Bourbaki

Exercise 11, § 3

Topological Vector Spaces, IV, p. 60

Bourbaki marks this one as harder than the rest.

Let (cmn)(c_{mn}) be a double sequence of numbers >0> 0 such that cm,n≤cm+1,nc_{m,n} \leq c_{m+1,n} and let EE be the space of sequences x=(xn)x = (x_n) of real numbers such that one has pm(x)=∑ncmn∣xn∣<+∞p_m(x) = \sum_n c_{mn}|x_n| < + \infty for every integer mm. We endow EE with the topology defined by the seminorms pmp_m, for which EE is a Fréchet space of countable type; the dual E′E' of EE is identified with the space of sequences x′=(xn′)x' = (x'_n) such that sup⁡ncmn−1∣xn′∣<+∞\sup_n c_{mn}^{-1}|x'_n| < + \infty for at least one mm, the canonical bilinear form ⟨x,x′⟩\langle x, x' \rangle being identified with ∑nxnxn′\sum_n x_n x'_n (IV, p. 47, Exercise 1, c)). It is assumed that there is no subsequence (nk)(n_k) such that there are a sequence (am)(a_m) of numbers ≥0\geq 0 and an index m0m_0 for which one has cm,nk≤amcm0,nkc_{m,n_k} \leq a_m c_{m_0,n_k} for all m≥m0m \geq m_0 and kk. Under these conditions, every weakly convergent sequence in E′E' is strongly convergent, and consequently (IV, p. 60, Exercise 10) EE is a Montel space. (Reason by contradiction; by making, if necessary, a transform of the form (xn)↦(anxn)(x_n) \mapsto (a_n x_n), reduce to the case where cm0n=1c_{m_0 n} = 1 for all nn and an index m0m_0, and where there exists a sequence (x(p))p≥0(x^{(p)})_{p \geq 0} weakly convergent to 0 in E′E', such that ∣xn(p)∣≤1|x^{(p)}_n| \leq 1 for every pair (p,n)(p, n), and a bounded set BB in EE, defined by pm(x)≤bmp_m(x) \leq b_m for all m≥0m \geq 0, and such that sup⁡x∈B∣⟨x,x(p)⟩∣≥2δ>0\sup_{x \in B} |\langle x, x^{(p)} \rangle| \geq 2\delta > 0 for every integer pp. Deduce from these assumptions that there exists a strictly increasing sequence (rq)(r_q) of integers, and a sequence (x(q))(x^{(q)}) of points of BB such that

∑k=rq+1rq+1∣xk(q)∣>δ \sum_{k=r_q+1}^{r_{q+1}} |x_k^{(q)}| > \delta

for each index qq. Now prove, by reasoning by contradiction, that for every qq, there exists at least one index sqs_q such that rq<sq≤rq+1r_q < s_q \leq r_{q+1} and that for every integer mm, one has cm,sq≤bm2m+2/δc_{m,s_q} \leq b_m 2^{m+2}/\delta, a result contrary to the assumption.)

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.