Uniqueness of extension of semi-Lipschitz functions on quasi-metric spaces

Costica Mustata
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Costica Mustata
“Tiberiu Popoviciu” Institute of Numerical Analysis, Romanian Academy, Romania

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C. Mustăţa, Uniqueness of extension of semi-Lipschitz functions on quasi-metric spaces, Bul. Ştiinţ. Univ. Baia Mare, Seria B, Fascicola matematică-informatică, 16 (2000) no. 2, pp. 207-212.

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Bul. Stiint. Univ. Baia Mare

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[1] Cobzas, S., Mustata, C., Norm Preserving Extension of Convex Lipschitz Functions , J. Approx. Theory, 24 (1978), 555-564.
[2] Fletcher, P., Lindgren, W.F., Quasi-Uniform Spaces, Dekker, York, 1982.
[3] McShane, J. A,, Extension of range of functions, Bull. Amer. Math. Soc. 40(1934) 837-842.
[4] Mustata, C., Best Approximation and Unique Extension of Lipschitz Functions, J. Approx. Theory 19(1977), 222-230.
[5] Mustata, C., On the Extension of Semi-Lipschitz Functions on Quasi-Metric Space (to appear).
[6] Romaguera, S., Sanchis, M., Semi-Lipschitz Functions and Best Approximation in Quasi-Metric Spaces, J. Approx. Theory 103 (2000) 292-301.
[7] Wells, J.H., Williams, L.R., Embeddings and Extensions in Analysis, Springer- Verlag, Berlin 1975.

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2000-Mustata-Bul.-Stiint.-Univ.-Baia-Mare-Uniqueness-of-the-extension-of-semi-Lipschitz

Dedicated to Maria S. Pop on her 60 th 60 th  60^("th ")60^{\text {th }}60th  anniversary

UNIQUENESS OF THE EXTENSION OF SEMI-LIPSCHITZ FUNCTIONS ON QUASI - METRIC SPACES

Costică MUSTĂTA

Let X X XXX be a nonvoid set and d : X × X → [ 0 , ∞ ) d : X × X → [ 0 , ∞ ) d:X xx X rarr[0,oo)d: X \times X \rightarrow[0, \infty)d:X×X→[0,∞) a function satisfying the following conditions:
(i) d ( x , y ) = 0 ⟺ x = y d ( x , y ) = 0 ⟺ x = y d(x,y)=0Longleftrightarrow x=yd(x, y)=0 \Longleftrightarrow x=yd(x,y)=0⟺x=y
(ii) d ( x , y ) ≤ d ( x , z ) + d ( z , y ) d ( x , y ) ≤ d ( x , z ) + d ( z , y ) d(x,y) <= d(x,z)+d(z,y)d(x, y) \leq d(x, z)+d(z, y)d(x,y)≤d(x,z)+d(z,y)
for all x , y , z ∈ X x , y , z ∈ X x,y,z in Xx, y, z \in Xx,y,z∈X. We call d d ddd a quasi-metric on X X XXX and the pair ( X , d X , d X,dX, dX,d ), a quasimetric space. Remark that the main difference with respect to a metric is the symmetry condition, d ( x , y ) = d ( y , x ) d ( x , y ) = d ( y , x ) d(x,y)=d(y,x)d(x, y)=d(y, x)d(x,y)=d(y,x), which is not satisfied by a quasimetric.
The conjugate of a quasi-metric d d ddd,denoted by d − 1 d − 1 d^(-1)d^{-1}d−1 is defined by
(1) d − 1 ( x , y ) = d ( y , x ) (1) d − 1 ( x , y ) = d ( y , x ) {:(1)d^(-1)(x","y)=d(y","x):}\begin{equation*} d^{-1}(x, y)=d(y, x) \tag{1} \end{equation*}(1)d−1(x,y)=d(y,x)
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X. Obviously, that the mapping d s : X × X → [ 0 , ∞ ) d s : X × X → [ 0 , ∞ ) d^(s):X xx X rarr[0,oo)d^{s}: X \times X \rightarrow[0, \infty)ds:X×X→[0,∞) defined by
(2) d s ( x , y ) = max { d ( x , y ) , d − 1 ( x , y ) } , x , y ∈ X (2) d s ( x , y ) = max d ( x , y ) , d − 1 ( x , y ) , x , y ∈ X {:(2)d^(s)(x","y)=max{d(x,y),d^(-1)(x,y)}","x","y in X:}\begin{equation*} d^{s}(x, y)=\max \left\{d(x, y), d^{-1}(x, y)\right\}, x, y \in X \tag{2} \end{equation*}(2)ds(x,y)=max{d(x,y),d−1(x,y)},x,y∈X
is a metric on X X XXX, i.e. d s d s d^(s)d^{s}ds satisfies the conditions (i), (ii) and the symmetry condition:
(iii) d s ( x , y ) = d s ( y , x ) , x , y ∈ X (iii) d s ( x , y ) = d s ( y , x ) , x , y ∈ X {:(iii)d^(s)(x","y)=d^(s)(y","x)","quad x","y in X:}\begin{equation*} d^{s}(x, y)=d^{s}(y, x), \quad x, y \in X \tag{iii} \end{equation*}(iii)ds(x,y)=ds(y,x),x,y∈X
A function f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R,defined on a quasi-metric space ( X , d X , d X,dX, dX,d ) is called semi-Lipschitz provided there exists a number K ≥ 0 K ≥ 0 K >= 0K \geq 0K≥0 such that
(3) f ( x ) − f ( y ) ≤ K d ( x , y ) (3) f ( x ) − f ( y ) ≤ K d ( x , y ) {:(3)f(x)-f(y) <= Kd(x","y):}\begin{equation*} f(x)-f(y) \leq K d(x, y) \tag{3} \end{equation*}(3)f(x)−f(y)≤Kd(x,y)
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X. A function f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R is called ≤ d − ≤ d − <= _(d^(-))\leq_{d^{-}}≤d−increasing if
(3a) d ( x , y ) = 0 ⟹ f ( x ) − f ( y ) ≤ 0 (3a) d ( x , y ) = 0 ⟹ f ( x ) − f ( y ) ≤ 0 {:(3a)d(x","y)=0Longrightarrow f(x)-f(y) <= 0:}\begin{equation*} d(x, y)=0 \Longrightarrow f(x)-f(y) \leq 0 \tag{3a} \end{equation*}(3a)d(x,y)=0⟹f(x)−f(y)≤0
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X.
