Extremal function pairs in asymmetric normed linear spaces

Topology and its Applications 166 (2014) 98–107

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Extremal function pairs in asymmetric normed linear spaces ✩ Olivier Olela Otafudu School of Mathematical Sciences, North-West University (Mafikeng campus), Mmabatho 2735, South Africa

a r t i c l e

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Article history: Received 15 October 2013 Received in revised form 22 January 2014 Accepted 20 February 2014 MSC: 54E15 54B20 54E35 54C60 54E55

a b s t r a c t In [4] Kemajou et al. constructed the injective hull in the category of T0 -quasi-metric spaces with nonexpansive maps that they called q-hyperconvex hull. In this paper, we study properties of functions pairs of the q-hyperconvex hull of asymmetric normed linear spaces. We show, for instance, that any point in the q-hyperconvex hull of an asymmetric normed linear space is convex. © 2014 Elsevier B.V. All rights reserved.

Keywords: Ampleness Asymmetric norm T0 -quasi-metric Bicompletion q-Hyperconvex q-Hyperconvex hull

1. Introduction In [4] Kemajou et al. studied the concept of hyperconvexity that is appropriate to the category of T0 -quasi-metric spaces and nonexpansive maps. They provided an explicit construction of the corresponding hull (called q-hyperconvex hull, or Isbell-convex hull) of a T0 -quasi-metric space. In this paper we study properties of functions pairs in the q-hyperconvex hull of an asymmetric normed linear space. We point out that every point in the q-hyperconvex hull of an asymmetric normed linear space is convex and we show that the translation of any point in the injective hull of an asymmetric normed linear space is extremal. We hope that the results of this paper will be applied for our further investigations of Banach properties of the q-hyperconvex hull of an asymmetric normed linear space. ✩

The author would like to thank the South African National Research Foundation for partial financial support. E-mail address: [email protected].

http://dx.doi.org/10.1016/j.topol.2014.02.006 0166-8641/© 2014 Elsevier B.V. All rights reserved.

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2. Preliminaries For the convenience of the reader and in order to fix our terminology we recall the following concepts. Definition 1. Let X be a set and let d : X × X → [0, ∞) be a function mapping into the set [0, ∞) of the nonnegative reals. Then d is called a quasi-pseudometric on X if (a) d(x, x) = 0 whenever x ∈ X, (b) d(x, z)  d(x, y) + d(y, z) whenever x, y, z ∈ X. We shall say that d is a T0 -quasi-metric provided that d also satisfies the following condition: For each x, y ∈ X, d(x, y) = 0 = d(y, x) implies that x = y. Remark 1. In some cases we need to replace [0, ∞) by [0, ∞] (where for a d attaining the value ∞ the triangle inequality is interpreted in the obvious way). In such a case we shall speak of an extended quasi-pseudometric. Remark 2. Let d be a quasi-pseudometric on a set X, then d−1 : X × X → [0, ∞) defined by d−1 (x, y) = d(y, x) whenever x, y ∈ X is also a quasi-pseudometric, called the conjugate quasi-pseudometric of d. As usual, a quasi-pseudometric d on X such that d = d−1 is called a pseudometric. Note that for any (T0 -)quasi-pseudometric d, ds = max{d, d−1 } = d ∨ d−1 is a pseudometric (metric). ˙ b = max{a − b, 0}. For any a, b ∈ R, we shall set a − The following definition can be found in [2] (compare [3]). Definition 2. Let X be a linear space over R and let .| : X → [0, ∞) be a function mapping into the set [0, ∞) of the nonnegative reals. Then .| is called an asymmetric norm on X if (AN1) x| = −x| = 0 =⇒ x = 0 for all x ∈ X, (AN2) αx| = αx| for all x ∈ X and α  0, (AN3) x + y|  x| + y| for all x, y ∈ X. The pair (X, .|) is called an asymmetric normed linear space. Remark 3. Let .| be an asymmetric norm on a linear space X over R, then |. : X → [0, ∞) defined by |x = −x| whenever x ∈ X is also an asymmetric norm, called the conjugate asymmetric norm of .|. As usual, an asymmetric norm .| on X such that .| = |. is called a norm. Furthermore, for any asymmetric norm .|, . = max{.|, |.} is a norm and (X, .) is a normed linear space. The asymmetric norm induces, in a natural way, a quasi-metric d.| on X defined by d.| (x, y) = x − y| for all x, y ∈ X. The following example is well-known, but important. Example 1. ([2, Example 1.1.3]) Let the set R of real numbers be equipped with the asymmetric norm v(x) = x+ = max{x, 0}. Then, for x ∈ R, v −1 (x) = x− = max{−x, 0} and v s = |x|. The topology τ (v) generated by v is called the upper topology of R; while the topology τ (v −1 ) generated by v −1 is called the lower topology of R.

