Open mapping theorem (complex analysis)
In complex analysis, the open mapping theorem states that if {\displaystyle U} is a domain of the complex plane {\displaystyle \mathbb {C} } and {\displaystyle f:U\to \mathbb {C} } is a non-constant holomorphic function, then {\displaystyle f} is an open map (i.e. it sends open subsets of {\displaystyle U} to open subsets of {\displaystyle \mathbb {C} }, 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 {\displaystyle f(x)=x^{2}} is not an open map, as the image of the open interval {\displaystyle (-1,1)} is the half-open interval {\displaystyle [0,1)}.
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 {\displaystyle f:U\to \mathbb {C} } is a non-constant holomorphic function and {\displaystyle U} is a domain of the complex plane. We have to show that every point in {\displaystyle f(U)} is an interior point of {\displaystyle f(U)}, i.e. that every point in {\displaystyle f(U)} has a neighborhood (open disk) which is also in {\displaystyle f(U)}.
Consider an arbitrary {\displaystyle w_{0}} in {\displaystyle f(U)}. Then there exists a point {\displaystyle z_{0}} in {\displaystyle U} such that {\displaystyle w_{0}=f(z_{0})}. Since {\displaystyle U} is open, we can find {\displaystyle d>0} such that the closed disk {\displaystyle B} around {\displaystyle z_{0}} with radius {\displaystyle d} is fully contained in {\displaystyle U}. Consider the function {\displaystyle g(z)=f(z)-w_{0}}. Note that {\displaystyle z_{0}} is a root of the function.
We know that {\displaystyle g(z)} is non-constant and holomorphic. The roots of {\displaystyle g} are isolated by the identity theorem, and by further decreasing the radius of the disk {\displaystyle B}, we can assure that {\displaystyle g(z)} has only a single root in {\displaystyle B} (although this single root may have multiplicity greater than 1).
The boundary of {\displaystyle B} is a circle and hence a compact set, on which {\displaystyle |g(z)|} is a positive continuous function, so the extreme value theorem guarantees the existence of a positive minimum {\displaystyle e}, that is, {\displaystyle e} is the minimum of {\displaystyle |g(z)|} for {\displaystyle z} on the boundary of {\displaystyle B} and {\displaystyle e>0}.
Denote by {\displaystyle D} the open disk around {\displaystyle w_{0}} with radius {\displaystyle e}. By Rouché's theorem, the function {\displaystyle g(z)=f(z)-w_{0}} will have the same number of roots (counted with multiplicity) in {\displaystyle B} as {\displaystyle h(z):=f(z)-w_{1}} for any {\displaystyle w_{1}} in {\displaystyle D}. This is because {\displaystyle h(z)=g(z)+(w_{0}-w_{1})}, and for {\displaystyle z} on the boundary of {\displaystyle B}, {\displaystyle |g(z)|\geq e>|w_{0}-w_{1}|}. Thus, for every {\displaystyle w_{1}} in {\displaystyle D}, there exists at least one {\displaystyle z_{1}} in {\displaystyle B} such that {\displaystyle f(z_{1})=w_{1}}. This means that the disk {\displaystyle D} is contained in {\displaystyle f(B)}.
The image of the ball {\displaystyle B}, {\displaystyle f(B)} is a subset of the image of {\displaystyle U}, {\displaystyle f(U)}. Thus {\displaystyle w_{0}} is an interior point of {\displaystyle f(U)}. Since {\displaystyle w_{0}} was arbitrary in {\displaystyle f(U)} we know that {\displaystyle f(U)} is open. Since {\displaystyle U} was arbitrary, the function {\displaystyle f} is open.
Applications
[edit ]See also
[edit ]References
[edit ]- Rudin, Walter (1966), Real & Complex Analysis, McGraw-Hill, ISBN 0-07-054234-1