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