����������� ������ 45 (1993), 49–53

UDC 517.982.2

���������� ������ ��� research paper

A NOTE ON A SUPPORT OF A LINEAR MAPPING

Lj. Čukić

Abstract. In this note a notion of the support of a linear mapping from C_{b}(T) into a locally convex space is introduced. Some of its properties are established.

Introduction

If E is a locally convex space and P ⊂ E′ is a weakly-∗-bounded set, then P is equicontinuous iff the linear mapping f from E into the Banach space C_{b}(P), defined by f(e)(p) = p(e) (p ∈ P, e ∈ E) is continuous. For the case E = (C_{b}(T), β_{t}), as we will see, some information concerning the continuity of the mapping f is provided by its support.

Preliminaries

All topological spaces considered here are assumed to be completely regular Hausdorff. If T is such a spce, then C_{b}(T) (resp. C(T)) denotes the space of bounded (resp. all) real-valued continuous functions on T. βT is the Stone-Čech compactification of T. For each x ∈ C_{b}(T) its continuous extension to βT is denoted by x^{β}. If x ∈ C(βT) and if A ⊂ βT, then x|A denotes the restriction of x to A. cl_{X} A is the closure of A ⊂ X.

We denote by ∥ ∥ supremum norm on C_{b}(T), and by B the unit ball { x ∈ C_{b}(T) : ∥x∥ ⩽ 1 }. M(T) is the Banach space dual to (C_{b}(T), ∥ ∥). If H ⊂ C_{b}(T) (or if H ⊂ M(T)), then H^{+} denotes the set { h ∈ H : h ⩾ 0 }. For such H, if h ∈ H, then h^{+} = sup{h, 0}, h^{−} = sup{−h, 0}, |h| = h^{+} + h^{−}.

Let t_{co} be the compact-open topology on C_{b}(T), i.e. t_{co} is the locally convex topology on C_{b}(T) defined by the family of seminorms p_{K}(x) = sup{ |x(t)| : t ∈ K }, K runs through the compact subsets of T. Then, the strict topology β_{t} on C_{b}(T) is the finest locally convex topology on C_{b}(T) coinciding with t_{co} on the unit ball B ([2],[6]). From definition of β_{t} immediately follows that if f is a linear mapping from C_{b}(T) into an LCS (a locally convex Hausdorff space) then f is β_{t}- continuous iff its restriction f|B is t_{co} - continuous. M_{t}(T) denotes the continuous dual of (C_{b}(T), β_{t}).

Results

A well-known theorem of Nachbin (see [5], III.1.2) says that if F ⊂ C(T) is absolutely convex and if εB ⊂ F for some ε > 0, then there is a minimal compact set K ⊂ βT with the property: if x ∈ C(T) and if x^{β}|K = 0, then x ∈ F. We prove the following variant of Nachbin’s theorm.

Theorem 1. If F ≠ {0} is a non-empty norm-closed absolutely convex subset of C_{b}(T), then there is a minimal compact set S(F) ⊂ βT with the property: if x ∈ C_{b}(T) and if x^{β}|S(F) = 0, then x ∈ F.

Proof. Let L = { L ⊂ βT : L is compact such that (∀x ∈ C_{b}(T))x^{β}|L = 0 ⇒ x ∈ F } and let M_{K} = { x ∈ C_{b}(T) : there exists an open G ⊃ K with x^{β}|G = 0 }, for compact K ⊂ βT. Then: (1) L ∈ L iff M_{L} ⊂ F; (2) if L_{1}, L_{2} ∈ L, then L_{1} ∩ L_{2} ∈ L; (3) S(F) = ⋂{ L : L ∈ L }. Proofs of (2) and (3) are the same as in [5], pp. 63–64. One half of (1) is trivial. To obtain the other half, suppose that M_{L} ⊂ F and x ∈ C_{b}(T), x^{β}|L = 0. Let y_{n}(t) = x^{β}(t) if |x^{β}(t)| < 1/n and y_{n}(t) = x^{β}(t)/(n|x^{β}(t)|) if |x^{β}(t)| ⩾ 1/n. Then y_{n} ∈ C(βT) and (x^{β} − y_{n})|G_{n} = 0, for G_{n} = { t ∈ βT : |x^{β}(t)| < 1/n }. From L ⊂ G_{n} it follows that (x^{β} − y_{n})|T ∈ F for each n = 1, 2,…. Then x ∈ F, because F is closed and ∥y_{n}|T ∥ ⩽ 1/n.

