= Solution
{c}
The <Liouville theorem> states that every bounded <entire function> is constant.
By the <Laurent theorem>, $f$ has a unique <Laurent series> $f(z)=\sum_{n=-\infty}^\infty c_nz^n$ on $R<|z|<S$. Split it into
$$
\boxed{g(z)=\sum_{n=0}^\infty c_nz^n,\qquad h(z)=\sum_{m=1}^\infty c_{-m}z^{-m}.}
$$
Here $g$ is <holomorphic> on $D_S$ and $h$ is <holomorphic> on $A_{R,\infty}$. To check their extended regions of convergence explicitly, fix $r\in(R,S)$ and put $M_r=\max_{|z|=r}|f(z)|$. The <Laurent coefficient> formula gives
$$
|c_n|\leq M_rr^{-n}\quad(n\geq0),\qquad |c_{-m}|\leq M_rr^m\quad(m\geq1).
$$
The first bound ensures convergence uniformly on every closed <disk> $|z|\leq r'<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\geq r''>r$; letting $r$ range down to $R$ covers $A_{R,\infty}$. Locally uniform convergence of these <power series>, in $z$ and $1/z$ respectively, proves the claimed <holomorphy>. On the original <annulus> their sum is $f$.
The second estimate also proves $h(z)\to0$ as $|z|\to\infty$: for any fixed $r\in(R,S)$ and $|z|>r$,
$$
|h(z)|\leq M_r\sum_{m=1}^\infty(r/|z|)^m=\frac{M_rr}{|z|-r}\longrightarrow0.
$$
For uniqueness under this condition, suppose $f=g_1+h_1=g_2+h_2$. Define $F=g_1-g_2$ on $D_S$ and $F=h_2-h_1$ on $A_{R,\infty}$. The two definitions agree on the overlap, and their <domains> cover $\mathbb C$, so they define an <entire function>. Moreover, $F(z)\to0$ at infinity. It is bounded outside a large <disk> by this limit, and inside by <continuity> and the <extreme value theorem>. The <Liouville theorem> makes $F$ constant, and its limit makes that constant zero. Hence $g_1=g_2$ and $h_1=h_2$. Without the limiting condition, adding a constant to $g$ and subtracting it from $h$ would preserve the decomposition.
Back to article page
