Let $X$$X$XX$X$ be a nonvoid set. A function $d:X\times X\to [0,\mathrm{\infty })$$d:X\times X\to [0,\mathrm{\infty })$d:X xx X rarr[0,oo)d: X \times X \rightarrow[0, \infty)$d:X\times X\to [0,\mathrm{\infty })$ satisfying the conditions:
(i) $d(x,y)=0⟺x=y$$d(x,y)=0⟺x=y$quad d(x,y)=0Longleftrightarrow x=y\quad d(x, y)=0 \Longleftrightarrow x=y$d(x,y)=0⟺x=y$,
(ii) $d(x,y)\le d(x,z)+d(z,y)$$d(x,y)\le d(x,z)+d(z,y)$quad d(x,y) <= d(x,z)+d(z,y)\quad d(x, y) \leq d(x, z)+d(z, y)$d(x,y)\le d(x,z)+d(z,y)$, for all $x,y,z\in X$$x,y,z\in X$x,y,z in Xx, y, z \in X$x,y,z\in X$ is called a quasi-metric on $X$$X$XX$X$ and the pair $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ is called quasi-metric space. The essential difference with respect to a metric on $X$$X$XX$X$ is that a quasi-metric does not satisfy 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)$.

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

Let $\theta \in X$$\theta \in X$theta in X\theta \in X$\theta \in X$ be the null element of the linear space $X$$X$XX$X$. A subset $Y$$Y$YY$Y$ of $X$$X$XX$X$ is called starshaped (with respect to $\theta$$\theta$theta\theta$\theta$ ) if it satisfies the condition:

If $Y$$Y$YY$Y$ is a starshaped subset of the linear space $X$$X$XX$X$ then a function $f:Y\to \mathbb{R}$$f:Y\to \mathbb{R}$f:Y rarrRf: Y \rightarrow \mathbb{R}$f:Y\to \mathbb{R}$ is called a starshaped function provided:

Obviously that the condition (1) implies $\theta \in Y$$\theta \in Y$theta in Y\theta \in Y$\theta \in Y$ and the condition (2) implies $f(\theta )\le 0$$f(\theta )\le 0$f(theta) <= 0f(\theta) \leq 0$f(\theta )\le 0$. In what follows we shall consider only starshaped functions on $Y$$Y$YY$Y$ which vanish at $\theta$$\theta$theta\theta$\theta$, i.e. $f(\theta )=0$$f(\theta )=0$f(theta)=0f(\theta)=0$f(\theta )=0$.

If ( $X,d$$X,d$X,dX, d$X,d$ ) is a quasi-metric linear space and $Y\subset X$$Y\subset X$Y sub XY \subset X$Y\subset X$ is a starshaped set, the quasi-metric $d$$d$dd$d$ is called starshaped on $Y$$Y$YY$Y$ if

Let $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ be a quasi-metric space and $Y\subset X,Y\ne \mathrm{\varnothing }$$Y\subset X,Y\ne \mathrm{\varnothing }$Y sub X,Y!=O/Y \subset X, Y \neq \emptyset$Y\subset X,Y\ne \mathrm{\varnothing }$. A function $f:Y\to \mathbb{R}$$f:Y\to \mathbb{R}$f:Y rarrRf: Y \rightarrow \mathbb{R}$f:Y\to \mathbb{R}$ is called semi-Lipschitz if it satisfies the condition

for all $x,y\in Y$$x,y\in Y$x,y in Yx, y \in Y$x,y\in Y$.
A number $K_Y\ge 0$$K_Y\ge 0$K_(Y) >= 0K_{Y} \geq 0$K_Y\ge 0$ for which (4) holds is called a semi-Lipschitz constant for $f$$f$ff$f$ (on $Y$$Y$YY$Y$ ).

One sees that

is the smallest semi-Lipschitz constant for the function $f$$f$ff$f$ on the set $Y$$Y$YY$Y$ (see $[6]$$[6]$[6][6]$[6]$ and [4, Th.1]).

Let

the set of all real-valued semi-Lipschitz functions defined on the quasi-metric space $(Y,d),Y\subseteq X$$(Y,d),Y\subseteq X$(Y,d),Y sube X(Y, d), Y \subseteq X$(Y,d),Y\subseteq X$.