Remark 2. If F is as in theorem 1 and if F is norm-bounded, then S(F) = βT. In fact, if t ∈ βT \ S(F), then there is x ∈ C_{b}(T) with x^{β}(t) = 1, x^{β}|S(F) = 0. Hence nx ∈ F, because nx^{β}|S(F) = 0 (n = 1, 2, …), i.e. F is not norm-bounded.

Definition 3. Let f ≠ 0 be a norm-continuous linear mapping from C_{b}(T) into an LCS E. The big support of f is bsupp f = S(f^{−1}(0)) and the support of f is supp f = bsupp f ∩ T.

Remark 4. If f is a norm-continuous linear functional on C_{b}(T), then f can be identified, via Alexandroff representation theorem ([6], 5.1) with the unique Baire measure μ on the minimal algebra which contains all zero sets from T. It is not difficult to see that supp f and supp μ coincide.

In the light of the preceeding remark, next result is not new, but we give a proof which is independent from the measure theory.

Proposition 5. Let f ∈ M^{+}(T) and f ≠ 0. Then:

(a) If x ∈ C_{b}^{+}(T) and f(x) = 0, then x^{β}| bsupp f = 0.

(b) The space bsupp f with the induced topology satisfies the countable chain condition.

Proof. (a) Let x^{β}(s) > 0 for some s ∈ bsupp f. Then there exist an open set G ⊂ βT and r > 0 with x^{β}(t) > r for all t ∈ G. We will prove that bsupp f is contained in βT \ G, which is impossible because s ∈ bsupp f ∩ G. Let y ∈ C_{b}(T), y^{β}|βT \ G = 0 and ∥y∥ < k. From x^{β}(t) > r(y^{±})^{β}(t)/k for all t ∈ βT and from non-negativity of f it follows that f(y^{±}) = 0. Then f(y) = f(y^{+}) − f(y^{−}) = 0. Hence bsupp f ⊂ βT \ G, by the minimality of bsupp.

(b) Let S = bsupp f and let the functional g_{0} on C_{b}^{+}(S) be defined by g_{0}(x) = f(x̄), where x̄ is any non-negative continuous extension of x ∈ C_{b}^{+}(S) on βT. The functional g_{0} is well-defined because each two such extensions coincide on S. It is trivial to see that g_{0} is a non-negative additive functional, and by [1, Chap.II, §2, Prop.2] there is a non-negative linear functional g on C_{b}(S) that extends g_{0}. By [4, V.5.5], g ∈ M^{+}(S).

Let { G_{α} : α ∈ A } be a family of non-empty pairwise disjoint open subsets of S and let t_{α} ∈ G_{α}. Then there are x_{α} ∈ C_{b}^{+}(S), x_{α} ⩽ 1, such that x_{α}(t_{α}) = 1 and x_{α}|S \ G_{α} = 0. From 0 ⩽ ∑_{Φ} x_{α} ⩽ 1 on S it follows that 0 ⩽ ∑_{Φ} g(x_{α}) ⩽ g(1) for all finite Φ ⊂ A. Then the set {α ∈ A : g(x_{α}) ⩾ g(1)/n } is finite for each n ∈ N. Countability of A then follows from the inequality g(x_{α}) > 0 (by (a)).

Theorem 6. Let E be a metrizable LCS, let (U_{n}) be its neighborhood basis of origin consisting of absolutely convex sets with 2U_{n+1} ⊂ U_{n}, and let f ≠ 0 be a norm-continuous linear mapping from C_{b}(T) into E. Then f is β_{t}-continuous if and only if there are compact sets L_{n} ⊂ T (n ∈ N) with the property that f(x) ∈ U_{n}, whenever x ∈ B^{+} and x|L_{n} = 0. Moreover, L_{n}’s may be chosen such that supp f = cl_{T} (⋃_{n=1}^{∞} L_{n}).