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3. q-Hyperconvex space hull of a T0 -quasi-metric space The following definition was given in [4]. A T0 -quasi-metric space (X, d) is said to be q-hyperconvex if for each family (xi )i∈I of points of X and families (ri )i∈I and (si )i∈I of nonnegative real numbers the  following conditions hold: if d(xi , xj )  ri + sj for all i, j ∈ I, then i∈I (Cd (xi , ri ) ∩ Cd−1 (xi , si )) = ∅, where Cd (x, r) := {y ∈ X: d(x, y)  r} denotes the closed ball of center x and radius r > 0. Let (X, d) be a T0 -quasi-metric space. We say that a function pair f = (f1 , f2 ) on (X, d), where fi : X → [0, ∞), is ample if d(x, y)  f2 (x) + f1 (y) whenever x, y ∈ X. Let Aq (X, d) be the set of all ample function pairs on (X, d). (If we do not have confusion about d we may ˙ g1 (x)) ∨ supx∈X (g2 (x) − ˙ f2 (x)). also write Aq (X).) For each f, g ∈ Aq (X) we set D(f, g) = supx∈X (f1 (x) − Then D is an extended T0 -quasi-metric on Aq (X) (by extended quasi-pseudometric we mean: a quasipseudometric in which we replace in its definition [0, ∞) by [0, ∞]). A function pair f with fi : X → [0, ∞) (i = 1, 2) is called minimal (or extremal) on (X, d) (among the ample function pairs on (X, d)) if it is ample and whenever g is ample on (X, d) and for each x ∈ X we have g1 (x)  f1 (x) and g2 (x)  f2 (x) (in this case we shall write g  f ), then g = f . By q (X, d) we shall denote the set of all minimal ample pairs on (X, d) equipped with the restriction of D to q (X, d) × q (X, d), which for convenience we shall denote by D. Then D is a (real valued) T0 -quasi-metric on q (X, d) × q (X, d) (see [4, Remark 6]). It was shown in [5, Remark 2] that any f ∈ Aq (X, d) belongs to q (X, d) if and only if   ˙ f1 (y) f2 (x) = sup d(x, y) − y∈X

and   ˙ f2 (y) f1 (x) = sup d(y, x) − y∈X

whenever x ∈ X. Furthermore, if f ∈ q (X, d) implies that f1 (x) − f1 (y)  d(y, x) and f2 (x) − f2 (y)  d(x, y) for all x, y ∈ X and if f, g ∈ q (X, d) then     ˙ g1 (x) = sup g2 (x) − ˙ f2 (x) D(f, g) = sup f1 (x) − x∈X

x∈X

whenever f, g ∈ q (X, d) (see [4, Lemma 7]). For all x ∈ X, we define the minimal function pair   fx (y) = d(x, y), d(y, x) whenever y is a point in the T0 -quasi-metric space. The map eX defined by eX (x) = fx whenever x ∈ X is an isometric embedding of (X, d) into (q (X, d), D) for more details (see [4, Lemma 1]). The pair (q (X, d), D) is called the q-hyperconvex hull of (X, d) and it is q-hyperconvex. It is well-known that D(f, fx ) = f1 (x) and D(fx , f ) = f2 (x) whenever x ∈ X and f ∈ q (X, d) [4, Lemma 8]. 4. Some first results Lemma 1. Let (X, .|) be an asymmetric normed linear space and let f = (f1 , f2 ) be an ample function pair in (X, .|). Then the function pair defined by f s = (f1s , f2s ), where for s ∈ X, fis (x) is defined by fi (x + s) for x ∈ X and i = 1, 2, is ample too.

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Proof. Let x, y ∈ X. Then x − y| = x − s + s − y|  f2 (x + s) + f1 (y + s) = f2s (x) + f1s (y).