If ( $X,d$$X,d$X,dX, d$X,d$ ) is a quasi-metric linear space and $Y$$Y$YY$Y$ is a subset of $X$$X$XX$X$ containing $\theta$$\theta$theta\theta$\theta$ then the set

is a semilinear space and the functional $\Vert \cdot {\Vert }_Y:S$$\Vert \cdot {\Vert }_Y:S$||*||_(Y):S\|\cdot\|_{Y}: S$\Vert \cdot {\Vert }_Y:S$ Lip ${}_{0}Y\to {\mathbb{R}}_{+}$${}_{0}Y\to {\mathbb{R}}_{+}$_(0)Y rarrR_(+)_{0} Y \rightarrow \mathbb{R}_{+}${}_{0}Y\to {\mathbb{R}}_{+}$defined by

is a quasi-norm on $SLi{p}_{0}X$$SLi{p}_{0}X$SLip_(0)XS L i p_{0} X$SLi{p}_{0}X$ (see [4] and [6]).
A semilinear space satisfies axioms similar to those defining a linear space, excepting the existence of the opposite element (the inverse with respect to ${}^{+}$${}^{+}$^(+){ }^{+}${}^{+}$) and that the multiplication is defined only for positive scalars. (see [6]).

The following problem is treated in [4]:

Let ( $X,d$$X,d$X,dX, d$X,d$ ) be a quasi-metric space, $Y\subset X$$Y\subset X$Y sub XY \subset X$Y\subset X$ and $f\in S$$f\in S$f in Sf \in S$f\in S$ Lip $Y$$Y$YY$Y$. One asks to find a function $F\in S\mathrm{Lip}X$$F\in S\mathrm{Lip}X$F in S Lip XF \in S \operatorname{Lip} X$F\in S\mathrm{Lip}X$ such that

One shows ([4, Th.2]) that for every $f\in S$$f\in S$f in Sf \in S$f\in S$ LipY there exists $F\in S$$F\in S$F in SF \in S$F\in S$ LipX satisfying (9). This result is similar to a result of Mc Shane from 1934 (see [2]) asserting that every real valued Lipschitz function defined on a subset of a metric space $X$$X$XX$X$ admits an extension to the whole space with the same Lipschitz constant.

In the present paper we study the extension problem for bounded starshaped semi-Lipschitz functions defined on sharshaped subsets of quasi-metric linear spaces:

For a quasi-metric linear space $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ with the quasi-metric $d$$d$dd$d$ starshaped, a starshaped subset $Y$$Y$YY$Y$ of $X$$X$XX$X$ and a bounded semi-Lipschitz starshaped function $f\in {\mathrm{SLip}}_{0}Y$$f\in {\mathrm{SLip}}_{0}Y$f inSLip_(0)Yf \in \operatorname{SLip}_{0} Y$f\in {\mathrm{SLip}}_{0}Y$ find a bounded starshaped function $F\in {\mathrm{SLip}}_{0}X$$F\in {\mathrm{SLip}}_{0}X$F inSLip_(0)XF \in \operatorname{SLip}_{0} X$F\in {\mathrm{SLip}}_{0}X$ such that

where $\Vert \cdot {\Vert }_{\mathrm{\infty }}$$\Vert \cdot {\Vert }_{\mathrm{\infty }}$||*||_(oo)\|\cdot\|_{\infty}$\Vert \cdot {\Vert }_{\mathrm{\infty }}$ denotes the sup-norm.
Observe that if $Y$$Y$YY$Y$ is a subspace of $X$$X$XX$X$ and $f\in {\mathrm{SLip}}_{0}Y$$f\in {\mathrm{SLip}}_{0}Y$f inSLip_(0)Yf \in \operatorname{SLip}_{0} Y$f\in {\mathrm{SLip}}_{0}Y$ is starshaped then it is possible that $f$$f$ff$f$ be unbounded on $Y$$Y$YY$Y$.

More exactly we have:
Lemma 1 Let $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ be a quasi-metric linear space, $Y$$Y$YY$Y$ a subspace of $X$$X$XX$X$ and $f\in$$f\in$f inf \in$f\in$ SLip ${}_{0}Y$${}_{0}Y$_(0)Y_{0} Y${}_{0}Y$ be starshaped. If there exists $x_0\in Y$$x_0\in Y$x_(0)in Yx_{0} \in Y$x_0\in Y$ such that $f\left(x_0\right)>0$$f\left(x_0\right)>0$f(x_(0)) > 0f\left(x_{0}\right)>0$f\left(x_0\right)>0$ then $f$$f$ff$f$ is unbounded on $Y$$Y$YY$Y$.

