§ 1. Bornology in a topological vector space

1. Bornologies

Definition 1

— A bornology on a set $E$ is a subset $\mathscr{B}$ of the set of all subsets of $E$ satisfying the following conditions (cf. GT, X, § 1.2, Remark 2). (B1) Every subset of a set of $\mathscr{B}$ belongs to $\mathscr{B}$ . (B2) Every finite union of a set of $\mathscr{B}$ belongs to $\mathscr{B}$ .

We say that $\mathscr{B}$ is covering if every element of $E$ is contained in a set which belongs to $\mathscr{B}$ , or, which is the same, if $\mathscr{B}$ is a cover of $E$ .

Example

— Let E be a metric space; the set of all bounded subsets of E (GT, IX, § 2, No. 2) is a covering bornology on E. Let G be the group of isometries of E; the set of all subsets M of G such that for every $x \in E$ , the set $M.x$ is a bounded subset of E, is a covering bornology on G.

If $\mathscr{B}$ is a bornology on a set E, a subset $\mathscr{B}_1$ of $\mathscr{B}$ is said to be a base of $\mathscr{B}$ if every set of $\mathscr{B}$ is contained in a set of $\mathscr{B}_1$ .

The intersection of a family of bornologies on E is a bornology; consequently for every subset $\mathfrak{S}$ of $\mathfrak{P}(E)$ , there exists a smallest bornology containing $\mathfrak{S}$ ; this bornology is said to be generated by $\mathfrak{S}$ and admits as a base the set of finite unions of sets of $\mathfrak{S}$ . If E and E' are two sets, and $\mathscr{B}$ (resp. $\mathscr{B}'$ ) a bornology on E (resp. E'), the product bornology is the bornology on $E \times E'$ which admits the sets $M \times M'$ as a base, where $M \in \mathscr{B}$ and $M' \in \mathscr{B}'$ .

Definition 2

Let $E$ be a vector space. A bornology $\mathscr{B}$ on $E$ is said to be convex, if for every $X \in \mathscr{B}$ and $t \in K$ , the homothetic set $tX$ and the convex balanced envelope $\Gamma(X)$ (II, p. 10) of $X$ belong to $\mathscr{B}$ .

If $X$ and $Y$ are two subsets of $E$ , we have

X + Y \subset 2\Gamma(X \cup Y)

\lambda X \subset t\Gamma(X) \quad \text{for} \quad |\lambda| \leq t .

Consequently, if $\mathscr{B}$ is a convex bornology on $E$ , if $A$ is a bounded subset of $K$ and if $X, Y$ belong to $\mathscr{B}$ , then $X + Y \in \mathscr{B}$ and $A.X \in \mathscr{B}$ .

2. Bounded subsets of a topological vector space

Definition 3

Let $E$ be a topological vector space. A subset $A$ of $E$ is said to be bounded if it is absorbed by every neighbourhood of $0$ in $E$ (I, p. 7, def. 4).

In order that $A$ be bounded, it is sufficient that $A$ be absorbed by every neighbourhood of a fundamental system of neighbourhoods of $0$ . Since there exists a fundamental system of balanced neighbourhoods of $0$ (I, p. 7, prop. 4), this is the same as saying that, for every neighbourhood $V$ of $0$ in $E$ , there exists $\lambda \in K$ such that $A \subset \lambda V$ .

Suppose the topology of $E$ is defined by a fundamental system $\Gamma$ of semi-norms (II, p. 3); then a subset $A$ of $E$ is bounded if and only if every semi-norm $p \in \Gamma$ is bounded on $A$ .

In particular, if $E$ is a semi-normed space, a subset $A$ of $E$ is bounded if and only if it is contained in a ball. In other words, if $E$ is normed this means that $A$ is bounded for the metric space structure of $E$ (GT, IX, § 2, No. 2).

Remark

— 1) If $E$ is a semi-normed space, the balls form a fundamental system of bounded neighbourhoods of $0$ in $E$ . Conversely, if $E$ is a locally convex topological vector space, and if there exists a bounded neighbourhood of $0$ in $E$ , this neighbourhood contains a convex balanced neighbourhood $W$ , and the gauge of $W$ is then a semi-norm defining the topology of $E$ .

