Bourbaki

Proposition 8

Tag 030S, permanent. Cite it as [Bourbaki, Tag 030S].

Théories spectrales, chapter IV, § 3. Distributions et distributions tempérées

The space D(U)\mathscr{D}(U) is dense in D′(U)\mathscr{D}'(U).

Let λ\lambda be a linear form on D′(U)\mathscr{D}'(U) which is zero on D(U)\mathscr{D}(U). Since D(U)\mathscr{D}(U) is reflexive, there exists a test function φ∈D(U)\varphi \in \mathscr{D}(U) such that λ(f)=⟨f,φ⟩\lambda (f) =\langle f, \varphi \rangle for every f∈D′(U)f\in \mathscr{D}'(U). Hence

0=λ(φ‾)=⟨φ,φ⟩=∫U∣φ∣2dμ 0 =\lambda (\overline{\varphi}) =\langle \varphi , \varphi \rangle =\int_U|\varphi |^2d\mu

whence φ=0\varphi = 0. The proposition then follows from the Hahn–Banach theorem (EVT, II, p. 49, cor. 3 (ii)).

For h∈C∞h\in \mathscr{C}^{\infty}(U), the transpose of the continuous linear mapping φ↦hφ\varphi \mapsto h\varphi of D(U)\mathscr{D}(U) into itself is a continuous linear mapping of D′(U)\mathscr{D}'(U) into itself, which is denoted by f↦hff\mapsto hf. This definition is justified because if ff is the distribution associated with a measure ν\nu on U, then hfhf is associated with the measure h⋅νh\cdot \nu. In fact, for every test function φ∈D\varphi \in \mathscr{D}(U), one computes

⟨hf,φ⟩=⟨f,hφ⟩=∫Uhφdν=∫Uφd(h⋅ν) \langle hf, \varphi \rangle =\langle f, h\varphi \rangle =\int_Uh\varphi d\nu =\int_U\varphi d(h\cdot \nu )

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.

Machine translation of the English, by gpt-5-mini, gpt-5.4, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.