Open mapping theorem

In mathematics, there are two theorems with the name "open mapping theorem". In both cases, they give conditions under which certain maps are open maps, i.e. they map open sets to open sets. They are significant results in their respective contexts since, unlike inverse images, direct images of functions are much less tractable in general.

Functional analysis

In functional analysis, the open mapping theorem, also known as the Banach-Schauder theorem, is a fundamental result which states: if A : X → Y is a surjective continuous linear operator between Banach spaces X and Y, then A is an open map (i.e. if U is an open set in X, then A(U) is open in Y).

The proof uses the Baire category theorem.

The open mapping theorem has two important consequences:

• If A : X → Y is a bijective continuous linear operator between the Banach spaces X and Y, then the inverse operator A^-1 : Y → X is continuous as well (this is called the bounded inverse theorem).
• If A : X → Y is a linear operator between the Banach spaces X and Y, and if for every sequence (x_n) in X with x_n → 0 and Ax_n → y it follows that y = 0, then A is continuous (Closed graph 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 function f(x) = x² is not open.

The theorem for example implies that a non-constant holomorphic function cannot map an open disk onto a portion of a line.

Proof

Blue dots represent zeros. Black spikes represent poles. The boundary of an open set is given by a dashed line. Note that all poles are exterior to the open set.

First assume $f$ is a non-constant holomorphic function and $U$ is a connected open subset of the complex plane. If every point in $f(U)$ is an interior point of $f(U)$ then $f(U)$ is open. Thus, if every point in $f(U)$ is contained in a disk which is contained in $f(U)$, then $f(U)$ is open.

Around every point in $U$, there is a relevant ball in $U$. Consider an arbitrary $z_{0}$ in $U$, and then consider its image point, $w_{0}=f(z_{0})$. Then $f(z_{0})-w_{0}=0$, making $z_{0}$ a root of $f(z)-w_{0}$. The function $f(z)-w_{0}$ may have another root at a distance $d_{1}$ from $z_{0}$. Additionally, the distance from $z_{0}$ to a point not in $U$ shall be written $d_{2}$. Any ball $B$ of radius less than the minimum of $d_{1}$ and $d_{2}$ will be contained in $U$, and at least one exists because $d_{1},d_{2}>0$.

Denote by $B_{2}$ the ball around $w_{0}$ with radius $e$ whose elements are written $w$. By Rouché's theorem or the Argument principle, the function $f(z)-w_{0}$ will have the same number of roots as $f(z)-w$ for any $w$ within a distance $e$ of $f(z_{0})$. Let $z_{1}$ be the root, or one of the roots of $f(z)-w$ just shown to exist. Thus, for every $w$ in $B_{2}$, there exists a $z_{1}$ in $B$ so that $f(z_{1})=w$, The image of B_2 is a subset of the image of B, which is a subset of $f(U)$.

Thus $w$ is an interior point of $f(U)$ for arbitrary $w$, and the theorem is proved.