Open mapping theorem (complex analysis): Difference between revisions

Revision as of 05:26, 3 June 2012

Revision as of 05:34, 3 June 2012

Line 13:

Consider an arbitrary <math>w_0</math> in <math>f(U)</math>. Then there exists a point <math>z_0</math> in ''U'' such that <math>w_0 = f(z_0)</math>. Since ''U'' is open, we can find <math>d>0</math> such that the closed disk <math>B</math> around <math>z_0</math> with radius ''d'' is fully contained in ''U''. Consider the function <math>g(z)=f(z)-w_0</math>. Note that <math>z_0</math> is a [[root of a function|root]] of the function.

We know that ''g''(''z'') is not constant and holomorphic. The reciprocal of any holomorphic ''g''(''z'') is [[meromorphic]] and has isolated poles. Thus the roots of holomorphic non-constant functions are isolated. In particular, the roots of ''g'' are isolated and by further decreasing the radius of the image disk ''d'', we can assure that ''g''(''z'') has only a single root in ''B'' (although this single root may have multiplicity greater than 1).

The boundary of ''B'' is a circle and hence a [[compact set]], and |''g''(''z'')| is a [[continuous function]], so the [[extreme value theorem]] guarantees the existence of a minimum. Let ''e'' be the minimum of |''g''(''z'')| for ''z'' on the boundary of ''B'', a positive number.