In mathematics, particularly in functional analysis and topology, the closed graph theorem is a result connecting the continuity of certain kinds of functions to a topological property of their graph. In its most elementary form, the closed graph theorem states that a linear function between two Banach spaces is continuous if and only if the graph of that function is closed.

The closed graph theorem has extensive application throughout functional analysis, because it can control whether a partially-defined linear operator admits continuous extensions. For this reason, it has been generalized to many circumstances beyond the elementary formulation above.

Preliminaries

The closed graph theorem is a result about linear map ${\displaystyle f:X\to Y}$ between two vector spaces endowed with topologies making them into topological vector spaces (TVSs). We will henceforth assume that ${\displaystyle X}$ and ${\displaystyle Y}$ are topological vector spaces, such as Banach spaces for example, and that Cartesian products, such as ${\displaystyle X\times Y,}$ are endowed with the product topology. The graph of this function is the subset

$${\displaystyle \operatorname {graph} {\!(f)}=\{(x,f(x)):x\in \operatorname {dom} f\},}$$

${\displaystyle \operatorname {dom} (f)\times Y=X\times Y,}$

${\displaystyle \operatorname {dom} f=X}$

${\displaystyle f:X\to Y}$

in ${\displaystyle X\times Y}$

${\displaystyle \operatorname {graph} f}$

${\displaystyle X\times Y}$

${\displaystyle f}$

${\displaystyle \operatorname {graph} f}$

${\displaystyle X\times Y.}$

A closed linear operator is a linear map whose graph is closed (it need not be continuous or bounded). It is common in functional analysis to call such maps "closed", but this should not be confused the non-equivalent notion of a "closed map" that appears in general topology.

Partial functions

It is common in functional analysis to consider partial functions, which are functions defined on a dense subset of some space ${\displaystyle X.}$ A partial function ${\displaystyle f}$ is declared with the notation ${\displaystyle f:D\subseteq X\to Y,}$ which indicates that ${\displaystyle f}$ has prototype ${\displaystyle f:D\to Y}$ (that is, its domain is ${\displaystyle D}$ and its codomain is ${\displaystyle Y}$) and that ${\displaystyle \operatorname {dom} f=D}$ is a dense subset of ${\displaystyle X.}$ Since the domain is denoted by ${\displaystyle \operatorname {dom} f,}$ it is not always necessary to assign a symbol (such as ${\displaystyle D}$) to a partial function's domain, in which case the notation ${\displaystyle f:X\rightarrowtail Y}$ or ${\displaystyle f:X\rightharpoonup Y}$ may be used to indicate that ${\displaystyle f}$ is a partial function with codomain ${\displaystyle Y}$ whose domain ${\displaystyle \operatorname {dom} f}$ is a dense subset of ${\displaystyle X.}$^[1] A densely defined linear operator between vector spaces is a partial function ${\displaystyle f:D\subseteq X\to Y}$ whose domain ${\displaystyle D}$ is a dense vector subspace of a TVS ${\displaystyle X}$ such that ${\displaystyle f:D\to Y}$ is a linear map. A prototypical example of a partial function is the derivative operator, which is only defined on the space ${\displaystyle D:=C^{1}([0,1])}$ of once continuously differentiable functions, a dense subset of the space ${\displaystyle X:=C([0,1])}$ of continuous functions.

Every partial function is, in particular, a function and so all terminology for functions can be applied to them. For instance, the graph of a partial function ${\displaystyle f}$ is (as before) the set ${\textstyle \operatorname {graph} {\!(f)}=\{(x,f(x)):x\in \operatorname {dom} f\}.}$ However, one exception to this is the definition of "closed graph". A partial function ${\displaystyle f:D\subseteq X\to Y}$ is said to have a closed graph (respectively, a sequentially closed graph) if ${\displaystyle \operatorname {graph} f}$ is a closed (respectively, sequentially closed) subset of ${\displaystyle X\times Y}$ in the product topology; importantly, note that the product space is ${\displaystyle X\times Y}$ and not ${\displaystyle D\times Y=\operatorname {dom} f\times Y}$ as it was defined above for ordinary functions.^{[note 1]}

