EXTENSIONS OF SEMI-LIPSCHITZ FUNCTIONS
ON QUASI-METRIC SPACES

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 extension theorem for real-valued Lipschitz functions defined on a metric space ([2], [4]).

1. Introduction

Let $X$ be a nonvoid set. A quasi-metric on $X$ is a function $d:X\times X\to [0,\mathrm{\infty })$ satisfying the conditions

(i) $d(x,y)$ $=d(y,x)=0⟺x=y;x,y\in X,$

(ii) $d(x,y)$ $\le d(x,z)+d(z,y),x,y,z\in X.$

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

The conjugate of quasi-metric $d$, denoted by $d^{-1}$ is defined by $d^{-1}(x,y)=d(y,x),$ $x,y\in X.$

Obviously the function $d^s:X\times X\to [0,\mathrm{\infty })$ defined by

$$d^s(x,y)=\max \{d(x,y),d^{-1}(x,y)\};x,y\in X$$

is a metric on $X.$

If the quasi-metric $d$ can take the value $+\mathrm{\infty }$, then it is called an extended quasi-metric.

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

(1) $$f\left(x\right)-f\left(y\right)\le K\cdot d(x,y),$$

for all $x,y\in X.$ The number $K\ge 0$ in (1) is called a semi-Lipschitz constant for $f.$

For a quasi-metric space $(X,d)$ the real-valued function $f:X\to ℝ$ is said to be ${\le }_d$-increasing if

(2) $$d(x,y)=0\text{implies }f\left(x\right)-f\left(y\right)\le 0,x,y\in X$$

or equivalently,

(3) $$f\left(x\right)-f\left(y\right)>0\text{ implies }d(x,y)>0,x,y\in X.$$

Note that every semi-Lipschitz function on quasi-metric space $(X,d)$ is ${\le }_d$-increasing (see (1)).

For a semi-Lipschitz function $f:X\to ℝ$, where $(X,d)$ is a quasi-metric space, denote by ${\Vert f\Vert }_d$ the constant:

(4) $${\Vert f\Vert }_d=sup\{\frac{\left(f\left(x\right)-f\left(y\right)\right)\vee 0}{d(x,y)}:d(x,y)>0,x,y\in X\}.$$

Theorem 1.

Let $(X,d)$ a quasi-metric space and $f:X\to ℝ$ a semi-Lipschitz function. Then ${\Vert f\Vert }_d$ defined by (4) is the smallest semi-Lipschitz constant for $f.$

Proof.

If $f:X\to ℝ$ is semi-Lipschitz, then $f$ is ${\le }_d$-increasing, and then $f\left(x\right)-f\left(y\right)>0$ implies $d(x,y)>0.$ It follows that

$$\frac{\left(f\left(x\right)-f\left(y\right)\right)\vee 0}{d(x,y)}=\frac{f\left(x\right)-f\left(y\right)}{d(x,y)}>0.$$

The inequalities $f\left(x\right)-f\left(y\right)\le 0$ and $d(x,y)>0$ imply

$$\frac{\left(f\left(x\right)-f\left(y\right)\right)\vee 0}{d(x,y)}=0.$$

Consequently ${\Vert f\Vert }_d\ge 0.$

For it follows $(f(x)-f(y))/d(x,y)\le {\Vert f\Vert }_d$ and obviously for $f\left(x\right)-f\left(y\right)\le 0$ we have $f\left(x\right)-f\left(y\right)\le 0\le {\Vert f\Vert }_d\cdot d(x,y).$

Consequently

$$f\left(x\right)-f\left(y\right)\le {\Vert f\Vert }_d\cdot d(x,y)$$

for all $x,y\in X.$

Now let $K\ge 0$ such that

$$f\left(x\right)-f\left(y\right)\le K\cdot d(x,y),\text{ for all }x,y\in X.$$

The function $f$ is ${\le }_d$-increasing, and then

Consequently ${\Vert f\Vert }_d\le K.$ ∎

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

(5)

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

2. Extensions of semi-Lipschitz functions

Let $Y\subset X$ 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 by $SLipY$ the set of all semi-Lipschitz functions defined on $Y$ and let

(6) $${\Vert f\Vert }_d=sup\{\frac{\left(f\left(x\right)-f\left(y\right)\right)\vee 0}{d(x,y)}:x,y\in Y,d(x,y)\ne 0\}$$

be the smallest semi-Lipschitz constant for $f\in SLipY.$

If $f\in SLipY,$ a function $F\in SLipX$ is called an extension (preserving the smallest semi-Lipschitz constant) of $f$ if:

(7) $${F|}_Y=f\text{ and }{\Vert F\Vert }_d={\Vert f\Vert }_d.$$

Denote by $E_Y\left(f\right)$ the set of all extensions of the function $f\in SLipY,$ i.e.