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2

Lemma 2. Let (X, .|) be an asymmetric normed linear space and let f = (f1 , f2 ) be an ample function pair in (X, .|). Then the function pair defined by ϕ(x) = (bf1 ( xb ), bf2 ( xb )) whenever x ∈ X and b > 0 is ample too. Proof. Let x, y ∈ X. Since f = (f1 , f2 ) is ample function pair, then        x y  1   f2 x + f1 y . x − y| =  − b b b b b Thus x − y|  bf2 ( xb ) + bf1 ( xb ) = ϕ2 (x) + ϕ1 (y). Therefore ϕ is ample. 2 Proposition 1. Let (X, .|) be an asymmetric normed linear space. (1) If f = (f1 , f2 ) ∈ q (X, .|) then function pair f s , defined in Lemma 1, where s ∈ X, is an element of q (X, .|). (2) Let b > 0. If f = (f1 , f2 ) ∈ q (X, .|) then function pair ϕ(x) = (bf1 ( xb ), bf2 ( xb )), x ∈ X, defined in Lemma 2, is an element of q (X, .|). Proof. (1) By Lemma 1 f s is ample. We need to show that f s is extremal where s ∈ X. Let g = (g1 , g2 ) be an ample function pair such that g1 (x)  f1s (x) and g2 (x)  f2s (x) whenever x ∈ X. Then g1 (y − s)  f1 (y) and g2 (y − s)  f2 (y) whenever y ∈ X and since f = (f1 , f2 ) is an extremal function pair on (X, .|) and from Lemma 1 g −s is still an ample function pair on (X, .|). By the minimality of f = (f1 , f2 ) we have g1 (y − s) = f1 (y) and g2 (y − s) = f2 (y) whenever y ∈ X. Therefore g1 (x) = f1s (x) and g2 (x) = f2s (x) whenever x ∈ X. Hence f s ∈ q (X, .|). (2) Consider f = (f1 , f2 ) an extremal function pair on (X, .|) and ϕ the ample function pair defined in Lemma 2. We have to show that ϕ is minimal. Let h = (h1 , h2 ) be an ample function pair on (X, .|) such that for b > 0, h1 (x)  bf1 ( xb ) and h2 (x)  bf2 ( xb ). Then 1b h1 (x)  f1 ( xb ) and 1b h2 (x)  f2 ( xb ) whenever x ∈ X. Hence 1b h1 (by)  f1 (y) and 1b h1 (by)  f1 (y) whenever y ∈ X. By Lemma 2 ( 1b h1 (by), 1b h2 (by)) is an ample function pair on (X, .|) and by the minimality of f = (f1 , f2 ) we have 1b h1 (by) = f1 (y) and 1b h2 (by) = f2 (y) whenever y ∈ X. Thus h1 (x) = bf1 ( xb ) and h2 (x) = bf2 ( xb ) whenever x ∈ X. 2 Proposition 2. Let (X, .|) be an asymmetric normed linear space. Then every element of q (X, .|) is convex. Moreover, if f = (f1 , f2 ) ∈ q (X, .|) then for each s ∈ X, the function pair ψ defined by ψ(x) = ˙ f1 (s), f2 (x + s) − ˙ f2 (s)) for any x ∈ X is still convex. (f1 (x + s) − Proof. Let f = (f1 , f2 ) ∈ q (X, .|). We have to show that f1 and f2 are convex. Suppose that f2 is not convex: there are x0 , y0 ∈ X and α ∈ (0, 1) such that αf2 (x0 ) + (1 − α)f2 (y0 ) < f2 (z0 ) where z0 = αx0 + (1 − α)y0 . Set g2 (z) = f2 (z) if z ∈ X, and g2 (z0 ) = αf2 (x0 ) + (1 − α)f2 (y0 ) if z = z0 . Clearly, (f1 , g2 ) < (f1 , f2 ). Now consider x ∈ X with x = z0 , we have     z0 − x| = αx0 + (1 − α)y0 − x = α(x0 − x) + (1 − α)(y0 − x)

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which implies that z0 − x|  αx0 − x| + (1 − α)y0 − x|. But x0 − x|  f2 (x0 ) + f1 (x) and y0 − x|  f2 (y0 ) + f1 (x). Then



