Applied Categorical Structures 8: 579–593, 2000. © 2000 Kluwer Academic Publishers. Printed in the Netherlands.

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p-Banach Spaces and p-Totally Convex Spaces. II RALF KEMPER Fachbereich Mathematik, Fernuniversität, D-58084 Hagen, Germany (Received: 30 October 1997; accepted: 8 April 1998) Abstract. We show that the results about the set S := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z ∼ β 1/p y + (1 − β)1/p z}, where x, y, z elements of a p-absolutely convex space D and ‘∼’ is a congruence relation on D are the best possible. Finally, we give an explicit construction of the left b p : Banp → TCp (resp. b p,fin : Vecp → ACp ). adjoint of the comparison functor Mathematics Subject Classifications (2000): 18C05, 18C20. Key words: p-Banach space, p-normed vector space, p-totally convex space, p-absolutely convex space, comparison functor, congruence relation, Eilenberg–Moore category, monad.

Introduction In [7] and [8] Pumplün and Röhrl introduced the category TC (resp. TCfin ) of (finitely) totally convex (t.c.) spaces, which are the Eilenberg–Moore algebras of the monad induced by the unit ball functor from the category of Banach spaces (resp. normed vector spaces) with linear contractions to the category of sets. In accordance with [1] we use the term ‘absolutely convex’ (a.c.) for the spaces Pumplün and Röhrl call ‘finitely totally convex’; AC denotes the category of absolutely convex spaces. In [4] the categories TCp of p-totally convex and ACp of p-absolutely convex spaces are introduced and investigated (0 < p < 1). The notion of p-absolutely convex spaces is a generalization of p-absolutely convex subsets of K-vector spaces and p-totally convex spaces are a generalization of p-totally convex subsets of topological K-vector spaces, K = R, C. For p = 1, TCp and ACp coincide with the categories TC and AC and the results proved by Pumplün and Röhrl in [7], Sections 1–3, are contained in [4]. Furthermore, a generalization of the central Theorem 4.1 in [7] for p-absolutely (resp. p-totally) convex spaces, p < 1, is proved in [4, 4.11]. In the present paper it is shown that this generalization, i.e. the results 4.11– 4.14 in [4] are the best possible in case p < 1. Furthermore, the results proved by Pumplün and Röhrl in [7], Sections 4 and 5, are generalized to p < 1, especially b p : Banp → TCp and the left adjoints Sp and Sp,fin of the comparison functors b p,fin : Vecp → ACp are explicitely constructed.

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For a p-absolutely convex space D a so-called ‘p-norm’ k k: D → R is defined by the Minkowski functional kxk := inf{|λ| | λ ∈ (K) and there is y ∈ D with x = λy} ([4, 4.2]). Let ‘∼’ be a congruence relation on D ∈ TCp (resp. D ∈ ACp ), x, y, z, x0 , y0 , z0 ∈ D, σ , τ , γ ∈ ]0, 1] with x = σ x0 , y = τy0 and z = γ z0 . If there is an α ∈ ]0, 1[ with αx ∼ αy, then for S := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z ∼ β 1/p y + (1 − β)1/p z} in case p < 1 we have the following central result ([4, 4.11]): p

p

(i) If kzk < 1 ≤ kxk 1−p + kyk 1−p , then " " 1 − kzkp 0,  ⊂ S. 1−p p p p 1−p 1−p + kyk − kzk kxk p

p

(ii) If kxk 1−p + kyk 1−p < 1, then S = [0, 1]. In 1.5 it will be shown that [4, 4.11(i)] is the best possible result. The same holds for [4, 4.12(i), 4.13 and 4.14]. For an absolutely convex space D the set S := {β ∈ [0, 1]|βx +(1−β)z ∼ βy +(1−β)z} ([1, 1.5]) resp. S := {β ∈ (K) | βx ∼ βy} ([7, 4.1]) does not depend on the (1-)norm of the elements. Contrary to this the pnorm plays an important role for p < 1. One gets results for elements with ‘small’ p-norm [4, 4.14] that correspond to the absolutely convex case ([2, 1.3, 1.5]). If the p-norms of x, y ∈ D are ‘big’ and kzk = 1 (z ∈ D), it will be shown in 1.11 that S := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z ∼ β 1/p y + (1 − β)1/p z} is, for p < 1, neither convex nor p-convex ([5, p. 101]). The following chapter about congruence relations is the continuation of [4, Section 4]. 1. Congruence Relations II DEFINITION 1.1. Let K := R, p < 1, and e1 , e2 , e3 be the unit-vectors in R3 . b p ◦ lp ({e1 , e2 , e3 }) = {(x, y, z) ∈ R3 | |x|p + |y|p + |z|p ≤ 1}. For Define A := ρ ∈ R a relation ‘∼ ρ ’ on the p-totally convex space A is defined in the following (x2 , y2 , z2 ) if and only if way: For (xi , yi , zi ) ∈ A (i = 1, 2) (x1 , y1 , z1 ) ∼ ρ (x1 , y1 , z1 ) = (x2 , y2 , z2 ) or the following two conditions are satisfied: (i) z1 = z2 and y1 + ρx1 = y2 + ρx2 . (ii) For all λ ∈ [0, 1] |λx1 + (1 − λ)x2 |p + |λy1 + (1 − λ)y2 |p + |λz1 + (1 − λ)z2 |p < 1. PROPOSITION 1.2. The above relation ‘∼ ρ ’ (ρ ∈ R) is a congruence relation on is a p-totally convex space. A, i.e. Aρ := A/∼ ρ Proof. Put ‘∼’ := ‘∼ ρ ’ for the sake of brevity. Obviously, ‘∼’ is reflexive and symmetric. Consider (xi , yi , zi ) ∈ A (i = 1, 2, 3) with (x1 , y1 , z1 ) ∼ (x2 , y2 , z2 ) and (x2 , y2 , z2 ) ∼ (x3 , y3 , z3 ). Without loss of generality one may assume

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P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

