View of Extensions of semi-Lipschitz functions on quasi-metric spaces

Rev. Anal. Num´er. Th´eor. Approx., vol. 30 (2001) no. 1, pp. 61–67 ictp.acad.ro/jnaat

EXTENSIONS OF SEMI-LIPSCHITZ FUNCTIONS ON QUASI-METRIC SPACES

COSTIC ˘A MUST ˘AT¸ A

Dedicated to the memory of Acad. Tiberiu Popoviciu

Abstract. The aim of this note is to prove an extension theorem for semi- Lipschitz real functions defined on quasi-metric spaces, similar to McShane ex- tension theorem for real-valued Lipschitz functions defined on a metric space ([2], [4]).

MSC 2000. 46A22, 26A16, 26A48.

1. INTRODUCTION

Let X be a nonvoid set. A quasi-metric on X is a function d:X×X → [0,∞) satisfying the conditions

d(x, y) =d(y, x) = 0⇐⇒x=y; x, y∈X, (i)

d(x, y)≤d(x, z) +d(z, y), x, y, z∈X.

(ii)

If d is a quasi-metric on X, then the pair (X, d) is called a quasi-metric space.

The conjugate of quasi-metricd, denoted by d⁻¹ is defined by d⁻¹(x, y) = d(y, x), x, y∈X.

Obviously the functiond^s:X×X→[0,∞) defined by d^s(x, y) = max

d(x, y), d⁻¹(x, y) ; x, y∈X is a metric on X.

“T. Popoviciu” Institute of Numerical Analysis, P.O. Box 68–1, 3400 Cluj-Napoca, Ro- mania, e-mail: [email protected].

quasi-metric.

Let (X, d) be a quasi-metric space. A function f :X → R is called semi- Lipschitz if there exists a constant K≥0 so that

(1) f(x)−f(y)≤K·d(x, y),

for all x, y∈X. The numberK ≥0 in (1) is called a semi-Lipschitz constant forf.

For a quasi-metric space (X, d) the real-valued functionf :X →R is said to be≤_d-increasing if

(2) d(x, y) = 0 implies f(x)−f(y)≤0, x, y∈X or equivalently,

(3) f(x)−f(y)>0 implies d(x, y)>0, x, y∈X.

Note that every semi-Lipschitz function on quasi-metric space (X, d) is≤_d- increasing (see (1)).

For a semi-Lipschitz function f : X → R, where (X, d) is a quasi-metric space, denote by kfk_d the constant:

(4) kfk_d= sup

(f(x)−f(y))∨0

d(x, y) :d(x, y)>0, x, y∈X

.

Theorem 1. Let (X, d) a quasi-metric space and f : X → R a semi- Lipschitz function. Then kfk_d defined by (4) is the smallest semi-Lipschitz constant forf.

Proof. If f : X → R is semi-Lipschitz, then f is ≤_d-increasing, and then f(x)−f(y)>0 impliesd(x, y)>0.It follows that

(f(x)−f(y))∨0

d(x, y) = f(x)−f(y) d(x, y) >0.

The inequalities f(x)−f(y)≤0 andd(x, y)>0 imply (f(x)−f(y))∨0

d(x, y) = 0.

Consequentlykfk_d≥0.

Forf(x)−f(y)<0 it follows (f(x)−f(y))/d(x, y) ≤ kfk_d and obviously forf(x)−f(y)≤0 we have f(x)−f(y)≤0≤ kfk_d·d(x, y).

Consequently

f(x)−f(y)≤ kfk_d·d(x, y) for all x, y∈X.

Now let K ≥0 such that

f(x)−f(y)≤K·d(x, y), for all x, y∈X.

The function f is≤_d-increasing, and then (f(x)−f(y))∨0

d(x, y) =

( _f(x)−f(y)

d(x,y) ≤K, iff(x)−f(y)>0, 0≤K, iff(x)−f(y)≤0,

Consequently kfk_d≤K.

For a quasi-metric (X, d) let us consider the set:

(5)

S Lip X = (

f :X→R|f is ≤_d-increasing, sup

d(x,y)6=0

(f(x)−f(y))∨0 d(x,y) <∞

) .

It is straightforward to see that S Lip X is exactly the set of all semi- Lipschitz functions on (X, d) (see [6]).

2. EXTENSIONS OF SEMI-LIPSCHITZ FUNCTIONS

Let Y ⊂ X where (X, d) is a quasi-metric space. Then (Y, d) is a quasi- metric space with the quasi-metric induced by d (denoted by d too). Let us denote byS Lip Y the set of all semi-Lipschitz functions defined onY and let (6) kfk_d= sup

(f(x)−f(y))∨0

d(x, y) :x, y∈Y, d(x, y)6= 0

be the smallest semi-Lipschitz constant for f ∈S Lip Y.

If f ∈ S Lip Y, a function F ∈ S Lip X is called an extension (preserving the smallest semi-Lipschitz constant) off if:

(7) F|_Y =f and kFk_d=kfk_d.

Denote by E_Y (f) the set of all extensions of the functionf ∈S Lip Y, i.e.

(8) E_Y (f) =

F ∈S Lip X :F|_Y =f and kFk_d=kfk_d

Theorem 2. Let (X, d) be a quasi-metric space andY a nonvoid subset of X. Then for every f ∈S Lip Y the setEY (f) is nonvoid.

Consider the function

(9) F(x) = inf

y∈Y {f(y) +kfk_dd(x, y)}, x∈X.

a)First we show that F is well defined.