The definition of ≤ d ≤ d <= _(d)\leq_{d}≤d - increasing function f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R is consistent for T 0 T 0 T_(0)T_{0}T0 - separated quasi-metric space ( X , d ) ( X , d ) (X,d)(X, d)(X,d) (see [6]). In this note the quasi-metric space ( X . d X . d X.dX . dX.d ) is T 1 T 1 T_(1)T_{1}T1 - separated (see the condition (i) and (ii)).
Since d ( x , y ) = 0 ⟺ x = y d ( x , y ) = 0 ⟺ x = y d(x,y)=0Longleftrightarrow x=yd(x, y)=0 \Longleftrightarrow x=yd(x,y)=0⟺x=y, it follows that f ( x ) ≤ f ( y ) f ( x ) ≤ f ( y ) f(x) <= f(y)f(x) \leq f(y)f(x)≤f(y) for any function f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R i.e. any real - valued function on a quasi - metric space X X XXX is ≤ d ≤ d <= _(d)\leq_{d}≤d - increasing.
Theorem 1 Let f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R be such that
(4) ‖ f ‖ d = sup { ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) : x , y ∈ X , d ( x , y ) > 0 } < ∞ (4) ‖ f ‖ d = sup ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) : x , y ∈ X , d ( x , y ) > 0 < ∞ {:(4)||f||_(d)=s u p{((f(x)-f(y))vv0)/(d(x,y)):x,y in X,d(x,y) > 0} < oo:}\begin{equation*} \|f\|_{d}=\sup \left\{\frac{(f(x)-f(y)) \vee 0}{d(x, y)}: x, y \in X, d(x, y)>0\right\}<\infty \tag{4} \end{equation*}(4)‖f‖d=sup{(f(x)−f(y))∨0d(x,y):x,y∈X,d(x,y)>0}<∞
Then f f fff satisfies the inequality
(5) f ( x ) − f ( y ) ≤ ‖ f ‖ d ⋅ d ( x , y ) , ∀ x , y ∈ X (5) f ( x ) − f ( y ) ≤ ‖ f ‖ d ⋅ d ( x , y ) , ∀ x , y ∈ X {:(5)f(x)-f(y) <= ||f||_(d)*d(x","y)","AA x","y in X:}\begin{equation*} f(x)-f(y) \leq\|f\|_{d} \cdot d(x, y), \forall x, y \in X \tag{5} \end{equation*}(5)f(x)−f(y)≤‖f‖d⋅d(x,y),∀x,y∈X
and ‖ f ‖ d ‖ f ‖ d ||f||_(d)\|f\|_{d}‖f‖d is the smallest constant for which the inequality (3) holds.
P roof. nce f f fff is ≤ d ≤ d <= _(d)\leq_{d}≤d - increasing (see (3a)) it follows that f ( x ) − f ( y ) > 0 f ( x ) − f ( y ) > 0 f(x)-f(y) > 0f(x)-f(y)>0f(x)−f(y)>0 implies d ( x , y ) > 0 d ( x , y ) > 0 d(x,y) > 0d(x, y)>0d(x,y)>0. But then
f ( x ) − f ( y ) d ( x , y ) > 0 and ‖ f ‖ d = sup d ( x , y ) > 0 ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) ≥ f ( x ) − f ( y d ( x , y ) f ( x ) − f ( y ) d ( x , y ) > 0  and  ‖ f ‖ d = sup d ( x , y ) > 0   ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) ≥ f ( x ) − f ( y d ( x , y ) (f(x)-f(y))/(d(x,y)) > 0" and "||f||_(d)=s u p_(d(x,y) > 0)((f(x)-f(y))vv0)/(d(x,y)) >= (f(x)-f(y)/(d(x,y))\frac{f(x)-f(y)}{d(x, y)}>0 \text { and }\|f\|_{d}=\sup _{d(x, y)>0} \frac{(f(x)-f(y)) \vee 0}{d(x, y)} \geq \frac{f(x)-f(y}{d(x, y)}f(x)−f(y)d(x,y)>0 and ‖f‖d=supd(x,y)>0(f(x)−f(y))∨0d(x,y)≥f(x)−f(yd(x,y)
implying
f ( x ) − f ( y ) ≤ ‖ f ‖ d ⋅ d ( x , y ) . f ( x ) − f ( y ) ≤ ‖ f ‖ d ⋅ d ( x , y ) . f(x)-f(y) <= ||f||_(d)*d(x,y).f(x)-f(y) \leq\|f\|_{d} \cdot d(x, y) .f(x)−f(y)≤‖f‖d⋅d(x,y).