(x1 , y1 , z1 ) 6= (x2 , y2 , z2 ) 6= (x3 , y3 , z3 ). (i) is trivially fulfilled. Next, for λ ∈ [0, 1], there exist n, m ∈ N with (n, m) = (1, 2) or (n, m) = (2, 3) and λx1 + (1 − λ)x3 = αxn + (1 − α)xm for some α ∈ [0, 1]. Because of (i) there is a c ∈ R with yi = −ρxi + c (i = 1, 2, 3), which implies λy1 + (1 − λ)y3 = λ(−ρx1 + c) + (1 − λ)(−ρx3 + c) = −ρ(λx1 + (1 − λ)x3 ) + c = −ρ(αxn + (1 − α)xm ) + c = α(−ρxn + c) + (1 − α)(−ρxm + c) = αyn + (1 − α)ym . This leads to |λx1 + (1 − λ)x3 |p + |λy1 + (1 − λ)y3 |p + |λz1 + (1 − λ)z3|p = |αxn + (1 − α)xm |p + |αyn + (1 − α)ym |p + |αzn + (1 − α)zm|p < 1, thus ‘∼’ is transitive. Let α ∈ p , (x1i , y1i , z1i ), (x2i , y2i , z2i ) ∈ A with (x1i , y1i , z1i ) ∼ (x2i , y2i , z2i ) (i ∈ N). We may assume (x1i0 , y1i0 , z1i0 ) 6= (x2i0 , y2i0 , z2i0 ) for some i0 ∈ supp α. For λ ∈ [0, 1], (ii) yields p X p X X X αi x1i + (1 − λ) αi x2i + λ αi y1i + (1 − λ) αi y2i + λ i

i

X p X i i + λ αi z1 + (1 − λ) αi z2 ≤

X

i

i

i

i

|αi |

p

(|λx1i

+ (1 − λ)x2i |p + |λy1i + (1 − λ)y2i |p +

i

+ |λz1i + (1 − λ)z2i |p ) <

X

|αi |p ≤ 1.

i

Hence ‘∼’ is a congruence relation on A.

2

DEFINITION 1.3. If πρ : A → Aρ (ρ ∈ R) is the canonical projection one simply writes (x, y, z) := πρ ((x, y, z)), if no misunderstandings are possible. LEMMA 1.4. If p < 1, then, for all elements (x, y, 0) ∈ A with |x|p + |y|p < 1, there exist a, b ∈ (R) with ab = 0 and (x, y, 0) ∼ 1 (a, b, 0). Proof. Denote ‘∼ 1 ’ by ‘∼’ and define the mappings g, h: [0, 1] → R by g(λ) := λp |x|p + |y + (1 − λ)x|p and h(λ) := λp |y|p + |x + (1 − λ)y|p (λ ∈ [0, 1]). In case x = 0 or y = 0 there is nothing to prove. So assume xy 6= 0. By 1.1, (x, y, 0) ∼ (0, x + y, 0) is equivalent with g(λ) < 1 for all λ ∈ [0, 1] and (x, y, 0) ∼ (x + y, 0, 0) is equivalent with h(λ) < 1 for all λ ∈ [0, 1]. First, let 0 < x ≤ y. g is differentiable in ]0, 1[ with g 0 (λ) = p(λp−1 x p − . From x+y ≥ 1, (y + (1 − λ)x)p−1 x). g 0 (λ) = 0 is equivalent with λ = x+y 2x 2x p p p p p g(0) = (x + y) ≤ x + y < 1 and g(1) = x + y < 1 one concludes g(λ) < 1 for λ ∈ [0, 1], hence (x, y, 0) ∼ (0, x + y, 0). If 0 < y ≤ x one gets h(λ) < 1 (λ ∈ [0, 1]), thus (x, y, 0) ∼ (x + y, 0, 0) by a symmetry argument. Consider now 0 < x and y < 0. In case x + y ≤ 0, for λ ∈ [0, 1], y + (1 − λ)x ≤ 0 and g 0 (λ) = p(λp−1 x p + (−(y + (1 − λ)x))p−1 x) > 0 holds. Because of g(1) = x p + (−y)p < 1 and g(0) < 1, g(λ) < 1 (λ ∈ [0, 1]) follows, hence

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RALF KEMPER

(x, y, 0) ∼ (0, x+y, 0). If x+y ≥ 0, 0 < −y, −x < 0 and (−y)+(−x) ≤ 0 holds, and by symmetry one gets h(λ) < 1 (λ ∈ [0, 1]), thus (x, y, 0) ∼ (x + y, 0, 0). If x < 0 < y and x + y ≥ 0, (x, y, 0) = −(−x, −y, 0) ∼ −(0, −(x + y), 0) = (0, x + y, 0) follows. Analogously, x + y ≤ 0 yields (x, y, 0) = −(−x, −y, 0) ∼ −(−(x + y), 0, 0) = (x + y, 0, 0). The assertion in the last case, x < 0, y < 0, follows immediately from the above considerations. 2 THEOREM 1.5. For p < 1, 0 < γ < 1, 1 ≤ r ≤ 2, K := R define σ := ( 2r )

1−p p

,

p 1−p

+ x := σ e1 , y := σ e2 , z := γ e3 ∈ A1 . Then kxk = kyk = σ , kzk = γ , kxk p kyk 1−p = r and for S := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z = β 1/p y + (1 − β)1/p z} " " 1 − kzkp holds. S = 0, p p (kxk 1−p + kyk 1−p )1−p − kzkp Proof. Put ‘∼’ := ‘∼ 1 ’ (1.1, 1.2). Obviously one has kzk ≤ γ . Assume that there exists an  with 0 <  < γ and z = (a, b, c) ((a, b, c) ∈ A1 ). Then γ = c = |c| ≤  follows (1.1), contradicting  < γ , i.e. kzk = γ . As kxk ≤ σ , assume that there exists an  with 0 <  < σ and x = (a, b, c). Then c = 0 because of 1.1. For any  0 ∈ ], σ [, x = (a, b, c) =  0 ( 0 a, 0 b, 0) and k( 0 a, 0 b, 0)k = k 0 (a, b, 0)k ≤ 0 < 1 follows. By 1.4 there are a 0 ∈ (R) or b0 ∈ (R) with x =  0 (a 0 , 0, 0) or x = 0  (0, b0 , 0). This implies σ =  0 a 0 =  0 |a 0 | ≤  0 < σ or σ =  0 b0 =  0 |b0 | ≤  0 < σ 1 and we get kxk = σ and similarly kyk = σ . If α := ( 12 ) p +1 and λ ∈ [0, 1], (λασ )p + ((1 − λ)ασ )p = α p σ p (λp + (1 − λ)p )  1 p ≤2 α p σ p ≤ 2α p < 1 2 holds, implying (ασ, 0, 0) = (0, ασ, 0), thus αx = αy. β 6= 0, β ∈ S is equivalent to (β 1/p σ, 0, (1 − β)1/p γ ) ∼ (0, β 1/p σ, (1 − β)1/p γ ). Define fβ : [0, 1] → R by fβ (λ) := βσ p (λp + (1 − λ)p ) + (1 − β)γ p (λ ∈ [0, 1], β ∈ ]0, 1]). Because of 1.1, β ∈ S is equivalent to fβ (λ) < 1 for λ ∈ [0, 1]. Hence β ∈ S implies p βr 1−p + (1 − β)γ p = βσ p (2( 12 )p ) + (1 − β)γ p = fβ ( 12 ) < 1. From kxk 1−p + p

p

kyk 1−p = 2σ 1−p = r ≥ 1 and kzk = γ < 1 we conclude β < 4.11(i)] the assertion is proved.

