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Let $f : \mathbb{C} \to \mathbb{C}$ be a continuous function holomorphic on $\mathbb{C} \setminus \mathbb{R}$. Show that $f$ is entire.

This question has been answered here, but I would like to clarify on the solution provided by Pp.. I understand that it reduces to the case where we examine triangles with an edge on the line (say $\Delta$). According to Pp..'s solution, he breaks the triangle down to smaller and smaller triangles. He claimed that, since each of the smaller triangles that intersect the line can get infinitely small, the path integral over each of this smaller triangle gets infinitely small. I agree with this, as we have the following: $$ \int_\phi f(z) \; \mathrm{d}z \leq \text{Length}(\phi) \cdot \sup_{x \in \phi^*}|f(z)| $$ where $\phi^*$ is the set of points which $\phi$ runs across.

However, while the path integral over each of the smaller triangles gets arbitrarily small, the number of such smaller triangles also gets arbitrary many. My question is, how can we ensure that the overall path integral (i.e. path integral over the original triangle $\Delta$) gets arbitrarily small as well?

Any help is appreciated.

Clement Yung
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    Split it such that you put a rectangle with one side on the line. Then you by making the side orthogonal to the line small, this contributions gets as small as you want. On the other hand the contributions from the sides parallel to the line have opposite sign, so you will get $\int (F(z)- F(z+\varepsilon z_0))$ and thus this gets small by uniform continuity. – Severin Schraven Apr 30 '20 at 09:49
  • @SeverinSchraven Why do the contributions from the sides parallel to the line have opposite signs? – Clement Yung Apr 30 '20 at 09:56
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    Because once you walk from left to right and the other time from right to left. If you now want to integrate over the same path, you make a substitution which yields exactly what I wrote above. – Severin Schraven Apr 30 '20 at 10:03
  • @SeverinSchraven Oh I understand now. Thank you! – Clement Yung Apr 30 '20 at 10:07
  • You are welcome :) – Severin Schraven Apr 30 '20 at 10:09

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