Letz∈Yand x∈X.For any y∈Y we have

f(y) +kfk_dd(x, y) =f(z) +kfk_dd(x, y)−(f(z)−f(y))

≥f(z) +kfk_dd(x, y)− kfk_dd(z, y)

=f(z)− kfk_d(d(z, y)−d(x, y)). The inequality d(z, y)−d(x, y)≤d(z, x) =d⁻¹(x, z) implies (10) f(y) +kfk_dd(x, y)≥f(z)− kfk_d·d⁻¹(x, z)

showing that for everyx∈Xthe set{f(y) +kfk_dd(x, y) :y∈Y}is bounded from above by f(z)− kfk_dd⁻¹(x, z),and the infimum (9) is finite.

b) We show now that F(y) =f(y) for all y∈Y.

Lety∈Y.Then

F(y)≤f(y) +kfk_dd(y, y) =f(y). For anyv∈Y we have

f(y)−f(v)≤ kfk_d·d(y, v) so that

f(v) +kfk_d·d(y, v)≥f(y) and

F(y) = inf{f(v) +kfk_dd(y, v) :v∈Y} ≥f(y). It follows F(y) =f(y).

c)We prove that kFk_d=kfk_d.

Since F|_Y =f, the definitions ofkFk_d andkfk_dyield kFk_d≥ kfk_d. Letx₁, x₂ ∈X and ε >0.Choosing y∈Y such that

F(x₁)≥f(y) +kfk_dd(x₁, y)−ε we obtain

F(x₂)−F(x₁)≤f(y) +kfk_dd(x₂, y)−(f(y) +kfk_d·d(x₁, y)−ε)

=kfk_d[d(x₂, y)−d(x₁, y)] +ε

≤ kfk_d·d(x₂, x₁) +ε.

Since ε >0 is arbitrary, it follows

F(x₂)−F(x₁)≤ kfk_d·d(x₂, x₁) for any x₁, x₂ ∈X andkFk_d≤ kfk_d.

d) The function F is ≤_d-increasing.

Indeed, let beu, v∈Xandd(u, v) = 0.We haved(u, y)≤d(u, v)+d(v, y). Consequently

d(u, y)≤d(v, y). Then

f(y) +kfk_dd(u, y)≤f(y) +kfk_dd(v, y). It follows that

F(u)≤F(v), and consequently d(u, v) = 0 impliesF(u)≤F(v).

It follows that F ∈E_Y^d(f) so thatE_Y^d (t)6=∅.

Remarks 1. 1⁰ Similarly, the function

(11) G(x) = sup

y∈Y

f(y)− kfk_dd⁻¹(x, y)

is ≤_d-increasing, andGbelongs to E_Y^d (f) too.

2⁰ The inequality

(12) G(x)≤F(x),

holds for everyx∈X.

Indeed, taking the infimum with respect to z∈Y and then the supremum with respect toy ∈Y in (10) we find

G(x) = sup

y∈Y

f(y)− kfk_dd⁻¹(x, y) ≤ inf

z∈Y {f(z) +kfk_dd(x, z)}=F(x). In fact, the following theorem holds:

Theorem 3. Let (X, d) be a quasi-metric space, Y a nonvoid subset of X and f ∈S Lip Y.

Then for any H ∈E_Y^d (f) we have

(13) G(x)≤H(x)≤F(x), x∈X.

H(x)−H(y)≤ kfk_dd(x, y) implying

H(x)≤H(y) +kfk_dd(x, y) =f(y) +kfk_d(x, y). Taking the imfimum with respect to y∈Y we get

H(x)≤ inf

y∈Y

f(y) +kfk_dd(x, y) =F(x).

The inequality H(x)≥G(x), x∈X can be proved similarly.

Corollary 4. A function f ∈S Lip Y has a unique extension in S Lip X if and only if the following relation

(14) inf

y∈Y {f(y) +kfk_dd(x, y)}= sup

y∈Y

{f(y)− kfkd(y, x)}, holds for every x∈X.

Example.

LetR be the real axis andd:R×R→[0,∞) the quasi-metric defined by d(x, y) =

x−y, if x≥y 1, if x < y.

Let Y be given by Y = [0,1] ⊂ R and f :Y → R, f(y) = 2y. Then f is semi-Lipschitz onY andkfk_d= 2.The extensionF defined by (9) is

F(x) =

2, if x <0 2x, if x≥0 and the extensionG defined by (11) is

G(x) =

2x, x≤1 0, x >1 Obviously,G(x)≤F(x), x∈R.

REFERENCES

[1] S. CobzasandC. Must˘at¸a,Norm preserving extension of convex Lipschitz functions, J. Approx. Theory,29(1978), 555–569.

[2] J. CzipserandL. Geh´er,Extension of functions satisfying a Lipschitz condition, Acta Math. Sci. Hungar.,6(1955), 213–220.

[3] P. FletcherandW. F. Lindgren,Quasi-Uniform Spaces, Dekker, New York, 1982.

[4] J. A. McShane,Extension of range of functions, Bull. Amer. Math. Soc.,40(1939), 837–842.

[5] C. Must˘at¸a,Best approximation and unique extension of Lipschitz functions, J. Ap- prox. Theory, 19(1977), 222–230.

[6] S. Romagueraand M. Sanchis,Semi-Lipschitz functions and best approximation in quasi-metric spaces, J. Approx. Theory,103(2000), 292–301.

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Received: August 8, 2000.