The space D(Rn) is contained in S(Rn), and the inclusion of D(Rn) in S(Rn) is continuous with dense image.
Let B⊂D(Rn) be a bounded subset, and let K be a compact subset of Rn such that B⊂CK∞(Rn). Let α∈Nn and k∈N. For every function φ∈B, one has
qα,k(φ)⩽(supx∈K∥x∥k)pα,K(φ),
therefore B is bounded in S(Rn). The continuity of the inclusion then follows from the fact that the spaces S(Rn) and D(Rn) are bornological.
Let us prove that D(Rn) is dense in S(Rn). Let B be the unit ball of Rn and let η∈D(Rn) be a test function with support contained in 2B such that 0⩽η⩽1 and η(x)=1 for every x∈B (lemma 1, a) of IV, p. 196).
Let φ∈S(Rn). For every integer m⩾1 and every x∈Rn, put ηm(x)=η(x/m). Finally define φm=ηmφ; we have φm∈D(Rn). Since ∂αηm=m−∣α∣(∂αη)(x/m) for all α∈Nn and x∈Rn, one deduces from formula (2) of IV, p. 210 that the sequence (φm) is bounded in S(Rn).
Let C be a compact subset of Rn. The sequence (φm)m⩾1 converges to φ in CK∞(Rn) since φm coincides with φ on C for every sufficiently large m. Thus the sequence (φm) converges to φ in C∞(Rn), and the corollary to proposition 10 of IV, p. 211 makes it possible to conclude that the sequence (φm) converges to φ in S(Rn).
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.