If f ( x ) − f ( y ) ≤ 0 f ( x ) − f ( y ) ≤ 0 f(x)-f(y) <= 0f(x)-f(y) \leq 0f(x)−f(y)≤0 then
( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) = 0 ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) = 0 ((f(x)-f(y))vv0)/(d(x,y))=0\frac{(f(x)-f(y)) \vee 0}{d(x, y)}=0(f(x)−f(y))∨0d(x,y)=0
implying f ( x ) − f ( y ) ≤ ‖ f ‖ d ⋅ d ( x , y ) f ( x ) − f ( y ) ≤ ‖ f ‖ d ⋅ d ( x , y ) f(x)-f(y) <= ||f||_(d)*d(x,y)f(x)-f(y) \leq\|f\|_{d} \cdot d(x, y)f(x)−f(y)≤‖f‖d⋅d(x,y).
Let now K ≥ 0 K ≥ 0 K >= 0K \geq 0K≥0 be such that
f ( x ) − f ( y ) ≤ K ⋅ d ( x , y ) f ( x ) − f ( y ) ≤ K ⋅ d ( x , y ) f(x)-f(y) <= K*d(x,y)f(x)-f(y) \leq K \cdot d(x, y)f(x)−f(y)≤K⋅d(x,y)
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X. Then f f fff is ≤ d ≤ d <= _(d)\leq_{d}≤d - increasing and
( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) = f ( x ) − f ( y ) d ( x , y ) ≤ K if f ( x ) − f ( y ) > 0 ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) = f ( x ) − f ( y ) d ( x , y ) ≤ K  if  f ( x ) − f ( y ) > 0 ((f(x)-f(y))vv0)/(d(x,y))=(f(x)-f(y))/(d(x,y)) <= K quad" if "quad f(x)-f(y) > 0\frac{(f(x)-f(y)) \vee 0}{d(x, y)}=\frac{f(x)-f(y)}{d(x, y)} \leq K \quad \text { if } \quad f(x)-f(y)>0(f(x)−f(y))∨0d(x,y)=f(x)−f(y)d(x,y)≤K if f(x)−f(y)>0
and
( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) = 0 ≤ K if f ( x ) − f ( y ) ≤ 0 . ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) = 0 ≤ K  if  f ( x ) − f ( y ) ≤ 0 . ((f(x)-f(y))vv0)/(d(x,y))=0 <= K quad" if "quad f(x)-f(y) <= 0.\frac{(f(x)-f(y)) \vee 0}{d(x, y)}=0 \leq K \quad \text { if } \quad f(x)-f(y) \leq 0 .(f(x)−f(y))∨0d(x,y)=0≤K if f(x)−f(y)≤0.
Consequently, ‖ f ‖ d ≤ K ‖ f ‖ d ≤ K ||f||_(d) <= K\|f\|_{d} \leq K‖f‖d≤K.
Denoting by SLip X X XXX the set of all real - valued semi - Lipschitz functions defined on a quasi - metric space ( X , d X , d X,dX, dX,d ) we have
(6) SLip X = { f : X → R , sup d ( x , y ) > 0 ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) < ∞ } (6)  SLip  X = f : X → R , sup d ( x , y ) > 0   ( f ( x ) − f ( y ) ) ∨ 0 d ( x , y ) < ∞ {:(6)" SLip "X={f:X rarrR,s u p_(d(x,y) > 0)((f(x)-f(y))vv0)/(d(x,y)) < oo}:}\begin{equation*} \text { SLip } X=\left\{f: X \rightarrow \mathbb{R}, \sup _{d(x, y)>0} \frac{(f(x)-f(y)) \vee 0}{d(x, y)}<\infty\right\} \tag{6} \end{equation*}(6) SLip X={f:X→R,supd(x,y)>0(f(x)−f(y))∨0d(x,y)<∞}
Let Y ⊂ X , Y ≠ ∅ Y ⊂ X , Y ≠ ∅ Y sub X,Y!=O/Y \subset X, Y \neq \emptysetY⊂X,Y≠∅, where ( X , d ) ( X , d ) (X,d)(X, d)(X,d) is a quasi - metric space. It follows that ( Y , d Y , d Y,dY, dY,d ) is a quasi - metric space, too, and let's denote by SLip Y Y YYY the set of all semi - Lipschitz functions on Y Y YYY.
