How to determine the uniqueness of the "Laurent representation" of a function?

Question

Let $A$ be the annulus $\{z:r_1<|z|<r_2\}$, where $r_1$ and $r_2$ are given positive numbers.

(a) Show that the Cauchy formula

$$ f(z)=\frac{1}{2\pi{i}}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)\frac{f(\zeta)}{\zeta-z}d\zeta $$

is valid under the following condition: $f \in H(A)$,

$$ r_1 +\varepsilon < |z| < r_2 - \varepsilon \quad $$

and

$$\gamma_1(t)=(r_1+\varepsilon)e^{-it}, \quad \gamma_2(t)=(r_2-\varepsilon)e^{it}\quad 0 \leq t \leq 2\pi$$

(b) Show by means of (a) that every $f\in H(A)$ can be decomposed into a sum $f=f_1+f_2$, where $f_1$ is holomorphic outside $\overline{D}(0;r_1)$ and $f_2 \in H(D(0;r_2))$; the decomposition is unique if we require that $f_1(z)\to0$ as $|z| \to \infty$.

This is an exercise comes from Rudin's Real and Complex Analysis with topic ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS. The rest part of the question goes to Laurent series by using this decomposition. My question is, how to determine the uniqueness of this decomposition?

If I'm not wrong, the first part of (b) is

$$f_1(z)=\frac{1}{2\pi{i}}\int_{\gamma_1}\frac{f(\zeta)}{\zeta-z}d\zeta$$

and

$$f_2(z)=\frac{1}{2\pi{i}}\int_{\gamma_2}\frac{f(\zeta)}{\zeta-z}d\zeta$$

and by some inequality stuff, it can be shown that $f_1(z) \to 0$ as $|z| \to \infty$. And I think to prove the uniqueness of this decomposition, I should directly try to prove the uniqueness of $f_1$, by assuming that there is another decomposition $(f_3,f_4)$, and prove that $f_3-f_1=0$ for all $z$ outside $\overline{D}(0;r_1)$ (let's call it $B$). But I found no way out.

I was considering Liouville's theorem, but unfortunately $f_3-f_1$ is not entire. Is it possible to define the function inside $B$? or should I consider $f_3(\frac{1}{z})-f_1(\frac{1}{z})$ inside $D(0;\frac{1}{r_1})$ (probably I should remove the origin here?). How would you determine the uniqueness? Appreciate in advance!

Answer 1

To show that the decompostion is unique given $f_1(z)\to0$ as $|z|\to\infty$, assume that $f = g_1 + g_2$, such that $g_1 \in H(\mathbb{C}\setminus \overline{D(0; r_1)})$, and $g_2\in H(D(0;r_2))$. Hence \begin{equation*} \begin{aligned} f_1 + f_2 &= g_1 + g_2 \\ f_1 - g_1 &= g_2 - f_2 \\ \end{aligned} \end{equation*} Then we can define \begin{equation*} \begin{aligned} h(z) = \begin{cases} f_1(z) - g_1(z) & z \in \mathbb{C}\setminus \overline{D(0; r_1)} \\ g_2(z) - f_2(z) & z \in D(0;r_2) \\ \end{cases} \end{aligned} \end{equation*} For $z \in A$, the two cases of $h(z)$ are equivalent, hence $h(z)$ is entire, and well defined. Then we have that as $|z|\to\infty$ then $f_1(z), g_1(z) \to 0$, hence $h(z)\to0$, hence $h(z)$ is bounded. Then by Liouville's Theorem, $h(z)$ is constant, hence $h(z) = 0, \ \forall z \in \mathbb{C}$. Hence for $z \in \mathbb{C}\setminus \overline{D(0; r_1)}$, $f_1(z) = g_1(z)$. Similarly for $z \in D(0;r_2)$, $f_2(z) = g_2(z)$.

Answer 2

Define

$g(z)=\left\{ \begin{array}{c l} f_3(z)-f_1(z) & (|z|>r_1)\\ f_2(z)-f_4(z) & (|z|<r_2) \end{array}\right.$

which is analytic on the whole plane by analyticity of $f_1,f_2,f_3,f_4$ as you have defined, then use Liouvelle's theorem about bounded holomorphic functions.