# Open mapping theorem (functional analysis)

In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theorem (named after Stefan Banach and Juliusz Schauder), is a fundamental result that states that if a bounded or continuous linear operator between Banach spaces is surjective then it is an open map.
A special case is also called the bounded inverse theorem (also called inverse mapping theorem or Banach isomorphism theorem), which states that a bijective bounded linear operator $T$ from one Banach space to another has bounded inverse $T^{-1}$.

## Statement and proof

Open mapping theorem — Let $T:E\to F$ be a surjective continuous linear map between Banach spaces (or more generally Fréchet spaces). Then $T$ is an open mapping (that is, if $U\subset E$ is an open subset, then $T(U)$ is open).
The proof here uses the Baire category theorem, and completeness of both $E$ and $F$ is essential to the theorem. The statement of the theorem is no longer true if either space is assumed to be only a normed vector space; see § Counterexample.
The proof is based on the following lemmas, which are also somewhat of independent interest. A linear map $f:E\to F$ between topological vector spaces is said to be nearly open if, for each neighborhood $U$ of zero, the closure ${\overline {f(U)}}$ contains a neighborhood of zero. The next lemma may be thought of as a weak version of the open mapping theorem.
Lemma — A linear map $f:E\to F$ between normed spaces is nearly open if the image of $f$ is non-meager in $F$. (The continuity is not needed.)
Proof: Shrinking $U$, we can assume $U$ is an open ball centered at zero. We have $f(E)=f\left(\bigcup _{n\in \mathbb {N} }nU\right)=\bigcup _{n\in \mathbb {N} }f(nU)$. Thus, some ${\overline {f(nU)}}$ contains an interior point $y$; that is, for some radius $r>0$,
    $B(y,r)\subset {\overline {f(nU)}}.$
Then for any $v$ in $F$ with $\|v\|<r$, by linearity, convexity and $(-1)U\subset U$,
    $v=v-y+y\in {\overline {f(-nU)}}+{\overline {f(nU)}}\subset {\overline {f(2nU)}}$,
which proves the lemma by dividing by $2n$.$\square$ (The same proof works if $E,F$ are pre-Fréchet spaces.)
The completeness on the domain then allows to upgrade nearly open to open.
Lemma (Schauder) — Let $f:E\to F$ be a continuous linear map between normed spaces.
If $f$ is nearly-open and if $E$ is complete, then $f$ is open and surjective.
More precisely, if $B(0,\delta )\subset {\overline {f(B(0,1))}}$ for some $\delta >0$ and if $E$ is complete, then
    $B(0,\delta )\subset f(B(0,1))$
where $B(x,r)$ is an open ball with radius $r$ and center $x$.
Proof: Let $y$ be in $B(0,\delta )$ and $c_{n}>0$ some sequence. We have: ${\overline {B(0,\delta )}}\subset {\overline {f(B(0,1))}}$. Thus, for each $\epsilon >0$ and $z$ in $F$, we can find an $x$ with $\|x\|<\delta ^{-1}\|z\|$ and $z$ in $B(f(x),\epsilon )$. Thus, taking $z=y$, we find an $x_{1}$ such that
    $\|y-f(x_{1})\|<c_{1},\,\|x_{1}\|<\delta ^{-1}\|y\|.$
Applying the same argument with $z=y-f(x_{1})$, we then find an $x_{2}$ such that
    $\|y-f(x_{1})-f(x_{2})\|<c_{2},\,\|x_{2}\|<\delta ^{-1}c_{1}$
where we observed $\|x_{2}\|<\delta ^{-1}\|z\|<\delta ^{-1}c_{1}$. Then so on. Thus, if $c:=\sum c_{n}<\infty$, we found a sequence $x_{n}$ such that $x=\sum _{1}^{\infty }x_{n}$ converges and $f(x)=y$. Also,
    $\|x\|\leq \sum _{1}^{\infty }\|x_{n}\|\leq \delta ^{-1}\|y\|+\delta ^{-1}c.$
Since $\delta ^{-1}\|y\|<1$, by making $c$ small enough, we can achieve $\|x\|<1$. $\square$ (Again the same proof is valid if $E,F$ are pre-Fréchet spaces.)
Proof of the theorem: By Baire's category theorem, the first lemma applies. Then the conclusion of the theorem follows from the second lemma. $\square$
In general, a continuous bijection between topological spaces is not necessarily a homeomorphism. The open mapping theorem, when it applies, implies the bijectivity is enough:
Corollary (Bounded inverse theorem) — A continuous bijective linear operator between Banach spaces (or Fréchet spaces) has continuous inverse. That is, the inverse operator is continuous.
Even though the above bounded inverse theorem is a special case of the open mapping theorem, the open mapping theorem in turns follows from that. Indeed, a surjective continuous linear operator $T:E\to F$ factors as
    $T:E{\overset {p}{\to }}E/\operatorname {ker} T{\overset {T_{0}}{\to }}F.$
Here, $T_{0}$ is continuous and bijective and thus is a homeomorphism by the bounded inverse theorem; in particular, it is an open mapping. As a quotient map for topological groups is open, $T$ is open then.
Because the open mapping theorem and the bounded inverse theorem are essentially the same result, they are often simply called Banach's theorem.

