Mergelyan’s theorem

Theorem (Mergelyan).

Let $K\mathrm{\subset }\mathrm{C}$ be a compact subset of the complex plane such that$\mathrm{C}\mathrm{\backslash \backslash }K$ (the complement of $K$) is connected, and let$f\mathrm{:}K\mathrm{\to }\mathrm{C}$ be a continuous function which is also holomorphicon the interior of $K\mathrm{.}$ Then $f$ is the uniform limit on $K$ of holomorphicpolynomials (polynomials in one complex variable).

So for any $ϵ>0$ one can find a polynomial $p(z)={\sum }_{j=1}^n{a}_j{z}^j$such that for all $z\in K.$

Do note that this theorem is not a weaker version of Runge’s theorem. Here, we do notneed $f$ to be holomorphic on a neighbourhood of $K,$ but just on the interior of $K.$ For example, if the interior of $K$ is empty, then $f$ just needs to be continuous on $K.$ Further, it could be that the closure of the interior of $K$might not be all of $K.$ Consider $K=D\cup [-10,10],$ where $D$is the closed unit disc. Then $K$ has two lines coming out of either end of the disc and $f$ needs to only be continuous there.

Also note that this theorem is distinct from the Stone-Weierstrass theorem. The point here is that the polynomials areholomorphic in Mergelyan’s theorem.