1−γ p r 1−p −γ p

and by [4, 2

Theorem 1.5 raises the question if any intervall [0, s], for arbitrary 0 < s < 1, appears as a S for a suitable D ∈ ACp . DEFINITION 1.6. For p < 1, K := R, ρ ∈ R, the relation ∧ρ on A (cf. 1.1) is defined by (x1 , y1 , z1 ) ∧ρ (x2 , y2 , z2 ) if and only if the following two conditions are fulfilled:

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P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

(i) z1 = z2 and y1 + ρx1 = y2 + ρx2 . (ii) For all λ ∈ [0, 1] |λx1 + (1 − λ)x2 |p + |λy1 + (1 − λ)y2 |p + |λz1 + (1 − λ)z2 |p ≤ 1. Modifying the proof of 1.2, one easily gets the PROPOSITION 1.7. ‘∧ρ ’ (ρ ∈ R) is a congruence relation on A. LEMMA 1.8. If p < 1 then, for all (x, y, 0) ∈ A, there exist a, b ∈ (R) with ab = 0 and (x, y, 0) ∧1 (a, b, 0). Proof. Write ‘∼’ := ‘∧1 ’, consider the mappings g, h: [0, 1] → R defined in 1.4, and replace g(λ) < 1 (resp. h(λ) < 1) in the proof of 1.4 by g(λ) ≤ 1 (resp. h(λ) ≤ 1) and the assertion follows. 2 Define Bρ := A/∧ρ (ρ ∈ R), and let πρ0 : A → Bρ be the canonical projection. If no misunderstandings are possible one simply writes (x, y, z) := πρ0 ((x, y, z)). THEOREM 1.9. For p < 1, 0 < γ < 1, 1 ≤ r ≤ 2, K := R define σ := ( 2r )

1−p p

,

p 1−p

+ x := σ e1 , y := σ e2 , z := γ e3 ∈ B1 . Then kxk = kyk = σ , kzk = γ , kxk p 1/p 1/p 1/p 1/p 1−p kyk = r and for S := {β ∈ [0, 1] | β x + (1 − β) z = β y + (1 − β) z} # " 1 − kzkp holds. S = 0, p p (kxk 1−p + kyk 1−p )1−p − kzkp Proof. Write ‘∼’ := ‘∧1 ’ for simplicity. The proof of the statements kxk = kyk = σ, kzk = γ is similar to the corresponding proof in 1.5. Also αx = 1 αy for α := ( 12 ) p +1 , follows as in 1.5. For β ∈ ]0, 1], β ∈ S is equivalent to (β 1/p σ, 0, (1 − β)1/p γ ) ∼ (0, β 1/p σ, (1 − β)1/p γ ). fβ : [0, 1] → R is defined by fβ (λ) := βσ p (λp + (1 − λ)p ) + (1 − β)γ p (λ ∈ [0, 1], β ∈ ]0, 1]). Because of 1.6, β ∈ S is equivalent to fβ (λ) ≤ 1 for λ ∈ [0, 1]. Hence, in particuar, β ∈ S implies βr 1−p +(1−β)γ p = βσ p (2( 12 )p )+(1−β)γ p = fβ ( 12 ) ≤ p p p 1−γ p 1. From kxk 1−p + kyk 1−p = 2σ 1−p = r ≥ 1 and kzk = γ < 1, β ≤ r 1−p −γ p p

1−γ follows. β0 := r 1−p yields fβ0 (λ) = β0 σ p (λp + (1 − λ)p ) + (1 − β0 )γ p ≤ 1 −γ p for each λ ∈ [0, 1], hence β0 ∈ S. This proves the assertion ([4, 4.11(i)]). 2

If D ∈ TCp (ACp ) and y ∈ D one defines the mapping ψy : D → {0, 1} (0 6= 1) by ψy (x) := 1 if there exist z ∈ D, β ∈ ]0, 1], γ ∈ K, |γ | = 1, with γ y = β 1/p x + (1 − β)1/p z, and ψy (x) := 0, otherwise (cf. [3, p. 3185]). A relation ‘∼’ on D is defined by a ∼ b if and only if a = b or ψy (a) = ψy (b) = 0. PROPOSITION 1.10. The relation ‘∼’ is a congruence relation on D. with ai ∼ Proof. Obviously, ‘∼’ is an equivalence relation. Let ai , bi ∈ D P bP i (i ∈ N) and α ∈ p (p,fin ). If for all i ∈ supp α ai = bi , then P i αi ai = α b holds. If there is an i ∈ supp α with a = 6 b assume ψ ( 0 i0 i0 y i i i i αi ai ) = 1. Then there exist z ∈ D, β ∈ ]0, 1], γ ∈ K, |γ | = 1 with

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RALF KEMPER

γ y = β 1/p

X i

αi ai + (1 − β)1/p z

 αi0 ai + = (β |αi0 |) |αi0 | 0 X  (1 − β)1/p β 1/p αi p 1/p +(1 − β|αi0 | ) ai + z . (1 − β|αi0 |p )1/p (1 − β|αi0 |p )1/p i6=i 

