Open mapping theorem (complex analysis)
In complex analysis, the open mapping theorem states that if is a domain of the complex plane and is a non-constant holomorphic function, then is an open map (i.e. it sends open subsets of to open subsets of , and we have invariance of domain.).
The open mapping theorem points to the sharp difference between holomorphy and real-differentiability. On the real line, for example, the differentiable function is not an open map, as the image of the open interval is the half-open interval .
The theorem for example implies that a non-constant holomorphic function cannot map an open disk onto a portion of any line embedded in the complex plane. Images of holomorphic functions can be of real dimension zero (if constant) or two (if non-constant) but never of dimension 1.
Proof
[edit]Assume is a non-constant holomorphic function and is a domain of the complex plane. We have to show that every point in is an interior point of , i.e. that every point in has a neighborhood (open disk) which is also in .
Consider an arbitrary in . Then there exists a point in such that . Since is open, we can find such that the closed disk around with radius is fully contained in . Consider the function . Note that is a root of the function.
We know that is non-constant and holomorphic. The roots of are isolated by the identity theorem, and by further decreasing the radius of the disk , we can assure that has only a single root in (although this single root may have multiplicity greater than 1).
The boundary of is a circle and hence a compact set, on which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle |g(z)|} is a positive continuous function, so the extreme value theorem guarantees the existence of a positive minimum , that is, is the minimum of for on the boundary of and .
Denote by the open disk around with radius . By Rouché's theorem, the function will have the same number of roots (counted with multiplicity) in as for any in . This is because , and for on the boundary of , . Thus, for every in , there exists at least one in such that . This means that the disk is contained in .
The image of the ball , is a subset of the image of , . Thus is an interior point of . Since was arbitrary in we know that is open. Since was arbitrary, the function is open.
Applications
[edit]See also
[edit]References
[edit]- Rudin, Walter (1966), Real & Complex Analysis, McGraw-Hill, ISBN 0-07-054234-1