Thus, if $E$ is locally convex and metrizable, and if its topology cannot be defined by a single norm, then there exists no distance on $E$ defining its topology and such that the bounded subsets for $d$ (GT, IX, § 2, No. 2) are the bounded subsets of $E$ . More precisely, for every distance $d$ on $E$ , which is translation invariant and which defines the topology of $E$ , the bounded subsets of $E$ are bounded for $d$ (III, p. 38, exerc. 3), but the converse is false.

2) Let $M$ be a vector subspace of $E$ endowed with the induced topology. In order that a subset of $M$ be bounded in $M$ , it is necessary and sufficient that it be bounded in $E$ .

3) Let $N$ be the intersection of all neighbourhoods of $0$ in $E$ , so that $\tilde{E} = E/N$ is the Hausdorff vector space associated with $E$ . Then $N$ is bounded; if $\pi : E \to \tilde{E}$ is the canonical homomorphism then a subset $B$ of $E$ is bounded if and only if $\pi(B)$ is bounded.

4) Let $E$ be a Hausdorff locally convex space; then for every $x \neq 0$ in $E$ , there exists a continuous semi-norm $p$ such that $p(x) \neq 0$ ; this semi-norm is not bounded on the real half-line $\mathbf{R}_+ . x$ generated by $x$ . Hence no non-null subspace of $E$ is bounded. In particular, a bounded subset does not contain any line.

Definition 4

Let $E$ be a locally convex space. A bornology $\mathscr{B}$ on $E$ is said to be adapted to $E$ , if it is convex, is composed of bounded subsets of $E$ and if the closure of every set of $\mathscr{B}$ belongs to $\mathscr{B}$ .

Proposition 1

Let $E$ be a locally convex space. The set of bounded subsets of $E$ is an adapted bornology.

We need to establish the following properties : a) If $B$ is a bounded subset of $E$ , every subset of $B$ is bounded. b) The union of two bounded subsets is bounded. c) Every set that is homothetic to a bounded set is bounded. d) The closed convex balanced envelope (II, p. 13) of a bounded subset is bounded. If $p$ is a continuous semi-norm on $E$ , the balls of $p$ are convex, balanced, closed and the set homothetic to a ball is a ball. Hence, if $p$ is bounded on two subsets $X$ and $Y$ of $E$ , it is also bounded on the closed convex balanced envelope of $X \cup Y$ , and on the sets homothetic to these. This establishes properties $b), c)$ and d), and a) is obvious.

Definition 5

Let $E$ be a locally convex space. The set of all bounded subsets of $E$ is called the canonical bornology of $E$ .

If $\mathscr{B}$ is a set of bounded subsets of $E$ , then there exists a smallest bornology $\tilde{\mathscr{B}}$ adapted to $E$ and containing $\mathscr{B}$ . The sets of $\tilde{\mathscr{B}}$ are those that are contained in a set homothetic to the closed convex balanced envelope of a finite union of sets of $\mathscr{B}$ .

Every adapted bornology is contained in the canonical bornology.

Proposition 2

In a locally convex space $E$ , every precompact set is bounded.* Let $A$ be a precompact subset of $E$ , and $V$ be a convex balanced neighbourhood of $0$ . There exists a finite sequence $(a_i)_{1 \leq i \leq n}$ of points of $A$ such that

A \subset \bigcup_{1 \leq i \leq n} (a_i + V).

Since

B = \{a_1, ..., a_n\}

is bounded, there exists a scalar

\lambda

such that

0 < \lambda < 1

and

\lambda B \subset V

; we have

\lambda A \subset \lambda B + \lambda V \subset V + V

, from which the proposition follows.

Corollary

In a locally convex space, the set of points of a Cauchy sequence is bounded.* In fact, this set is precompact (GT, II, § 4, No. 2).

Remark 5

— In general the bounded subsets of a locally convex space $E$ are not all precompact (for example, if $E$ is an infinite dimensional normed space, its unit ball is not compact (I, p. 15, th. 3)). However, this is so if $E$ is a weak space (II, p. 42) : for the Hausdorff topological vector space associated with E is then isomorphic to a subspace of a product K^l whose bounded subsets are precompact (cf. III, p. 4, cor. 2). For other examples, see IV, p. 18.

Proposition 3

— Let A be a subset of a locally convex space E. Suppose that A is bounded; then for every sequence (xn) of points of A and for every sequence (\lambda_n) of scalars tending to 0, the sequence (\lambda_n x_n) tends to 0. Conversely, if there exists a sequence (\lambda_n) of non-zero scalars such that for every sequence (x_n) of points of A, the sequence (\lambda_n x_n) is bounded, then A is bounded.

