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 $f : X \to Y$ between Banach spaces is a closed operator if and only if it is a bounded operator and the domain of the operator is $X$. 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
$\operatorname = \.$
However, one exception to this is the definition of "closed graph". A function $f : D \subseteq X \to Y$ is said to have a closed graph if $\operatorname f$ is a closed subset of $X \times Y$ in the product topology; importantly, note that the product space is $X \times Y$ and $D \times Y = \operatorname f \times Y$ 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 $\operatorname f$ is a closed subset of $D \times Y.$ If $\operatorname f$ is a closed subset of $X \times Y$ then it is also a closed subset of $\operatorname (f) \times Y$ although the converse is not guaranteed in general.

Definition: If and are topological vector spaces (TVSs) then we call a linear map a closed linear operator if its graph is closed in .

Closable maps and closures

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

If $f : D \subseteq X \to Y$ is a closable linear operator then a or an 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 $\operatorname f$ in $X \times Y$ is equal to the closure of $\operatorname f\big\vert_C$ 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 $\operatorname : (X, \tau) \to (X, \nu)$ a closed discontinuous linear operator.
* Consider the derivative operator $A = \frac$ where $X = Y = C($, b is the Banach space of all continuous functions on an interval $$, b If one takes its domain $D(f)$ to be $C^1($, b, then $f$ is a closed operator, which is not bounded. On the other hand, if $D(f)$ is the space $C^\infty($, 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 $C^1($, b.

Basic properties

The following properties are easily checked for a linear operator between Banach spaces:

* If is closed then is closed where is a scalar and is the identity function;
* If is closed, then its kernel (or nullspace) is a closed vector subspace of ;
* If is closed and injective then its inverse is also closed;
* A linear operator admits a closure if and only if for every and every pair of sequences and in both converging to in , such that both and converge in , one has .

References

*
*
*

Linear operators