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.

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 ${\displaystyle \lim _{i\to \infty }f\left(x_{i}\right)=\lim _{i\to \infty }f\left(z_{i}\right).}$

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]

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.

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