1/p

0

(Here the bracket is to be read as one p (p,fin )-operation.) This implies α ψy ( |αii0 | ai0 ) = 1, thus ψy (ai0 ) = 1, contradicting ai0 ∼ bi0 , ai0 6= bi0 . Hence, 0 P P ( α a ) = 0 and, by symmetry, ψ ( necessarily ψ y i i y i i αi bi ) = 0. This proves P P 2 i αi ai ∼ i αi bi and ‘∼’ is a congruence relation. THEOREM 1.11. For p < 1 and M ⊂ [0, 1[ with 0 ∈ M there exist a p-totally p p convex space C and x, y, z ∈ C with kzk = kxk 1−p + kyk 1−p = 1 (resp. kzk = p p 1, kxk 1−p + kyk 1−p = 2) such that for S := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z = β 1/p y + (1 − β)1/p z} S = M holds. b p ◦ lp ({(0, 1), (1, 0)}) and put Proof. First let M 6= [0, 1[. Define D := x := (1, 0), yν := (0, ν), z := (0, 1) ∈ D with ν ∈ {−1, 0}. For α ∈ M c := [0, 1[ \M define cα := α 1/p x + (1 − α)1/p z. The relation ‘∼ α ’ on D is defined by a ∼ α b if and only if a = b or ψcα (a) =Tψcα (b) = 0. is a congruence relation on D. Denote By 1.10, the relation ∼ := α∈M c ∼ α D/∼ by C and by [a] the set {b ∈ D | a ∼ b} (a ∈ D). Because of M c ⊂ ]0, 1[, cα = α 1/p x + (1 − α)1/p z, ψcα (x) = ψcα (z) = 1 (α ∈ M c ), |[x]| = |[z]| = 1 and k[x]k = k[z]k = 1 follows. From −cα = α 1/p (−1, 0) + (1 − α)1/p (0, −1) and M c ⊂ ]0, 1[ one gets ψcα ((0, −1)) = 1 (α ∈ M c ), hence |[(0, −1)]| = 1 and p p k[y−1 ]k = k[(0, −1)]k = 1. This implies k[z]k = k[x]k 1−p + k[y0 ]k 1−p = 1 and p p k[x]k 1−p + k[y−1 ]k 1−p = 2. Put Sν := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z ∼ β 1/p yν + (1 − β)1/p z}. For all α ∈ M c ψcα (cα ) = 1 holds, thus α 1/p x + (1 − α)1/p z = cα 6∼ (0, 0). For every β ∈ M\{0} |β 1/p ν + (1 − β)1/p 1| < 1 holds, and kcα k = 1 implies (0, 0) ∼ (0, β 1/p ν + (1 − β)1/p 1) = β 1/p yν + (1 − β)1/p z ([4, 4.5]). This yields M c ∩ Sν = ∅. Now let β ∈ M\{0}. Assume that there exists an α0 ∈ M c with ψcα0 (β 1/p x + (1 − β)1/p z) = 1. Then there are (r, s) ∈ D, γ ∈ K with |γ | = 1 and  ∈ ]0, 1] with γ cα0 =  1/p (β 1/p x + (1 − β)1/p z) + (1 − )1/p (r, s) = ((β)1/p + (1 − )1/p r, ((1 − β))1/p + (1 − )1/p s). |γ | = 1, kcα0 k = 1 implies kγ cα0 k = 1, hence |r|p + |s|p = k(r, s)kp = 1.  = 1 1/p leads to γ cα0 = (β 1/p , (1 − β)1/p ), hence γ α0 = β 1/p and γ = 1, implying c α0 = β ∈ M ∩ M = ∅. Therefore  < 1 holds and one gets 1 = kγ cα0 k =

585

P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

|(β)1/p + (1 − )1/p r|p + |((1 − β))1/p + (1 − )1/p s|p ≤ β + (1 − )|r|p + (1−β)+(1−)|s|p = 1, which implies r = s = 0, contradicting |r|p +|s|p = 1. Therefore, for all α ∈ M c , β ∈ M\{0} ψcα (β 1/p x +(1−β)1/p z) = 0, consequently β 1/p x + (1 − β)1/p z ∼ (0, 0) ∼ β 1/p yν + (1 − β)1/p z and β ∈ Sν . Obviously, 0 ∈ Sν holds, and M ⊂ Sν is proved. The assumption 1 ∈ Sν , i.e. x ∼ yν implies the contradiction (1, 0) = x = yν = (0, ν) because of ψcα (x) = 1 (α ∈ M c ). Hence 1 6∈ Sν and Sν = M (ν = −1, 0) in case M 6= [0, 1[ holds. For M = [0, 1[ define C := L ([7, p. 985]), L the totally convex Linton space, and put z := 1, x := 1, y := 0 ∈ C (resp. z := 1, x := 1, y := −1 ∈ C), and the assertion is proved. 2 THEOREM 1.12. Let p < 1 and 1 < r ≤ 2. Then there exist a real p-totally p p convex space D, x, y ∈ D with kxk 1−p + kyk 1−p = r, ( 12 )1/p x = ( 12 )1/p y and there are zn ∈ D with kzn k = 1, such that for the sets Sn := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p zn = β 1/p y + (1 − β)1/p zn } Sn 6= {0} (n ∈ N), and limn→∞ sup Sn = 0 holds. 1−p Proof. Put σ := ( 2r ) p and D := A1 ∈ TCp (1.1, 1.2). Let x := σ (1, 0, 0), y := p p σ (0, 1, 0) ∈ A1 . By 1.5, kxk = kyk = σ , which implies kxk 1−p + kyk 1−p = r. Let zi ∈ (R) (i = 1, 2, 3) with 0 < z1 , z2 < 0 and |z1 |p + |z2 |p + |z3 |p = 1. Then we have z := (z1 , z2 , z3 ) ∈ A1 and kzk = 1 holds. Let ‘∼’ := ‘∼ 1 ’ (1.1) and S := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p z = β 1/p y + (1 − β)1/p z} = {β ∈ [0, 1] | (β 1/p σ + (1 − β)1/p z1 , (1 − β)1/p z2 , (1 − β)1/p z3 ) ∼ ((1 − β)1/p z1 , β 1/p σ + (1 − β)1/p z2 , (1 − β)1/p z3 )}. For every β ∈ ]0, 1] β ∈ S is equivalent with the following condition (∗) because of β 1/p σ 6= 0: (∗) For all λ ∈ [0, 1] |(1 − λ)β 1/p σ + (1 − β)1/p z1 |p + |λβ 1/p σ + (1 − β)1/p z2 |p + + (1 − β)|z3 |p < 1. First we assert S 6= {0}. For every λ ∈ [0, 1] and 0 < β, λβ 1/p σ +(1−β)1/p z2 < 0 is equivalent with λ < λ0 , where λ0 := ( β1 − 1)1/p |zσ2 | . λ0 > 1 holds if and only if

| β < |z2|z|p2+σ p . For a fixed β0 with 0 < β0 < defined by p

|z2 |p |z2 |p +σ p

a function g: [0, 1] → R is

 1/p 1/p g(t) := (1 − t)β0 σ + (1 − β0 )1/p z1 +   p 1/p + − tβ0 σ + (1 − β0 )1/p z2 + (1 − β0 )|z3 |p g is differentiable in ]0, 1[ with

(t ∈ [0, 1]).

