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

 

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

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

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

See also

• Open mapping theorem (functional analysis)

References

• Rudin, Walter (1966), Real & Complex Analysis, McGraw-Hill, ISBN 0-07-054234-1