Lemma

Let F be a Fréchet space, $F'$ its strong dual, $F''$ its bidual. If every subset of $F''$, bounded for $\sigma(F'', F')$, is contained in the closure (for $\sigma(F'', F')$) of a bounded subset of F, then $F'$ is the direct limit of a sequence of Banach spaces.

For, let $(V_n)$ be a decreasing fundamental sequence of convex, balanced and closed neighborhoods of 0 in F. For every integer $n$, let $G_n$ be the linear subspace of $F'$ generated by the polar $V_n^\circ$ of $V_n$. In $G_n$, $V_n^\circ$ is an absorbent convex set, therefore its gauge $p_n$ is a norm on $G_n$; moreover, $V_n^\circ$ is a complete subset of the strong dual $F'$ (TVS, III, §3, No. 8, Prop. 11); thus $G_n$, equipped with the norm $p_n$, is a Banach space (GT, III, §3, No. 5, Cor. 2 of Prop. 9). We are going to show that the strong topology on $F'$ is the direct limit of these Banach space topologies on the $G_n$, or again that, for a strongly closed, balanced, convex subset U of $F'$ to be a strong neighborhood of 0, it is necessary and sufficient that it absorb each of the $V_n^\circ$. It is obvious that this condition is necessary; to see that it is sufficient, it will suffice to prove that U contains a barrel of $F'$. Indeed, its polar $U^\circ$ in $F''$ will then be bounded for $\sigma(F'', F')$, hence, by hypothesis, will be contained in the closure (for $\sigma(F'', F')$) of a bounded subset B of F, from which one can conclude that U (which is closed for $\sigma(F', F'')$) contains the strong neighborhood $B^\circ$ of 0 (TVS, II, §6, No. 3, Th. 1 and §8, No. 4).

By hypothesis, for every integer $n$ there exists a number $\lambda_n > 0$ such that $\lambda_n V_n^\circ \subset \frac{1}{2}U$; let $A_n$ be the convex envelope of the union of the $\lambda_i V_i^\circ$ for $i \leq n$. Then $A_n \subset \frac{1}{2}U$ for every $n$; let W be the union of the $A_n$; W is a convex, balanced, absorbent set contained in $\frac{1}{2}U$, and it will suffice to show that its strong closure (which is a barrel) is contained in U.

Thus, let $x'$ be a point of $F'$ not belonging to U. Since each of the $V_n^\circ$ is compact for $\sigma(F', F)$, the same is true of the $A_n$ (TVS, II, §2, No. 6, Prop. 15), and since $x' \notin 2A_n$, there exists an element $x_n$ belonging to the polar of $A_n$ in F such that $\langle x', x_n \rangle = 2$ (TVS, II, §5, No. 3, Prop. 4). The sequence $(x_n)$ is bounded in F: for, every $y' \in F'$ belongs to some $V_k^\circ$, consequently $|\langle y', x_n \rangle| \leq \lambda_k^{-1}$ for $n \geq k$, whence our assertion (TVS, IV, §1, No. 1, Prop. 1). Let C be a bounded, balanced convex set in F containing all the $x_n$; $C^\circ$ is then a neighborhood of 0 in $F'$, and the polar $C^{\circ\circ}$ of $C^\circ$ in $F''$ is compact for $\sigma(F'', F')$ (TVS, III, §3, No. 4, Cor. 2 of Prop. 4 and No. 5, Prop. 7). Thus one sees that the sequence $(x_n)$ has a cluster point $x''$ in $F''$ for $\sigma(F'', F')$; obviously $\langle x', x'' \rangle = 2$ and, on the other hand, $x''$ belongs to the polar of $A_n$ in $F''$ for every $n$, hence to the polar $W^\circ$ of W in $F''$. From this, one concludes that $x' \notin W^{\circ\circ}$, hence is not in the closure of W for $\sigma(F', F'')$ (TVS, II, §6, No. 3, Th. 1 and §8, No. 4), therefore a fortiori for the strong topology, which completes the proof.

Exercises