Let f∈C∞(Rn). Suppose that for every α in Nn, the function ∂αf has polynomial growth. The linear mapping of the space S(Rn) into itself defined by φ↦fφ is continuous.
For every φ in S(Rn), the function fφ belongs to C∞(Rn). By hypothesis, for every α∈Nn, there exist an integer kα⩾0 and a real Cα such that ∣∂αf(x)∣⩽Cα(1+∥x∥)kα for every x in Rn. Let φ∈S(Rn). Let α∈Nn and k∈N. By Prop. 1 of IV, p. 196, there follows
(α)kβα−β
qα,k(fφ)⩽∑ sup ∥x∥∣∂f(x)∂φ(x)∣
βx∈Rn
0⩽β⩽α
(α)kkα−β
⩽∑Cβ sup ∥x∥(1+∥x∥)β∣∂φ(x)∣.
βx∈Rn
0⩽β⩽α
Let β∈Nn be such that 0⩽β⩽α. For every x∈Rn, one has
∥x∥k(1+∥x∥)kβ∣∂α−βφ(x)∣⩽ sup∥x∥⩽12kβ∣∂α−βφ(x)∣
+ supx∈Rn2kβ∥x∥k+kβ∣∂α−βφ(x)∣ whence finally
(α)k()
qα,k(fφ)⩽∑2βCβqα−β,0(φ)+qα−β,k+kβ(φ)
β
0⩽β⩽α
which implies the proposition.
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.