Proof. For every $x\in Y,x\ne 0$$x\in Y,x\ne 0$x in Y,x!=0x \in Y, x \neq 0$x\in Y,x\ne 0$, the function $h:(0,\mathrm{\infty })\to \mathbb{R}$$h:(0,\mathrm{\infty })\to \mathbb{R}$h:(0,oo)rarrRh:(0, \infty) \rightarrow \mathbb{R}$h:(0,\mathrm{\infty })\to \mathbb{R}$ defined by $h(t)=f(tx)/t$$h(t)=f(tx)/t$h(t)=f(tx)//th(t)=f(t x) / t$h(t)=f(tx)/t$ is non-increasing. Indeed, if $0<t_1<t_2$$0<t_1<t_2$0 < t_(1) < t_(2)0<t_{1}<t_{2}$0<t_1<t_2$ then

In particular, for $t>1$$t>1$t > 1t>1$t>1$ and $x_0\in Y$$x_0\in Y$x_(0)in Yx_{0} \in Y$x_0\in Y$ with $f\left(x_0\right)>0$$f\left(x_0\right)>0$f(x_(0)) > 0f\left(x_{0}\right)>0$f\left(x_0\right)>0$ we have

so that

for every $t>1$$t>1$t > 1t>1$t>1$, which shows that $f$$f$ff$f$ is unbounded on $Y$$Y$YY$Y$.
The following theorem answers positively the question on the extension of bounded starshaped semi-Lipschitz functions.

Theorem 2 Let ( $X,d$$X,d$X,dX, d$X,d$ ) be a quasi-metric linear space and $Y$$Y$YY$Y$ a starshaped subset of $X$$X$XX$X$. Suppose that the quasi-metric $d$$d$dd$d$ is starshaped on $Y$$Y$YY$Y$.

Let $f\in S$$f\in S$f in Sf \in S$f\in S$ Lip ${}_{0}Y$${}_{0}Y$_(0)Y_{0} Y${}_{0}Y$ be bounded and starshaped. In order to exist a bounded starshaped function $F\in SLi{p}_{0}X$$F\in SLi{p}_{0}X$F in SLip_(0)XF \in S L i p_{0} X$F\in SLi{p}_{0}X$ such that

it is necessary and sufficient that $f(y)\le 0$$f(y)\le 0$f(y) <= 0f(y) \leq 0$f(y)\le 0$ for all $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$.
Proof. Sufficiency suppose that $f(y)\le 0$$f(y)\le 0$f(y) <= 0f(y) \leq 0$f(y)\le 0$ for all $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$. The function

is starshaped on $X$$X$XX$X$ and satisfies the conditions

(see [4,Th.2] and [5, Th.8).
Indeed, let $z\in Y$$z\in Y$z in Yz \in Y$z\in Y$ and $x\in X$$x\in X$x in Xx \in X$x\in X$. For any $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$ we have

The inequality $d(z,y)-d(x,y)\le d(z,x)$$d(z,y)-d(x,y)\le d(z,x)$d(z,y)-d(x,y) <= d(z,x)d(z, y)-d(x, y) \leq d(z, x)$d(z,y)-d(x,y)\le d(z,x)$ implies

showing that for every $x\in X$$x\in X$x in Xx \in X$x\in X$ the set $\{f(y)+\Vert f{\Vert }_{Y}d(x,y):y\in Y\}$$\left\{f(y)+\Vert f{\Vert }_{Y}d(x,y):y\in Y\right\}${f(y)+||f||_(Y)d(x,y):y in Y}\left\{f(y)+\|f\|_{Y} d(x, y): y \in Y\right\}$\{f(y)+\Vert f{\Vert }_{Y}d(x,y):y\in Y\}$ is bounded from above by $f(z)-\Vert f{\Vert }_{Y}d(z,x)$$f(z)-\Vert f{\Vert }_{Y}d(z,x)$f(z)-||f||_(Y)d(z,x)f(z)-\|f\|_{Y} d(z, x)$f(z)-\Vert f{\Vert }_{Y}d(z,x)$, and the infimum (11) is finite.

We show now that $H(y)=f(y)$$H(y)=f(y)$H(y)=f(y)H(y)=f(y)$H(y)=f(y)$ for all $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$.
Let $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$. Then

For any $v\in Y$$v\in Y$v in Yv \in Y$v\in Y$ we have

so that

and

It follows $H(y)=f(y)$$H(y)=f(y)$H(y)=f(y)H(y)=f(y)$H(y)=f(y)$.
We prove that $\Vert H{\Vert }_Y=\Vert f{\Vert }_Y$$\Vert H{\Vert }_Y=\Vert f{\Vert }_Y$||H||_(Y)=||f||_(Y)\|H\|_{Y}=\|f\|_{Y}$\Vert H{\Vert }_Y=\Vert f{\Vert }_Y$.
Since ${H|}_Y=f$${\left.H\right|}_Y=f$H|_(Y)=f\left.H\right|_{Y}=f${H|}_Y=f$, the definitions of $\Vert H{\Vert }_Y$$\Vert H{\Vert }_Y$||H||_(Y)\|H\|_{Y}$\Vert H{\Vert }_Y$ and $\Vert f{\Vert }_Y$$\Vert f{\Vert }_Y$||f||_(Y)\|f\|_{Y}$\Vert f{\Vert }_Y$ yield $\Vert H{\Vert }_Y\ge \Vert f{\Vert }_Y$$\Vert H{\Vert }_Y\ge \Vert f{\Vert }_Y$||H||_(Y) >= ||f||_(Y)\|H\|_{Y} \geq\|f\|_{Y}$\Vert H{\Vert }_Y\ge \Vert f{\Vert }_Y$.
Let $x_1,x_2\in X$$x_1,x_2\in X$x_(1),x_(2)in Xx_{1}, x_{2} \in X$x_1,x_2\in X$ and $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0$\epsilon >0$. Choosing $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$ such that