Suppose that A is bounded. If (\lambda_n) is a sequence of scalars tending to 0, and V is a neighbourhood of 0, we have \lambda_n A \subset V whenever n is large enough, and the first assertion follows.

Conversely, if A is not bounded and if (\lambda_n) is a sequence of scalars \neq 0, then there exists a continuous semi-norm p and a sequence (x_n) of points of A, such that $p(x_n) \geq \frac{n}{|\lambda_n|}$ . We have then that $p(\lambda_n x_n) \geq n$ , and the sequence (\lambda_n x_n) is not bounded.

Corollary

— *A subset A of E is bounded if and only if every countable subset of A is bounded.

3. Image under a continuous mapping

Proposition 4

— *Let E and F be two locally convex spaces and f : E \to F a continuous mapping. Assume that f(0) = 0 and that there exists a real number m \geq 0 such that f(\lambda x) = \lambda^m f(x) for every \lambda > 0. Then, if A is a bounded subset of E, f(A) is bounded in F.

In fact, if V is a neighbourhood of 0 in F, then $f^{-1}(V)$ is a neighbourhood of 0 in E. If A is bounded in E, there exists \lambda > 0 such that A \subset \lambda f^{-1}(V) and this implies that $f(A) \subset \lambda^m V$ .

Corollary 1

— *Let E and F be two locally convex spaces, and u : E \to F be a continuous linear mapping. If A is a bounded subset of E, then u(A) is bounded in F.

Corollary 2

— Let E = \prod{i \in I} E_i be the product of a family of locally convex spaces. Then a subset of E is bounded if and only if all its projections are bounded.

More generally :

Corollary 3

— Let E be a vector space, (Fi)_{i \in I} a family of locally convex spaces and f_i a linear mapping from E into F_i (for i \in I). Suppose E is assigned the coarsest topology (locally convex) for which all the f_i are continuous (II, p. 26). Then, for a subset A of E to be bounded, it is necessary and sufficient that f_i(A) is bounded in F_i for all i \in I.

In fact, if A is bounded, so are the f_i(A) (cor. 1). Conversely, if the f_i(A) are bounded and if p is a continuous semi-norm on E, then there exists a finite subset J of I and a family $(q_j)_{j \in J}$ , where $q_j$ is a continuous semi-norm on $F_j$ , such that $p \leq \sup_{j \in J} (q_j \circ f_j)$ consequently $p$ is bounded on $A$ .

Corollary 4

Let $E_i$ ( $1 \leq i \leq n$ ) and $F$ be locally convex spaces, and $f$ be a continuous multilinear map from $\prod_{i=1}^n E_i$ into $F$ . If $B_i$ is a bounded subset of $E_i$ , for $1 \leq i \leq n$ , then $f(\prod_{i=1}^n B_i)$ is bounded in $F$ .

Corollary 5

Let $E$ and $F$ be two locally convex spaces and $u : E \to F$ be a continuous polynomial mapping. If $A$ is a bounded subset of $E$ , then $u(A)$ is bounded.

4. Bounded subsets in certain inductive limits

Proposition 5

Let $(E_i)_{i \in I}$ be a family of Hausdorff locally convex spaces, and let $E$ be the topological direct sum of this family (II, p. 29). In order that a subset $B$ of $E$ be bounded, it is necessary and sufficient that there exists a finite subset $J$ of $I$ such that $\operatorname{pr}_i(B)$ is bounded in $E_i$ for $i \in J$ and $\operatorname{pr}_i(B) \subset \{0\}$ for all $i \notin J$ .

Let $J$ be a finite subset of $I$ . Since the topology of $E$ induces the product topology on $\prod_{j \in J} E_j$ (II, p. 30, prop. 7 and p. 31, prop. 8), it follows from III, p. 4, cor. 2 that the condition is sufficient.

Conversely, let $B$ be a bounded subset of $E$ . Then $\operatorname{pr}_i(B)$ is bounded for all $i$ (III, p. 4, cor. 1). Therefore it is enough to prove that there exists a finite subset $J$ of $I$ such that $\operatorname{pr}_i(B) \subset \{0\}$ for all $i \notin J$ . If not, then there exists an infinite sequence $(i_n)$ of distinct elements of $I$ and an infinite sequence $(x_n)$ of elements of $B$ such that $\operatorname{pr}_{i_n}(x_n) \neq 0$ . Since $E_{i_n}$ is Hausdorff, there exists a continuous semi-norm $p_n$ on $E_{i_n}$ such that $p_n(\operatorname{pr}_{i_n}(x_n)) \geq n$ . Hence $p = \sum_{n \geq 1} p_n \circ \operatorname{pr}_{i_n}$ is a continuous semi-norm on $E$ and $p$ is not bounded on $B$ , which is a contradiction.

