Closed linear operator

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

The closed graph theorem says a linear operator between Banach spaces is a closed operator if and only if it is a bounded operator. Hence, a closed linear operator that is used in practice is typically only defined on a dense subspace of a Banach space.

Definition

It is common in functional analysis to consider partial functions, which are functions defined on a subset of some space $X.$ A partial function $f$ is declared with the notation $f:D\subseteq X\to Y,$ which indicates that $f$ has prototype $f:D\to Y$ (that is, its domain is $D$ and its codomain is $Y$ )

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 $f$ is 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 $f:D\subseteq X\to Y$ is said to have a closed graph if $\operatorname {graph} f$ is a closed subset of $X\times Y$ in the product topology; importantly, note that the product space is $X\times Y$ and not $D\times Y=\operatorname {dom} f\times Y$ as it was defined above for ordinary functions. In contrast, when $f:D\to Y$ is considered as an ordinary function (rather than as the partial function $f:D\subseteq X\to Y$ ), then "having a closed graph" would instead mean that $\operatorname {graph} f$ is a closed subset of $D\times Y.$ If $\operatorname {graph} f$ is a closed subset of $X\times Y$ then it is also a closed subset of $\operatorname {dom} (f)\times Y$ although the converse is not guaranteed in general.

Definition: If $X$ and $Y$ are topological vector spaces (TVSs) then we call a linear map $f : D (f) \subseteq X \to Y$ a closed linear operator if its graph is closed in $X \times Y$ .

Closable maps and closures

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

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

Examples

A bounded operator is a closed operator. Here are examples of closed operators that are not bounded.

If $(X,\tau )$ is a Hausdorff TVS and $\nu$ is a vector topology on $X$ that is strictly finer than $\tau ,$ then the identity map $\operatorname {Id} :(X,\tau )\to (X,\nu )$ a closed discontinuous linear operator.^[1]
Consider the derivative operator $A={\frac {d}{dx}}$ where $X=Y=C([a,b])$ is the Banach space of all continuous functions on an interval $[a,b].$ If one takes its domain $D(f)$ to be $C^{1}([a,b]),$ then $f$ is a closed operator, which is not bounded.^[2] On the other hand, if $D(f)$ is the space $C^{\infty }([a,b])$ of smooth functions scalar valued functions then $f$ will no longer be closed, but it will be closable, with the closure being its extension defined on $C^{1}([a,b]).$

Basic properties

The following properties are easily checked for a linear operator $f : D (f) \subseteq X \to Y$ between Banach spaces:

If $A$ is closed then $A - λ Id D (f)$ is closed where $λ$ is a scalar and $Id D (f)$ is the identity function;
If $f$ is closed, then its kernel (or nullspace) is a closed vector subspace of $X$ ;
If $f$ is closed and injective then its inverse $f -1$ is also closed;
A linear operator $f$ admits a closure if and only if for every $x \in X$ and every pair of sequences $x • = (x i) \infty i =1$ and $y • = (y i) \infty i =1$ in $D (f)$ both converging to $x$ in $X$ , such that both $f (x •) = (f (x i)) \infty i =1$ and $f (y •) = (f (y i)) \infty i =1$ converge in $Y$ , one has $lim i \to \infty fx i = lim i \to \infty fy i$ .

References

^ Narici & Beckenstein 2011, p. 480.
^ Kreyszig, Erwin (1978). Introductory Functional Analysis With Applications. USA: John Wiley & Sons. Inc. p. 294. ISBN 0-471-50731-8.

Dolecki, Szymon; Mynard, Frédéric (2016). Convergence Foundations Of Topology. New Jersey: World Scientific Publishing Company. ISBN 978-981-4571-52-4. OCLC 945169917.
Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.

[FOOTNOTENariciBeckenstein2011480-1] Narici & Beckenstein 2011, p. 480.

[2] Kreyszig, Erwin (1978). Introductory Functional Analysis With Applications. USA: John Wiley & Sons. Inc. p. 294. ISBN 0-471-50731-8.

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