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My question is motivated by this post.

Let $f : \mathbb{C} \to \mathbb{C}$ be continuous and $f|_{\mathbb{C} \backslash S^1} :\mathbb{C} \backslash S^1 \to \mathbb{C}$ be holomorphic. I am trying to show that $f$ is entire, i.e. holomorphic everywhere.

My attempt: As the hint in the aforementioned post suggested, we'll try to use Morera: Let $\Omega := \mathbb{C}$ be the open set in Morera. Let $a,b,c \in \mathbb{C}$ and let $T$ be the convex hull of $a,b,c$. Then, we want to show that $\int_{\partial T}f(z)\,dz=0$ for all triangles $\partial T$.

If $\partial T \cap S^1 = \emptyset$, then $\int_{\partial T}f(z)\,dz=0$ anyway. So assume $\partial T \cap S^1 \neq \emptyset$. Then, we see that $\#(\partial T \cap S^1) \leq 6$, since each edge of $\partial T$ has at most two points of intersection with $S^1$. Let $I := \partial T \cap S^1$. Assume w.l.o.g. $\#I=1$ and that this one point of intersection lies in the middle of the edge connecting $a$ and $b$. Then, because $I$ is just a single point, we can write \begin{align} \int_{\partial T}f(z)\,dz &= \int_{\partial T \backslash I}f(z)\,dz = 0, \end{align} where the last equality follows from Goursat. This is the part where I'm confused. My idea was that this would work similarly to Riemann integrals over $\mathbb{R}$, where it is possible to integrate a function $g : [0,1]\to\mathbb{R}$ if $g$ has finitely many points where $g$ is not differentiable. So it seemed like it would be okay to integrate over $\partial T\backslash I$, because it is almost a closed curve, on which $f$ is differentiable.

Am I missing something, or does anybody have any suggestions as to how I can finish my proof more rigorously?

jasnee
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    Note that by elementary geometry the intersection cannot have one point unless that is a vertex and the triangle sits inside or outside fully and the result is easy by continuity; in general one decomposes the triangle in pieces inside and outside the circle with a few points on the boundary and then we get a sum of integrals on closed curves all inside or outside except for some arcs on the boundary (the extra curve segments cancel out in the sum as they appear in opposite orientation pairs), so one needs to prove the result in that case where it follows by continuity – Conrad Jul 27 '21 at 13:20
  • @Conrad Thanks, that was helpful. However, isn't it possible for the intersection to have one point that is not a vertex of the triangle? For instance, if one of the sides of the triangle lies tangential to $S^1$ and the rest of the triangle lies outside the circle. – jasnee Jul 27 '21 at 13:25
  • That's true of course but like with the vertex case this case is easy by continuity; the more difficult case is when the triangle has vertices both inside and outside circle where the smoothness of the circle is essential (in jargon the circle has analytic measure or capacity zero in the plane - there are Jordan curves of nonzero analytic measure where the quoted result is false like the Osgood curve) – Conrad Jul 27 '21 at 14:39

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I hope it helps you :). I am going to use a general theorem of Cauchy: If $C$ is a closed simple curve such that $f$ is holomorphic over it and its interior then $\int_Cf(z)dz=0$. enter image description here We take the triangle such that it intersects the circle $S^1$ and break it with $\alpha$ and $\beta$ where $\alpha$ is the part of the triangle above the circle and joined with a circular curve ($e^{i\theta}\delta$ where $\delta>1$ and $\theta\in[\theta_1,\theta_2]$), $\beta$ is the part of the triangle below the circle joined with a circular curve ($e^{i\theta}\epsilon$ where $0<\epsilon<1$ and $\theta\in[\theta_2,\theta_1]$). Then if we integrate $$\int_\alpha fdz+\int_\beta fdz=\int_{l_1} fdz+\int_{l_2} fdz+\int_{l_3} fdz+\int_{l_4} fdz+\int_{l_5} fdz +\int_{\theta_1}^{\theta_2}f(e^{i\theta}\delta)-f(e^{i\theta}\epsilon)d\theta$$ where $1>\epsilon>0$ tends to 1 and $\delta>1$ tends to 1 and $l_1,l_2,l_3,l_4,l_5$ are the lines of the triangle. Then since $f$ is holomorphic in $\mathbb{C}\backslash S^1$, using the theorem of Cauchy, then $$\int_\alpha fdz+\int_\beta fdz=0.$$ Since $f$ is continuous $$\lim_{\delta,\epsilon\rightarrow 1} \int_{\theta_1}^{\theta_2}f(e^{i\theta}\delta)-f(e^{i\theta}\epsilon)d\theta=0.$$ So finally $$\int_T fdz=\lim_{\delta,\epsilon\rightarrow 1} \left(\int_\alpha fdz+\int_\beta fdz\right)=0.$$

Luz
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