Open mapping theorem (complex analysis)

In complex analysis, the open mapping theorem states that if $U$ is a domain of the complex plane $\mathbb{ℂ}$ and $f:U\to \mathbb{ℂ}$ is a non-constant holomorphic function, then $f$ is an open map (i.e. it sends open subsets of $U$ to open subsets of $\mathbb{ℂ}$, 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 $f(x)=x^2$ 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

File:Openmappingtheorem.png

Black dots represent zeros of $g(z)$. Black annuli 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 disk is $B$, centered at $z_0$.

Assume $f:U\to \mathbb{ℂ}$ 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 $B$, $|g(z)|\ge e>|w_0-w_1|$. 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)

References

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