(8) $$E_Y\left(f\right)=\{F\in SLipX:{F|}_Y=f\text{ and }{\Vert F\Vert }_d={\Vert f\Vert }_d\}$$

Theorem 2.

Let $(X,d)$ be a quasi-metric space and $Y$ a nonvoid subset of $X.$ Then for every $f\in SLipY$ the set $E_Y\left(f\right)$ is nonvoid.

Proof.

Let $f\in SLipY$ and the constant ${\Vert f\Vert }_d$ defined by (6).

Consider the function

(9) $$F\left(x\right)=\underset{y\in Y}{inf}\{f\left(y\right)+\Vert {f\parallel }_{d}d(x,y)\},x\in X.$$

a) First we show that $F$ is well defined.

Let $z\in Y$and $x\in X.$For any $y\in Y$ we have

$f\left(y\right)+{\Vert f\Vert }_{d}d(x,y)$ $=f\left(z\right)+{\Vert f\Vert }_{d}d(x,y)-\left(f\left(z\right)-f\left(y\right)\right)$

$\ge f\left(z\right)+{\Vert f\Vert }_{d}d(x,y)-{\Vert f\Vert }_{d}d(z,y)$

$=f\left(z\right)-{\Vert f\Vert }_d\left(d(z,y)-d(x,y)\right).$

The inequality $d(z,y)-d(x,y)\le d(z,x)=d^{-1}(x,z)$ implies

(10) $$f\left(y\right)+{\Vert f\Vert }_{d}d(x,y)\ge f\left(z\right)-{\Vert f\Vert }_d\cdot d^{-1}(x,z)$$

showing that for every $x\in X$ the set $\{f\left(y\right)+\Vert {f\parallel }_{d}d(x,y):y\in Y\}$ is bounded from above by $f\left(z\right)-{\Vert f\Vert }_d{d}^{-1}(x,z),$ and the infimum (9) is finite.

b) We show now that $F\left(y\right)=f\left(y\right)$ for all $y\in Y.$

Let $y\in Y.$ Then

$$F\left(y\right)\le f\left(y\right)+{\Vert f\Vert }_{d}d(y,y)=f\left(y\right).$$

For any $v\in Y$ we have

$$f\left(y\right)-f\left(v\right)\le {\Vert f\Vert }_d\cdot d(y,v)$$

so that

$$f\left(v\right)+{\Vert f\Vert }_d\cdot d(y,v)\ge f\left(y\right)$$

and

$$F\left(y\right)=inf\{f\left(v\right)+\Vert {f\parallel }_{d}d(y,v):v\in Y\}\ge f\left(y\right).$$

It follows $F\left(y\right)=f\left(y\right).$

c) We prove that ${\Vert F\Vert }_d={\Vert f\Vert }_d.$

Since ${F|}_Y=f,$ the definitions of ${\Vert F\Vert }_d$ and ${\Vert f\Vert }_d$ yield ${\Vert F\Vert }_d\ge {\Vert f\Vert }_d.$

Let $x_1,x_2\in X$ and $\epsilon >0.$ Choosing $y\in Y$ such that

$$F\left(x_1\right)\ge f\left(y\right)+{\Vert f\Vert }_{d}d(x_1,y)-\epsilon$$

we obtain

$F\left(x_2\right)-F\left(x_1\right)$ $\le f\left(y\right)+{\Vert f\Vert }_{d}d(x_2,y)-\left(f\left(y\right)+{\Vert f\Vert }_d\cdot d(x_1,y)-\epsilon \right)$

$={\Vert f\Vert }_d\left[d(x_2,y)-d(x_1,y)\right]+\epsilon$

$\le {\Vert f\Vert }_d\cdot d(x_2,x_1)+\epsilon .$

Since $\epsilon >0$ is arbitrary, it follows

$$F\left(x_2\right)-F\left(x_1\right)\le {\Vert f\Vert }_d\cdot d(x_2,x_1)$$

for any $x_1,x_2\in X$ and ${\Vert F\Vert }_d\le {\Vert f\Vert }_d.$

d) The function $F$ is ${\le }_d$-increasing.