Proposition 6

Let $E$ be a locally convex space which is the strict inductive limit of an increasing sequence $(E_n)$ of closed vector subspaces of $E$ (II, p. 33). A subset $B$ of $E$ is bounded if and only if it is contained in one of the subspaces $E_n$ , and is bounded in this subspace.

The condition is sufficient, since the topology induced on $E_n$ by that of $E$ is precisely the given topology of $E_n$ (II, p. 32, prop. 9). To see that the condition is necessary, it is enough (III, p. 4, prop. 3) to prove that if a sequence $(x_m)$ of points of $E$ is not contained in any of the subspaces $E_n$ , then it cannot tend to 0. By extracting a subsequence of the sequence $(x_m)$ , we can assume that there exists a strictly increasing sequence $(n_k)$ of integers such that, for every index $k$ , we have $x_k \notin E_{n_k}$ and $x_k \in E_{n_{k+1}}$ . Then there exists (II, p. 33, lemma 2) an increasing sequence $(V_k)$ of convex sets such that $V_k$ is a neighbourhood of 0 in $E_{n_k}$ , $V_{k+1} \cap E_{n_k} = V_k$ and such that $x_k \notin V_{k+1}$ for every index $k$ . The union $V$ of the $V_k$ is then a neighbourhood of 0 in $E$ , and we have $x_k \notin V$ for all $k$ . This proves that the sequence $(x_k)$ does not tend to 0.

The conclusion of prop. 6 is not necessarily true for a space E which is the inductive limit of a non-denumerable directed set of closed subspaces of E (cf. III, p. 38, exerc. 7).

Proposition 7

— Let $(E_n)_{n \geq 0}$ be a sequence of Hausdorff locally convex spaces, and for every $n$ , let $u_n : E_n \to E_{n+1}$ be an injective linear mapping which is compact (i.e. such that there exists a neighbourhood of 0 in $E_n$ whose image under $u_n$ is relatively compact; this implies that $u_n$ is continuous). Let $E$ be the inductive limit of the system $(E_n, u_n)$ (II, p. 29), and let $v_n$ be the canonical mapping from $E_n$ into $E$ . Then the locally convex space $E$ is Hausdorff. Moreover, for every subset $A$ of $E$ , the following conditions are equivalent : (i) $A$ is bounded; (ii) there exists an integer $n$ such that $A$ is the image under $v_n$ of a bounded subset of $E_n$ ; (iii) $A$ is relatively compact. We identify $E_n$ with a vector subspace of $E$ (endowed with a topology finer than the induced topology).

Lemma 1

Under the hypothesis of prop. 7, the topology of $E$ is the finest topology for which all the mappings $v_n : E_n \to E$ are continuous.

We need to prove that, if $U$ is a subset of $E$ such that $U \cap E_n$ is open in $E_n$ for every $n$ , then $U$ is open in $E$ ; in other words, we must prove that, for every $x \in U$ , there exists a convex balanced set $V$ such that $x + V \subset U$ and that $V \cap E_n$ is a neighbourhood of 0 in $E_n$ for every large enough $n$ (II, p. 27, prop. 5). For every $n$ , let $W_n$ be a convex balanced neighbourhood of 0 in $E_n$ such that the closure $H_n$ of $W_n$ in $E_{n+1}$ is compact. Let $x \in U$ and let $n_0$ be an integer such that $x \in E_{n_0}$ . We shall construct, by induction, a sequence $(\varepsilon_n)_{n \geq 0}$ of scalars $> 0$ such that $x + \sum_{n_0 \leq i \leq n} \varepsilon_i H_i$ is contained in $U$ for $n \geq n_0$ . Suppose that the $\varepsilon_i$ for $i < n$ have been constructed. If $n = n_0$ , set $V_{n-1} = \{0\}$ ; if not, set

V_{n-1} = \sum_{n_0 \leq i \leq n-1} \varepsilon_i H_i.