we obtain

Since $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0$\epsilon >0$ is arbitrary it follows

for any $x_1,x_2\in X$$x_1,x_2\in X$x_(1),x_(2)in Xx_{1}, x_{2} \in X$x_1,x_2\in X$ and $\Vert H{\Vert }_Y\le \Vert f{\Vert }_Y$$\Vert H{\Vert }_Y\le \Vert f{\Vert }_Y$||H||_(Y) <= ||f||_(Y)\|H\|_{Y} \leq\|f\|_{Y}$\Vert H{\Vert }_Y\le \Vert f{\Vert }_Y$. Then $\Vert H{\Vert }_X=\Vert f{\Vert }_Y$$\Vert H{\Vert }_X=\Vert f{\Vert }_Y$||H||_(X)=||f||_(Y)\|H\|_{X}=\|f\|_{Y}$\Vert H{\Vert }_X=\Vert f{\Vert }_Y$.
We shall show that $H$$H$HH$H$ is also starshaped on $X$$X$XX$X$. To this end let $x\in X,z\in Y$$x\in X,z\in Y$x in X,z in Yx \in X, z \in Y$x\in X,z\in Y$ and $\alpha \in [0,1]$$\alpha \in [0,1]$alpha in[0,1]\alpha \in[0,1]$\alpha \in [0,1]$. We have

Taking the infimum with respect to $z\in Y$$z\in Y$z in Yz \in Y$z\in Y$ we get

for all $x\in X$$x\in X$x in Xx \in X$x\in X$ and all $\alpha \in [0,1]$$\alpha \in [0,1]$alpha in[0,1]\alpha \in[0,1]$\alpha \in [0,1]$, showing that the function $H$$H$HH$H$ defined by (11) is a starshaped extension of $f$$f$ff$f$.

Consider the function

Since ${F|}_Y={H|}_Y=f,\Vert H{\Vert }_X=\Vert f{\Vert }_Y$${\left.F\right|}_Y={\left.H\right|}_Y=f,\Vert H{\Vert }_X=\Vert f{\Vert }_Y$F|_(Y)=H|_(Y)=f,||H||_(X)=||f||_(Y)\left.F\right|_{Y}=\left.H\right|_{Y}=f,\|H\|_{X}=\|f\|_{Y}${F|}_Y={H|}_Y=f,\Vert H{\Vert }_X=\Vert f{\Vert }_Y$ and $\Vert H{\Vert }_X\ge \Vert F{\Vert }_X$$\Vert H{\Vert }_X\ge \Vert F{\Vert }_X$||H||_(X) >= ||F||_(X)\|H\|_{X} \geq\|F\|_{X}$\Vert H{\Vert }_X\ge \Vert F{\Vert }_X$, it follows that $\Vert F{\Vert }_X=\Vert f{\Vert }_Y$$\Vert F{\Vert }_X=\Vert f{\Vert }_Y$||F||_(X)=||f||_(Y)\|F\|_{X}=\|f\|_{Y}$\Vert F{\Vert }_X=\Vert f{\Vert }_Y$. Therefore $F$$F$FF$F$ is an extension of $f$$f$ff$f$ with the same semi-Lipschitz constant.

Let $x\in X$$x\in X$x in Xx \in X$x\in X$. If $H(x)\le 0$$H(x)\le 0$H(x) <= 0H(x) \leq 0$H(x)\le 0$ then

for any $\alpha \in [0,1]$$\alpha \in [0,1]$alpha in[0,1]\alpha \in[0,1]$\alpha \in [0,1]$ so that

If $H(x)>0$$H(x)>0$H(x) > 0H(x)>0$H(x)>0$ then $F(x)=0$$F(x)=0$F(x)=0F(x)=0$F(x)=0$. Let $\alpha \in (0,1)$$\alpha \in (0,1)$alpha in(0,1)\alpha \in(0,1)$\alpha \in (0,1)$. If $H(\alpha x)>0$$H(\alpha x)>0$H(alpha x) > 0H(\alpha x)>0$H(\alpha x)>0$ then