### Transpose formulation

Here is a formulation of the open mapping theorem in terms of the transpose of an operator.
Theorem — Let $X$ and $Y$ be Banach spaces, let $B_{X}$ and $B_{Y}$ denote their open unit balls, and let $T:X\to Y$ be a bounded linear operator. If $\delta >0$ then among the following four statements we have $(1)\implies (2)\implies (3)\implies (4)$ (with the same $\delta$)
* $\delta \left\|y'\right\|\leq \left\|T'y'\right\|$ for all $y'\in Y'$ = continuous dual of $Y$;
* $\delta B_{Y}\subset {\overline {T\left(B_{X}\right)}}$;
* $\delta B_{Y}\subset {T\left(B_{X}\right)}$;
* $T$ is surjective.
Furthermore, if $T$ is surjective then (1) holds for some $\delta >0.$
Proof: The idea of 1. $\Rightarrow$ 2. is to show: $y\notin {\overline {T(B_{X})}}\Rightarrow \|y\|>\delta ,$ and that follows from the Hahn–Banach theorem. 2. $\Rightarrow$ 3. is exactly the second lemma in § Statement and proof. Finally, 3. $\Rightarrow$ 4. is trivial and 4. $\Rightarrow$ 1. easily follows from the open mapping theorem. $\square$
Alternatively, 1. implies that $T'$ is injective and has closed image and then by the closed range theorem, that implies $T$ has dense image and closed image, respectively; i.e., $T$ is surjective. Hence, the above result is a variant of a special case of the closed range theorem.

### Quantative formulation

Terence Tao gives the following quantitative formulation of the theorem:
Theorem — Let $T:E\to F$ be a bounded operator between Banach spaces. Then the following are equivalent:
* $T$ is open.
* $T$ is surjective.
* There exists a constant $C>0$ such that, for each $f$ in $F$, the equation $Tu=f$ has a solution $u$ with $\|u\|\leq C\|f\|$.
* 3. holds for $f$ in some dense subspace of $F$.
Proof: 2. $\Rightarrow$ 1. is the usual open mapping theorem.
1. $\Rightarrow$ 4.: For some $r>0$, we have $B(0,2)\subset T(B(0,r))$ where $B$ means an open ball. Then ${\frac {f}{\|f\|}}=T\left({\frac {u}{\|f\|}}\right)$ for some ${\frac {u}{\|f\|}}$ in $B(0,r)$. That is, $Tu=f$ with $\|u\|<r\|f\|$.
4. $\Rightarrow$ 3.: We can write $f=\sum _{0}^{\infty }f_{j}$ with $f_{j}$ in the dense subspace and the sum converging in norm. Then, since $E$ is complete, $u=\sum _{0}^{\infty }u_{j}$ with $\|u_{j}\|\leq C\|f_{j}\|$ and $Tu_{j}=f_{j}$ is a required solution. Finally, 3. $\Rightarrow$ 2. is trivial. $\square$

## Counterexample

The open mapping theorem may not hold for normed spaces that are not complete. A quickest way to see this is to note that the closed graph theorem, a consequence of the open mapping theorem, fails without completeness. But here is a more concrete counterexample. Consider the space X of sequences x : N → R with only finitely many non-zero terms equipped with the supremum norm. The map T : X → X defined by
    $Tx=\left(x_{1},{\frac {x_{2}}{2}},{\frac {x_{3}}{3}},\dots \right)$
is bounded, linear and invertible, but T$^{−1}$ is unbounded. This does not contradict the bounded inverse theorem since X is not complete, and thus is not a Banach space. To see that it's not complete, consider the sequence of sequences x$^{(n)}$ ∈ X given by
    $x^{(n)}=\left(1,{\frac {1}{2}},\dots ,{\frac {1}{n}},0,0,\dots \right)$
converges as n → ∞ to the sequence x$^{(∞)}$ given by
    $x^{(\infty )}=\left(1,{\frac {1}{2}},\dots ,{\frac {1}{n}},\dots \right),$
which has all its terms non-zero, and so does not lie in X.
The completion of X is the space $c_{0}$ of all sequences that converge to zero, which is a (closed) subspace of the ℓ$_p$ space ℓ$_∞$(N), which is the space of all bounded sequences. However, in this case, the map T is not onto, and thus not a bijection. To see this, one need simply note that the sequence
    $x=\left(1,{\frac {1}{2}},{\frac {1}{3}},\dots \right),$
is an element of $c_{0}$, but is not in the range of $T:c_{0}\to c_{0}$. Same reasoning applies to show $T$ is also not onto in $l^{\infty }$, for example $x=\left(1,1,1,\dots \right)$ is not in the range of $T$.