Then

V_{n-1}

is compact in

E_n

, and a fortiori in

E_{n+1}

. Since

U \cap E_{n+1}

is open in

E_{n+1}

, there exists a scalar

\varepsilon_n > 0

such that

x + V_n = x + V_{n-1} + \varepsilon_n H_n

is contained in

U

(GT, II, § 4, No. 3, cor.). Let

V = \bigcup_{n \geq n_0} V_n

. Then

V

is convex and balanced; for

n \geq n_0

, we have

V \cap E_n \supset \varepsilon_n H_n \cap E_n \supset \varepsilon_n W_n

, hence

V \cap E_n

is a neighbourhood of 0 in

E_n

. This completes the proof of the lemma.

The set $U = E - \{0\}$ is such that the set $U \cap E_n = E_n - \{0\}$ is open in $E_n$ for every $n$ , hence $U$ is open in $E$ , which proves that $E$ is Hausdorff (GT, III, § 1, No. 3, prop. 2). It is clear that property (ii) implies (iii) and that (iii) implies (i). Finally we show that (i) implies (ii). For this it is enough to show that if a subset $A$ of $E$ is not absorbed by any of the sets $\sum_{0 \leq i \leq n} H_i$ , then $A$ is not bounded. But then there exists a sequence $(x_n)_{n \geq 1}$ of points of $A$ such that, for every $n$ , we have $x_n \notin n^2 \sum_{0 \leq i \leq n} H_i$ .

Then the set of the $x_n / n$ is closed by virtue of lemma 1, since its intersection with $E_m$ is discrete for every integer $m$ . The complement of the set of the $x_n / n$ is then an open neighbourhood of 0 which does not absorb the sequence $(x_n)$ , hence $A$ is not bounded.

Remark 1

With the above notations, let $F_n$ be the vector space generated by $H_n$ , with a norm equal to the gauge of $H_n$ . We shall see (III, p. 8, cor.) that $F_n$ is a Banach space. The injection from $F_n$ into $E_{n+1}$ is compact, hence a fortiori also the injection $w_n$ from $F_n$ into $F_{n+1}$ . Further, $E$ is the inductive limit of the inductive system $(F_n, w_n)$ of Banach spaces. For, a convex balanced neighbourhood $V$ of 0 in $E$ is such that $V \cap E_n$ absorbs $H_{n-1}$ for all $n \geq 1$ , and conversely, if a convex balanced set $W$ in $E$ is such that $W \cap E_n$ absorbs $H_{n-1}$ , then $W \cap E_{n-1}$ contains a set homothetic to $W_{n-1}$ for all $n \geq 1$ , and hence $W$ is a neighbourhood of 0 in $E$ .

Remark 2

Let $F$ be a locally convex space, $k$ an integer $\geq 0$ and $f : E^k \to F$ a multilinear mapping. For $f$ to be continuous, it is necessary and sufficient that the restriction of $f$ to $E_n^k$ is continuous for every $n$ . We verify immediately that $E^k$ has the final locally convex topology for the family of linear mappings $v_n \times \cdots \times v_n : E_n \times \cdots \times E_n \to E \times \cdots \times E$ (II, p. 28, cor. 2 and p. 30, prop. 7) and that $u_n \times \cdots \times u_n$ is a compact injective linear mapping from $(E_n)^k$ into $(E_{n+1})^k$ . It is now enough to apply lemma 1.

5. The spaces $E_A$ (A bounded)

Let $E$ be a locally convex space and $A$ be a convex balanced subset of $E$ . We recall that $E_A$ denotes the vector space generated by $A$ , with $p_A$ the gauge of $A$ , as the semi-norm (II, p. 26, Example 3). We verify immediately that the canonical injection of $E_A$ into $E$ is continuous if and only if $A$ is bounded. If, in addition, $E$ is Hausdorff, a bounded set $A$ does not contain a line (III, p. 2, Remark 4) and so $p_A$ is a norm (II, loc. cit.).

We shall say that a uniform space $X$ is semi-complete if every Cauchy sequence in $X$ is convergent. A complete uniform space is semi-complete; but the converse is not always true (GT, II, § 4, exerc. 4); however, a metrizable semi-complete space is complete (GT, IX, § 2, No. 6, prop. 9).

Proposition 8

Let $E$ be a Hausdorff locally convex space and let $A$ be a closed, balanced, bounded and convex subset of $E$ . Let $(x_n)$ be a Cauchy sequence in $E_A$ . Then this sequence converges in $E_A$ if and only if it converges in $E$ .