586

RALF KEMPER

p−1  1/p 1/p g 0 (t) = −pβ0 σ (1 − t)β0 σ + (1 − β0 )1/p z1 + p−1    1/p . + − tβ0 σ + (1 − β0 )1/p z2 Thus g 0 (t) < 0 for all t ∈ ]0, 1[ and g(0) = (β0 σ + (1 − β0 )1/p z1 )p + (1 − p β0 )(|z2 |p +|z3 |p ) < β0 σ p +(1−β0 )z1 +(1−β0 )(|z2 |p +|z3 |p ) = β0 σ p +(1−β0 ) ≤ 1 holds implying 0 ≤ g(t) < 1 for all t ∈ [0, 1]. Thus β0 fulfills (∗) and S 6= {0} follows.  1/p    1 1/p    γ 1/p 1 1/p 1/p x= γ x + (1 − γ ) z − (1 − γ )1/p z 2 2 2  1/p   1/p    1 1 γ 1/p y + (1 − γ )1/p z − (1 − γ )1/p z = 2 2  γ 1/p y (γ ∈ S\{0}) = 2 1/p

yields ( 12 )1/p x = ( 12 )1/p y because of [4, 4.14]. Now, for β ∈ ]0, 1[ put t 0 := 2p z

p

0 0 1 + ( β1 − 1)1/p zσ1 and s(z1 ) := 2p zp +σ p . Obviously, 0 < t holds and t < 1 is 1 equivalent with s(z1 ) < β. Put z2 := −z1 < 0. If βP∈ S with s(z1 ) < β, then 0 < t 0 < 1 follows and (∗) is fulfilled for λ = t 0 . 1 = 3i=1 |zi |p and 0 < z1 imply r 1−p − |z3 |p holds. Application of (∗) to λ = t 0 leads |z3 | < 1 < r, hence 0 < 1−|z3 |p to β < s(z3 ) := r 1−p −|z |p by an elementary computation. Consequently, β ∈ S 3 implies β ≤ max{s(z1 ), s(z3 )}. Now, for every n ∈ N, we choose an element zn := (z1,n, z2,n , z3,n ) ∈ D with P 0 < z1,n < 1, 0 ≤ z3,n < 1, z2,n := −z1,n , 3i=1 |zi,n |p = 1; furthermore, the sequences (z1,n )n∈N and (z3,n )n∈N are supposed to be convergent with limn→∞ z1,n = 0 and limn→∞ z3,n = 1. For the sets Sn := {β ∈ [0, 1] | β 1/p x + (1 − β)1/p zn = β 1/p y + (1 − β)1/p zn } we have Sn 6= {0} (n ∈ N) and ( 12 )1/p x = ( 12 )1/p y holds. Because of limn→∞ z1,n = 0 and limn→∞ z3,n = 1, limn→∞ s(z1,n ) = 0 and limn→∞ s(z3,n ) = 0 follows. This leads to limn→∞ max{s(z1,n ), s(z3,n )} = 0. As seen above, β ∈ Sn implies β ≤ max{s(z1,n ), s(z3,n )} (n ∈ N) hence we get 2 limn→∞ sup Sn = 0. 1 2

ki )i∈I be a family of p-absolutely convex subspaces of D ∈ ACp Let (Di , k T and put C := i∈I Di . In case p = 1 kxkC = sup{kxki |i ∈ I } holds for all x ∈ C ([7, 6.6]). For p < 1 one has the following PROPOSITION 1.13. For p < 1 sup{kxki |i ∈ I } ≤ kxkC (x ∈ C) holds, but in general (for K := R) sup{kxki | i ∈ I } 6= kxkC for some x ∈ C. Proof. Let ini : C ,→ Di be the inclusion of C in Di (i ∈ I ). For every x ∈ C kxki = kini (x)ki ≤ kxkC (i ∈ I ) holds ([4, 4.3(ii)]), thus sup{kxki | i ∈ I } ≤ ◦

kxkC . Put D := A1 ∈ TCp (1.1, 1.2). For all x ∈ (R), (x, 0, 0) ∼ 1 (0, x, 0) is

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P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

equivalent by 1.1 with |x|p (λp + (1 − λ)p ) < 1 for all λ ∈ [0, 1]. The mapping g: [0, 1] → R, defined by g(t) := t p + (1 − t)p (t ∈ [0, 1]) has an absolute maximum in t = 12 with g( 12 ) = ( 12 )p−1 . Hence (x, 0, 0) ∼ 1 (0, x, 0) is equivalent ◦

with |x| < ωp := ( 12 ) p −1 for all x ∈ (R). b p (R)(0, 1, 0) and C2 := b p (R)(1, 0, 0) are p-totally convex subC1 := 1

spaces of D, and z := (0, ( 12 )1/p , 0) = (( 12 )1/p , 0, 0) ∈ C1 ∩ C2 holds. By the ◦

◦

above considerations we get C1 ∩ C2 = (R)(0, ωp , 0). For every γ ∈ (R) with z = γ (0, ωp , 0) Definition 1.1 yields 12 ωp = ( 12 )1/p = γ ωp , hence γ = 12 . Now, a 2 simple computation shows kzkC1 ∩C2 = 12 > sup{kzkCi | i = 1, 2} = ( 12 )1/p . DEFINITION 1.14. Let D be a p-absolutely convex space and x, y ∈ D. (i) The sets ]x, y[ := {β 1/p x +(1−β)1/p y | 0 < β < 1}, [x, y[ := {β 1/p x +(1− β)1/p y | 0 < β ≤ 1}, ]x, y] := {β 1/p x + (1 − β)1/p y | 0 ≤ β < 1}, [x, y] := {β 1/p x + (1 − β)1/p y | 0 ≤ β ≤ 1} are the so-called open (right open, left open, closed) segment between x and y. (ii) ∂D := {z ∈ D | kzk = 1} is the boundary of D (cf. [9, p. 1473]). ◦

(iii) D := D\∂D is the interior of D ([8, 10.1]). PROPOSITION 1.15. Let D be a p-absolutely convex space and x, y ∈ D. Then the following statements hold: ◦

(i) For p = 1 ]x, y[ ⊂ D or ]x, y[ ⊂ ∂D. ◦ (ii) For p < 1, x0 , y0 ∈ [x, y[ or x0 , y0 ∈ ]x, y] implies ]x0 , y0 [ ∈ D . Proof. (i) Let a ∈ ]x, y[ with kak < 1. Then there exists an α with 0 < α < 1 and a = αx + (1 − α)y. For every b ∈ ]x, y[ there exists a β with 0 < β < 1 and ∈ [0, 1[ one gets γ x + (1 − γ )a = b = βx + (1 − β)y. In case β ≥ α for γ := β−α 1−α β−α 1−β x+ 1−α (αx+(1−α)y) = βx+(1−β)y = b and kbk ≤ γ kxk+(1−γ )kak < 1 1−α ∈ ]0, 1[. Then γ y + (1 − γ )a = holds by [4, 4.5]. In case β < α put γ := α−β α α−β β y + (αx + (1 − α)y) = βx + (1 − β)y = b follows, and, by [4, 4.5], kbk < 1. α α ◦