The following extension problem arises naturally: for f ∈ S L i p Y f ∈ S L i p Y f in SLipYf \in S L i p Yf∈SLipY find F ∈ F ∈ F inF \inF∈ SLip X X XXX such that
(7) F | Y = f and ‖ F ‖ d = ‖ f ‖ d (7) F Y = f  and  ‖ F ‖ d = ‖ f ‖ d {:(7)F|_(Y)=f" and "||F||_(d)=||f||_(d):}\begin{equation*} \left.F\right|_{Y}=f \text { and }\|F\|_{d}=\|f\|_{d} \tag{7} \end{equation*}(7)F|Y=f and ‖F‖d=‖f‖d
The answer is affirmative. In [5] it was shown that the functions
(8) F ( x ) = inf y ∈ Y [ f ( y ) + ‖ f ‖ d ⋅ d ( x , y ) ] , x ∈ X (8) F ( x ) = inf y ∈ Y   f ( y ) + ‖ f ‖ d ⋅ d ( x , y ) , x ∈ X {:(8)F(x)=i n f_(y in Y)[f(y)+||f||_(d)*d(x,y)]","x in X:}\begin{equation*} F(x)=\inf _{y \in Y}\left[f(y)+\|f\|_{d} \cdot d(x, y)\right], x \in X \tag{8} \end{equation*}(8)F(x)=infy∈Y[f(y)+‖f‖d⋅d(x,y)],x∈X
(9) G ( x ) = sup y ∈ Y [ f ( y ) − ‖ f ‖ d ⋅ d − 1 ( x , y ) ] , x ∈ X (9) G ( x ) = sup y ∈ Y   f ( y ) − ‖ f ‖ d ⋅ d − 1 ( x , y ) , x ∈ X {:(9)G(x)=s u p_(y in Y)[f(y)-||f||_(d)*d^(-1)(x,y)]","x in X:}\begin{equation*} G(x)=\sup _{y \in Y}\left[f(y)-\|f\|_{d} \cdot d^{-1}(x, y)\right], x \in X \tag{9} \end{equation*}(9)G(x)=supy∈Y[f(y)−‖f‖d⋅d−1(x,y)],x∈X
satisfy the equalities
F | Y = G | Y = f and ‖ F ‖ d = ‖ G ‖ d = ‖ f ‖ d . F Y = G Y = f  and  ‖ F ‖ d = ‖ G ‖ d = ‖ f ‖ d . F|_(Y)=G|_(Y)=f quad" and "quad||F||_(d)=||G||_(d)=||f||_(d).\left.F\right|_{Y}=\left.G\right|_{Y}=f \quad \text { and } \quad\|F\|_{d}=\|G\|_{d}=\|f\|_{d} .F|Y=G|Y=f and ‖F‖d=‖G‖d=‖f‖d.
In other words, for any f ∈ S L i p Y f ∈ S L i p Y f in SLipYf \in S L i p Yf∈SLipY the set
(10) E Y d ( f ) := { H ∈ S Lip X : H | Y = f and ‖ H ‖ d = ‖ f ‖ d } (10) E Y d ( f ) := H ∈ S Lip X : H Y = f  and  ‖ H ‖ d = ‖ f ‖ d {:(10)E_(Y)^(d)(f):={H in S Lip X:H|_(Y)=f quad" and "quad||H||_(d)=||f||_(d)}:}\begin{equation*} E_{Y}^{d}(f):=\left\{H \in S \operatorname{Lip} X:\left.H\right|_{Y}=f \quad \text { and } \quad\|H\|_{d}=\|f\|_{d}\right\} \tag{10} \end{equation*}(10)EYd(f):={H∈SLipX:H|Y=f and ‖H‖d=‖f‖d}
of all extensions of f f fff which preserve the smallest Lipschitz constant is nonvoid.
Concerning the unicity of the extension ( card E Y d ( f ) = 1 card E Y d ( f ) = 1 cardE_(Y)^(d)(f)=1\operatorname{card} E_{Y}^{d}(f)=1cardEYd(f)=1 ) one can prove:
Theorem 2 Let ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a quasi-metric space, Y ⊂ X Y ⊂ X Y sub XY \subset XY⊂X and f ∈ S L i p Y f ∈ S L i p Y f in SLipYf \in S L i p Yf∈SLipY. Then
a) For every H ∈ E Y d ( f ) H ∈ E Y d ( f ) H inE_(Y)^(d)(f)H \in E_{Y}^{d}(f)H∈EYd(f) the following inequalities hold:
(11) G ( x ) ≤ H ( x ) ≤ F ( x ) , x ∈ X (11) G ( x ) ≤ H ( x ) ≤ F ( x ) , x ∈ X {:(11)G(x) <= H(x) <= F(x)","quad x in X:}\begin{equation*} G(x) \leq H(x) \leq F(x), \quad x \in X \tag{11} \end{equation*}(11)G(x)≤H(x)≤F(x),x∈X
where the functions F , G F , G F,GF, GF,G are defined by (8), (9);
b) card E Y d ( f ) = 1 E Y d ( f ) = 1 E_(Y)^(d)(f)=1E_{Y}^{d}(f)=1EYd(f)=1 if and only if
(12) sup y ∈ Y [ f ( y ) − ‖ f ‖ d d − 1 ( x , y ) ] = inf y ∈ Y [ f ( y ) + ‖ f ‖ d d ( x , y ) ] (12) sup y ∈ Y   f ( y ) − ‖ f ‖ d d − 1 ( x , y ) = inf y ∈ Y   f ( y ) + ‖ f ‖ d d ( x , y ) {:(12)s u p_(y in Y)[f(y)-||f||_(d)d^(-1)(x,y)]=i n f_(y in Y)[f(y)+||f||_(d)d(x,y)]:}\begin{equation*} \sup _{y \in Y}\left[f(y)-\|f\|_{d} d^{-1}(x, y)\right]=\inf _{y \in Y}\left[f(y)+\|f\|_{d} d(x, y)\right] \tag{12} \end{equation*}(12)supy∈Y[f(y)−‖f‖dd−1(x,y)]=infy∈Y[f(y)+‖f‖dd(x,y)]
for all x ∈ X x ∈ X x in Xx \in Xx∈X.
