Densely defined operator

Let X,Y be topological vector spaces. A densely defined linear operator T from X to Y is a linear operator of type T: D(T) \to Y, such that D(T) is a dense subset of X. In other words, T is a partial function whose domain is dense in X. Sometimes this is abbreviated as T : X \to Y when the context makes it clear that T might not be defined for all of X. == Properties ==

The Hausdorff property ensures sequential convergence is unique. The metrizability property ensures that sequentially closed sets are closed. In functional analysis, these conditions typically hold, as most spaces under consideration are Fréchet space, or stronger than Fréchet. In particular, Banach spaces are Fréchet. ==Examples==

Sequence Let X=\ell^{2}(\mathbb N) be the Hilbert space of square-summable sequences, with orthonormal basis (e_n)_{n\ge 1}. Define the diagonal operator A : D(A)\to \ell^{2}, \qquad (Ax)_n := n\,x_n, with domain D(A):=\left\{x=(x_n)_{n\ge 1}\in \ell^{2}:\sum_{n=1}^{\infty} n^{2} | x_n|^{2}Then D(A) is dense in \ell^2 because the finitely supported sequences c_{00}\subset D(A), and c_{00} is dense in \ell^2. The operator A is closed and unbounded, since \|Ae_n\|_2=n. There exists a bounded inverse: A^{-1}:\ell^{2}\to D(A),\qquad (A^{-1}y)_n:=\frac{y_n}{n},\qquad \|A^{-1}\|=\sup_{n\ge 1}\frac{1}{n}=1. Hence A:D(A)\to \ell^{2} is bijective with bounded inverse, so 0\in\rho(A) and, by the Neumann series argument, the resolvent set of A contains the open unit disk \{\,\lambda\in\mathbb C:\ |\lambda|. In fact, the spectrum of A (that is, the complement of its resolvent set) is precisely the set of positive integers, since for any \lambda \not\in \{1, 2, \dots\}, the diagonal formula (A-\lambda I)^{-1}y=\bigl(\tfrac{y_n}{n-\lambda}\bigr)_{n\ge 1} defines a bounded operator \ell^2\to D(A). Thus, A is a densely defined, closed, unbounded operator with bounded inverse and nontrivial, unbounded spectrum. Differentiation Consider the space C^0([0, 1]; \R) of all real-valued, continuous functions defined on the unit interval; let C^1([0, 1]; \R) denote the subspace consisting of all continuously differentiable functions. Equip C^0([0, 1]; \R) with the supremum norm \|\,\cdot\,\|_\infty; this makes C^0([0, 1]; \R) into a real Banach space. The differentiation operator D given by (\mathrm{D} u)(x) = u'(x) is a linear operator defined on the dense linear subspace C^1([0, 1]; \R) \subset C^0([0, 1]; \R), therefore it is a operator densely defined on C^0([0, 1]; \R). The operator \mathrm{D} is an example of an unbounded linear operator, since u_n (x) = e^{- n x} \quad \text{ has } \quad \frac{\left\|\mathrm{D} u_n\right\|_{\infty}}{\left\|u_n\right\|_\infty} = n. This unboundedness causes problems if one wishes to somehow continuously extend the differentiation operator D to the whole of C^0([0, 1]; \R). Paley–Wiener The Paley–Wiener integral is a standard example of a continuous extension of a densely defined operator. In any abstract Wiener space i : H \to E with adjoint j := i^* : E^* \to H, there is a natural continuous linear operator (in fact it is the inclusion, and is an isometry) from j\left(E^*\right) to L^2(E, \gamma; \R), under which j(f) \in j\left(E^*\right) \subseteq H goes to the equivalence class [f] of f in L^2(E, \gamma; \R). It can be shown that j\left(E^*\right) is dense in H. Since the above inclusion is continuous, there is a unique continuous linear extension I : H \to L^2(E, \gamma; \R) of the inclusion j\left(E^*\right) \to L^2(E, \gamma; \R) to the whole of H. This extension is the Paley–Wiener map. ==See also==