The canonical injection from $E_A$ into $E$ is continuous. Hence, if $(x_n)$ converges in $E_A$ , it converges in $E$ . Conversely, suppose $(x_n)$ converges to $y$ in $E$ . There exists an increasing sequence of integers $(n_k)$ such that $p_A(x_m - x_n) \leq 2^{-k-1}$ if $m \geq n_k$ and $n \geq n_k$ . Therefore the sequence $(x_{n_k} + 2^{-k}A)$ is decreasing. Since $A$ is closed in $E$ , we have $y \in \bigcap_k (x_{n_k} + 2^{-k}A)$ , which proves that $(x_{n_k})$ , hence $(x_n)$ , converges to $y$ in $E_A$ .

Corollary

— If A is semi-complete (in particular, complete) then EA is a Banach space.* In fact, a Cauchy sequence in E_A is also a Cauchy sequence for the topology of E and is contained in a set homothetic to A, hence converges in E.

6. Complete bounded sets and quasi-complete spaces

Definition 6

— A locally convex space E is said to be quasi-complete if every closed and bounded subset of E is complete (for the uniform structure induced by that of E). A complete locally convex space is quasi-complete, but the converse is not always true. For example, if E is an infinite dimensional Hilbert space, or more generally, an infinite dimensional reflexive Banach space, then E with its weakened topology is quasi-complete but not complete (II, p. 51, prop. 9). A quasi-complete space is semi-complete, since every Cauchy sequence is contained in a closed and bounded subset (III, p. 3, corollary and prop. 1). In particular, a locally convex metrizable and quasi-complete space is complete. In a Hausdorff quasi-complete space, the closure and the closed convex balanced envelope of a precompact subset are compact; in fact, they are precompact (II, p. 25, prop. 3), and complete being closed and bounded (III, p. 3, prop. 2).

Proposition 9

— (i) A closed vector subspace of a quasi-complete locally convex space is quasi-complete. (ii) The product of quasi-complete locally convex spaces is quasi-complete. (iii) The topological direct sum of quasi-complete locally convex spaces is quasi-complete. (iv) A locally convex space which is the strict inductive limit of a sequence of closed quasi-complete subspaces is quasi-complete. Assertion (i) follows from Remark 2 (III, p. 2), (ii) from III, p. 4, cor. 2, (iii) from prop. 5 (III, p. 5) and (iv) from prop. 6 (III, p. 5).

We shall see later that the quotient space of a quasi-complete locally convex space by a closed vector space is not necessarily quasi-complete (IV, p. 63, exerc. 10).

Proposition 10

— Let E be a locally convex space, M a vector subspace of E such that every point of E is in the closure of a bounded subset of M. Then every continuous linear mapping f from M into a Hausdorff quasi-complete locally convex space F uniquely extends to a continuous linear mapping from E into F. The hypothesis implies that M is dense in E, hence f extends uniquely to a continuous linear mapping $\hat{f}$ from E into the completion $\hat{F}$ of F (GT, III, § 3, No. 4, corollary). But every $x \in F$ lies in the closure of a bounded subset B of M; hence $\hat{f}(x)$ is in the closure of $f(B)$ in $\hat{F}$ . But $f(B)$ is bounded in F, hence its closure in F is complete, and coincides with its closure in $\hat{F}$ . This proves that $\hat{f}(x) \in F$ .

7. Examples

a) Let X be a topological space. Let $\mathscr{R}(X)$ , the vector space of numerical (finite) functions on X be assigned the topology of compact convergence (GT, X, § 1, No. 3): this is the coarsest topology for which the restriction mappings $\mathscr{R}(X) \to \mathscr{R}(H)$ are continuous (where H runs through the family of compact subsets of X and where $\mathscr{R}(H)$ is assigned the topology of uniform convergence). Cor. 3 of III, p. 4 shows that a subset A of $\mathscr{R}(X)$ is bounded if and only if, for every compact subset H of X, the set of restrictions to H of functions belonging to A is uniformly bounded.

