An analytic function $f(z)$ on $U$ can be decomposed as $f(z)=f_1(z)+f_2(z)$

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?

• If $f$ is continuous on the closure you can apply the Cauchy Integral Formula on $C_1\oplus C_2$. – David C. Ullrich May 18 at 16:42
• @DavidC.Ullrich Yes, I know this. But how to proceed? – Bach May 19 at 0:15

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.

• How to conclude that the decomposition is unique up to an additive constant? – Bach May 19 at 1:19
• What if it is not.. – reuns May 19 at 1:20
• 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) – reuns May 19 at 1:34
• No. Where is the problem with $f_1+e^z,f_2-e^z$ ? – reuns May 19 at 1:53
• 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. – Martin R Jun 25 at 12:16