# Mergelyan’s theorem

###### Theorem (Mergelyan).

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

So for any $\u03f5>0$ 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 not
need $f$ 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 are holomorphic in Mergelyan’s theorem.

## References

- 1 John B. Conway. . Springer-Verlag, New York, New York, 1978.
- 2 Walter Rudin. . McGraw-Hill, Boston, Massachusetts, 1987.

Title | Mergelyan’s theorem |
---|---|

Canonical name | MergelyansTheorem |

Date of creation | 2013-03-22 14:23:59 |

Last modified on | 2013-03-22 14:23:59 |

Owner | jirka (4157) |

Last modified by | jirka (4157) |

Numerical id | 7 |

Author | jirka (4157) |

Entry type | Theorem |

Classification | msc 30E10 |

Related topic | RungesTheorem |