P roof. t H ∈ E Y d ( f ) H ∈ E Y d ( f ) H inE_(Y)^(d)(f)H \in E_{Y}^{d}(f)H∈EYd(f). Then we have for every x ∈ X x ∈ X x in Xx \in Xx∈X and y ∈ Y y ∈ Y y in Yy \in Yy∈Y :
H ( x ) − H ( y ) ≤ ‖ f ‖ d d ( x , y ) H ( y ) − H ( x ) ≤ ‖ f ‖ d d ( y , x ) = ‖ f ‖ d ⋅ d − 1 ( x , y ) . H ( x ) − H ( y ) ≤ ‖ f ‖ d d ( x , y ) H ( y ) − H ( x ) ≤ ‖ f ‖ d d ( y , x ) = ‖ f ‖ d ⋅ d − 1 ( x , y ) . {:[H(x)-H(y) <= ||f||_(d)d(x","y)],[H(y)-H(x) <= ||f||_(d)d(y","x)=||f||_(d)*d^(-1)(x","y).]:}\begin{aligned} & H(x)-H(y) \leq\|f\|_{d} d(x, y) \\ & H(y)-H(x) \leq\|f\|_{d} d(y, x)=\|f\|_{d} \cdot d^{-1}(x, y) . \end{aligned}H(x)−H(y)≤‖f‖dd(x,y)H(y)−H(x)≤‖f‖dd(y,x)=‖f‖d⋅d−1(x,y).
The first inequality implies
H ( x ) ≤ H ( y ) + ‖ f ‖ d ⋅ d ( x , y ) = f ( y ) + ‖ f ‖ d d ( x , y ) H ( x ) ≤ H ( y ) + ‖ f ‖ d ⋅ d ( x , y ) = f ( y ) + ‖ f ‖ d d ( x , y ) H(x) <= H(y)+||f||_(d)*d(x,y)=f(y)+||f||_(d)d(x,y)H(x) \leq H(y)+\|f\|_{d} \cdot d(x, y)=f(y)+\|f\|_{d} d(x, y)H(x)≤H(y)+‖f‖d⋅d(x,y)=f(y)+‖f‖dd(x,y)
and, taking the infimum with respect to y ∈ Y y ∈ Y y in Yy \in Yy∈Y, we have
H ( x ) ≤ F ( x ) , x ∈ X . H ( x ) ≤ F ( x ) , x ∈ X . H(x) <= F(x),quad x in X.H(x) \leq F(x), \quad x \in X .H(x)≤F(x),x∈X.
Similarly, we get
H ( x ) ≥ H ( y ) − ‖ f ‖ d d − 1 ( x , y ) = f ( y ) − ‖ f ‖ d d − 1 ( x , y ) . H ( x ) ≥ H ( y ) − ‖ f ‖ d d − 1 ( x , y ) = f ( y ) − ‖ f ‖ d d − 1 ( x , y ) . H(x) >= H(y)-||f||_(d)d^(-1)(x,y)=f(y)-||f||_(d)d^(-1)(x,y).H(x) \geq H(y)-\|f\|_{d} d^{-1}(x, y)=f(y)-\|f\|_{d} d^{-1}(x, y) .H(x)≥H(y)−‖f‖dd−1(x,y)=f(y)−‖f‖dd−1(x,y).
Taking the supremum with respect to y ∈ Y y ∈ Y y in Yy \in Yy∈Y one obtains
H ( x ) ≥ G ( x ) , x ∈ X . H ( x ) ≥ G ( x ) , x ∈ X . H(x) >= G(x),quad x in X.H(x) \geq G(x), \quad x \in X .H(x)≥G(x),x∈X.
The assertion b) is a direct consequence of the inequalities (11).
Remark. 1 0 1 0 1^(0)1^{0}10. If the function f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R is constant on X X XXX then ‖ f ‖ d = 0 ‖ f ‖ d = 0 ||f||_(d)=0\|f\|_{d}=0‖f‖d=0, and the equality (12) holds.
