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 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 a line.

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 $z_{0}$ in $U$. Since U is open, we can find $d>0$ such that the closed disk $B$ around z₀ 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. Consider the image point, $w_{0}=f(z_{0})$. Then $f(z_{0})-w_{0}=0$, making $z_{0}$ a root of the function $g(z)=f(z)-w_{0}$.

We know that g(z) is not constant, and by further decreasing d, we can assure that g(z) has only a single root in B. (The roots of holomorphic non-constant functions are isolated.) Let e be the minimum of |g(z)| for z on the boundary of B, a positive number. (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 this minimum.) 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 at least one $z_{1}$ in $B$ so that $f(z_{1})=w$. This means that the disk D is contained in f(B), which is a subset of $f(U)$.