 Solution

The Liouville theorem states that every bounded entire function is constant.

By the Laurent theorem, $f$ has a unique Laurent series $f(z)={\sum }_{n=-\infty }^{\infty }{c}_n{z}^n$ on $R<|z|<S$. Split it intoHere $g$ 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 givesThe first bound ensures convergence uniformly on every closed disk $|z|\le r^{\prime }<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\ge r^{\prime \prime }>r$; 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 $h(z)\to 0$ as $|z|\to \infty$: for any fixed $r\in (R,S)$ and $|z|>r$,

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, $F(z)\to 0$ 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.