*b) (Spaces of infinitely differentiable functions.) Let $n \geqslant 1$ be an integer. For every open set U in $\mathbf{R}^n$ , let $\mathscr{C}^\infty(U)$ denote the vector space of infinitely differentiable functions on U (VAR, R, 2.3). Let f be in $\mathscr{C}^\infty(U)$ . For every multi-index $\alpha = (\alpha_1, ..., \alpha_n)$ in $\mathbf{N}^n$ , $\partial^\alpha f$ denotes the partial derivative $\partial^{|\alpha|} f / \partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}$ ; this is a continuous function on U (VAR, R, 2.3 and 2.4). For every integer $m \geqslant 0$ , and every compact subset H of U, set

p_{m,H}(f) = \sup_{\substack{|\alpha| \leqslant m \\ x \in H}} |\partial^\alpha f(x)| .

Then $p_{m,H}$ is a semi-norm on $\mathscr{C}^\infty(U)$ .

Let $\mathscr{C}^\infty(U)$ be assigned the topology defined by the semi-norms $p_{m,H}$ . This is the coarsest of the topologies for which the mappings $f \to \partial^\alpha f$ from $\mathscr{C}^\infty(U)$ into $\mathscr{R}(U)$ are continuous, where $\mathscr{R}(U)$ is assigned the topology of compact convergence. There exists an increasing sequence of compact subsets $(H_n)_{n \geqslant 0}$ of U whose interiors cover U; the family of semi-norms $p_{m,H_n}$ defines the topology of $\mathscr{C}^\infty(U)$ , which then becomes a locally convex metrizable space. The space $\mathscr{C}^\infty(U)$ is complete; in other words, it is a Fréchet space (II, p. 24): in fact, let $(f_k)$ be a Cauchy sequence in $\mathscr{C}^\infty(U)$ ; for every $\alpha \in \mathbf{N}^n$ , the sequence $(\partial^\alpha f_k)$ converges in the complete space $\mathscr{R}(U)$ (TG, X, § 1, No. 5, th. 1) to a continuous function $g_\alpha$ . By induction on $|\alpha|$ , we deduce from th. 1 of FVR, II, p. 2 that $g_\alpha = \partial^\alpha g_0$ for every $\alpha \in \mathbf{N}^n$ . In other words, the sequence $(f_k)$ converges to $g_0$ in $\mathscr{C}^\infty(U)$ .

Let A be a subset of $\mathscr{C}^\infty(U)$ . In order that A be bounded, it is necessary and sufficient that the number $\sup_{f \in A} p_{m,H}(f)$ is finite for every integer $m \geqslant 0$ and for every compact subset H of U; this condition means that for every $\alpha \in \mathbf{N}^n$ , the set of functions $\partial^\alpha f|H$ for $f \in A$ is uniformly bounded for every compact $H \subset U$ .

Let $H \subset U$ be compact. We denote by $\mathscr{C}_H^\infty(U)$ the subspace of $\mathscr{C}^\infty(U)$ consisting of those functions whose support lies in H. The space $\mathscr{C}_c^\infty(U)$ of infinitely differentiable functions with compact support in U is the directed increasing union of subspaces $\mathscr{C}_H^\infty(U)$ where H runs through the family of compact subsets of U. Each space $\mathscr{C}_H^\infty(U)$ will be assigned the topology induced by that of $\mathscr{C}^\infty(U)$ , and $\mathscr{C}_c^\infty(U)$ with the corresponding inductive limit topology. If the sets $H_n$ are such that their interiors form a cover of U, then the space $\mathscr{C}_c^\infty(U)$ is the strict inductive limit of the Fréchet spaces $\mathscr{C}_{H_n}^\infty(U)$ ; it is therefore complete (II, p. 32, prop. 9) and every bounded subset of $\mathscr{C}_c^\infty(U)$ is contained in one of the subspaces $\mathscr{C}_{H_n}^\infty(U)$ (III, p. 5, prop. 6). \*

c) (Gevrey’s spaces.) Let I be a compact interval in $\mathbf{R}$ . For every integer $n \geqslant 0$ , $D^n f$ denotes the $n$ th derivative of a numerical function $f$ defined on I (whenever this derivative exists). Let $s \geqslant 1$ and $M \geqslant 0$ be two real numbers. Let $\mathscr{G}_{s,M}(I)$ denote the vector space of those infinitely differentiable functions $f$ on I (FVR, I, p. 28) for which the sequence $(|D^n f|/M^n(n!)^s)_{n \geqslant 0}$ is bounded in the space $\mathscr{C}(I)$ of all continuous functions on I (with the topology of uniform convergence). The space $\mathscr{G}_{s,M}(I)$ is a Banach space with the norm

\|f\|_{s,M} = \sup_{n \geqslant 0, x \in I} |D^n f(x)|/M^n(n!)^s .