z0 − x|  α f2 (x0 ) + f1 (x) + (1 − α) f2 (y0 ) + f1 (x)  αf2 (x0 ) + (1 − α)f2 (y0 ) + f1 (x) < g2 (z0 ) + f1 (x) whenever z0 , x ∈ X. It follows that (f1 , g2 ) is ample and we have reached a contradiction to the minimality of (f1 , f2 ). By similar arguments, one shows that f1 is convex. Moreover f = (f1 , f2 ) is convex. Namely that the function ψ defined, for fixed s ∈ X, by   ψ(x) = f1 (x + s) − f1 (s), f2 (x + s) − f2 (s) , x ∈ X is convex. The convexity of ψ1 (·) = f1s (·) − f1 (s) is obvious since any translate a + h of a convex function h is convex too. Similarly by the same argument ψ2 is convex too. Therefore ψ = (ψ1 , ψ2 ) is a convex function pair. 2 Lemma 3. (Compare [4, Lemma 3].) Let f = (f1 , f2 ) ∈ q (X, .|). Then f1 (x) − f1 (y)  |x − y whenever x, y ∈ X and f2 (x) − f2 (y)  x − y| whenever x, y ∈ X. (Hence f1 is nonexpansive map on (X, |.) and f2 is nonexpansive on (X, .|)), when considered as maps into ([0, ∞), v), where v is the asymmetric norm in Example 1. 5. More properties of extremal function pairs In this section d is the T0 -quasi-metric induced by .|. The following lemma will be crucial to define algebraic properties in the injective hull of an asymmetric normed linear space for further investigations. Lemma 4. (Compare [1, Lemma 2.2].) Let (X, .|) be an asymmetric normed linear space. If we identify x ∈ X with the distance function pair from x,i.e., with eX (x)(y) = (x − y|, y − x|), then the algebraic operations on X are transformed by eX in the following way: for x1 , x2 ∈ X and λ  0 and y ∈ X

  eX (x1 + x2 )(y) = eX (x1 ) ⊕ eX (x2 ) (y) := eX (x1 + x2 )1 (y), eX (x1 + x2 )2 (y) with   eX (x1 + x2 )1 (y) := inf d(x1 , z) + d(x2 , y − z) z∈X

and   eX (x1 + x2 )2 (y) = inf d−1 (x1 , z) + d−1 (x2 , y − z) . z∈X

Furthermore

  eX (λx)(y) = λ • eX (x) (y) := eX (λx)1 (y), eX (λx)2 (y)

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with    y  x − eX (λx)1 (y) = λ  λ

   y eX (λx)2 (y) = λx −  λ

and

if λ > 0

and eX (λx)1 (y) = y|

and

eX (λx)2 (y) = |y

if λ = 0.

Proof. By definition of eX we have   eX (x1 + x2 )1 (y) = x1 + x2 − y|  x1 − z| + x2 − (y − z). For z = x1 , we have the equality   eX (x1 + x2 )1 (y) = inf d(x1 , z) + d(x2 , y − z) . z∈X

Similarly we have   eX (x1 + x2 )2 (y) = inf d−1 (x1 , z) + d−1 (x2 , y − z) . z∈X

For x, y ∈ X and λ  0, then    y  x − eX (λx)1 (y) = λx − y| = λ  λ

if λ > 0

and by similar argument we have    y if λ > 0. eX (λx)2 (y) = λx −  λ For λ = 0, eX (λx)1 (y) = y|

and eX (λx)2 (y) = |y.

2

Lemma 5. (Compare [1, Lemma 2.3].) Let (X, .|) be an asymmetric normed linear space. For f = (f1 , f2 ) ∈ q (X, .|), the function pair f = (f1 , f2 ) defined, for every h ∈ q (X, .|) by     ˙ f2 (z) f1 (h) = sup Dd eX (z), h − z∈X

and     ˙ f1 (z) f2 (h) = sup Dd h, eX (z) − z∈X

is the only extremal extension function pair of f to q (X, .|). Moreover Dd (f, g) = Dd (f, g) whenever f, g ∈ q (X, .|). Proof. Indeed f = (f1 , f2 ) = (Dd (f, h), Dd (h, f )) implies that f is an extremal function pair on q (X, .|). Furthermore, we have     f1 ex (y) = Dd f, eX (y) = f1 (y) and

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    f2 ex (y) = Dd eX (y), f = f2 (y) whenever y ∈ X. Let g = (g1 , g2 ) be an extremal extension of f ∈ q (X, .|). We have g1 (h) =

      ˙ g2 (u)  sup Dd h, eX (z) − ˙ f2 (z) = f1 (h) Dd (h, u) −

sup

z∈X

u∈q (X,.|)

and g2 (h) =

      ˙ g1 (u)  sup Dd eX (z), h − ˙ f1 (z) = f2 (h). Dd (u, h) −

sup

z∈X

u∈q (X,.|)

Then f = g, by the minimality of g.

