Abstract
Authors
“Tiberiu Popoviciu” Institute of Numerical Analysis, Romanian Academy, Romania
Keywords
Paper coordinates
C. Mustăţa, The extension of starshaped bounded Lipschitz functions, Anal. Numer. Théor. Approx. 9 (1980) 1, 93-99
About this paper
Journal
Revue d’Analyse Numer. Theor.Approximation
Publisher Name
Publishing Romanian Academy
Print ISSN
2457-6794
Online ISSN
2501-059X
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[1] Cobzas, S., Mustata, C., Norm Preserving Extension of Convex Lipschitz Functions, J. Approx. Theory 24 (1978), 555-564.
[2] Holmes, R.B., A Course on Optimization and Best Approximation, Lectures Notes in Math. No. 257, Springer Verlag, Berlin-Heidelberg-New York, 1972.
[3] Johnson, J.A., Banach Spaces of Lipschitz Functions and Vector-Valued Lipschitz Functions, Trans. Amer. Math. Soc. 148 (1970), 147-169.
[4] McShane, R.J., Extension of Range of Functions, Bull. Amer. Math. Soc. 40 (1934), 837-842.
[5] Musata, C., Best Approximation and Unique Extension of Lipschitz Functions, H. Approx. Theory 19 (1977), 222-230.
[6] Mustata, C., A Characterization of Chebyshevian Subspace of Y^{}-Type, Mathematica – Revue Anal. Num. Teor. Approx., L’Analyse Num. Teor. approx. 6, 1 (1977), 51-56.
[7] Mustata, C., Norm Preserving Extension of Starshaped Lipschitz Functions, Mathematica 19 (42) 2 (1977), 183-187.
[8] Phelps, R.R., Uniqueness of Hahn-Banach Extension and Unique Best Approximation, Trans. Amer. Math. Soc. 95 (1960), 238-255.
Paper (preprint) in HTML form
THE EXTENSION OF STARSHAPED BOUNDED LIPSCHITZ FUNCTIONS
- Let XX be a normed space and YY a nonvoid subset of XX. The set YY is called starshaped (with respect to theta in X\theta \in X ) if alpha y in Y\alpha y \in Y for all alpha in[0,1]\alpha \in[0,1] and y in Yy \in Y. A real function ff defined on YY is called starshaped if
{:(1.1)f(alpha y) <= alpha f(y)",":}\begin{equation*}
f(\alpha y) \leq \alpha f(y), \tag{1.1}
\end{equation*}
for all alpha in[0,1]\alpha \in[0,1] and y in Yy \in Y. From (1.1) follows f(theta) <= 0f(\theta) \leq 0. In the following we consider only starshaped functions vanishing at theta.Y\theta . Y will denote always a starshaped set.
A function f:Y rarr Rf: Y \rightarrow R is called Lipschitz on YY if there exists K >= 0K \geq 0 such that
{:(1.2)|f(x)-f(y)| <= K*||x-y||:}\begin{equation*}
|f(x)-f(y)| \leq K \cdot\|x-y\| \tag{1.2}
\end{equation*}
for all x,y in Yx, y \in Y. Denote by Lip_(0)Y\operatorname{Lip}_{0} Y ( Lip_(0)X\operatorname{Lip}_{0} X ) the space of all Lipschitz functions on YY (respectively on XX ) vanishing at theta in Y sub X\theta \in Y \subset X ([5]) and by BLip_(0)Y\mathrm{BLip}_{0} Y ( BLip_(0)X\mathrm{BLip}_{0} X ) their subspaces formed of bounded functions on YY (respectively on XX ). Denote also
{:[(1.3)BS_(Y)={f:f inBLip_(0)Y,f," is starshaped "}],[(1.4)BS_(X)={F:F inBLip_(0)X,F" is starshaped "}]:}\begin{align*}
& \mathrm{BS}_{Y}=\left\{f: f \in \mathrm{BLip}_{0} Y, f, \text { is starshaped }\right\} \tag{1.3}\\
& \mathrm{BS}_{X}=\left\{F: F \in \mathrm{BLip}_{0} X, F \text { is starshaped }\right\} \tag{1.4}
\end{align*}
The sets BS_(Y)\mathrm{BS}_{Y} and BS_(X)\mathrm{BS}_{X} are convex cones in BLip_(0)Y\mathrm{BLip}_{0} Y and in BLip_(0)X\mathrm{BLip}_{0} X, respectively, i.e. f+gf+g and lambda f\lambda f are in BS_(Y)\mathrm{BS}_{Y} (in BS_(X)\mathrm{BS}_{X} ) for all f,gf, g in BS_(Y)\mathrm{BS}_{Y} (in BS_(X)\mathrm{BS}_{X} ) and lambda >= 0\lambda \geq 0.
