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 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 $U$ is a connected open subset 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 $f(U)$ is contained in a disk which is contained 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. Since U is connected and f is not constant on U, we then know that f is not constant on B because of Analytic continuation. 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 not constant and holomorphic. The reciprocal of any holomorphic g(z) is meromorphic and has isolated poles. Thus the roots of holomorphic non-constant functions are isolated. In particular, 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.

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 $D$ the 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 in B as $f(z)-w$ for any $w$ within a distance $e$ of $w_{0}$. Thus, for every $w$ in $D$, there exists one (and only one) $z_{1}$ in $B$ so that $f(z_{1})=w$. 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 the image of an open set by a holomorphic function $f(U)$. 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

References

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