Let k∈N and α∈Nn. For every function φ∈S(Rn) and every real number T>0, one has
kα 1
(3) qα,k(φ)⩽T sup ∣∂φ(x)∣+qα,k+1(φ).
∥x∥⩽T T
In fact, one has
qα,k(φ)⩽ sup∥x∥⩽T∥x∥k∣∂αφ(x)∣+ sup∥x∥>T∥x∥k∣∂αφ(x)∣
⩽Tk sup∥x∥⩽T∣∂αφ(x)∣+T1 sup∥x∥>T∥x∥k+1∣∂αφ(x)∣.
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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