Bourbaki

Proposition 17

Tag 031E, permanent. Cite it as [Bourbaki, Tag 031E].

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

The spaces D(Rn)\mathscr{D}(\mathbf{R}^n) and S(Rn)\mathscr{S}(\mathbf{R}^n) are dense in S′(Rn)\mathscr{S}'(\mathbf{R}^n).

It is enough to prove that D(Rn)\mathscr{D}(\mathbf{R}^n) is dense in S′(Rn)\mathscr{S}'(\mathbf{R}^n). Let λ\lambda be a continuous linear form on S′(Rn)\mathscr{S}'(\mathbf{R}^n) which is zero on D(Rn)\mathscr{D}(\mathbf{R}^n). Since the space S(Rn)\mathscr{S}(\mathbf{R}^n) is reflexive, there exists a function φ∈S(Rn)\varphi \in \mathscr{S}(\mathbf{R}^n) such that λ(f)=⟨f,φ⟩\lambda (f) =\langle f, \varphi \rangle for every f∈S′(Rn)f\in \mathscr{S}'(\mathbf{R}^n). Hence

0=λ(ψ)=⟨ψ,φ⟩=∫Rnψφdμ 0 =\lambda (\psi ) =\langle \psi , \varphi \rangle =\int_{\mathbf{R}^n}\psi \varphi d\mu

for every ψ∈D(Rn)\psi \in \mathscr{D}(\mathbf{R}^n). The measure φ⋅μ\varphi \cdot \mu on Rn\mathbf{R}^n is therefore zero (prop. 6 of IV, p. 206), whence φ=0\varphi = 0. The proposition then follows from the Hahn–Banach theorem (EVT, II, p. 46, cor. 1).

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.

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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.