Consider on R R R\mathbb{R}R de quasi-metric
d ( x , y ) = { x − y , x ≥ y 0 , x < y d ( x , y ) = x − y , x ≥ y 0 , x < y d(x,y)={[x-y",",x >= y],[0",",x < y]:}d(x, y)=\left\{\begin{array}{cc} x-y, & x \geq y \\ 0, & x<y \end{array}\right.d(x,y)={x−y,x≥y0,x<y
and let Y = [ 0 , 1 ] Y = [ 0 , 1 ] Y=[0,1]Y=[0,1]Y=[0,1] and f ( y ) = 2 y , y ∈ Y f ( y ) = 2 y , y ∈ Y f(y)=2y,y in Yf(y)=2 y, y \in Yf(y)=2y,y∈Y.Then ‖ f ‖ d = 2 ‖ f ‖ d = 2 ||f||_(d)=2\|f\|_{d}=2‖f‖d=2 and the extremal extensions F , G F , G F,GF, \mathrm{G}F,G are
F ( x ) = { 2 x < 0 2 x x ≥ 0 and G ( x ) = { 2 x x ≤ 1 0 x > 1 F ( x ) = 2 x < 0 2 x x ≥ 0  and  G ( x ) = 2 x x ≤ 1 0 x > 1 F(x)={[2,x < 0],[2x,x >= 0]quad" and "quad G(x)={[2x,x <= 1],[0,x > 1]:}F(x)=\left\{\begin{array}{cc} 2 & x<0 \\ 2 x & x \geq 0 \end{array} \quad \text { and } \quad G(x)=\left\{\begin{array}{cc} 2 x & x \leq 1 \\ 0 & x>1 \end{array}\right.\right.F(x)={2x<02xx≥0 and G(x)={2xx≤10x>1
which are distinct.
2 0 2 0 2^(0)2^{0}20. By Theorem 2, if f ∈ S L i p Y f ∈ S L i p Y f in SLipYf \in S L i p Yf∈SLipY has a unique extension then the equality (12) holds and, since
inf y ∈ Y [ f ( y ) + ‖ f ‖ d ⋅ d ( x , y ) ] ≥ inf y ∈ Y f ( y ) + ‖ f ‖ d ⋅ d ( x , Y ) sup y ∈ Y [ f ( y ) − ‖ f ‖ d ⋅ d − 1 ( x , y ) ] ≤ sup y ∈ Y f ( y ) − ‖ f ‖ d ⋅ d − 1 ( x , Y ) inf y ∈ Y   f ( y ) + ‖ f ‖ d ⋅ d ( x , y ) ≥ inf y ∈ Y   f ( y ) + ‖ f ‖ d ⋅ d ( x , Y ) sup y ∈ Y   f ( y ) − ‖ f ‖ d ⋅ d − 1 ( x , y ) ≤ sup y ∈ Y   f ( y ) − ‖ f ‖ d ⋅ d − 1 ( x , Y ) {:[i n f_(y in Y)[f(y)+||f||_(d)*d(x,y)] >= i n f_(y in Y)f(y)+||f||_(d)*d(x","Y)],[s u p_(y in Y)[f(y)-||f||_(d)*d^(-1)(x,y)] <= s u p_(y in Y)f(y)-||f||_(d)*d^(-1)(x","Y)]:}\begin{aligned} \inf _{y \in Y}\left[f(y)+\|f\|_{d} \cdot d(x, y)\right] & \geq \inf _{y \in Y} f(y)+\|f\|_{d} \cdot d(x, Y) \\ \sup _{y \in Y}\left[f(y)-\|f\|_{d} \cdot d^{-1}(x, y)\right] & \leq \sup _{y \in Y} f(y)-\|f\|_{d} \cdot d^{-1}(x, Y) \end{aligned}infy∈Y[f(y)+‖f‖d⋅d(x,y)]≥infy∈Yf(y)+‖f‖d⋅d(x,Y)supy∈Y[f(y)−‖f‖d⋅d−1(x,y)]≤supy∈Yf(y)−‖f‖d⋅d−1(x,Y)
where
d ( x , Y ) = inf { d ( x , y ) : y ∈ Y } d ( x , Y ) = inf { d ( x , y ) : y ∈ Y } d(x,Y)=i n f{d(x,y):y in Y}d(x, Y)=\inf \{d(x, y): y \in Y\}d(x,Y)=inf{d(x,y):y∈Y}
and
d − 1 ( x , Y ) = inf { d ( y , x ) : y ∈ Y } d − 1 ( x , Y ) = inf { d ( y , x ) : y ∈ Y } d^(-1)(x,Y)=i n f{d(y,x):y in Y}d^{-1}(x, Y)=\inf \{d(y, x): y \in Y\}d−1(x,Y)=inf{d(y,x):y∈Y}
we obtain the inequality
(13) d ( x , Y ) + d − 1 ( x , Y ) ≤ 1 ‖ f ‖ d ( sup y ∈ Y f ( y ) − inf y ∈ Y f ( y ) ) (13) d ( x , Y ) + d − 1 ( x , Y ) ≤ 1 ‖ f ‖ d sup y ∈ Y   f ( y ) − inf y ∈ Y   f ( y ) {:(13)d(x","Y)+d^(-1)(x","Y) <= (1)/(||f||_(d))(s u p_(y in Y)f(y)-i n f_(y in Y)f(y)):}\begin{equation*} d(x, Y)+d^{-1}(x, Y) \leq \frac{1}{\|f\|_{d}}\left(\sup _{y \in Y} f(y)-\inf _{y \in Y} f(y)\right) \tag{13} \end{equation*}(13)d(x,Y)+d−1(x,Y)≤1‖f‖d(supy∈Yf(y)−infy∈Yf(y))
Theorem 3 Let ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a quasi-metric space and Y ⊂ X , Y ≠ X Y ⊂ X , Y ≠ X Y sub X,Y!=XY \subset X, Y \neq XY⊂X,Y≠X, containing at least one cluster point. If each function f ∈ S L i p Y f ∈ S L i p Y f in SLipYf \in S L i p Yf∈SLipY has a unique extension then Y ¯ = X Y ¯ = X bar(Y)=X\bar{Y}=XY¯=X.