Thus ]x, y[ ⊂ D or ]x, y[ ⊂ ∂D. (ii) Because of the assumption there exist β, γ ∈ [0, 1] with (β, γ ) 6= (1, 0), (0, 1) and x0 = β 1/p x + (1 − β)1/p y, y0 = γ 1/p x + (1 − γ )1/p y. Let a ∈ ]x0 , y0 [ thus a = α 1/p x0 + (1 − α)1/p y0 for some α ∈ ]0, 1[. From [4, 4.5] one gets kakp = kα 1/p (β 1/p x + (1 − β)1/p y)+ + (1 − α)1/p (γ 1/p x + (1 − γ )1/p y)kp = k((αβ)1/p + ((1 − α)γ )1/p )x + ((α(1 − β))1/p + + ((1 − α)(1 − γ ))1/p )ykp ≤ ((αβ)1/p + ((1 − α)γ )1/p )p + ((α(1 − β))1/p + + ((1 − α)(1 − γ ))1/p )p .

588

RALF KEMPER

Since in all cases (0 < αβ and 0 < (1 − α)γ ) or (0 < α(1 − β) and 0 < (1 − α)(1 − γ )) holds, kakp < αβ + (1 − α)γ + α(1 − β) + (1 − α)(1 − γ ) = 1 ◦ ◦ 2 follows, i.e. a ∈ D . This implies ]x0 , y0 [ ⊂ D . Certain subsets of the boundary of an absolutely convex space and the interior of such spaces are convex spaces ([10, 1.8; 9, Section 2]). In case p < 1, D ∈ 1 ACp , for every y ∈ ∂D, ( 12 )1/p y + ( 12 )1/p y = ωp y 6∈ ∂D, where ωp := ( 12 ) p −1 , holds, so that the boundary of D is not accessible by the methods of [9, Section 2]. Furthermore, 1.15(ii), applied to the p-absolutely convex space D :=Pp,fin , p < p 1, shows (with 0p,fin := {α = (αi )i∈N ∈ [0, 1]N | |supp α| < ∞ and i αi = 1}) that there do not exist ‘p-convex spaces’ in the sense of [4, 2.4]. LEMMA 1.16. For D ∈ ACp let ‘∼’ be a congruence relation on D. Then, for all ωp a, b ∈ D, and all α ∈ 2 (K)\{0} ([4, 4.9]), αa − αb ∼ 0 implies βa ∼ βb for ◦

all β ∈ ωp (K). Proof. From (( 14 )1/p α)b = ( 14 )1/p 0+( 14 )1/p (αb) ∼ ( 14 )1/p (αa−αb)+( 14 )1/p (αb) 2 = (( 14 )1/p α)a and α 6= 0 the assertion follows by [4, 4.14]. b If no misunderstandings are possible, we will simply write (K) instead of b b b

p (K), resp. fin (K) instead of p,fin (K) in the following. b THEOREM 1.17 (cf. [7, 4.4]). The p-totally (p-absolutely) convex space (K) b fin (K)) permits for p = 1, K = R, C, precisely three, for p < 1, K = R, ( precisely four, and for p < 1, K = C, infinitely many congruence relations. b Proof. First let p ≤ 1, K = R, C and ‘∼’ be a congruence relation on (K) ∈ 1 1/p 1 1/p b TCp . For a, b ∈ (K) with a ∼ b and a 6= b, ( 2 ) a − ( 2 ) b ∈ (K)\{0} follows and (( 12 )1/p a − ( 12 )1/p b)1 = ( 12 )1/p a − ( 12 )1/p b ∼ ( 12 )1/p a − ( 12 )1/p a = 0 implies S 6= {0} for the set S := {λ ∈ (K) | λ1 ∼ λ0}. By [4, 4.17(ii)], applied ◦ b to (K)/∼, we have S = (K) or S = (K). S = (K) means x ∼ y for all ◦

b b x, y ∈ (K). If S = (K), for every x ∈ (K) with |x| = 1, obviously x 6∼ 0 holds. Hence, for p = 1 a ∼ b with |a| = |b| = 1 leads to 1 = 1b b ∼ b1 a = ab and 1=

b a 2a b

+ 12 1 ∼

b 1 + 12 1 2a

◦

= 12 ( ab + 1). This implies a = b, since 12 ( ab + 1) ∈ (K) ◦

otherwise. Thus the equivalence classes in case S = (K) for p = 1 are precisely ◦ b

(K) and {x} (x ∈ (K) with |x| = 1). b Now, for every p ≤ 1 the relation ‘≈’ on (K), defined by x ≈ y if and ◦

b only if x = y or x, y ∈ (K) obviously is an equivalence relation on (K). b Let xi , yi ∈ (K)Pwith xi ≈Pyi (i ∈ N), α ∈ p . In case xi = yi for all αi xi = i αi yP i ∈ supp α, we get i P i . So, let i0 ∈ supp α with xi0 6= yi0 . Because p 1/p of kxi0 k, kyi0 k
P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

589

b relation on (K). This means that there are precisely three congruence relations b on (K) for p ≤ 1 and at last three for p < 1. b Now let p < 1, K = R, C, and let ‘∼’ be an equivalence relation on (K) ◦ b with x ∼ y for all x, y ∈ (K), and the following property: For all a, b ∈ (K) and all α ∈ (K) with |α| = 1, a ∼ b implies αa ∼ αb. We show that ‘∼’ is b b with ai P ∼ bi (i ∈ a congruence relation on (K). For this let ai , bi ∈ (K) is an i0 ∈ supp α with 0 < |αi0 | < 1, P| i αi ai | ≤ N) p . If there P P and α ∈ P p |α | |a | ≤ |α | < i i i i i P i |αi | ≤ 1 follows and by symmetry | i αi bi | < 1, P hence i αi ai ∼ i αi bi . This is also trivially fulfilled for α = 0, and P in case i ∈ supp α by assumption α a ∼ α b , hence |αi0 | = 1 for some i0 i0 i0 i0 i αi ai = P0 b αi0 ai0 ∼ αi0 bi0 = i αi bi . Consequently, ‘∼’ is a congruence relation on (K). b In case p < 1, K = R, the relation ‘∼’ on (R) defined by a ∼ b if and ◦ b is, as seen above, the last only if a, b ∈ (R) or |a| = |b| = 1 (a, b ∈ (R)) b possibility for a congruence relation on (R). Obviously, ‘∼’ is an equivalence b b relation on (R) and a ∼ b implies αa ∼ αb for all a, b ∈ (R), α ∈ (R) b with |α| = 1. By the above result, ‘∼’ is a congruence relation on (R), hence b there are precisely four congruence relations on (R), p < 1. b Obviously, the relation ‘∼’ on (C), defined by a ∼ b if and only if a, b ∈ ◦ b