Proof. ⟹ The restriction f|B is t_{co}-continuous. Then there are an increasing sequence of compact sets K_{n} ⊂ T and a decreasing sequence ε_{n} of positive numbers with the property: if x ∈ B and p_{K_{n}}(x) < ε_{n}, then f(x) ∈ U_{n}. We will first prove that ⋃_{n=1}^{∞} K_{n} ∩ bsupp f ≠ ∅. Suppose the contrary. Then, there are x_{n} ∈ B^{+} such that x_{n}|K_{n} = 0, x_{n}| bsupp f = 1. There is u ∈ B such that f(u) ≠ 0. From (ux_{n})^{β}|K_{n} = 0, (ux_{n})^{β}| bsupp f = u^{β}| bsupp f it follows that f(u) = f(ux_{n}) ∈ U_{n} for all n, which is in contradiction with f(u) ≠ 0.

Hence, there is k ∈ N such that K_{n} ∩ bsupp f is non-empty for all n ⩾ k. Let L_{n} = K_{n+k} ∩ bsupp f, δ_{n} = ε_{n+k} and let x ∈ B^{+}, x|L_{n} = 0. If K_{n+k} ⊂ G_{n} = { t ∈ βT : x^{β}(t) < δ_{n} } then f(x) ∈ U_{n+k} ⊂ U_{n}. If K_{n+k} ⊄ G_{n}, then from L_{n} ⊂ G_{n} it follows that there is y ∈ B^{+} with y^{β}|K_{n+k} ∩ (T \ G_{n}) = 0, y^{β}| bsupp f = 1. Since K_{n+k} = (L_{n} ∪ (K_{n+k} \ (T \ G_{n}))) ∪ (K_{n+k} ∩ (G_{n} \ L_{n})), then p_{K_{n+k}}(xy) < δ_{n}.

From this and from the fact that x^{β} and (xy)^{β} coincide on bsupp f it follows that f(x) = f(xy) ∈ U_{n+k} ⊂ U_{n}.

For the equality supp f = cl_{T} (⋃_{n=1}^{∞} L_{n}), only inclusion bsupp f ⊂ cl_{βT} (⋃_{n=1}^{∞} L_{n}) needs a proof. If z ∈ C_{b}(T) and z^{β}| cl_{βT} (⋃_{n=1}^{∞} L_{n}) = 0, then (z^{±}/∥z∥) ∈ B^{+} and (z^{±}/∥z∥)|L_{n} = 0. It follows that f(z^{±}) ∈ ∥z∥U_{n} for all n, i.e. f(z) = 0. By the minimality of bsupp f, the proof is finished.

⟸ Since f is norm-continuous, we may choose positive numbers a_{n} < 1 so that 4f(a_{n}B) ⊂ U_{n+1} for each n. We will show that f(V_{n} ∩ B) ⊂ U_{n}, where V_{n} is the set { x ∈ C_{b}(T) : p_{L_{n+2}}(x) < a_{n} }. Let x ∈ V_{n} ∩ B and let

y^{+}(t) = \{ \begin{matrix}x^{+}(t), & if x^{+}(t) < a_{n}, \\ a_{n}, & if x^{+}(t) ⩾ a_{n}, \end{matrix} y^{−}(t) = \{ \begin{matrix}x^{−}(t), & if x^{−}(t) < a_{n}, \\ a_{n}, & if x^{−}(t) ⩾ a_{n}. \end{matrix}

Then x^{±} − y^{±} ∈ B^{+}, y^{±} ∈ a_{n}B, (x^{±} − y^{±})|L_{n+2} = 0, and so f(x^{±} − y^{±}) ∈ U_{n+2} and 4f(y^{±}) ∈ U_{n+1}. From this it follows that f(x) = f(x^{+}) − f(x^{−}) = (f(x^{+} − y^{+}) + f(y^{+})) − (f(x^{−} − y^{−}) + f(y^{−})) ∈ 2U_{n+2} + \frac{1}{2}U_{n+1} ⊂ U_{n}, which completes the proof of the theorem.

