarily this point is a vertex of $\Delta$.
- $|\Delta\cap L|=2$, then there is a segment of $L$ inside the area bounded by $\Delta$.
- $|\Delta\cap L|=\mathfrak c$, then there is a whole segment of $\Delta$ contained in $L$.
Let a homotopy $H\in C^1([0,1]^2,U)$ such that each $H([0,1],t)$ is triangular. Now I want to show that if $f$ is continuous then the map $t\mapsto\int_{\Delta_t}f(z)\, dz$ is also continuous, where $\Delta_t:=H([0,1],t)$. Then $$ \left|\int_{\Delta_{t_1}} f(z)\, dz-\int_{\Delta_{t_2}}f(z)\, dz\right|=\left|\int_0^1(f\circ H)(s,t_1)\partial_1H(s,t_1)\, ds-\int_0^1(f\circ H)(s,t_2)\partial_1H(s,t_2)\, ds\right|\\ =\left|\int_0^1\big((f\circ H)(s,t_1)\partial_1H(s,t_1)-(f\circ H)(s,t_2)\partial_1H(s,t_2)\big)\, ds\right|\\ \le\max_{s\in[0,1]}|(f\circ H)(s,t_1)\partial_1H(s,t_1)-(f\circ H)(s,t_2)\partial_1H(s,t_2)|\tag1 $$ Now observe that the function $g:=(f\circ H)(\cdot,\cdot)\partial_1 H(\cdot,\cdot)$ is uniformly continuous (because it is a continuous function with compact domain) so for every chosen $\epsilon>0$ there is a $\delta>0$ such that $$ |(s,t_1)-(s,t_2)|=|t_1-t_2|<\delta\implies |g(s,t_1)-g(s,t_2)|<\epsilon\tag2 $$ Consequently the map $t\mapsto\int_{\Delta_t}f(z)\, dz$ is continuous. Thus if we choose $H$ such that $\Delta_t\subset U\setminus L$ for all $t\in[0,1)$ and $\Delta_1\subset U$ then $$ \int_{\Delta_1}f(z)\, dz=\lim_{t\to 1}\int_{\Delta_t}f(z)\, dz=\lim_{t\to 1}0=0\tag3 $$ Now observe that this hold for every triangular path on $U$ of kind $1$ and $3$ because the interior of the region bounded by these triangles belongs completely to $U\setminus L$, so we can define such $H$ in any case. By last observe that a triangle of kind $2$ can be decomposed in two or three triangular paths of kind $1$ and $3$.
Then $(3)$ holds for every $\Delta_1\subset U$ so, by Morera's theorem, $f$ is analytic in $U$.$\Box$
Your proof is not correct. You only considered triangles such that one of its vertices belong to $L$. That's far from being the general case.