(C) or |a| = |b| = 1 (a, b ∈ (C)) is for p < 1 a congruence relation on b b

(C). Finally, we have to show that (C), for p < 1, permits infinitely many b congruence relations. Take any m ∈ N. A relation ‘∼’ on (C) is defined by a ∼ b r b if and only if there exists a r ∈ N0 with a = ζm b (a, b ∈ (C)). Obviously, ‘∼’ is b b an equivalence relation on (C). Let α ∈ (C) with |α| = 1, and a, b ∈ (C) r r with a ∼ b. Then there is a r ∈ N0 with a = ζm b. This implies αa = ζm (αb), b hence αa ∼ αb. Hence, ‘∼’ is a congruence relation on (C). For all m, n ∈ N with m < n there does not exist a s ∈ N0 with ζn = ζms . Thus we get infinitely b b fin (K) ∈ ACp is many congruence relations on (C) for p < 1. The proof for similar. 2 6= 1), i ∈ N, COROLLARY 1.18. If one defines, for p < 1 and xi ∈ {0, 1} (0 P P α x := 1, if there is an i ∈ N with |α | = 1 and x = 1, and 0 i0 i0 i i i i αi xi := 0 otherwise, then {0, 1} becomes a p-totally convex space. DEFINITION 1.19 (cf. [7, p. 985]). Let p ≤ 1, K = R, C. The relation ‘∼’ on ◦ b b

(K) defined by x ∼ y if and only if x, y ∈ (K) or x = y (x, y ∈ (K)) b b is the is a congruence relation on (K) (cf. 1.17). The quotient LK := (K)/∼ so-called Linton space with |LR | = 3 and |LC | = ∞. COROLLARY 1.20 (cf. [4, 1.4, 1.6]). The functors p : Banp → Set and p,fin : Vecp → Set are not monadic. COROLLARY 1.21. Let D 6= {0} be an absolutely convex space. Then, for any x ∈ D\{0}, and α, β ∈ K with |α| = |β| = 1, αx = βx implies α = β.

590

RALF KEMPER

b Proof. Obviously, the mapping f : (K) → D, defined by f (λ) := λx (λ ∈ b b

(K)) is an AC-morphism. Denote by ‘∼’ the congruence relation on (K) induced by f . Let α, β ∈ (K) with |α| = |β| = 1 and αx = βx. Then α ∼ β and, since x 6= 0, 1 6∼ 0 holds. ( α2 − β2 )x = 12 (αx) − 12 (βx) = 0 implies S 6= {0} for the set S := {γ ∈ (K) | γ x = 0}, if α 6= β. Because of 1 6∼ 0 we then get ◦

S = (K) ([4, 4.17(ii)]). Now, 1.17 yields α = β.

2

Corollary 1.21 does not hold for p < 1, since in {0, 1} (1.18) we have (−1)·1 = 1 · 1. COROLLARY 1.22 ([7, 4.7]). Let D 6= {0} be a complex absolutely convex space. Then card(D) ≥ card(C). b Proof. Let x ∈ D\{0}. The mapping f : (C) → D defined by f (λ) := λx is b injective on ∂ (C) (1.21). This implies card(D) ≥ card(C). 2 COROLLARY 1.23 ([7, p. 986]). Let D be a real absolutely convex space with card (D) < ∞. Then card (D) is odd. Proof. Let x ∈ D\{0}. By 1.21, x 6= −x holds, and we are finished. 2 Corollary 1.18 shows that 1.23 does not hold for p < 1. b p : Banp → TCp 2. The Left Adjoint of b p has a left adjoint Sp . We will now construct It is known (cf., e.g. [6, 3.7]) that Sp explicitly. For this let D be a p-totally convex space. On the set K × D one defines an equivalence relation ‘∼’ by putting (λ0 , d0 ) ∼ (λ1 , d1 ) if and only if there is a λ ≥ max{|λ0 |, |λ1 |} with λλ0 d0 = λλ1 d1 . The equivalence class of (λ, x) is denoted by (λ, x), and the set of equivalence classes of K × D by Sp (D). There is a canonical mapping σD0 : D → Sp (D), defined by σD0 (x) := (1, x) (x ∈ D). Next one introduces scalar-multiplication and addition on Sp (D) by µ(λ, x) := (µλ, x) ((λ, x) ∈ K × D, µ ∈ K) resp. (λ0 , d0 ) + (λ1 , d1 ) := (λ, λλ0 d0 + λλ1 d1 ) where λ > (|λ0 |p + |λ1 |p )1/p . Obviously, the scalar-multiplication and, by a simple computation, the addition on Sp (D) is well-defined and Sp (D) is a K-vector space. Furthermore, for (λ, d) ∈ Sp (D) one puts k(λ, d)k := inf{|µ| | µ ∈ K and (λ, d) = µ(1, y) for some y ∈ D}. Then one has the following PROPOSITION 2.1. For D ∈ TCp , Sp (D) is a p-normed K-vector space. Proof. k(λ, d)k ≥ 0 is trivial. For (λ, d) ∈ K × D with k(λ, d)k = 0, for µ := 1 1/p ( 2 ) there is a y ∈ D with (λ, d) = (µ2 , y) = (µ, µy), hence k(µ, µy)k = 0. Let ε ∈ ]0, ωp [. Then there exists a z ∈ D with (µ, µy) = (εµ, z). Consequently, z = µσ (εz), implying µy = εz because of there is a σ > µ with µσ (µy) = εµ σ