Remark 7. If E is a non-metrizable LCS, then supp f need not be the closure of a σ-compact subset of T, as the following example shows. Let T be the discrete space, card T = c. Then T is a realcomplete ([3,11.D.(a)]) metrizable space. By [5,III.3.5 and III.4.3] E = (C(T), t_{co}) is a bornological barrellled complete LCS. The inclusion mapping i from C_{b}(T) into E is β_{t}-continuous and from remark 2 it follows that supp i = T. Each compact subset of T is finite, hence cl_{T} (⋃_{n=1}^{∞} L_{n}) = ⋃_{n=1}^{∞} L_{n} ≠ T for all compact L_{n}’s.

Remark 8. In the proof of theorem 6 we showed also that cl_{T} ⋃_{n=1}^{∞} L_{n} is dense in bsupp f. Hence, supp f is dense in bsupp f.

The next lemma is well-known and we omit the proof.

Lemma 9. Let f_{n} ∈ M_{t}^{+}(T), ∥f_{n}∥ ⩽ 1 and let f = ∑_{n=1}^{∞} 2^{−n}f_{n}. Then f ∈ M_{t}^{+}(T), ∥f∥ ⩽ 1 and supp f = cl_{T} (⋃_{n=1}^{∞} supp f_{n}).

Theorem 10. Let f ≠ 0 be a weakly continuous linear mapping from (C_{b}(T), β_{t}) into an LCS E. Then:

(a) If F ⊂ E′ is weakly-∗-dense in E′, then supp f = cl_{T} (⋃_{w∈F} supp(wf)).

(b) supp f is dense in bsupp f.

(c) If E′ is weakly-∗-separable, then there is μ ∈ M_{t}^{+}(T), ∥μ∥ ⩽ 1 such that supp f = supp μ.

(d) If E′ is weakly-∗-separable, then supp f satisfies the countable chain condition.

Proof. (a) From f^{−1}(0) ⊂ (wf)^{−1}(0) and the theorem 1 it follows that bsupp(wf) ⊂ bsupp f for each w ∈ F. On the other hand, if x^{β}| cl_{βT} (⋃_{w∈F} supp(wf)) = 0, then by the remark 8, x^{β}| bsupp(wf) = 0. This gives that wf(x) = 0 for all w ∈ F. Hence f(x) = 0. From the theorem 1 it follows that bsupp f ⊂ cl_{βT} (⋃_{w∈F} supp(wf)).

(b) Immediately follows from (a).

(c) Let { w_{n} : n ∈ N } be weakly-∗-dense in E′. Then supp f = cl_{T} (⋃_{n=1}^{∞} supp(w_{n}f)), by (a). If μ = ∑_{n∈N_{1}} 2^{−n}(|w_{n}f|/∥w_{n}f∥), where N_{1} = { n : w_{n}f ≠ 0 }, then μ ∈ M_{t}^{+}(T) and supp f = supp μ, by the lemma 9.

(d) From (b) and (c) it follows that supp f is dense in bsupp μ. Then, by the proposition 5, from [3, 2.J.(d)] it follows that supp f satisfies the countable chain condition.

Remark 11. Assertions in (c), (d) are not true if we omit the separability condition, even if E is a Banach space. For example, let T be the compact space βN \ N and let f be the identity mapping on (C(T), ∥ ∥). Then supp f = T, but T does not satisfy the countable chain condition [3, 3.6.Example 2].

Applications of our results will be given in a subsequent paper.

REFERENCES

[1] N. Bourbaki, Integration, Chap. I–IV, 2-eme ed., Hermann, Paris 1965

[2] J. B. Cooper, Saks spaces and applications to functional analisys, North-Holland, Amsterdam-New York-Oxford, 1978

[3] R. Engelking, General Topology, PWN, Warszawa, 1977

[4] H. Schaefer, Topological Vector Spaces, Macmillan, New York, 1966

[5] J. Schmets, Espaces de fonctions continues, Springer, Berlin-Heidelberg-New York, 1976

[6] R. F. Wheeler, A survey of Baire measures and strict topologies, Expo. Math. 2 (1983), 97–190

(received 03 06 1993)

Department of Mathematics, Faculty of Civil Engineering, Bul. Revolucije 73, 11000 Beograd, Yugoslavia