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This is exercise 1, page 101 of Michael Taylor's Introduction to Complex Analysis, which you may find here.

Statement: Let Ω ⊂ C be a connected domain. Suppose γ is a smooth curve in Ω, and Ω \ γ has two connected pieces, say Ω±. Assume g is continuous on Ω, and holomorphic on Ω+ and on Ω−. Show that g is holomorphic on Ω. Hint. Verify the hypotheses of Morera’s theorem.


My proof idea: We assume without loss of generality in the following that $\gamma$ is defined on $[0,1]$ (if it is not we may reparametrize it).

We aim to verify the hypotheses of Morera's theorem. To this end consider a rectangle $R \subseteq \Omega$. Clearly there is nothing to do if $R \subseteq \Omega_+$$R \subseteq \Omega_-$; $g$ is holomorphic on these regions so the contour integral is $0$ as a consequence of Cauchy's integral theorem.

Suppose then that $R$ intersects $\gamma$ and let $O$ be some compact set that contains $\gamma$. In the following let $P_n = \{t_{k,n}\}$ be the standard partition of $[0,1]$: $t_{k,n} = \frac{k}{n}$ for $0 \le k \le n$. The path $\gamma$ is smooth, and so in particular it is Lipschitz so we have for all $k$: \begin{equation*} \vert \gamma(t_{k+1, n}) - \gamma(t_{k, n}) \vert \le C \vert t_{k+1, n} - t_{k, n} \vert \le C \frac{1}{n}. \end{equation*} In particular this means that the path segment $\gamma \colon t \in [t_{k, n}, t_{k+1, n}]$ is contained in a rectangle $R_{k,n}$ with sides of length inferior or equal to $\frac{C}{n}$.

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Write \begin{align*} R_+ &= \left ( R \setminus \left ( \bigcup_{k=0}^n R_{k,n} \right ) \right ) \cap \Omega_+ \\ R_- &= \left ( R \setminus \left ( \bigcup_{k=0}^n R_{k,n} \right ) \right ) \cap \Omega_-. \end{align*} So $R_+$ is the part of $R$ lying in $\Omega_+$ from which we have removed the rectangles $R_{k,n}$, and vice-versa for $R_-$. With a little effort of imagination we see that: \begin{equation*} \int_{\partial R} g = \int_{\partial R_+} g + \int_{\partial R_-} + \sum_{k = 0}^{n} \int_{\partial R_{k,n}} g. \end{equation*}

The first two integrals are $0$ as a consequence of the Cauchy integral theorem. Then we may bound the sum in the following manner: \begin{align*} \vert \sum_{k = 0}^{n} \int_{\partial R_{k,n}} g \vert &\le \sum_{k = 0}^{n} \sup_{\partial R_{k,n}} \vert g \vert \cdot \ell (\partial R_{k,n}) \\ &\le \sum_{k = 0}^{n} \sup_{\partial R_{k,n}} \vert g \vert \cdot \frac{4C}{n}. \end{align*}

But here I meet a big obstacle; there are n terms in my sum that are all of the shape 1/n, so this does not let me conclude that everything converges to $0$...

I'm a bit stumped on how to proceed and if this is even the right idea so I would really appreciate any pointers.

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    $\begingroup$ Your drawing is somewhat misleading, because $\gamma$ can never touch the boundary of $\Omega$. This is because $\gamma([0,1])$ is a compact subset contained in $\Omega$ and so the distance to the boundary is $>0$. $\endgroup$ Commented Oct 12 at 6:51
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    $\begingroup$ If $\gamma$ is a polygonal path only moving vertically and horizontally, then $\int_R g(z) dz =0$ follows from Theorem 5.9 of the linked notes. In general it might be possible to conclude the general case by approximating $\gamma$ with such a polygonal path. $\endgroup$ Commented Oct 12 at 8:30
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    $\begingroup$ You want to use the cancellation of integrating along a path and along another path that is nearly the same, but going in the opposite direction. In your bound you are adding both contributions, instead of letting them subtract. Try this. Write $g$ on the sides of your rectangles as $g$ on the curve plus $\epsilon$. Then integrate along the sides of the rectangles and close the loop coming back along the curve. Do let two integrals of $g$ along the curve to cancel, and bound the integral of $\epsilon$ as you did. $\endgroup$ Commented Oct 12 at 9:28
  • $\begingroup$ Thank you so much, beginner! I don't know how I didn't see that... $\endgroup$ Commented Oct 12 at 20:14

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