Open mapping theorem (complex analysis) explained

In complex analysis, the open mapping theorem states that if

U is a domain of the complex plane C and f: U\to C is a non-constant holomorphic function, then f is an open map (i.e. it sends open subsets of U to open subsets of 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} is not an open map, as the image of the open interval ( -1 , 1 ) is the half-open interval [ 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\to C is a non-constant holomorphic function and U is a domain of the complex plane. We have to show that every point in f(U) is an interior point of f(U) , i.e. that every point in f(U) has a neighborhood (open disk) which is also in f(U) .

Consider an arbitrary

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

We know that

g(z) is non-constant and holomorphic. The roots of g are isolated by the identity theorem, and by further decreasing the radius of the disk B , 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, on which |g(z)| is a positive continuous function, so the extreme value theorem guarantees the existence of a positive minimum e , that is, e is the minimum of |g(z)| for z on the boundary of B and e> 0 .

Denote by

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

The image of the ball

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

Applications

• Maximum modulus principle
• Rouché's theorem
• Schwarz lemma

See also

• Open mapping theorem (functional analysis)