Mergelyan's theorem

Remarks

The conditions on $f$ are necessary, since a uniform limit of polynomials on $K$ is continuous on $K$ and holomorphic on its interior. The condition on $K$ is also necessary. On the unit circle, $f(z) = 1/z$ is continuous and the interior is empty, yet $f$ is not a uniform limit of polynomials. Indeed $\oint p(z) \, dz = 0$ for polynomials while $\oint z^{-1} \, dz = 2\pi i$.

Two special cases were known before. If $f$ is holomorphic on a neighbourhood of $K$, the result is Runge's theorem. If $K$ is the closed unit disc, it suffices to approximate $f$ by $f(rz)$ with $r < 1$ and then truncate its Taylor series. The general proof is much harder; it uses estimates on Cauchy integrals that are uniform over the shape of $K$.

Example: let $K = [0, 1]$, a segment with empty interior and connected complement. The theorem says every continuous function on $[0, 1]$ is a uniform limit of complex polynomials, which recovers the Weierstrass approximation theorem.

Going deeper

For a student

Mergelyan's theorem completes the theory of polynomial approximation on compact sets with connected complement. It contains the Weierstrass approximation theorem and its complex analogue for sets with empty interior. For sets with holes, the question of approximation by rational functions is harder and is described by analytic capacity (Vitushkin's theory). On unbounded closed sets the corresponding statement is Arakelyan's theorem.