Bourbaki

Proposition 3

Tag 030H, permanent. Cite it as [Bourbaki, Tag 030H].

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

Let N be a norm on Rn,r\mathbf{R}^n,r a real number and p∈[1,+∞[p\in [1,+\infty [.

a) The function (1+N)r(1 + N)^r belong to Lp(Rn,μ)\mathscr{L}^p(\mathbf{R}^n, \mu) if and only if rp<−nrp <-n;

a′)') The restriction to Zn\mathbf{Z}^n of the function (1+N)r(1 + N)^r belong to ℓp(Zn)\ell^p(\mathbf{Z}^n) if and only if rp<−nrp <-n;

b) Let V be a bounded measurable neighbourhood of 0 in Rn\mathbf{R}^n. The function NrN^r belong to Lp(Rn−V,μ)\mathscr{L}^p(\mathbf{R}^n-V, \mu) if and only if rp<−nrp <-n.

By virtue of the equivalence of norms on Rn\mathbf{R}^n (EVT I, p. 14, th. 2 and TG, IX, p. 32, prop. 8), we may suppose that the norm N is given by N(x)=N(x) = sup∣xi∣|x_i| for x=(xi)∈Rnx= (x_i)\in \mathbf{R}^n. Let B denote the unit ball of Rn\mathbf{R}^n for this norm.

Let V be a bounded measurable neighbourhood of 0 in Rn\mathbf{R}^n. The function NrN^r is continuous and bounded on (B- VV)∪(V\cup (V- B), which shows that assertion b) is valid if and only if it is so when V = B, which we shall henceforth suppose. Since the function (1+N)r(1 + N)^r is continuous and bounded on B, and satisfies the inequalities Nr⩽(1+N)r⩽2rNrN^r\leqslant (1 + N)^r\leqslant 2^rN^r on Rn−\mathbf{R}^n- B, we see that assertions a) and b) are equivalent.

Let us prove b) when V = B. We may suppose that p=1p= 1. Since the case where r>0r >0 is elementary, suppose that r⩽0r\leqslant 0. For every integer j⩾1j\geqslant 1, the set

Cj={x∈Rn∣2j−1⩽N(x)<2j} C_j=\{x\in \mathbf{R}^n|2^{j-1}\leqslant N(x)<2^j\}

is of measure 2nj(2n−1)2^{nj}(2^n-1). The sets (Cj)j⩾1(C_j)_{j\geqslant 1} form a partition of Rn−\mathbf{R}^n- B. According to INT, V, p. 4, § 1, No.o1^o1, cor., one therefore has

∫∗Nrdμ=∑∫∗Nrdμ \int^*N^rd\mu=\sum\int^*N^rd\mu
Rn−Bj⩾1Cj\mathbf{R}^{n-}Bj\geqslant 1C_j

⩽∑j⩾12n+nj2r(j−1)=2n−r∑j⩾12(n+r)j\leqslant \sum_{j\geqslant 1}2^{n+nj}2^{r(j-1)}= 2^{n-r}\sum_{j\geqslant 1}2^{(n+r)j} which is finite if r<−nr <-n. On the other hand, one has (loc. cit.)

∫∗Nrdμ⩾∑2rj2nj(2n−1)=(2n−1)∑2(r+n)j \int^*N^rd\mu\geqslant \sum 2^{rj}2^{nj}(2^n-1) = (2^n-1)\sum 2^{(r+n)j}
Rn−Bj⩾1j⩾1\mathbf{R}^{n-}Bj\geqslant 1j\geqslant 1

which is infinite if r⩾−nr\geqslant -n.

One proves a′)') similarly by considering the sets Cj∩ZnC_j\cap \mathbf{Z}^n which cover Zn−{0}\mathbf{Z}^n-\{0\} and satisfy

Card(Cj∩Zn_j\cap \mathbf{Z}^n) =(2j+1−1)n−(2j−1)n= (2^{j+1}-1)^n-(2^j-1)^n,

which belongs to the interval [(1−2−n)(2j+1−1)n,2n(j+1)][(1-2^{-n})(2^{j+1}-1)^n,2^{n(j+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.

Cited by

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Cites

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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.