In functional analysis, a branch of mathematics, a closed linear operator or often closed operator is a linear operator whose graph is closed. It is a basic example of an unbounded operator.

The closed graph theorem says a closed linear operator between Banach spaces is continuous; thus is a bounded operator. Hence, a closed linear operator that is used in practice is typically not defined on a Banach space.

We 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 $${\displaystyle \operatorname {graph} {\!(f)}=\{(x,f(x)):x\in \operatorname {dom} f\},}$$ of ${\displaystyle \operatorname {dom} (f)\times Y=X\times Y,}$ where ${\displaystyle \operatorname {dom} f=X}$ denotes the function's domain. The map ${\displaystyle f:X\to Y}$ is said to have a closed graph (in ${\displaystyle X\times Y}$) if its graph ${\displaystyle \operatorname {graph} f}$ is a closed subset of product space ${\displaystyle X\times Y}$ (with the usual product topology). Similarly, ${\displaystyle f}$ is said to have a sequentially closed graph if ${\displaystyle \operatorname {graph} f}$ is a sequentially closed subset of ${\displaystyle X\times Y.}$

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]}

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 ${\displaystyle X\times Y}$).

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

• 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. ^[3] 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.

• 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. ^[4]
• 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. ^[5] 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]

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