Jump to content
Wikipedia The Free Encyclopedia

Open mapping theorem (complex analysis)

From Wikipedia, the free encyclopedia

In complex analysis, the open mapping theorem states that if U {\displaystyle U} {\displaystyle U} is a domain of the complex plane C {\displaystyle \mathbb {C} } {\displaystyle \mathbb {C} } and f : U C {\displaystyle f:U\to \mathbb {C} } {\displaystyle f:U\to \mathbb {C} } is a non-constant holomorphic function, then f {\displaystyle f} {\displaystyle f} is an open map (i.e. it sends open subsets of U {\displaystyle U} {\displaystyle U} to open subsets of C {\displaystyle \mathbb {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}} {\displaystyle f(x)=x^{2}} is not an open map, as the image of the open interval ( 1 , 1 ) {\displaystyle (-1,1)} {\displaystyle (-1,1)} is the half-open interval [ 0 , 1 ) {\displaystyle [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

[edit ]
Black dots represent zeros of g ( z ) {\displaystyle g(z)} {\displaystyle g(z)}. Black annuli represent poles. The boundary of the open set U {\displaystyle U} {\displaystyle U} is given by the dashed line. Note that all poles are exterior to the open set. The smaller red disk is B {\displaystyle B} {\displaystyle B}, centered at z 0 {\displaystyle z_{0}} {\displaystyle z_{0}}.

Assume f : U C {\displaystyle f:U\to \mathbb {C} } {\displaystyle f:U\to \mathbb {C} } is a non-constant holomorphic function and U {\displaystyle U} {\displaystyle U} is a domain of the complex plane. We have to show that every point in f ( U ) {\displaystyle f(U)} {\displaystyle f(U)} is an interior point of f ( U ) {\displaystyle f(U)} {\displaystyle f(U)}, i.e. that every point in f ( U ) {\displaystyle f(U)} {\displaystyle f(U)} has a neighborhood (open disk) which is also in f ( U ) {\displaystyle f(U)} {\displaystyle f(U)}.

Consider an arbitrary w 0 {\displaystyle w_{0}} {\displaystyle w_{0}} in f ( U ) {\displaystyle f(U)} {\displaystyle f(U)}. Then there exists a point z 0 {\displaystyle z_{0}} {\displaystyle z_{0}} in U {\displaystyle U} {\displaystyle U} such that w 0 = f ( z 0 ) {\displaystyle w_{0}=f(z_{0})} {\displaystyle w_{0}=f(z_{0})}. Since U {\displaystyle U} {\displaystyle U} is open, we can find d > 0 {\displaystyle d>0} {\displaystyle d>0} such that the closed disk B {\displaystyle B} {\displaystyle B} around z 0 {\displaystyle z_{0}} {\displaystyle z_{0}} with radius d {\displaystyle d} {\displaystyle d} is fully contained in U {\displaystyle U} {\displaystyle U}. Consider the function g ( z ) = f ( z ) w 0 {\displaystyle g(z)=f(z)-w_{0}} {\displaystyle g(z)=f(z)-w_{0}}. Note that z 0 {\displaystyle z_{0}} {\displaystyle z_{0}} is a root of the function.

We know that g ( z ) {\displaystyle g(z)} {\displaystyle g(z)} is non-constant and holomorphic. The roots of g {\displaystyle g} {\displaystyle g} are isolated by the identity theorem, and by further decreasing the radius of the disk B {\displaystyle B} {\displaystyle B}, we can assure that g ( z ) {\displaystyle g(z)} {\displaystyle g(z)} has only a single root in B {\displaystyle B} {\displaystyle B} (although this single root may have multiplicity greater than 1).

The boundary of B {\displaystyle B} {\displaystyle B} is a circle and hence a compact set, on which | g ( z ) | {\displaystyle |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} {\displaystyle e}, that is, e {\displaystyle e} {\displaystyle e} is the minimum of | g ( z ) | {\displaystyle |g(z)|} {\displaystyle |g(z)|} for z {\displaystyle z} {\displaystyle z} on the boundary of B {\displaystyle B} {\displaystyle B} and e > 0 {\displaystyle e>0} {\displaystyle e>0}.

Denote by D {\displaystyle D} {\displaystyle D} the open disk around w 0 {\displaystyle w_{0}} {\displaystyle w_{0}} with radius e {\displaystyle e} {\displaystyle e}. By Rouché's theorem, the function g ( z ) = f ( z ) w 0 {\displaystyle 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} {\displaystyle B} as h ( z ) := f ( z ) w 1 {\displaystyle h(z):=f(z)-w_{1}} {\displaystyle h(z):=f(z)-w_{1}} for any w 1 {\displaystyle w_{1}} {\displaystyle w_{1}} in D {\displaystyle D} {\displaystyle D}. This is because h ( z ) = g ( z ) + ( w 0 w 1 ) {\displaystyle h(z)=g(z)+(w_{0}-w_{1})} {\displaystyle h(z)=g(z)+(w_{0}-w_{1})}, and for z {\displaystyle z} {\displaystyle z} on the boundary of B {\displaystyle B} {\displaystyle B}, | g ( z ) | e > | w 0 w 1 | {\displaystyle |g(z)|\geq e>|w_{0}-w_{1}|} {\displaystyle |g(z)|\geq e>|w_{0}-w_{1}|}. Thus, for every w 1 {\displaystyle w_{1}} {\displaystyle w_{1}} in D {\displaystyle D} {\displaystyle D}, there exists at least one z 1 {\displaystyle z_{1}} {\displaystyle z_{1}} in B {\displaystyle B} {\displaystyle B} such that f ( z 1 ) = w 1 {\displaystyle f(z_{1})=w_{1}} {\displaystyle f(z_{1})=w_{1}}. This means that the disk D {\displaystyle D} {\displaystyle D} is contained in f ( B ) {\displaystyle f(B)} {\displaystyle f(B)}.

The image of the ball B {\displaystyle B} {\displaystyle B}, f ( B ) {\displaystyle f(B)} {\displaystyle f(B)} is a subset of the image of U {\displaystyle U} {\displaystyle U}, f ( U ) {\displaystyle f(U)} {\displaystyle f(U)}. Thus w 0 {\displaystyle w_{0}} {\displaystyle w_{0}} is an interior point of f ( U ) {\displaystyle f(U)} {\displaystyle f(U)}. Since w 0 {\displaystyle w_{0}} {\displaystyle w_{0}} was arbitrary in f ( U ) {\displaystyle f(U)} {\displaystyle f(U)} we know that f ( U ) {\displaystyle f(U)} {\displaystyle f(U)} is open. Since U {\displaystyle U} {\displaystyle U} was arbitrary, the function f {\displaystyle f} {\displaystyle f} is open.

Applications

[edit ]

See also

[edit ]

References

[edit ]

AltStyle によって変換されたページ (->オリジナル) /