Open mapping theorem (complex analysis)

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

References

* {{citation, first=Walter, last=Rudin, authorlink=Walter Rudin, title=Real & Complex Analysis, publisher=McGraw-Hill, year=1966, isbn=0-07-054234-1

Theorems in complex analysis
Articles containing proofs