Let N be the euclidean norm on Rn. Let k be an integer ⩾0. The Sobolev space Hk(Rn) is the space of f∈S′(Rn) such that (1+Nk)F(f) belongs to L2(Rn).
We proceed by induction on k. When k=0, the result is a consequence of the Plancherel theorem (II, p. 215, th. 1). Suppose that k=1. For f∈S′(Rn), one has (1+N)Ff∈L2(Rn) if and only if f∈L2(Rn) and NFf∈L2(Rn). Moreover, NF∈L2(Rn) if and only if, for every α∈Nn such that ∣α∣=1, one has XαFf∈L2(Rn). Since one has F(∂αf)=2iπXαFf, this condition means that F(∂αf)∈L2(Rn) for every α with ∣α∣=1, that is, ∂αf∈L2(Rn) for every α with ∣α∣=1. It follows that the assertion is true for k=1.
Suppose now that k⩾ 2 and that the assertion concerning Hℓ(Rn) is valid for every positive integer ℓ⩽k−1. Let f∈S′(Rn). By definition, one has f∈Hk(Rn) if and only if f∈L2(Rn) and, for every β∈Nn such that ∣β∣⩽1, the distribution ∂βf belongs to Hk−1(Rn). This is equivalent, by the induction hypothesis, to f∈L2(Rn) and (1+Nk−1)F(∂βf)∈L2(Rn) for every β∈Nn such that ∣β∣⩽1. Since F(∂βf)=(2iπX)∣β∣F(f), the condition f∈Hk(Rn) is equivalent to saying that Ff∈L2(Rn) and (1+Nk−1)XβFf∈L2(Rn) for β∈Nn such that ∣β∣⩽1.
The inequalities
1+Nk⩽1+Nk−1β∣β∈∣N⩽1n∑∣Xβ∣⩽1+n1/2Nk⩽(1+n1/2)(1+Nk)
then imply that f∈Hk(Rn) if and only if (1+Nk)F∈L2(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.
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