§ 2. Bidual. Reflexive spaces

1. Bidual

Definition 1

— *Let $E$ be a locally convex space and $E'_b$ its strong dual. The dual of the locally convex space $E'_b$ is called the bidual of $E$ and is denoted by $E''$ .

For every $x \in E$ , let $\tilde{x}$ be the linear form $x' \mapsto \langle x, x' \rangle$ on $E'$ ; it is continuous for the weak topology $\sigma(E', E)$ , hence a fortiori, for the strong topology on $E'$ ; therefore $\tilde{x} \in E''$ for all $x \in E$ . The map $c_E : x \mapsto \tilde{x}$ from $E$ into $E''$ is a linear mapping, said to be canonical.

Proposition 1

— *The kernel of $c_E : E \to E''$ is the closure of 0 in $E$ . If $E$ is Hausdorff, $c_E$ is injective.

By construction, the kernel of $c_E$ is the intersection of the kernels of the continuous linear forms on $E$ , i.e. the closure of $\{0\}$ in $E$ (II, p. 24, cor. 1).

When $E$ is Hausdorff, we identify $E$ with a subspace of $E''$ , by means of the mapping $c_E$ .

BIDUAL. REFLEXIVE SPACES

TVS IV.15

The strong topology on $E''$ is the $\mathscr{S}$ -topology, where $\mathscr{S}$ is the family of all strongly bounded subsets of $E'$ . Since every equicontinuous subset of $E'$ is strongly bounded (III, p. 22, prop. 9), the initial topology on $E$ is coarser than the topology obtained by taking the inverse image under $c_E$ of the strong topology on $E''$ ; it can be strictly coarser (IV, p. 52, exerc. 1). However:

Proposition 2

— Suppose that the space $E$ is bornological or barrelled. The initial topological on $E$ is the inverse image under $c_E$ of the strong topology on $E''$ .

For, every subset of $E'$ which is strongly bounded is equicontinuous (III, p. 22, prop. 10 and III, p. 24).

Proposition 3

— Let $E$ be a locally convex Hausdorff space. In order that the strong dual $E'_b$ of $E$ be barrelled, it is necessary and sufficient that every subset of $E''$ which is bounded for $\sigma(E'', E')$ , is contained in the closure, for $\sigma(E'', E')$ , of a bounded subset of $E$ .

The equicontinuous subsets of $E''$ are the subsets contained in the bipolar (for the duality between $E''$ and $E'$ ) of a bounded subset of the subspace $E$ of $E''$ . It is now enough to apply the theorem of bipolars (II, p. 45, cor. 3) and the definition of a barrelled space (III, p. 24).

Remark

Let $E$ be a locally convex Hausdorff space, $E'$ its dual and $E''$ its bidual. We have $E \subset E'' \subset {E'}^*$ , where ${E'}^*$ is the algebraic dual of $E'$ . If $B$ is a bounded subset of $E$ , its closure $\overline{B}$ in ${E'}^*$ endowed with $\sigma({E'}^*, E')$ is contained in $E''$ : for, the polar $U = B^\circ$ of $B$ in $E'$ is a neighbourhood of 0 in $E'_b$ , and we have $\overline{B} \subset U^\circ \subset E''$ .

2. Semi-reflexive spaces

Definition 2

— Let $E$ be a locally convex space. We say that $E$ is semi-reflexive if the canonical mapping $c_E$ from $E$ into $E''$ is bijective.

This implies that $E$ is Hausdorff, and that every linear form on $E'$ , which is continuous for the strong topology $\beta(E', E)$ , is of the form $x' \mapsto \langle x, x' \rangle$ with $x \in E$ , i.e. continuous for the weak topology $\sigma(E', E)$ .

Theorem 1

— A locally convex Hausdorff space is semi-reflexive if and only if every bounded subset of $E$ is relatively compact for the weakened topology $\sigma(E, E')$ . If $E$ is semi-reflexive, the strong dual $E'_b$ of $E$ is barrelled.

The second assertion follows from prop. 3 (IV, p. 15), and the identity between bounded subsets for the initial topology and for the weakened topology of $E$ (III, p. 27, cor. 3).

To say that $E$ is semi-reflexive means that the topology on $E'_b$ is compatible with the duality between $E$ and $E'$ , in other words, by Mackey's theorem (IV, p. 2, th. 1) that the topology on $E'_b$ is coarser than $\tau(E', E)$ (and in fact is identical with it); by definition (IV, p. 2), this means that every closed, convex and bounded subset of $E$ is compact for $\sigma(E, E')$ , and this is equivalent to saying that every bounded subset of E is relatively compact for $\sigma(E, E')$ , because the closed convex envelope of a bounded subset of E is bounded (III, p. 3, prop. 1).

