Exercise 11, § 3
Topological Vector Spaces, IV, p. 60
Bourbaki marks this one as harder than the rest.
Let be a double sequence of numbers such that and let be the space of sequences of real numbers such that one has for every integer . We endow with the topology defined by the seminorms , for which is a Fréchet space of countable type; the dual of is identified with the space of sequences such that for at least one , the canonical bilinear form being identified with (IV, p. 47, Exercise 1, c)). It is assumed that there is no subsequence such that there are a sequence of numbers and an index for which one has for all and . Under these conditions, every weakly convergent sequence in is strongly convergent, and consequently (IV, p. 60, Exercise 10) is a Montel space. (Reason by contradiction; by making, if necessary, a transform of the form , reduce to the case where for all and an index , and where there exists a sequence weakly convergent to 0 in , such that for every pair , and a bounded set in , defined by for all , and such that for every integer . Deduce from these assumptions that there exists a strictly increasing sequence of integers, and a sequence of points of such that
for each index . Now prove, by reasoning by contradiction, that for every , there exists at least one index such that and that for every integer , one has , a result contrary to the assumption.)
Solution
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