Closable maps and closures

A linear operator ${\displaystyle f:D\subseteq X\to Y}$ is closable in ${\displaystyle X\times Y}$ if there exists a vector subspace ${\displaystyle E\subseteq X}$ containing ${\displaystyle D}$ and a function (resp. multifunction) ${\displaystyle F:E\to Y}$ whose graph is equal to the closure of the set ${\displaystyle \operatorname {graph} f}$ in ${\displaystyle X\times Y.}$ Such an ${\displaystyle F}$ is called a closure of ${\displaystyle f}$ in ${\displaystyle X\times Y}$, is denoted by ${\displaystyle {\overline {f)),}$ and necessarily extends ${\displaystyle f.}$

If ${\displaystyle f:D\subseteq X\to Y}$ is a closable linear operator then a core or an essential domain of ${\displaystyle f}$ is a subset ${\displaystyle C\subseteq D}$ such that the closure in ${\displaystyle X\times Y}$ of the graph of the restriction ${\displaystyle f{\big \vert }_{C}:C\to Y}$ of ${\displaystyle f}$ to ${\displaystyle C}$ is equal to the closure of the graph of ${\displaystyle f}$ in ${\displaystyle X\times Y}$ (i.e. the closure of ${\displaystyle \operatorname {graph} f}$ in ${\displaystyle X\times Y}$ is equal to the closure of ${\displaystyle \operatorname {graph} f{\big \vert }_{C))$ in ${\displaystyle X\times Y}$).

Characterizations of closed graphs (general topology)

Throughout, let ${\displaystyle X}$ and ${\displaystyle Y}$ be topological spaces and ${\displaystyle X\times Y}$ is endowed with the product topology.

Function with a closed graph

If ${\displaystyle f:X\to Y}$ is a function then it is said to have a closed graph if it satisfies any of the following are equivalent conditions:

1. (Definition): The graph ${\displaystyle \operatorname {graph} f}$ of ${\displaystyle f}$ is a closed subset of ${\displaystyle X\times Y.}$
2. For every ${\displaystyle x\in X}$ and net ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I))$ in ${\displaystyle X}$ such that ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle X,}$ if ${\displaystyle y\in Y}$ is such that the net ${\displaystyle f\left(x_{\bullet }\right)=\left(f\left(x_{i}\right)\right)_{i\in I}\to y}$ in ${\displaystyle Y}$ then ${\displaystyle y=f(x).}$^[2]
   • Compare this to the definition of continuity in terms of nets, which recall is the following: for every ${\displaystyle x\in X}$ and net ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I))$ in ${\displaystyle X}$ such that ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle X,}$ ${\displaystyle f\left(x_{\bullet }\right)\to f(x)}$ in ${\displaystyle Y.}$
   • Thus to show that the function ${\displaystyle f}$ has a closed graph, it may be assumed that ${\displaystyle f\left(x_{\bullet }\right)}$ converges in ${\displaystyle Y}$ to some ${\displaystyle y\in Y}$ (and then show that ${\displaystyle y=f(x)}$) while to show that ${\displaystyle f}$ is continuous, it may not be assumed that ${\displaystyle f\left(x_{\bullet }\right)}$ converges in ${\displaystyle Y}$ to some ${\displaystyle y\in Y}$ and instead, it must be proven that this is true (and moreover, it must more specifically be proven that ${\displaystyle f\left(x_{\bullet }\right)}$ converges to ${\displaystyle f(x)}$ in ${\displaystyle Y}$).

and if ${\displaystyle Y}$ is a Hausdorff compact space then we may add to this list:

3. ${\displaystyle f}$ is continuous.^[3]

and if both ${\displaystyle X}$ and ${\displaystyle Y}$ are first-countable spaces then we may add to this list:

4. ${\displaystyle f}$ has a sequentially closed graph in ${\displaystyle X\times Y.}$

Function with a sequentially closed graph

If ${\displaystyle f:X\to Y}$ is a function then the following are equivalent:

1. ${\displaystyle f}$ has a sequentially closed graph in ${\displaystyle X\times Y.}$
2. Definition: the graph of ${\displaystyle f}$ is a sequentially closed subset of ${\displaystyle X\times Y.}$
3. For every ${\displaystyle x\in X}$ and sequence ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle X}$ such that ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle X,}$ if ${\displaystyle y\in Y}$ is such that the net ${\displaystyle f\left(x_{\bullet }\right):=\left(f\left(x_{i}\right)\right)_{i=1}^{\infty }\to y}$ in ${\displaystyle Y}$ then ${\displaystyle y=f(x).}$^[2]

Basic properties of maps with closed graphs

Suppose ${\displaystyle f:D(f)\subseteq X\to Y}$ is a linear operator between Banach spaces.