For $M \leqslant M'$ , we have $\mathscr{G}_{s,M}(I) \subset \mathscr{G}_{s,M'}(I)$ , and

\|f\|_{s,M'} \leqslant \|f\|_{s,M}

for every $f \in \mathscr{G}_{s,M}(I)$ . Let $\mathscr{G}_s(I)$ denote the union of the spaces $\mathscr{G}_{s,M}(I)$ and endow it with the inductive limit topology of the topologies of $\mathscr{G}_{s,M}(I)$ .

Let $M < M'$ and let B be the unit ball (closed) in $\mathscr{G}_{s,M}(I)$ . We will prove that B is a compact subset of the Banach space $\mathscr{G}_{s,M'}(I)$ . It is clear that B is closed in $\mathscr{G}_{s,M'}(I)$ and so it is enough to prove that B is precompact in $\mathscr{G}_{s,M'}(I)$ . Let $\varepsilon > 0$ and let N be a positive integer such that $(M/M')^N \leqslant \varepsilon/2$ . Let $k$ be a positive integer; the set of all functions $D^{k+1} f$ , as $f$ ranges over B, is bounded in $\mathscr{C}(I)$ , hence the set of all functions $D^k f$ , as $f$ ranges over B, is relatively compact in $\mathscr{C}(I)$ : this follows from the theorem of finite increments (FVR, I, p. 23, cor. 1) and from Ascoli’s theorem (GT, X, § 2, No. 5). We define a norm on $\mathscr{G}_{s,M}(I)$ by

q(f) = \sup_{0 \leqslant n \leqslant N, x \in I} |D^n f(x)|/M^n(n!)^s .

The above argument shows that B is precompact for the topology associated with the norm $q$ ; in other words, there exists a finite subset C of B such that for every $f \in B$ , there exists $g \in C$ such that $q(f - g) \leqslant \varepsilon$ . Finally, for every $n > N$ , we have

|D^n f(x) - D^n g(x)|/M^n(n!)^s \leqslant 2(M/M')^n \leqslant \varepsilon ,

from which we get $\|f - g\| \leqslant \varepsilon$ . This proves that B is precompact in $\mathscr{G}_{s,M'}(I)$ .

The space $\mathscr{G}_s(I)$ is the inductive limit of the spaces $\mathscr{G}_{s,k}(I)$ as $k$ ranges over $\mathbf{N}$ ; by prop. 7 (III, p. 6) every bounded subset of $\mathscr{G}_s(I)$ is contained in one of the spaces $\mathscr{G}_{s,k}(I)$ and is relatively compact in this space.

\* d) (Spaces of holomorphic functions.) Let $n \geqslant 1$ be an integer. For every open subset U of $\mathbf{C}^n$ , $\mathscr{H}(U)$ denotes the space of functions holomorphic in U, and is assigned the topology of compact convergence in U. For every compact subset L of $\mathbf{C}^n$ , $\mathscr{H}(L)$ denotes the space of germs of holomorphic functions in a neighbourhood of L; we endow this space with the finest locally convex topology for which the canonical mappings $\pi_U : \mathscr{H}(U) \to \mathscr{H}(L)$ are continuous, where U ranges over the set of open neighbourhoods of L.

For every integer $m \geq 1$ , let $U_m$ be the set of points of $\mathbf{C}^n$ which are at a distance $< 1/m$ from L. It can be shown that the canonical mapping $\pi_{U_m}$ from $\mathscr{H}(U_m)$ into $\mathscr{H}(L)$ is injective, and that the restriction mapping from $\mathscr{H}(U_m)$ into $\mathscr{H}(U_p)$ is compact for $p \geq m$ . We can then apply prop. 7 (III, p. 6). Let A be a bounded subset of $\mathscr{H}(L)$ ; then there exists an integer $m \geq 1$ such that A consists of germs of functions in a neighbourhood of L, belonging to a bounded set B in $\mathscr{H}(U_m)$ . Moreover, a mapping $\phi$ from $\mathscr{H}(L)$ into a topological space T is continuous if and only if the mapping $\phi \circ \pi_U$ from $\mathscr{H}(U)$ into T is continuous for every open neighbourhood U of L. \*

Exercises

See the exercises for § 1.