Corollary

— *Let E be a locally convex semi-reflexive space. Every closed vector subspace M of E is semi-reflexive; moreover, the strong topology on $E'/M^\circ$ (considered as the dual of M) is the quotient of the strong topology on E'.

Let B be a bounded subset of M. Since B is bounded in E, and the weakened topology $\sigma(M, M')$ on M is induced by $\sigma(E, E')$ (IV, p. 10, prop. 11), the closure of B in M endowed with $\sigma(M, M')$ is compact. Hence, by th. 1, M is semi-reflexive. The last assertion of the corollary follows from prop. 10 of IV, p. 9, applied to the set $\mathscr{S}$ of all closed, convex and bounded subsets of E.

Remark 1

Suppose E is semi-reflexive. Every subset of E which is convex, closed and bounded for the initial topology is compact for the topology $\sigma(E, E')$ (IV, p. 1, prop. 1). *On the other hand, the unit sphere (with the equation $\|x\| = 1$ ) of an infinite dimensional hilbertian space E is closed and bounded for the initial topology, but is not closed for the weakened topology, even if E is semi-reflexive. \

Remark 2

By remark 3 of IV, p. 5, we can reformulate th. 1 as follows : the Hausdorff space E is semi-reflexive if and only if it is quasi-complete for its weakened topology. If it is semi-reflexive, then it is quasi-complete for its initial topology (IV, p. 5, Remark 2).

Remark 3

Under the hypotheses of the above corollary, the space E/M is not necessarily semi-reflexive (IV, p. 63, exerc. 10).

3. Reflexive spaces

Definition 3

— *A locally convex space E is said to be reflexive if the canonical mapping $c_E$ from E into $E''$ is a topological vector space isomorphism from E onto the strong dual of $E'_b$ .

In particular, a reflexive space is semi-reflexive, hence Hausdorff.

Proposition 4

— *The strong dual of a reflexive space is reflexive.

This follows immediately from def. 3.

Theorem 2

— *In order that a locally convex Hausdorff space E be reflexive, it is necessary and sufficient that it is barrelled and that every bounded subset of E is relatively compact for the weakened topology $\sigma(E, E')$ .

By th. 1 (IV, p. 15), this is the same as saying that E is reflexive if and only if it is semi-reflexive and barrelled.

If E is reflexive, $E'_b$ is reflexive (prop. 4) and consequently E is barrelled (IV, p. 15, th. 1). Conversely, if E is semi-reflexive and barrelled, $c_E$ is a bijection and is bicontinuous by IV, p. 15, prop. 2, hence E is reflexive.

Remark 1

Let E be an infinite dimensional real hilbertian space. Let F denote the space E endowed with the weakened topology. The spaces E and F have the same dual E', and E is a reflexive Banach space (V, p. 17). Consequently, F is semi-reflexive.

However, the strong topology and the weakened topology on E are distinct, hence F is not reflexive.

Remark 2

Let E be a reflexive space and M a closed vector subspace of E. It may happen that neither M nor E/M are reflexive spaces (IV, p. 63, exerc. 10). \* For the case of normed spaces, see prop. 7 of IV, p. 17. \

4. The case of normed spaces

Let E be a normed space. The strong topology on the dual E' of E is defined by the norm

(1)

\|x'\| = \sup_{x \in E, \|x\| \leq 1} |\langle x, x' \rangle|,

and the strong dual of E is a Banach space (III, p. 24, cor. 2). Then the bidual E'' of E is also a Banach space, for the norm defined by

(2)

\|x''\| = \sup_{x' \in E', \|x'\| \leq 1} |\langle x', x'' \rangle|.

By prop. 8, (i) of IV, p. 7, the canonical linear mapping $c_E : E \to E''$ is an isometry. Henceforth, we shall identify E with a normed subspace of its bidual E''.

Proposition 5

— *Let E be a normed space, E' its dual and E'' its bidual. The (closed) unit ball in E'' is the closure of the unit ball B in E for the weak topology $\sigma(E'', E')$ .

By formulas (1) and (2), the unit ball in E'' is the bipolar $B^{\circ \circ}$ of B. Prop. 5 then follows from the theorem of bipolars (II, p. 45, cor. 3).

Remark

— A Banach space E is closed in its bidual E'' for the strong topology, but is dense for the weak topology (prop. 5).