• If ${\displaystyle A}$ is closed then ${\displaystyle A-s\operatorname {Id} _{D(f)))$ is closed where ${\displaystyle s}$ is a scalar and ${\displaystyle \operatorname {Id} _{D(f)))$ is the identity function.
• If ${\displaystyle f}$ is closed, then its kernel (or nullspace) is a closed vector subspace of ${\displaystyle X.}$
• If ${\displaystyle f}$ is closed and injective then its inverse ${\displaystyle f^{-1))$ is also closed.
• A linear operator ${\displaystyle f}$ admits a closure if and only if for every ${\displaystyle x\in X}$ and every pair of sequences ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty ))$ and ${\displaystyle z_{\bullet }=\left(z_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle D(f)}$ both converging to ${\displaystyle x}$ in ${\displaystyle X,}$ such that both ${\displaystyle f\left(x_{\bullet }\right)=\left(f\left(x_{i}\right)\right)_{i=1}^{\infty ))$ and ${\displaystyle f\left(z_{\bullet }\right)=\left(f\left(z_{i}\right)\right)_{i=1}^{\infty ))$ converge in ${\displaystyle Y,}$ one has ${\displaystyle \lim _{i\to \infty }f\left(x_{i}\right)=\lim _{i\to \infty }f\left(z_{i}\right).}$

Examples and counterexamples

Continuous but not closed maps

• Let ${\displaystyle X}$ denote the real numbers ${\displaystyle \mathbb {R} }$ with the usual Euclidean topology and let ${\displaystyle Y}$ denote ${\displaystyle \mathbb {R} }$ with the indiscrete topology (where ${\displaystyle Y}$ is not Hausdorff and that every function valued in ${\displaystyle Y}$ is continuous). Let ${\displaystyle f:X\to Y}$ be defined by ${\displaystyle f(0)=1}$ and ${\displaystyle f(x)=0}$ for all ${\displaystyle x\neq 0.}$ Then ${\displaystyle f:X\to Y}$ is continuous but its graph is not closed in ${\displaystyle X\times Y.}$^[2]
• If ${\displaystyle X}$ is any space then the identity map ${\displaystyle \operatorname {Id} :X\to X}$ is continuous but its graph, which is the diagonal ${\displaystyle \operatorname {graph} \operatorname {Id} =\{(x,x):x\in X\},}$ is closed in ${\displaystyle X\times X}$ if and only if ${\displaystyle X}$ is Hausdorff.^[4] In particular, if ${\displaystyle X}$ is not Hausdorff then ${\displaystyle \operatorname {Id} :X\to X}$ is continuous but not closed.
• If ${\displaystyle f:X\to Y}$ is a continuous map whose graph is not closed then ${\displaystyle Y}$ is not a Hausdorff space.

Closed but not continuous maps

• If ${\displaystyle (X,\tau )}$ is a Hausdorff TVS and ${\displaystyle \nu }$ is a vector topology on ${\displaystyle X}$ that is strictly finer than ${\displaystyle \tau ,}$ then the identity map ${\displaystyle \operatorname {Id} :(X,\tau )\to (X,\nu )}$ a closed discontinuous linear operator.^[5]
• Consider the derivative operator ${\displaystyle A={\frac {d}{dx))}$ where ${\displaystyle X=Y=C([a,b]).}$is the Banach space of all continuous functions on an interval ${\displaystyle [a,b].}$ If one takes its domain ${\displaystyle D(f)}$ to be ${\displaystyle C^{1}([a,b]),}$ then ${\displaystyle f}$ is a closed operator, which is not bounded.^[6] On the other hand, if ${\displaystyle D(f)}$ is the space ${\displaystyle C^{\infty }([a,b])}$ of smooth functions scalar valued functions then ${\displaystyle f}$ will no longer be closed, but it will be closable, with the closure being its extension defined on ${\displaystyle C^{1}([a,b]).}$
• Let ${\displaystyle X}$ and ${\displaystyle Y}$ both denote the real numbers ${\displaystyle \mathbb {R} }$ with the usual Euclidean topology. Let ${\displaystyle f:X\to Y}$ be defined by ${\displaystyle f(0)=0}$ and ${\displaystyle f(x)={\frac {1}{x))}$ for all ${\displaystyle x\neq 0.}$ Then ${\displaystyle f:X\to Y}$ has a closed graph (and a sequentially closed graph) in ${\displaystyle X\times Y=\mathbb {R} ^{2))$ but it is not continuous (since it has a discontinuity at ${\displaystyle x=0}$).^[2]
• Let ${\displaystyle X}$ denote the real numbers ${\displaystyle \mathbb {R} }$ with the usual Euclidean topology, let ${\displaystyle Y}$ denote ${\displaystyle \mathbb {R} }$ with the discrete topology, and let ${\displaystyle \operatorname {Id} :X\to Y}$ be the identity map (i.e. ${\displaystyle \operatorname {Id} (x):=x}$ for every ${\displaystyle x\in X}$). Then ${\displaystyle \operatorname {Id} :X\to Y}$ is a linear map whose graph is closed in ${\displaystyle X\times Y}$ but it is clearly not continuous (since singleton sets are open in ${\displaystyle Y}$ but not in ${\displaystyle X}$).^[2]

