Closed operator: Definition from Answers.com

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In mathematics, specifically in functional analysis, closed linear operators are an important class of linear operators on Banach spaces. They are more general than bounded operators, and therefore not necessarily continuous, but they still retain nice enough properties that one can define the spectrum and (with certain assumptions) functional calculus for such operators. Many important linear operators which fail to be bounded turn out to be closed, such as the derivative and a large class of differential operators.

Let $B$ denote a Banach space. A linear operator

$A\colon\mathcal{D}(A)\subset B\to B$

is closed if for every sequence $\{x_n\}_{n\in \mathbb{N}}$ in $\mathcal{D}(A)$ converging to $x\in B$ such that $Ax_n\to y\in B$ as $n\to\infty$ one has $x\in\mathcal{D}(A)$ and $A x = y .$ Equivalently, $A$ is closed if its graph is closed in the direct sum $B\oplus B.$

Given a linear operator $A$ , not necessarily closed, if the closure of its graph in $B\oplus B$ happens to be the graph of some operator, that operator is called the closure of $A$ , and we say that $A$ is closable. Denote the closure of $A$ by $\overline{A}.$ It follows easily that $A$ is the restriction of $\overline{A}$ to $\mathcal{D}(A).$

A core of a closable operator is a subset $\mathcal{C}$ of $\mathcal{D}(A)$ such that the closure of the restriction of $A$ to $\mathcal{C}$ is $\overline{A}.$

The following properties are easily checked:

Any closed linear operator defined on the whole space $B$ is bounded. This is the closed graph theorem;
If $A$ is closed then $A - λ I$ is closed where $λ$ is a scalar and $I$ is the identity function;
If $A$ is closed and injective, then its inverse $A - 1$ is also closed;
An operator $A$ admits a closure if and only if for every pair of sequences ${x n}$ and ${y n}$ in $\mathcal{D}(A)$ converging to $x$ and $y$ , respectively, such that both ${A x n}$ and ${A y n}$ converge, it holds $\lim_n Ax_n = \lim_n Ay_n$ if $x = y$ .

As an example, consider the derivative operator

$A f = f'\,$

where the Banach space B is the space C[a, b] of all continuous functions on an interval [a, b]. If one takes its domain $\mathcal{D}(A)$ to be the largest set possible, that is, $\mathcal{D}(A)=C^{1}[a, b],$ then A is a closed operator, which is not bounded.

If one takes $\mathcal{D}(A)$ to be instead the set of all infinitely differentiable functions, A will no longer be closed, but it will be closable, with the closure being its maximal extension defined on $C 1 [a, b].$

This article incorporates material from Closed operator on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

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