Bourbaki

Proposition 4

Tag 07U0, permanent. Cite it as [Bourbaki, Tag 07U0].

Topological Vector Spaces, chapter IV, § 3. Dual of a Fréchet space

— Let E be a reflexive Fréchet space. The strong dual Eb′E'_b of E is the inductive limit of a sequence of Banach spaces.

Let (Vn)n∈N(V_n)_{n \in \mathbf{N}} be a decreasing sequence of convex, balanced and closed neighbourhoods of 0 in E, such that every neighbourhood of 0 in E contains one of the VnV_n. Let AnA_n be the polar of VnV_n in E′E'. Then AnA_n is convex, balanced and compact for σ(E′,E)\sigma(E', E); by III, p. 8, corollary the space EAn′E'_{A_n} is a Banach space. We shall prove that Eb′E'_b is the inductive limit of the spaces EAn′E'_{A_n}; in other words, that every convex and balanced subset U of E′E' which absorbs each of the AnA_n is a neighbourhood of 0 in Eb′E'_b. For every n∈Nn \in \mathbf{N}, choose a real number λn>0\lambda_n > 0 such that λnAn⊂U\lambda_n A_n \subset U. Let BnB_n be the convex envelope of the set ⋃0≤i≤nλiAi\bigcup_{0 \leq i \leq n} \lambda_i A_i; put V=⋃nBnV = \bigcup_n B_n, then V⊂UV \subset U. For every n∈Nn \in \mathbf{N}, the set BnB_n is convex, balanced and compact for σ(E′,E)\sigma(E', E) (II, p. 14, prop. 15).

Now we shall show that 12Voo⊂V\frac{1}{2} V^{oo} \subset V. Let x∈Eb′−Vx \in E'_b - V; for every n∈Nn \in \mathbf{N}, we have x∉Bnx \notin B_n, and since BnB_n is closed for σ(E′,E)\sigma(E', E) there exists an element yny_n in Bn∘B_n^\circ such that \langle y_n, x \rangle = 1 (\text{II, p. 38, prop. 4}). Since E is reflexive, every bounded subset of E is relatively compact for σ(E,E′)\sigma(E, E') (IV, p. 16, th. 2). By the definition of BnB_n, we have

λiyn∈Vifor all n≥i, \lambda_i y_n \in V_i \quad \text{for all } n \geq i,
hence the sequence (yn)(y_n) is bounded. Let yy be a limit point of (yn)(y_n) for the topology σ(E,E′)\sigma(E, E'). We have y∈V∘=⋂nBn∘y \in V^\circ = \bigcap_n B_n^\circ and ⟨y,x⟩=1\langle y, x \rangle = 1. Hence x∉12V∘∘x \notin \frac{1}{2}V^{\circ\circ}, and so we have the inclusion 12V∘∘⊂V\frac{1}{2}V^{\circ\circ} \subset V and a fortiori, 12V∘∘⊂U\frac{1}{2}V^{\circ\circ} \subset U.

Since every bounded subset of Eb′E'_b is contained in one of the sets AnA_n, the set V=⋃nBnV = \bigcup_n B_n absorbs every bounded subset of Eb′E'_b. Consequently, V∘V^\circ is bounded in E, hence 12V∘∘\frac{1}{2}V^{\circ\circ} is a neighbourhood of 0 in Eb′E'_b. A fortiori, U is a neighbourhood of in Eb′E'_b.

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

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Cites

Transcribed from the printed volume. Not translated.