In order that a normed space be reflexive, it is necessary and sufficient that it is semi-reflexive; for, the initial topology of E is always induced by the strong topology of E''. Th. 1 (IV, p. 15) then implies the following result :

Proposition 6

— *In order that a normed space E be reflexive, it is necessary and sufficient that the unit ball in E be compact for the weakened topology $\sigma(E, E')$ .

We observe that a reflexive normed space is complete hence a Banach space, and that its dual is a reflexive Banach space by prop. 4 of IV, p. 16.

Proposition 7

— *Let E be a reflexive Banach space and M a closed vector subspace of E. Then M and E/M are reflexive Banach spaces.

Let E' be the dual of E and $M^\circ$ the orthogonal of M in E'. As a normed space, we can identify the space $E'/M^\circ$ with the dual $M'$ of M (IV, p. 9, prop. 10). Since M is semi-reflexive (IV, p. 16, corollary), it is reflexive, hence so is $E'/M^\circ$ ; similarly $M^\circ$ is reflexive, as also its dual $E/M^{\circ \circ} = E/M$ .

Example

— 1) Let $\ell^\infty(\mathbf{N})$ denote the Banach space of bounded sequences $x = (x_n)_{n \in \mathbf{N}}$ of scalars, with the norm

(3)

\| \mathbf{x} \| = \sup_{n \in \mathbf{N}} |x_n| \quad (\text{I, p. 4}) .

Let $c_0(\mathbf{N})$ be the closed vector subspace of $\ell^\infty(\mathbf{N})$ consisting of sequences tending to 0. Finally, let $\ell^1(\mathbf{N})$ be the vector space of summable sequences, endowed with the norm

(4)

\| \mathbf{x} \|_1 = \sum_{n \in \mathbf{N}} |x_n| .

We can show (IV, p. 47, exerc. 1) that the dual of $c_0(\mathbf{N})$ can be identified with $\ell^1(\mathbf{N})$ in such a way that we have

(5)

\langle \mathbf{x}, \mathbf{x}' \rangle = \sum_{n \in \mathbf{N}} x_n x'_n

for all $\mathbf{x} \in c_0(\mathbf{N})$ and $\mathbf{x}' \in \ell^1(\mathbf{N})$ . Similarly the dual of $\ell^1(\mathbf{N})$ can be identified with $\ell^\infty(\mathbf{N})$ in such a way that we have the relation (5) for all $\mathbf{x} \in \ell^1(\mathbf{N})$ and all $\mathbf{x}' \in \ell^\infty(\mathbf{N})$ . Hence $\ell^\infty(\mathbf{N})$ is the bidual of $c_0(\mathbf{N})$ , and this latter space is not reflexive.

\* 2) Every hilbertian space is a reflexive Banach space (V, p. 17). \* 3) Let X be a Hausdorff topological space and $\mu$ a complex measure on X. For every real number $p > 1$ , the Banach space $L^p(X, \mu)$ is reflexive, and its dual can be identified with $L^q(X, \mu)$ with $p^{-1} + q^{-1} = 1$ (INT, V, 2nd edition, § 5, No. 8 and IX, § 1, No. 10).

5. Montel spaces

Definition 4

A locally convex Hausdorff and barrelled space in which every bounded subset is relatively compact is called a Montel space.

Example 1

Every finite dimensional Hausdorff space is a Montel space. A normed space which is a Montel space is locally compact, hence is finite dimensional (I, p. 15, th. 3).

Example 2

With the notations and hypothesis of prop. 7 of III, p. 6, the space E, being the inductive limit of Banach spaces, is barrelled (III, p. 25); moreover, every bounded subset of E is relatively compact (III, p. 6, prop. 7). In other words, E is a Montel space. In particular, Gevrey spaces (III, p. 10) are Montel spaces. \* This is true for the space $\mathscr{H}(K)$ consisting of germs of functions analytic in a neighbourhood of a compact subset K of $\mathbf{C}^n$ (III, p. 10).

Example 3

Every strict inductive limit E of a sequence $(E_n)$ of Montel spaces (II, p. 33) such that $E_n$ is closed in $E_{n+1}$ for all n, is a Montel space; in fact, E is Hausdorff (II, p. 32, prop. 9 (i)), barrelled (III, p. 25, cor. 3) and every bounded subset of E is contained in one of the $E_n$ (III, p. 5, prop. 6) hence is relatively compact in $E_n$ , and consequently also in E.