P P P\mathbf{P}P roof. t y 0 ∈ Y t y 0 ∈ Y ty_(0)in Y\mathrm{t} y_{0} \in Yty0∈Y be a cluster point of the set Y Y YYY and let y n ∈ Y ∖ { y 0 } y n ∈ Y ∖ y 0 y_(n)in Y\\{y_(0)}y_{n} \in Y \backslash\left\{y_{0}\right\}yn∈Y∖{y0}, n = 1 , 2 … n = 1 , 2 … n=1,2dotsn=1,2 \ldotsn=1,2…, be such that lim n → ∞ d ( y n , y 0 ) = 0 lim n → ∞   d y n , y 0 = 0 lim_(n rarr oo)d(y_(n),y_(0))=0\lim _{n \rightarrow \infty} d\left(y_{n}, y_{0}\right)=0limn→∞d(yn,y0)=0.
Claim: There exists x 0 ∈ X x 0 ∈ X x_(0)in Xx_{0} \in Xx0∈X such that d ( x 0 , y 0 ) > 0 d x 0 , y 0 > 0 d(x_(0),y_(0)) > 0d\left(x_{0}, y_{0}\right)>0d(x0,y0)>0 and d ( x 0 , y n ) > 0 d x 0 , y n > 0 d(x_(0),y_(n)) > 0d\left(x_{0}, y_{n}\right)>0d(x0,yn)>0, n = 1 , 2 , … n = 1 , 2 , … n=1,2,dotsn=1,2, \ldotsn=1,2,….
Indeed, if contrary, then for every x ∈ X , d ( x , y 0 ) = 0 x ∈ X , d x , y 0 = 0 x in X,d(x,y_(0))=0x \in X, d\left(x, y_{0}\right)=0x∈X,d(x,y0)=0 or d ( x , y n ) = 0 d x , y n = 0 d(x,y_(n))=0d\left(x, y_{n}\right)=0d(x,yn)=0 for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N.In the first case x = y 0 ∈ Y x = y 0 ∈ Y x=y_(0)in Yx=y_{0} \in Yx=y0∈Y and in the second x = y n ∈ Y x = y n ∈ Y x=y_(n)in Yx=y_{n} \in Yx=yn∈Y.It follows Y = X Y = X Y=XY=XY=X, a contradiction.
Consider the function f : X → R f : X → R f:X rarrRf: X \rightarrow \mathbb{R}f:X→R defined by
f ( x ) = d ( x , y 0 ) − d ( x 0 , y 0 ) , x ∈ X f ( x ) = d x , y 0 − d x 0 , y 0 , x ∈ X f(x)=d(x,y_(0))-d(x_(0),y_(0)),x in Xf(x)=d\left(x, y_{0}\right)-d\left(x_{0}, y_{0}\right), x \in Xf(x)=d(x,y0)−d(x0,y0),x∈X
We have
f ( y 0 ) = d ( y 0 , y 0 ) − d ( x 0 , y 0 ) = − d ( x 0 , y 0 ) < 0 f ( y n ) = d ( y n , y 0 ) − d ( x 0 , y 0 ) > − d ( x 0 , y 0 ) f y 0 = d y 0 , y 0 − d x 0 , y 0 = − d x 0 , y 0 < 0 f y n = d y n , y 0 − d x 0 , y 0 > − d x 0 , y 0 {:[f(y_(0))=d(y_(0),y_(0))-d(x_(0),y_(0))=-d(x_(0),y_(0)) < 0],[f(y_(n))=d(y_(n),y_(0))-d(x_(0),y_(0)) > -d(x_(0),y_(0))]:}\begin{aligned} & f\left(y_{0}\right)=d\left(y_{0}, y_{0}\right)-d\left(x_{0}, y_{0}\right)=-d\left(x_{0}, y_{0}\right)<0 \\ & f\left(y_{n}\right)=d\left(y_{n}, y_{0}\right)-d\left(x_{0}, y_{0}\right)>-d\left(x_{0}, y_{0}\right) \end{aligned}f(y0)=d(y0,y0)−d(x0,y0)=−d(x0,y0)<0f(yn)=d(yn,y0)−d(x0,y0)>−d(x0,y0)
for all n = 1 , 2 , … n = 1 , 2 , … n=1,2,dotsn=1,2, \ldotsn=1,2,…. Define the sequence of functions φ n : f ( X ) → [ 0 , 1 ] φ n : f ( X ) → [ 0 , 1 ] varphi_(n):f(X)rarr[0,1]\varphi_{n}: f(X) \rightarrow[0,1]φn:f(X)→[0,1] by