591

P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

kµyk < ωp and kεzk < ωp ([4, 4.14]). This leads to kµyk = 0 and, by [4, 4.15], to µy = 0, hence (λ, d) = 0. Obviously, k−k is homogenous. The proof of the p-triangle inequality is a simple computation. 2 LEMMA 2.2. The following statements hold for every D ∈ TCp . (i) kσD0 (x)k ≤ kxk (x ∈ D). P P (ii) For α ∈ p,fin , xi ∈ D (i ∈ N) σD0 ( i αi xi ) = i αi σD0 (xi ) holds. (iii) For α ∈ p , (xi )i∈N ∈ D N and any n ∈ N one has X  X  X  n ∞ σD0 αi xi = αi σD0 (xi ) + σD0 αi xi . i

i=1

i=n+1

Proof. (i) x = λx0 implies σD0 (x) = λσD0 (x0 ), hence kσD0 (x)k ≤ kxk. (ii) This follows from the definition of the addition and the scalar-multiplication on Sp (D). P P (iii) Put x := ni=1 αi xi and y := ∞ i=n+1 αi xi . Then one gets by (ii) X  X   X  n ∞  1  0 0 1 0 0 αi xi = 2σD x + y = σD αi xi + σD αi xi σD 2 2 i i=1 i=n+1  X  n ∞ X 0 0 αi σD (xi ) + σD αi xi . = i=1

i=n+1

2 ◦

LEMMA 2.3. For D ∈ TCp p (Sp (D)) ⊂ σD0 (D) ⊂ p (Sp (D)) holds. ◦

Proof. The second inclusion follows from 2.2(i). For (λ, d) ∈ p (Sp (D)) there ◦

are µ ∈ (K), y ∈ D with (λ, d) = (µ, y) = σD0 (µy) ∈ σD0 (D).

2

PROPOSITION 2.4. For any D ∈ TCp , Sp (D) is a p-Banach space. Proof. P By 2.1, Sp (D) is a p-normed K-vector space. Let zi ∈ Sp (D) (i ∈ N) with i kzi kp < ∞. Clearly, we may P assume that zi 6= 0 (i P∈ N). Then one has αi := 2kzi k > 0 (i ∈ N), α := 2( i kzi kp )1/p < ∞ and i ( ααi )p = 1. By 2.3, P there are di ∈ D with α1i zi = σD0 (di ) (i ∈ N). Put d := i ααi di ∈ D. Then, due to 2.2(iii), [4, 4.5],

p  ∞  p n ∞ 

X X

0 X αi

αi p 1

0

σ

= ≤ zi d

σD (d) − i

D

α α α and thus

P∞

i=1

i=1 zi

=

i=n+1

ασD0 (d).

i=n+1

Therefore, Sp (D) is a p-Banach space.

2

b p ◦ Sp (D) denote σ 0 with the codomain restricted to bp ◦ Let σD : D → D Sp (D) (D ∈ TCp ). Then one has

592

RALF KEMPER

b p ◦ Sp (D) is in TCp . LEMMA 2.5. For any D ∈ TCp σD : D → Proof. Due to 2.2(i), (iii), [4, 4.5], one has for α ∈ p , xi ∈ D (i ∈ N),

p  ∞

  X  p n ∞ X X

0 X



≤

σD σ − α x α σ (x ) = α x |αi |p , i i i D i i i

D

i

i=1

i=n+1

i=n+1

n ∈ N, implying the assertion.

2

b p : Banp → TCp has as a left adjoint the functor THEOREM 2.6. The functor Sp with the object function D 7→ Sp (D). The unit of this adjunction is σD (D ∈ TCp ). b p (B) (B ∈ Banp ), a Proof. Obviously, for a TCp -morphism ψ: D → b p (ψ b: Sp (D) → B with ψ = b) ◦ σD is uniquely deterBanp -morphism ψ mined. Each z ∈ Sp (D) can be written as z = λσD (x) (λ ∈ K, x ∈ D). If 0 λσD (x) = λ0 σD (x 0 ), then, for some γ > max{|λ|, |λ0 |}, γλ x = λγ x 0 holds. This imb: Sp (D) → B, defined by ψ b(z) := λψ(x) plies λψ(x) = λ0 ψ(x 0 ), and therefore ψ b for z = λσD (x) is well-defined. By a simple computation, ψ is a Banp -morphism b p (ψ b) ◦ σD thus finishing the proof. b fulfills ψ = 2 and ψ For D ∈ ACp define Sp (D) just as in the case of D ∈ TCp ; this is possible since only finite operations enter into the definition of Sp (D). Put Np (D) := {z ∈ Sp (D) | kzk = 0}. Since k − k is a p-seminorm on Sp (D), Np (D) is a subvector space of Sp (D). Define Sp,fin (D) := Sp (D)/Np (D) and let πD : Sp (D) → b p,fin ◦ Sp,fin (D) Sp,fin (D) be the canonical projection. Denote by σD,fin : D → 0 b the restriction of πD ◦ σD to p,fin ◦ Sp,fin (D). πD is an isometry and one has the following three propositions: PROPOSITION 2.7. For D ∈ ACp the following hold: (i) kσD0 (x)k ≤ kxk (x ∈ D). P P (ii) For xi ∈ D (i ∈ N), α ∈ p,fin one has σD0 ( i αi xi ) = i αi σD0 (xi ). ◦

PROPOSITION 2.8. For any p-absolutely convex space D, p,fin (Sp,fin (D)) ⊂ σD,fin (D) ⊂ p,fin (Sp,fin (D)) holds. b p,fin (Sp,fin (D)) is an ACp PROPOSITION 2.9. For D ∈ ACp , σD,fin : D → morphism. b p,fin : Vecp → ACp has as a left adjoint the THEOREM 2.10. The functor functor Sp,fin : ACp → Vecp with the object function D 7→ Sp,fin (D). The unit of this adjunction is σD,fin (D ∈ ACp ). b p,fin (V ) be an ACp -morphism, V ∈ Vecp . For z ∈ Proof. Let ψ: D → Sp (D) there exist λ ∈ K, x ∈ D with z = λσD0 (x), hence πD (z) = λ(πD ◦ b: Sp,fin (D) → V with ψ = σD0 )(x) = λσD,fin (x). Therefore, a Vecp -morphism ψ

P -BANACH SPACES AND P -TOTALLY CONVEX SPACES. II

593

b p,fin (ψ b)◦σD,fin is uniquely determined. Define ψ b: Sp,fin (D) → V by ψ b(πD (z)) :=

b is well-defined λψ(x) for z = λσD0 (x) (λ ∈ K, x ∈ D). By the proof of 2.6, ψ b and K-linear. For z as above one has kψ (πD (z))k = kλψ(x)k ≤ |λ|kxk ≤ |λ|, b is a Vecp -morphism with ψ = b(πD (z))k ≤ kzk = kπD (z)k. Thus ψ hence kψ b p,fin (ψ b) ◦ σD,fin , finishing the proof.

2

References 1. 2. 3. 4.

5. 6. 7. 8. 9. 10.

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