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