Open mapping theorem (complex analysis)

In complex analysis, the open mapping theorem states that if U is a connected open subset of the complex plane C and f : U → 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).

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) = x² 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 real 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

 

Blue dots represent zeros of g(z). Black spikes represent poles. The boundary of the open set U is given by the dashed line. Note that all poles are exterior to the open set. The smaller red circle is the set B constructed in the proof.

Assume f:U → C is a non-constant holomorphic function and ${\displaystyle U}$  is a connected open subset 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 ${\displaystyle f(U)}$  is contained in a disk which is contained in ${\displaystyle f(U)}$ .

Consider an arbitrary ${\displaystyle w_{0}}$  in ${\displaystyle f(U)}$ . Then there exists a point ${\displaystyle z_{0}}$  in U such that ${\displaystyle w_{0}=f(z_{0})}$ . Since U is open, we can find ${\displaystyle d>0}$  such that the closed disk ${\displaystyle B}$  around ${\displaystyle z_{0}}$  with radius d is fully contained in 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 g(z) is not constant and holomorphic. The roots of g are isolated and by further decreasing the radius of the image disk d, 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, and |g(z)| is a continuous function, so the extreme value theorem guarantees the existence of a minimum. Let e be the minimum of |g(z)| for z on the boundary of B, a positive number.

Denote by ${\displaystyle D}$  the disk around ${\displaystyle w_{0}}$  with radius 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 B as ${\displaystyle f(z)-w}$  for any ${\displaystyle w}$  within a distance ${\displaystyle e}$  of ${\displaystyle w_{0}}$ . Thus, for every ${\displaystyle w}$  in ${\displaystyle D}$ , there exists at least one ${\displaystyle z_{1}}$  in ${\displaystyle B}$  so that ${\displaystyle f(z_{1})=w}$ . This means that the disk D is contained in ${\displaystyle f(B)}$ .

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

Applications

Maximum modulus principle

References

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