2

Remark 4. If we want to define addition (⊕) and multiplication (•) operations on q (X, .|). Then, we would like to point out that eX must be a linear function pair, eX (0) = (eX (0)1 , eX (0)2 ) has to be the neutral element for ⊕ and consequently the asymmetric norm of q (X, .|), if it exists, must be defined by   f | = Dd f, eX (0) = f1 (0)

  and |f  = Dd eX (0), f = f2 (0).

(1)

Therefore f  = max{f1 (0), f2 (0)}. Proposition 3. Let (X, .|) be an asymmetric normed linear space. Let q (X, .|) be equipped with the asymmetric norm defined in (1) and for any g ∈ q (X, .|) define a function pair ψ by      ˙ g1 (s) , sup g2 (x + s) − ˙ g2 (s) ψ(x) = sup g1 (x + s) − s∈X

s∈X

whenever x ∈ X.

Then ψ1 (x) = |x,

ψ2 (x) = x|

whenever x ∈ X.

Proof. Let g ∈ q (X, .|). Consider the function pair defined by      ˙ g1 (s) , sup g2 (x + s) − ˙ g2 (s) ψ(x) = sup g1 (x + s) − s∈X

s∈X

whenever x ∈ X.

We know that ψ is convex by Proposition 2. Furthermore, if x, y ∈ X, then   ˙ g1 (s) ψ1 (x + y) = sup g1 (x + y − s) − s∈X

implies     ˙ g1 (y + s) + sup g1 (y + s) − ˙ g1 (s) ψ1 (x + y)  sup g1 (x + y − s) − s∈X

s∈X

so that ψ1 (x + y)  ψ1 (x) + ψ1 (y). Analogously we have ψ2 (x + y)  ψ2 (x) + ψ2 (y).

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Thus ψ1 and ψ2 are subadditive functions with ψ1 (0) = 0 and ψ2 (0) = 0. We claim that ψ1 (px) = pψ1 (x) and ψ2 (px) = pψ2 (x) whenever x ∈ X and p > 0. Let us show that ψ1 (px) = pψ1 (x). The equality for ψ2 follows by symmetry. Let a be a real number such that 0 < a < 1, we have   ψ1 (ax) = ψ1 ax + (1 − a)0  aψ1 (x) + (1 − a)ψ1 (0) = aψ1 (x). Now consider p > 0, then there exists n ∈ N such that 1.n > p. Hence for any x ∈ X:   ψ1 (px) = ψ1 n(p/n)x  nψ1 (p/nx)  n(p/n)ψ1 (x) = pψ1 (x). Moreover, this is an equality, since:     ψ1 (x) = ψ1 (1/p)px  (1/p)ψ1 (px)  (1/p)p ψ1 (x) = ψ1 (x). Since g = (g1 , g2 ) ∈ q (X, .|), then g is ample. By taking s = 0 in the definition of ψ2 , we have

˙ g2 (0) = max 0, g2 (x) − g2 (0) , ψ2 (x)  g2 (x) −

x ∈ X.

The ampleness of g = (g1 , g2 ) implies that   g2 (x) − g2 (0) = g2 (x) + g1 (0) − g2 (0) − g1 (0)  x| − g2 (0) + g1 (0) . Hence   ˙ g2 (0) + g1 (0)  ψ2 (x). x| − ˙ (g2 (0) + g1 (0))  ψ2 ( xp ), for any p > 0 and x| − ˙ p(g2 (0) + g1 (0))  ψ2 (x). Therefore Thus  xp | − x|  ψ2 (x),

whenever x ∈ X.

It follows that g2 (x + s) − g2 (s)  x|, for all x, s ∈ X, since g2 is nonexpansive by Lemma 3. Moreover,   ˙ g2 (s)  x|. ψ2 (x) = sup g2 (x + s) − s∈X

Therefore ψ2 (x) = x| whenever x ∈ X. By similar arguments one shows that ψ1 (x) = |x whenever x ∈ X. 2 Proposition 4. Let (X, .|) be an asymmetric normed linear space. Let q (X, .|) be equipped with the asymmetric norm defined in (1). For any f, g ∈ q (X, .|) define a function pair ϕ by   ˙ g1 (s) ϕ1 (x) = sup f1 (x − s) − s∈X

and   ˙ g2 (s) ϕ2 (x) = sup f2 (x − s) − s∈X

whenever x ∈ X. Then the function pair ϕ = (ϕ1 , ϕ2 ) is ample.

