I'm trying to solve
Show that if $f:\mathbb{C}\to\mathbb{C}$ is a continuous function such that f is analytic off $[−1, 1]$ then $f$ is an entire function.
This is a solution of Andreas Kleefeld:
Solution. By Morera’s Theorem, it suffice to show that for every triangular path $T$ in $\mathbb{C}$, we have $\int_{T}f=0$, since we already assume $f$ is continuous. There are several cases to consider:
Case 1: A triangle does not intersect $I=[-1,1]$. In this case, we obtain by Cauchy’s Theorem, that $\int_{T}f=0$,
since we can find an open neighborhood $G$ containing $T$ (is closed and rectifiable) such that $f\in A(G)$ (by
assumption) and $n(T;w) = 0 \forall w \in\mathbb{C}\setminus G$ (by Theorem 4.4 p. 82).Case 2: A triangle touches $I$ exactly at one point $P$. This single point of intersection $P$ is a removable
singularity, since $f$ is continuous. Again apply Cauchy’s Theorem to obtain
$\int_{T}f=0$. (Procedure: Translate
the triangle by $\pm \epsilon i$ depending if it lies above or below the $x$-axis. By Cauchy’s Theorem $\int_{T}f=0$ over the
translated triangle and let $\epsilon\to 0$ since $f$ is continuous).Case 3: One edge of the triangle touches $I$. Like in case 2 we can translate the triangle by $\pm \epsilon i$ We can
also argue this way: Let $G$ be an open neighborhood containing the triangle $T$. Let $\left\{T_n\right\}$ be a sequence of
triangles that are not intersecting $I$, but whose limit is the given $T$. Then by the continuity of $f$ , we get
$$\int_{T}f=\lim_{n}\int_{T_n}f=\lim_{n}0=0$$
where the first equality follows by Lemma 2.7 and the second one by case 1.Case 4: A triangle $T$ is cut into 2 parts by $I$. Then, we can always decompose the triangle $T$ in three parts.
Two triangles are of the kind explained in case 3 and one is of the kind explained in case 2 (a sketch mighthelp). Thus, $\int_{T}f=0$ again.Case 5: A triangle $T$ contains parts of $I$ (also here a sketch might help). In this instance, we can decompose
the triangle into 5 parts. Two triangles are of the kind explained in case 3 and the other three triangles are
of the kind explained in case 2. Hence, $\int_{T}f=0$
Summary: Since $\int_{T}f=0$ for every triangular path in $\mathbb{C}$ and $f$ is continuous, we get by Morera’s Theorem:
$f\in A(\mathbb{C})$, that is $f$ is an entire function.
I can't see the sketch of case 5.What would that decomposition look like graphically? I can't break it down into 5 triangles.
Best Answer
I think you're missing the case where the "part" of $I$ in $T$ is the entirety of $I$:
This proof of Andreas Kleefeld also allows degenerate triangles in his decomposition of $T$ which I do not find helpful (i.e., he means at most 5 triangles)