Editing Distribution (mathematics) (section)

===Preliminaries: Transpose of a linear operator===
{{anchor|Transpose of a linear operator}}
{{Main|Transpose of a linear map}}

Operations on distributions and spaces of distributions are often defined using the [[Transpose of a linear map|transpose]] of a linear operator. This is because the transpose allows for a unified presentation of the many definitions in the theory of distributions and also because its properties are well-known in [[functional analysis]].<ref>{{harvnb|Strichartz|1994|loc=§2.3}}; {{harvnb|Trèves|2006}}.</ref> For instance, the well-known [[Hermitian adjoint]] of a linear operator between [[Hilbert space]]s is just the operator's transpose (but with the [[Riesz representation theorem]] used to identify each Hilbert space with its [[Strong dual space|continuous dual space]]). In general, the transpose of a continuous linear map <math>A : X \to Y</math> is the linear map 
<math display=block>{}^{t}A : Y' \to X' \qquad \text{ defined by } \qquad {}^{t}A(y') := y' \circ A,</math> 
or equivalently, it is the unique map satisfying <math>\langle y', A(x)\rangle = \left\langle {}^{t}A (y'), x \right\rangle</math> for all <math>x \in X</math> and all <math>y' \in Y'</math> (the prime symbol in <math>y'</math> does not denote a derivative of any kind; it merely indicates that <math>y'</math> is an element of the continuous dual space <math>Y'</math>). Since <math>A</math> is continuous, the transpose <math>{}^{t}A : Y' \to X'</math> is also continuous when both duals are endowed with their respective [[Strong dual space|strong dual topologies]]; it is also continuous when both duals are endowed with their respective [[Weak* topology|weak* topologies]] (see the articles [[Polar topology#Polar topologies and topological vector spaces|polar topology]] and [[Dual system#Weak topology|dual system]] for more details).

In the context of distributions, the characterization of the transpose can be refined slightly. Let <math>A : \mathcal{D}(U) \to \mathcal{D}(U)</math> be a continuous linear map. Then by definition, the transpose of <math>A</math> is the unique linear operator <math>{}^tA : \mathcal{D}'(U) \to \mathcal{D}'(U)</math> that satisfies:
<math display=block>\langle {}^{t}A(T), \phi \rangle = \langle T, A(\phi) \rangle \quad \text{ for all } \phi \in \mathcal{D}(U) \text{ and all } T \in \mathcal{D}'(U).</math>

Since <math>\mathcal{D}(U)</math> is dense in <math>\mathcal{D}'(U)</math> (here, <math>\mathcal{D}(U)</math> actually refers to the set of distributions <math>\left\{D_\psi : \psi \in \mathcal{D}(U)\right\}</math>) it is sufficient that the defining equality hold for all distributions of the form <math>T = D_\psi</math> where <math>\psi \in \mathcal{D}(U).</math> Explicitly, this means that a continuous linear map <math>B : \mathcal{D}'(U) \to \mathcal{D}'(U)</math> is equal to <math>{}^{t}A</math> if and only if the condition below holds:
<math display=block>\langle B(D_\psi), \phi \rangle = \langle {}^{t}A(D_\psi), \phi \rangle \quad \text{ for all } \phi, \psi \in \mathcal{D}(U)</math>
where the right-hand side equals <math>\langle {}^{t}A(D_\psi), \phi \rangle = \langle D_\psi, A(\phi) \rangle = \langle \psi, A(\phi) \rangle = \int_U \psi \cdot A(\phi) \,dx.</math>