= Solution
Runge's polynomial approximation theorem says that if $K\subset\mathbb C$ is compact, $\mathbb C\setminus K$ is connected, and $f$ is holomorphic on a neighborhood of $K$, then for every $\varepsilon>0$ there is a polynomial $p$ such that
$$
\sup_K|p-f|<\varepsilon.
$$
Let
$$
S=\{z:|z|=1,\ \operatorname{Re}z\leq0\}.
$$
This arc is compact, its complement is connected, and $1/z$ is holomorphic on a neighborhood of it. Applying Runge with error $1/n$ gives a polynomial $q_n$ satisfying
$$
\sup_{z\in S}|q_n(z)-z^{-1}|<\frac1n,
$$
which is the requested uniform approximation.
For the pointwise construction, for $n\geq2$ set
$$
\begin{aligned}
K_n^+&=\{z:|z|\leq1-1/n,\ \operatorname{Re}z\geq1/n\},\\
K_n^-&=\{z:|z|\leq1-1/n,\ \operatorname{Re}z\leq-1/n\},\\
K_n^0&=\{iy:|y|\leq1-1/n\},
\qquad K_n=K_n^+\cup K_n^-\cup K_n^0.
\end{aligned}
$$
The three pieces are disjoint compact sets and $\mathbb C\setminus K_n$ is connected. Define a function on a neighborhood of $K_n$ to equal $1$, $-1$, and $0$ on neighborhoods of $K_n^+$, $K_n^-$, and $K_n^0$, respectively. It is holomorphic because those neighborhoods may be chosen disjoint. Runge supplies a polynomial $P_n$ such that
$$
|P_n-1|<1/n\ \hbox{on }K_n^+,
\quad |P_n+1|<1/n\ \hbox{on }K_n^-,
\quad |P_n|<1/n\ \hbox{on }K_n^0.
$$
Every fixed point with $|z|<1$ eventually belongs to the appropriate one of these three sets, according to the sign of its real part. These inequalities therefore give precisely the asserted pointwise limits.
Back to article page