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Proof. Consider f = (f1 , f2 ), g = (g1 , g2 ) ∈ q (X, .|). The function pair ϕ(x) = (ϕ1 (x), ϕ2 (x)), x ∈ X, ˙ g1 (s)) and ϕ2 (x) = sups∈X (f2 (x − s) − ˙ g2 (s)). where ϕ1 (x) = sups∈X (f1 (x − s) − We are going to show that ϕ = (ϕ1 , ϕ2 ) is ample. For any x, y ∈ X, by definition of ϕ = (ϕ1 , ϕ2 ), we have     ˙ g2 (s2 ) + sup f1 (y − s1 ) − ˙ g1 (s1 ) ϕ2 (x) + ϕ1 (y) = sups2 ∈X f2 (x − s2 ) − s1 ∈X

so that ˙ g2 (s2 ) + f1 (y − s1 ) − ˙ g1 (s1 ). ϕ2 (x) + ϕ1 (y)  f2 (x − s2 ) − Hence



ϕ2 (x) + ϕ1 (y)  max 0, f2 (x − s2 ) − g2 (s2 ) + max 0, f1 (y − s1 ) − g1 (s1 ) . Moreover

ϕ2 (x) + ϕ1 (y)  max 0, f2 (x − s2 ) − g2 (s2 ) + f1 (y − s1 ) − g1 (s1 ) . Furthermore,   ˙ g2 (s2 ) + g1 (s1 ) ϕ2 (x) + ϕ1 (y)  f2 (x − s2 ) + f1 (y − s1 ) −

(2)

  ˙ g2 (s2 ) + g1 (s1 )  f2 (x − s2 ) + f1 (y − s1 ) − ˙ g2 (s2 ) − ˙ g1 (s1 ). f2 (x − s2 ) + f1 (y − s1 ) −

(3)

and

Since f = (f1 , f2 ) is an ample function pair, we have f2 (x − s2 ) + f1 (y − s1 )  x − y + s1 − s2 | so that f2 (x − s2 ) + f1 (y − s1 ) − g2 (s2 )  x − y + s1 − s2 | − g2 (s2 ) hence ˙ g2 (s2 )  x − y + s1 − s2 | − ˙ g2 (s2 ). f2 (x − s2 ) + f1 (y − s1 ) − Moreover,   ˙ g2 (s2 )  sup x − y + s1 − s2 | − ˙ g2 (s2 ) = g1 (x − y + s1 ) f2 (x − s2 ) + f1 (y − s1 ) − s2 ∈X

which implies ˙ g2 (s2 ) − ˙ g1 (s1 )  g1 (x − y + s1 ) − ˙ g1 (s1 ) f2 (x − s2 ) + f1 (y − s1 ) − consequently,   ˙ g2 (s2 ) − ˙ g1 (s1 )  sup g1 (x − y + s1 ) − ˙ g1 (s1 ) . f2 (x − s2 ) + f1 (y − s1 ) − s1 ∈X

(4)

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By (2), (3) and (4) we have   ˙ g1 (s1 ) = ψ1 (x − y) = x − y|. ϕ2 (x) + ϕ1 (y)  sups1 ∈X g1 (x − y + s1 ) − Therefore ϕ is an ample function pair. 2 Problem 1. Let (X, .|) be an asymmetric normed linear space. Under which condition is ϕ defined in Proposition 4 extremal? Problem 2. Let (X, .|) be an asymmetric normed linear space. How can the operation ⊕ be defined in q (X, .|) such that it is commutative? Acknowledgement The author would like to thank the referee for comments and suggestions which have improved the paper. References [1] F. Cianciaruso, E. De Pascale, Discovering the algebraic structure on the metric injective envelope of a real Banach space, Topology Appl. 78 (1997) 285–292. [2] S. Cobzas, Functional Analysis in Asymmetric Normed Spaces, Frontiers in Mathematics, Springer, Basel, 2013. [3] M. Grabiec, Y. Je Cho, V. Radu, On Nonsymmetric Topological and Probabilistic Structures, Nova Science Publ., New York, 2006. [4] E. Kemajou, H.-P.A. Künzi, O.O. Otafudu, The Isbell-hull of a di-space, Topology Appl. 159 (2012) 2463–2475. [5] H.-P.A. Künzi, M. Sanchis, The Katêtov construction modified for a T0 -quasi-metric space, Topology Appl. 159 (2012) 711–720.