Closed graph theorems

Between Banach spaces

The operator is required to be everywhere-defined, that is, the domain ${\displaystyle D(T)}$ of ${\displaystyle T}$ is ${\displaystyle X.}$ This condition is necessary, as there exist closed linear operators that are unbounded (not continuous); a prototypical example is provided by the derivative operator on ${\displaystyle C([0,1]),}$ whose domain is a strict subset of ${\displaystyle C([0,1]).}$

The usual proof of the closed graph theorem employs the open mapping theorem. In fact, the closed graph theorem, the open mapping theorem and the bounded inverse theorem are all equivalent. This equivalence also serves to demonstrate the importance of ${\displaystyle X}$ and ${\displaystyle Y}$ being Banach; one can construct linear maps that have unbounded inverses in this setting, for example, by using either continuous functions with compact support or by using sequences with finitely many non-zero terms along with the supremum norm.

Complete metrizable codomain

The closed graph theorem can be generalized from Banach spaces to more abstract topological vector spaces in the following ways.

Between F-spaces

There are versions that does not require ${\displaystyle Y}$ to be locally convex.

This theorem is restated and extend it with some conditions that can be used to determine if a graph is closed:

Complete pseudometrizable codomain

Every metrizable topological space is pseudometrizable. A pseudometrizable space is metrizable if and only if it is Hausdorff.

Codomain not complete or (pseudo) metrizable

An even more general version of the closed graph theorem is

Borel graph theorem

The Borel graph theorem, proved by L. Schwartz, shows that the closed graph theorem is valid for linear maps defined on and valued in most spaces encountered in analysis.^[13] Recall that a topological space is called a Polish space if it is a separable complete metrizable space and that a Souslin space is the continuous image of a Polish space. The weak dual of a separable Fréchet space and the strong dual of a separable Fréchet-Montel space are Souslin spaces. Also, the space of distributions and all Lp-spaces over open subsets of Euclidean space as well as many other spaces that occur in analysis are Souslin spaces. The Borel graph theorem states:

An improvement upon this theorem, proved by A. Martineau, uses K-analytic spaces.

A topological space ${\displaystyle X}$ is called a ${\displaystyle K_{\sigma \delta ))$ if it is the countable intersection of countable unions of compact sets.

A Hausdorff topological space ${\displaystyle Y}$ is called K-analytic if it is the continuous image of a ${\displaystyle K_{\sigma \delta ))$ space (that is, if there is a ${\displaystyle K_{\sigma \delta ))$ space ${\displaystyle X}$ and a continuous map of ${\displaystyle X}$ onto ${\displaystyle Y}$).

Every compact set is K-analytic so that there are non-separable K-analytic spaces. Also, every Polish, Souslin, and reflexive Fréchet space is K-analytic as is the weak dual of a Frechet space. The generalized Borel graph theorem states:

Related results

If ${\displaystyle F:X\to Y}$ is closed linear operator from a Hausdorff locally convex TVS ${\displaystyle X}$ into a Hausdorff finite-dimensional TVS ${\displaystyle Y}$ then ${\displaystyle F}$ is continuous.^[15]