Unbounded operator

In mathematics, more specifically functional analysis and operator theory, the notion of unbounded operator provides an abstract framework for dealing with differential operators, unbounded observables in quantum mechanics, and other cases.

The term "unbounded operator" can be misleading, since

• "unbounded" should sometimes be understood as "not necessarily bounded";
• "operator" should be understood as "linear operator" (as in the case of "bounded operator");
• the domain of the operator is a linear subspace, not necessarily the whole space;
• this linear subspace is not necessarily closed; often (but not always) it is assumed to be dense;
• in the special case of a bounded operator, still, the domain is usually assumed to be the whole space.

In contrast to bounded operators, unbounded operators on a given space do not form an algebra, nor even a linear space, because each one is defined on its own domain.

The term "operator" often means "bounded linear operator", but in the context of this article it means "unbounded operator", with the reservations made above. The given space is assumed to be a Hilbert space. Some generalizations to Banach spaces and more general topological vector spaces are possible.

Short history

The theory of unbounded operators developed in the late 1920s and early 1930s as part of developing a rigorous mathematical framework for quantum mechanics. The theory's development is due to John von Neumann and Marshall Stone. Von Neumann introduced using graphs to analyze unbounded operators in 1932.

Definitions and basic properties

Let X, Y be Banach spaces. An unbounded operator (or simply operator) T : D(T) → Y is a linear map T from a linear subspace D(T) ⊆ X—the domain of T—to the space Y. Contrary to the usual convention, T may not be defined on the whole space X.

An operator T is said to be closed if its graph Γ(T) is a closed set. (Here, the graph Γ(T) is a linear subspace of the direct sum X ⊕ Y, defined as the set of all pairs (x, Tx), where x runs over the domain of T .) Explicitly, this means that for every sequence {x_n} of points from the domain of T such that x_n → x and Tx_n → y, it holds that x belongs to the domain of T and Tx = y. The closedness can also be formulated in terms of the graph norm: an operator T is closed if and only if its domain D(T) is a complete space with respect to the norm:

$\|x\|_{T}={\sqrt {\|x\|^{2}+\|Tx\|^{2}}}.$

An operator T is said to be densely defined if its domain is dense in X. This also includes operators defined on the entire space X, since the whole space is dense in itself. The denseness of the domain is necessary and sufficient for the existence of the adjoint (if X and Y are Hilbert spaces) and the transpose; see the sections below.

If T : X → Y is closed, densely defined and continuous on its domain, then its domain is all of X.

A densely defined operator T on a Hilbert space H is called bounded from below if T + a is a positive operator for some real number a. That is, ⟨Tx|x⟩ ≥ −a ||x||² for all x in the domain of T (or alternatively ⟨Tx|x⟩ ≥ a ||x||² since a is arbitrary). If both T and −T are bounded from below then T is bounded.

Example

Let C([0, 1]) denote the space of continuous functions on the unit interval, and let C¹([0, 1]) denote the space of continuously differentiable functions. We equip $C([0,1])$ with the supremum norm, $\|\cdot \|_{\infty }$, making it a Banach space. Define the classical differentiation operator d/dx : C¹([0, 1]) → C([0, 1]) by the usual formula:

$\left({\frac {d}{dx}}f\right)(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}},\qquad \forall x\in [0,1].$

Every differentiable function is continuous, so C¹([0, 1]) ⊆ C([0, 1]). We claim that d/dx : C([0, 1]) → C([0, 1]) is a well-defined unbounded operator, with domain C¹([0, 1]). For this, we need to show that ${\frac {d}{dx}}$ is linear and then, for example, exhibit some $\{f_{n}\}_{n}\subset C^{1}([0,1])$ such that $\|f_{n}\|_{\infty }=1$ and $\sup _{n}\|{\frac {d}{dx}}f_{n}\|_{\infty }=+\infty$.

This is a linear operator, since a linear combination a f  + bg of two continuously differentiable functions  f , g is also continuously differentiable, and

$\left({\tfrac {d}{dx}}\right)(af+bg)=a\left({\tfrac {d}{dx}}f\right)+b\left({\tfrac {d}{dx}}g\right).$

The operator is not bounded. For example,

${\begin{cases}f_{n}:[0,1]\to [-1,1]\\f_{n}(x)=\sin(2\pi nx)\end{cases}}$

satisfy

$\left\|f_{n}\right\|_{\infty }=1,$

but

$\left\|\left({\tfrac {d}{dx}}f_{n}\right)\right\|_{\infty }=2\pi n\to \infty$

as $n\to \infty$.

The operator is densely defined, and closed.

The same operator can be treated as an operator Z → Z for many choices of Banach space Z and not be bounded between any of them. At the same time, it can be bounded as an operator X → Y for other pairs of Banach spaces X, Y, and also as operator Z → Z for some topological vector spaces Z. As an example let I ⊂ R be an open interval and consider

$\frac{d}{dx}:\left(C^1(I),{\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->\Vert <!--\; \Vert \; -->}_{C^1}\right)\to <!--\; \to \; -->(C(I)$