Example 4

\* Let U be an open set in $\mathbf{R}^n$ and let $\mathscr{C}^\infty(U)$ be the Fréchet space of infinitely differentiable functions on U (III, p. 9). We shall prove that this is a Montel space. Since $\mathscr{C}^\infty(U)$ is a Fréchet space, it is barrelled (III, p. 25, corollary). Let B be a bounded subset of $\mathscr{C}^\infty(U)$ and let K be a compact subset of U. For every $\alpha \in \mathbf{N}^n$ let $H_{\alpha, K}$ be the set of restrictions to K of the functions $\partial^\alpha f$ , as $f$ runs through B. Let $\alpha \in \mathbf{N}^n$ ; for every $\beta \in \mathbf{N}^n$ such that $|\beta| = |\alpha| + 1$ , the set $H_{\alpha, K}$ is bounded in $\mathscr{C}(K)$ since B is bounded in $\mathscr{C}^\infty(U)$ ; by VAR, R., No. 2.2.3, the set $H_{\alpha, K}$ is equicontinuous, hence (GT, X, § 2, No. 5) relatively compact in $\mathscr{C}(K)$ . But the topology of $\mathscr{C}^\infty(U)$ is the coarsest among the topologies for which all the maps $f \mapsto \partial^\alpha f|K$ from $\mathscr{C}^\infty(U)$ into $\mathscr{C}(K)$ are continuous, therefore B is relatively compact in $\mathscr{C}^\infty(U)$ (GT, I, § 4, No. 1, prop. 3 and § 9, No. 5, corollary).

Similarly, the space $\mathscr{C}_0^\infty(U)$ of all infinitely differentiable functions with compact support in U (III, p. 9) is a Montel space. For, $\mathscr{C}_0^\infty(U)$ is the strict inductive limit of a sequence $\mathscr{C}_{H_n}^\infty(U)$ of Fréchet spaces (III, p. 9), and it is enough to see that each of the spaces $\mathscr{C}_{H_n}^\infty(U)$ is a Montel space (Example 3). But a bounded and closed subset of $\mathscr{C}_0^\infty(U)$ is closed and bounded in $\mathscr{C}^\infty(U)$ , hence compact in $\mathscr{C}^\infty(U)$ , and consequently in $\mathscr{C}_{H_n}^\infty(U)$ . \

Proposition 8

Let E be a Montel space and $\mathfrak{T}$ a filter on E, which converges to a point $x_0$ in E for the weakened topology. If $\mathfrak{T}$ is a countable base, or contains a bounded set, then $\mathfrak{T}$ , converges to $x_0$ for the initial topology also.

Assume first that there exists a bounded set $\overline{B}$ in $\mathfrak{T}$ . The closure $\overline{B}$ of B for the initial topology of E is bounded; in addition, $\overline{B}$ is compact because E is a Montel space. The topology on $\overline{B}$ induced by $\sigma(E, E')$ is Hausdorff and coarser than the topology induced by the initial topology; they therefore coincide (GT, I, § 9, No. 4). This prove the proposition for this case.

Next assume that $\mathfrak{T}$ has a countable base. It is enough (GT, I, § 6, No. 8, prop. 11) to consider the case of a sequence $(x_n)_{n \geq 1}$ tending to $x_0$ for $\sigma(E, E')$ . Let B be the set of all $x_n$ for $n \geq 0$ . This set is bounded for $\sigma(E, E')$ , hence also for the initial topology (III, p. 27, cor. 3). Thus we have reduced to the first case of the proof.

Every Montel space is reflexive: this follows from def. 4 and from th. 2 of IV, p. 16. Further:

Proposition 9

The strong dual of a Montel space is a Montel space.

Let E be a Montel space and $E'_b$ its strong dual. Since E is reflexive, $E'_b$ is barrelled (IV, p. 15, th. 1). Since every bounded subset of E is relatively compact the strong topology on $E'$ coincides with the topology of compact convergence. Let B be a bounded subset of $E'_b$ ; it is bounded for the weak topology $\sigma(E', E)$ , hence is equicontinuous because E is barrelled. Then Ascoli’s theorem (GT, X, § 2, No. 4, cor. and § 2, No. 5, cor. 1) implies that the closure of B for $\sigma(E', E)$ is compact for the topology of compact convergence; therefore B is relatively compact in $E'_b$ .

Proposition 10

Every metrizable Montel space satisfies the first axiom of countability.

Let E be a metrizable Montel space. We know (II, p. 5) that E can be identified with a subspace of the product $F = \prod_{n \in \mathbf{N}} F_n$ of a sequence of normed spaces, and we can even assume that we have $\mathrm{pr}_n(E) = F_n$ for all $n \in \mathbf{N}$ . If each of the metrizable spaces $F_n$ satisfies the first axiom of countability, then so does F (GT, IX, § 2, No. 8), hence also E.