Indeed, let be $u,v\in X$ and $d(u,v)=0.$ We have $d(u,y)\le d(u,v)+d(v,y).$ Consequently

$$d(u,y)\le d(v,y).$$

Then

$$f\left(y\right)+{\Vert f\Vert }_{d}d(u,y)\le f\left(y\right)+{\Vert f\Vert }_{d}d(v,y).$$

It follows that

$$F\left(u\right)\le F\left(v\right),$$

and consequently $d(u,v)=0$ implies $F\left(u\right)\le F\left(v\right).$

It follows that $F\in E_Y^d\left(f\right)$ so that $E_Y^d\left(t\right)\ne \mathrm{\varnothing }.$ ∎

Remarks 1.

1${}^{\text{0}}$ Similarly, the function

(11) $$G\left(x\right)=\underset{y\in Y}{sup}\{f\left(y\right)-\Vert {f\parallel }_d{d}^{-1}(x,y)\}$$

is ${\le }_d$-increasing, and $G$ belongs to $E_Y^d\left(f\right)$ too.

2${}^{\text{0}}$ The inequality

(12) $$G\left(x\right)\le F\left(x\right),$$

holds for every $x\in X.$

Indeed, taking the infimum with respect to $z\in Y$ and then the supremum with respect to $y\in Y$ in (10) we find

$$G\left(x\right)=\underset{y\in Y}{sup}\{f\left(y\right)-\Vert {f\parallel }_d{d}^{-1}(x,y)\}\le \underset{z\in Y}{inf}\{f\left(z\right)+\Vert {f\parallel }_{d}d(x,z)\}=F\left(x\right).$$

In fact, the following theorem holds:

Theorem 3.

Let $(X,d)$ be a quasi-metric space, $Y$ a nonvoid subset of $X$ and $f\in SLipY.$

Then for any $H\in E_Y^d\left(f\right)$ we have

(13) $$G\left(x\right)\le H\left(x\right)\le F\left(x\right),x\in X.$$

Proof.

Let $H\in E_Y^d\left(f\right).$ For arbitrary $x\in X$ and $y\in Y$ we have

$$H\left(x\right)-H\left(y\right)\le {\Vert f\Vert }_{d}d(x,y)$$

implying

$$H\left(x\right)\le H\left(y\right)+{\Vert f\Vert }_{d}d(x,y)=f\left(y\right)+{\Vert f\Vert }_d(x,y).$$

Taking the imfimum with respect to $y\in Y$ we get

$$H\left(x\right)\le \underset{y\in Y}{inf}\{f\left(y\right)+\Vert {f\parallel }_{d}d(x,y)\}=F(x).$$

The inequality $H\left(x\right)\ge G\left(x\right),$ $x\in X$ can be proved similarly. ∎

Corollary 4.

A function $f\in SLipY$ has a unique extension in $SLipX$ if and only if the following relation

(14) $$\underset{y\in Y}{inf}\{f\left(y\right)+\Vert {f\parallel }_{d}d(x,y)\}=\underset{y\in Y}{sup}\{f\left(y\right)-\Vert f\parallel d(y,x)\},$$

holds for every $x\in X.$

Example.

Let $ℝ$ be the real axis and $d:ℝ\times ℝ\to [0,\mathrm{\infty })$ the quasi-metric defined by

Let $Y$ be given by $Y=[0,1]\subset ℝ$ and $f:Y\to ℝ,$ $f\left(y\right)=2y.$ Then $f$ is semi-Lipschitz on $Y$ and ${\Vert f\Vert }_d=2.$ The extension $F$ defined by (9) is

and the extension $G$ defined by (11) is

Obviously, $G\left(x\right)\le F\left(x\right),$ $x\in ℝ.$

Received: August 8, 2000.