φ n ( t ) = { 1 if t < f ( y 0 ) f ( y n ) − t f ( y n ) − f ( y 0 ) t ∈ [ f ( y 0 ) , f ( y n ) ] 0 t > f ( y n ) φ n ( t ) = 1       if  t < f y 0 f y n − t f y n − f y 0      t ∈ f y 0 , f y n 0      t > f y n varphi_(n)(t)={[1," if "t < f(y_(0))],[(f(y_(n))-t)/(f(y_(n))-f(y_(0))),t in[f(y_(0)),f(y_(n))]],[0,t > f(y_(n))]:}\varphi_{n}(t)= \begin{cases}1 & \text { if } t<f\left(y_{0}\right) \\ \frac{f\left(y_{n}\right)-t}{f\left(y_{n}\right)-f\left(y_{0}\right)} & t \in\left[f\left(y_{0}\right), f\left(y_{n}\right)\right] \\ 0 & t>f\left(y_{n}\right)\end{cases}φn(t)={1 if t<f(y0)f(yn)−tf(yn)−f(y0)t∈[f(y0),f(yn)]0t>f(yn)
The function Ψ n = φ n ∘ f : X → R , n = 1 , 2 , … Ψ n = φ n ∘ f : X → R , n = 1 , 2 , … Psi_(n)=varphi_(n)@f:X rarrR,n=1,2,dots\Psi_{n}=\varphi_{n} \circ f: X \rightarrow \mathbb{R}, n=1,2, \ldotsΨn=φn∘f:X→R,n=1,2,…, satisfy
‖ Ψ n ‖ d ≥ ( φ n ( f ( y 0 ) ) − φ n ( f ( y n ) ) ) ∨ 0 d ( y 0 , y n ) = 1 d ( y 0 , y n ) → ∞ Ψ n d ≥ φ n f y 0 − φ n f y n ∨ 0 d y 0 , y n = 1 d y 0 , y n → ∞ ||Psi_(n)||_(d) >= ((varphi_(n)(f(y_(0)))-varphi_(n)(f(y_(n))))vv0)/(d(y_(0),y_(n)))=(1)/(d(y_(0),y_(n)))rarr oo\left\|\Psi_{n}\right\|_{d} \geq \frac{\left(\varphi_{n}\left(f\left(y_{0}\right)\right)-\varphi_{n}\left(f\left(y_{n}\right)\right)\right) \vee 0}{d\left(y_{0}, y_{n}\right)}=\frac{1}{d\left(y_{0}, y_{n}\right)} \rightarrow \infty‖Ψn‖d≥(φn(f(y0))−φn(f(yn)))∨0d(y0,yn)=1d(y0,yn)→∞
for n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞.
By the inequality (12)
d ( x , Y ) + d − 1 ( x , Y ) ≤ 1 − 0 ‖ Ψ n ‖ d → 0 d ( x , Y ) + d − 1 ( x , Y ) ≤ 1 − 0 Ψ n d → 0 d(x,Y)+d^(-1)(x,Y) <= (1-0)/(||Psi_(n)||_(d))rarr0d(x, Y)+d^{-1}(x, Y) \leq \frac{1-0}{\left\|\Psi_{n}\right\|_{d}} \rightarrow 0d(x,Y)+d−1(x,Y)≤1−0‖Ψn‖d→0
for n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞,showing that Y Y YYY is dense in X X XXX, with respect to the quasi-metric d d ddd and with respect to d − 1 d − 1 d^(-1)d^{-1}d−1, as well.

References

[1] Cobzas, S., Mustăta, C., Norm Preserving Extension of Convex Lipschitz Functions, J.A.T. 24(1978), 555-564.
[2] Fletcher, P., Lindgren, W.F., "Quasi-Uniform Spaces" Dekken, New York, 1982.
[3] McShane, J.A., Extension of range of functions, Bull.Amer.Math.Soc. 40(1934) 837-842.
[4] Mustăţa, C., Best Approximation and Unique Extension of Lipschitz Functions, J.A.T. 19(1977), 222-230.
[5] Mustăţa, C., On the Extension of Semi-Lipschitz Functions on QuasiMetric Space (to appear).
[6] Romaguera, S., Sanchis, M., Semi-Lipschitz Functions and Best Approximation in Quasi-Metric Spaces, J.A.T. 103 (2000) 292-301.
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Received 19.10.2000
"T. Popoviciu" Institute of Numerical Analysis str. Gh. Bilaşcu nr. 37
C.P. 68, O.P. 1
3400 Cluj-Napoca
2000

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