We argue by reductio ad absurdum. Assume for example that $F_0$ does not satisfy the first axiom of countability. Let $B_0$ be the unit ball (closed) in $F_0$ ; this is a metric space which does not satisfy the first axiom of countability. We shall use the following lemma :

Lemma 1

Suppose the metric space X does not satisfy the first axiom of countability. Then there exists a real number $\varepsilon > 0$ and an uncountable subset A in X such that $d(x, y) \geq \varepsilon$ for all distinct $x, y$ in A.

For every integer $n \geq 1$ , let $\mathfrak{F}_n$ be the set (ordered by inclusion) of subsets D of X such that $d(x, y) \geq \frac{1}{n}$ for distinct $x, y$ in D. The set $\mathfrak{F}_n$ is of finite character, hence possesses a maximal element $D_n$ (S, III, § 4, No. 5). Then for all $y \in X$ , there exists a point $x$ in $D_n$ such that $d(x, y) < \frac{1}{n}$ , by virtue of the maximal character of $D_n$ .

Put $D = \bigcup_n D_n$ ; the set D is then dense in X, and since X does not satisfy the first axiom of countability, D is not countable, and so one of the $D_n$ is not countable.

Q.E.D.

By lemma 1, applied to $B_0$ , there exists an uncountable subset $A_0$ of $F_0$ and a number $\varepsilon > 0$ such that $\|x\| \leq 1$ and $\|x - y\| \geq \varepsilon$ for distinct $x, y$ in $A_0$ . We have $\mathrm{pr}_0(E) = F_0$ and hence there exists a subset A in E such that $\mathrm{pr}_0$ induces a bijection from A onto $A_0$ .

Lemma 2

There exists a sequence $(x_m)_{m \geq 0}$ consisting of distinct elements of A, which is bounded in E.

We shall construct a sequence $(x_m)_{m \geq 0}$ of points of A by induction; and a decreasing sequence $(C_m)_{m \geq 0}$ of subsets of A satisfying the following conditions : a) None of the sets $C_m$ is countable. b) For every $n \geq 0$ , the set $\mathrm{pr}_k(C_m)$ is bounded in $F_k$ for $0 \leq k \leq m$ . c) For every $m \geq 0$ , we have $x_m \in C_m - C_{m+1}$ .

We put $C_0 = A$ . Suppose the sets $C_m$ for $0 \leq m \leq n$ have been defined, so as to satisfy a) and b) for $0 \leq m \leq n$ , and also the points $x_m$ in $C_m - C_{m+1}$ for $0 \leq m < n$ . For every integer $r \geq 1$ , let $C_{n,r}$ be the set of all $x \in C_n$ such that

r - 1 \leq \| \mathrm{pr}_{n+1}(x) \| < r .

Since $C_n$ is not countable, there exists an integer $r \geq 1$ such that $C_{n,r}$ is not countable. We choose a point $x_n$ in $C_{n,r}$ and put $C_{n+1} = C_{n,r} - \{ x_n \}$ . Evidently $C_{n+1} \subset C_n$ and $x_n \in C_n - C_{n+1}$ , the set $C_{n+1}$ is not countable and $\mathrm{pr}_k(C_{n+1})$ is bounded in $F_n$ for $0 \leq k \leq n + 1$ .

We have $x_m \in C_m$ , and so $x_m \in C_n$ where $m \geq n$ . The projection of the sequence $(x_m)_{m \geq 0}$ on $F_n$ is therefore bounded for all $n \geq 0$ ; in other words, the sequence $(x_m)_{m \geq 0}$ is bounded in E, and this establishes lemma 2.

Q.E.D.

With the notations of lemma 2, the bounded sequence $(x_m)_{m \geq 0}$ has a limit point $y$ in E. Therefore the sequence $(\mathrm{pr}_0(x_m))_{m \geq 0}$ has a limit point $\mathrm{pr}_0(y)$ in $F_0$ , but this contradicts the construction of $A_0$ .

Corollary

— Let E be a metrizable Montel space. Then there exists a countable dense set in the strong dual of E.

On the dual E' of E, the strong topology is identical with that of compact convergence, since E is a Montel space. It is now enough to apply cor. 1 of prop. 6 of III, p. 18.

We can show that the strong dual of a metrizable Montel space E is not metrizable if E is infinite dimensional (IV, p. 57, exerc. 1).

Exercises

See the exercises for § 2.