Bourbaki

Lemma (i)

Tag 0BTJ, permanent. Cite it as [Bourbaki, Tag 0BTJ].

General Topology, chapter IV, § 7. Infinite sums and products of real numbers

If (ai)1≤i≤p(a_i)_{1 \leq i \leq p} is a finite sequence of numbers >0> 0, then

∏i=1p(1+ai)≥1+∑i=1pai. \prod_{i=1}^p (1 + a_i) \geq 1 + \sum_{i=1}^p a_i.
(ii) If also ai<1a_i < 1 for each ii, then
∏i=1p(1−ai)≥1−∑i=1pai. \prod_{i=1}^p (1 - a_i) \geq 1 - \sum_{i=1}^p a_i.

These inequalities are clear if p=1p = 1, and are proved by induction on pp. If

∏i=1p−1(1+ai)≥1+∑i=1p−1ai, \prod_{i=1}^{p-1} (1 + a_i) \geq 1 + \sum_{i=1}^{p-1} a_i,
we have
∏i=1p(1+ai)≥(1+ap)(1+∑i=1p−1ai)=1+∑i=1pai+ap⋅∑i=1p−1ai≥1+∑i=1pai. \prod_{i=1}^p (1 + a_i) \geq (1 + a_p) \left( 1 + \sum_{i=1}^{p-1} a_i \right) = 1 + \sum_{i=1}^p a_i + a_p \cdot \sum_{i=1}^{p-1} a_i \geq 1 + \sum_{i=1}^p a_i.
Again, if
∏i=1p−1(1−ai)≥1−∑i=1p−1ai, \prod_{i=1}^{p-1} (1 - a_i) \geq 1 - \sum_{i=1}^{p-1} a_i,
we have
∏i=1p(1−ai)≥(1−ap)(1−∑i=1p−1ai)=1−∑i=1pai+ap⋅∑i=1p−1ai≥1−∑i=1pai. \prod_{i=1}^p (1 - a_i) \geq (1 - a_p) \left( 1 - \sum_{i=1}^{p-1} a_i \right) = 1 - \sum_{i=1}^p a_i + a_p \cdot \sum_{i=1}^{p-1} a_i \geq 1 - \sum_{i=1}^p a_i.

Having proved the lemma, we note that if the family (1+ui)(1 + u_i) is multipliable then so are the families (1+ui+)(1 + u_i^+) and (1−ui−)(1 - u_i^-), since R∗\mathbf{R}^* is a complete group (Chapter III, § 5, no. 3, Proposition 2); and conversely, if the families (1+ui+)(1 + u_i^+) and (1−ui−)(1 - u_i^-) are multipliable, then so is (1+ui)(1 + u_i) (Chapter III, § 5, no. 3, Proposition 3). We need therefore consider only the cases in which all the uiu_i are ≥0\geq 0 and in which they are all ≥0\geq 0.

Suppose first that ui≥0u_i \geq 0 for all ii. If the family (1+ui)(1 + u_i) is multipliable, then for each ε>0\varepsilon > 0 there is a finite subset JJ of the index set in II such that, for each finite subset HH of II disjoint from JJ, we have 1≤∏i∈H(1+ui)≤1+ε1 \leq \prod_{i \in H} (1 + u_i) \leq 1 + \varepsilon; by (2) it follows that ∑i∈Hui≤ε\sum_{i \in H} u_i \leq \varepsilon, which shows that (ui)(u_i) is summable in R\mathbf{R} by virtue of Cauchy’s criterion (Chapter III, § 5, no. 2, Theorem 1).

Conversely, suppose that (ui)(u_i) is summable in R\mathbf{R}. For each ε\varepsilon such that 0<ε<10 < \varepsilon < 1, there is a finite subset JJ of II such that, for each finite subset HH of II disjoint from JJ, we have 0≤∑i∈Hui≤ε0 \leq \sum_{i \in H} u_i \leq \varepsilon. By (3), we have therefore ∏i∈H(1−ui)≥1−ε\prod_{i \in H} (1 - u_i) \geq 1 - \varepsilon; but since 1+u≤11−u1 + u \leq \frac{1}{1 - u} for every number uu such that 0≤u<10 \leq u < 1, it follows that

1≤∏i∈H(1+ui)≤11−ε, 1 \leq \prod_{i \in H} (1 + u_i) \leq \frac{1}{1 - \varepsilon},

and this shows that (1+ui)(1 + u_i) is multipliable (Cauchy’s criterion).

The proof is similar when all the uiu_i are ≤0\leq 0. To show that (ui)(u_i) is summable if (1+ui)(1 + u_i) is multipliable, use formula (2) and the inequality 1−u≤11+u1 - u \leq \frac{1}{1 + u} (0≤u<10 \leq u < 1); to show that (1+ui)(1 + u_i) is multipliable if (ui)(u_i) is summable, use formula (3).

In Chapter V (§ 4), the topological study of the group R∗\mathbf{R}^* will enable us to give another criterion for multipliability of a family in R∗\mathbf{R}^* with the help of the logarithmic function; in a later volume we shall establish the equivalence of this criterion with the one above, using the differential properties of the logarithm.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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