So for any one can find a polynomial such that for all
Do note that this theorem is not a weaker version of Runge’s theorem. Here, we do not need to be holomorphic on a neighbourhood of but just on the interior of For example, if the interior of is empty, then just needs to be continuous on Further, it could be that the closure of the interior of might not be all of Consider where is the closed unit disc. Then has two lines coming out of either end of the disc and 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.
- 1 John B. Conway. . Springer-Verlag, New York, New York, 1978.
- 2 Walter Rudin. . McGraw-Hill, Boston, Massachusetts, 1987.
|Date of creation||2013-03-22 14:23:59|
|Last modified on||2013-03-22 14:23:59|
|Last modified by||jirka (4157)|