Theorem 3 (Krein)

Let E be a Hausdorff and quasi-complete locally convex space, and let $\mathscr{T}$ be a topology on E compatible with the duality between E and E'. Let A be a subset of E which is compact for $\mathscr{T}$. Then the closed convex balanced envelope C of A is compact for $\mathscr{T}$.

We shall first make several reductions. A) The set C is precompact for $\mathscr{T}$ (II, p. 25, prop. 3), and A is compact for $\sigma(E, E')$. On account of prop. 1 (IV, p. 32), it is enough to prove that C is compact for $\sigma(E, E')$, and so we have reduced to the case where $\mathscr{T} = \sigma(E, E')$. B) Since C is precompact and closed for $\sigma(E, E')$, it is bounded and closed for the initial topology of E (III, p. 3, prop. 2 and IV, p. 1, prop. 1); hence it is complete since E is quasi-complete. In other words, C is the closed convex balanced envelope of $A$ in the completion $\hat{E}$ of $E$. Since the topology $\sigma(\hat{E}, E')$ induces $\sigma(E, E')$ on $E$, we have reduced to the case when $E$ is complete.

C) Let $\Gamma$ be the convex balanced envelope of $A$. Then $C$ is the closure of $\Gamma$ for $\sigma(E, E')$. By Eberlein’s theorem (IV, p. 35, th. 1), it is enough to prove that every sequence $(x_n)_{n \in \mathbf{N}}$ of points of $\Gamma$ has a limit point for $\sigma(E, E')$ in $E$. But $x_n$ belongs to the convex balanced envelope of a finite subset $B_n$ of $A$. Let $F$ be the closed vector subspace of $E$ generated by the countable set $B = \bigcup_n B_n$. Then $F$ is complete, the topology $\sigma(F, F')$ on $F$ is induced by $\sigma(E, E')$ and we have $x_n \in F$ for all $n \in \mathbf{N}$. Hence it is enough to prove that $(x_n)_{n \in \mathbf{N}}$ has a limit point for $\sigma(F, F')$, which gives the reduction to the case when there exists a countable dense set in $E$.

Let $A$ be assigned the topology induced by $\sigma(E, E')$, which makes it a compact space. We define a linear mapping $u : E' \to \mathscr{C}(A)$ by

Let $(x'_n)_{n \in \mathbf{N}}$ be an equicontinuous sequence in $E'$, converging to 0 for $\sigma(E', E)$. Then the sequence of functions $u(x'_n)$ is bounded in $\mathscr{C}(A)$ and converges simply to 0. For every $\mu \in \mathscr{C}(A)'$, we have $\lim_{n \to \infty} \mu(u(x'_n)) = 0$ by lemma 2 (IV, p. 37). By the criterion given in the remark in III, p. 21, the linear form $\mu \circ u$ on $E'$ is then continuous for $\sigma(E', E)$ for every $\mu \in \mathscr{C}(A)'$. Hence there exists a linear mapping $v : \mathscr{C}(A)' \to E$ satisfying the relation

It is clear that $v$ is continuous if $\mathscr{C}(A)'$ is assigned the topology $\sigma(\mathscr{C}(A)', \mathscr{C}(A))$ and $E$ the topology $\sigma(E, E')$.

The unit ball (closed) $B$ of the Banach space $\mathscr{C}(A)$ is compact for the topology $\sigma(\mathscr{C}(A)', \mathscr{C}(A))$ (III, p. 17, cor. 3). Consequently, $v(B)$ is a convex balanced and compact subset of $E$ for $\sigma(E, E')$. For every $a \in A$, the continuous linear form $\varepsilon_a : f \mapsto f(a)$ on $\mathscr{C}(A)$ belongs to $B$, and we have $v(\varepsilon_a) = a$ by formulas (7) and (8). Hence, $A \subset v(B)$, and so $C \subset v(B)$. This proves that $C$ is compact for $\sigma(E, E')$.

Q.E.D.