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Let $C_1$ and $C_2$ be simple closed curves in $\mathbb C$ and assume that $C_2$ is in the interior of $C_1$. Let $U$ be the region bounded by $C_1$ and $C_2$.

Prove that an analytic function $f(z)$ on $U$ can be decomposed as $$ f(z)=f_1(z)+f_2(z) ,$$ where $f_1(z)$ is analytic in the interior of $C_1$ and $f_2(z)$ is analytic in the exterior of $C_2$(including $\infty$). Moreover the decomposition is unique up to an additive constant.


My Attempt:

If $C_1$ and $C_2$ are circles then we know that $f_1(z)$ are the terms in the Laurent series with non-negative power, and $f_2(z)$ are those with negative power. Otherwise, we can imitate the important technique to prove the Laurent series locally so that we can still treat it as is the circle(since within a small angle, it preserves the inequalities and we can write $\frac{1}{\zeta-z}=\frac{1}{\zeta-z_0+z_0-z}=\frac{1}{1+\frac{z_0-z}{\zeta-z_0}}\cdot\frac{1}{\zeta-z_0}$ and use the Talor series of $\frac 1{1-z}$). Then since the closed curve is also compact, we can get finitely many expansions of $f(z)$ at different parts of the curve. Finally, since the Laurent series is unique locally, then we can glue each part to get a Laurent series of $f(z)$ in $U$. Does this work?

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  • $\begingroup$ If $f$ is continuous on the closure you can apply the Cauchy Integral Formula on $C_1\oplus C_2$. $\endgroup$ – David C. Ullrich May 18 at 16:42
  • $\begingroup$ @DavidC.Ullrich Yes, I know this. But how to proceed? $\endgroup$ – Bach May 19 at 0:15
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Shift $C_1,C_2$ slightly so that $f$ is analytic on an open containing $U$ and them.

From the Cauchy integral formula (the one for simply connected domains, adding an edge to make $U$ simply connected, edge which disappears since traversed in two opposite directions)

For $z$ in between the two simple closed curves $$f(z) = \frac{1}{2i\pi}\int_{C_1} \frac{f(s)}{s-z}ds-\frac{1}{2i\pi}\int_{C_2} \frac{f(s)}{s-z}ds$$ It is immediate that this is your decomposition.

When $C_1,C_2$ are circles expanding $\frac{1}{s-z}$ in geometric series in $s/z,z/s$ is how you get the Laurent series.

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  • $\begingroup$ How to conclude that the decomposition is unique up to an additive constant? $\endgroup$ – Bach May 19 at 1:19
  • $\begingroup$ What if it is not.. $\endgroup$ – reuns May 19 at 1:20
  • $\begingroup$ Replace $f_1,f_2$ by $f_1+e^z,f_2-e^z$. Where is the problem ? (also $\frac{1}{2i\pi}\int_{C_2} \frac{f(s)}{s-z}ds$ is a Laurent series) $\endgroup$ – reuns May 19 at 1:34
  • $\begingroup$ No. Where is the problem with $f_1+e^z,f_2-e^z$ ? $\endgroup$ – reuns May 19 at 1:53
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    $\begingroup$ The uniqueness works as for the Laurent decomposition: Assume that $f = f_1 + f_2 = g_1 + g_2$ are two such representations. Then $h = f_1 - g_1 = g_2 - f_2$ in $U$. The LHS is holomorphic inside $C_2$ and the RHS is holomorphic outside $C_1$. Therefore $h$ can be extended to an entire function, and since it is holomorphic at infinity, $h$ is constant. $\endgroup$ – Martin R Jun 25 at 12:16

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