## Consequences

The open mapping theorem has several important consequences:
* If $T:X\to Y$ is a bijective continuous linear operator between the Banach spaces $X$ and $Y,$ then the inverse operator $T^{-1}:Y\to X$ is continuous as well (this is called the bounded inverse theorem).
* If $T:X\to Y$ is a linear operator between the Banach spaces $X$ and $Y,$ and if for every sequence $\left(x_{n}\right)$ in $X$ with $x_{n}\to 0$ and $Tx_{n}\to y$ it follows that $y=0,$ then $T$ is continuous (the closed graph theorem).
* Given a bounded operator $T:E\to F$ between normed spaces, if the image of $T$ is non-meager and if $E$ is complete, then $T$ is open and surjective and $F$ is complete (to see this, use the two lemmas in the proof of the theorem).
* An exact sequence of Banach spaces (or more generally Fréchet spaces) is topologically exact.
* The closed range theorem, which says an operator (under some assumption) has closed image if and only if its transpose has closed image (see closed range theorem#Sketch of proof).
The open mapping theorem does not imply that a continuous surjective linear operator admits a continuous linear section. What we have is:
* A surjective continuous linear operator between Banach spaces admits a continuous linear section if and only if the kernel is topologically complemented.
In particular, the above applies to an operator between Hilbert spaces or an operator with finite-dimensional kernel (by the Hahn–Banach theorem). If one drops the requirement that a section be linear, a surjective continuous linear operator between Banach spaces admits a continuous section; this is the Bartle–Graves theorem.

## Generalizations

Local convexity of $X$ or $Y$  is not essential to the proof, but completeness is: the theorem remains true in the case when $X$ and $Y$ are F-spaces. Furthermore, the theorem can be combined with the Baire category theorem in the following manner:
Open mapping theorem for continuous maps — Let $A:X\to Y$ be a continuous linear operator from a complete pseudometrizable TVS $X$ onto a Hausdorff TVS $Y.$ If $\operatorname {Im} A$ is nonmeager in $Y$ then $A:X\to Y$ is a (surjective) open map and $Y$ is a complete pseudometrizable TVS. Moreover, if $X$ is assumed to be hausdorff (i.e. a F-space), then $Y$ is also an F-space.
(The proof is essentially the same as the Banach or Fréchet cases; we modify the proof slightly to avoid the use of convexity,)
Furthermore, in this latter case if $N$ is the kernel of $A,$ then there is a canonical factorization of $A$ in the form $X\to X/N{\overset {\alpha }{\to }}Y$ where $X/N$ is the quotient space (also an F-space) of $X$ by the closed subspace $N.$ The quotient mapping $X\to X/N$ is open, and the mapping $\alpha$ is an isomorphism of topological vector spaces.
An important special case of this theorem can also be stated as
Theorem — Let $X$ and $Y$ be two F-spaces. Then every continuous linear map of $X$ onto $Y$ is a TVS homomorphism, where a linear map $u:X\to Y$ is a topological vector space (TVS) homomorphism if the induced map ${\hat {u}}:X/\ker(u)\to Y$ is a TVS-isomorphism onto its image.
On the other hand, a more general formulation, which implies the first, can be given:
Open mapping theorem — Let $A:X\to Y$ be a surjective linear map from a complete pseudometrizable TVS $X$ onto a TVS $Y$ and suppose that at least one of the following two conditions is satisfied:
* $Y$ is a Baire space, or
* $X$ is locally convex and $Y$ is a barrelled space,
If $A$ is a closed linear operator then $A$ is an open mapping. If $A$ is a continuous linear operator and $Y$ is Hausdorff then $A$ is (a closed linear operator and thus also) an open mapping.
Nearly/Almost open linear maps
A linear map $A:X\to Y$ between two topological vector spaces (TVSs) is called a nearly open map (or sometimes, an almost open map) if for every neighborhood $U$ of the origin in the domain, the closure of its image $\operatorname {cl} A(U)$ is a neighborhood of the origin in $Y.$ Many authors use a different definition of "nearly/almost open map" that requires that the closure of $A(U)$ be a neighborhood of the origin in $A(X)$ rather than in $Y,$ but for surjective maps these definitions are equivalent. A bijective linear map is nearly open if and only if its inverse is continuous. Every surjective linear map from locally convex TVS onto a barrelled TVS is nearly open. The same is true of every surjective linear map from a TVS onto a Baire TVS.
Open mapping theorem — If a closed surjective linear map from a complete pseudometrizable TVS onto a Hausdorff TVS is nearly open then it is open.
Theorem — If $A:X\to Y$ is a continuous linear bijection from a complete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is a Baire space, then $A:X\to Y$ is a homeomorphism (and thus an isomorphism of TVSs).
Webbed spaces are a class of topological vector spaces for which the open mapping theorem and the closed graph theorem hold.