For f inBS_(Y)(f inBLip_(0)Y)f \in \mathrm{BS}_{Y}\left(f \in \mathrm{BLip}_{0} Y\right) put
{:(1.5))||f||_(d)=s u p{|f(y_(1))-f(y_(2))|//||y_(1)-y_(2)||:y_(1),y_(2)in Y,y_(1)!=y_(2)}:}\begin{equation*}
\|f\|_{d}=\sup \left\{\left|f\left(y_{1}\right)-f\left(y_{2}\right)\right| /\left\|y_{1}-y_{2}\right\|: y_{1}, y_{2} \in Y, y_{1} \neq y_{2}\right\} \tag{1.5)}
\end{equation*}
the Lipschitz norm of ff, and
{:(1.6)||f||_(oo)=s u p{|f(y)|:y in Y}",":}\begin{equation*}
\|f\|_{\infty}=\sup \{|f(y)|: y \in Y\}, \tag{1.6}
\end{equation*}
the uniform norm of ff.
For F inBLip_(0)XF \in \mathrm{BLip}_{0} X, the Lipschitz and the uniform norms are defined similarly.
2. It is well known (see e.g. [4]) that a Lipschitz function ff defined on a nonvoid subset YY of a metric space XX has a norm preserving Lipschitz extension FF on XX, i.e. F|_(Y)=f\left.F\right|_{Y}=f and ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d}. In [1] it was shown that if Y is a nonvoid convex subset of a normed space XX, then every convex Lipschitz function on YY has a convex norm preserving Lipschitz extension FF on XX. In a similar result was established for starshaped Lipschitz function [7].
For F inBLip_(0)XF \in \mathrm{BLip}_{0} X, the Lipschitz and the uniform norms are defined similarly.
2. It is well known (see e.g. [4]) that a Lipschitz function ff defined on a nonvoid subset YY of a metric space XX has a norm preserving Lipschitz extension FF on XX, i.e. F|_(Y)=f\left.F\right|_{Y}=f and ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d}. In [1] it was shown that if Y is a nonvoid convex subset of a normed space XX, then every convex Lipschitz function on YY has a convex norm preserving Lipschitz extension FF on XX. In a similar result was established for starshaped Lipschitz function [7].
This Note is concerned with the problem of extension of bounded starshaped Lipschitz functions, i.e. for f inBS_(Y)f \in \mathrm{BS}_{Y} find F inBS_(X)F \in \mathrm{BS}_{X} such that F|_(Y)=f,||F||_(d)=||f||_(d)\left.F\right|_{Y}=f,\|F\|_{d}=\|f\|_{d} and ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty}. The function FF is called briefly an extension of ff. We consider the problem of existence and unicity of such an extension.
Remark, that similar problem for bounded convex Lipschitz functions has a trivial answer: if YY is a convex subset of XX, such that theta in Y\theta \in Y, then the nul function is the only bounded convex (Lipschitz) function on YY which has a bounded convex Lipschitz extension on XX. This follows from the fact that the constant functions are only bounded convex functions on XX. Indeed, if F:X rarr RF: X \rightarrow R is a nonconstant there XX. XX unction, then there exist two points x_(0),x_(1)in Xx_{0}, x_{1} \in X sich that F(x_(0))!=F(x_(1))F\left(x_{0}\right) \neq F\left(x_{1}\right), say F(x_(0))<<F(x_(1))F\left(x_{0}\right)< <F\left(x_{1}\right). But then, since the function varphi:(0,oo)rarr R\varphi:(0, \infty) \rightarrow R, defined by
varphi(t)=(F(x_(0)+t(x_(1)-x_(0)))-F(x_(0)))/(t),t > 0\varphi(t)=\frac{F\left(x_{0}+t\left(x_{1}-x_{0}\right)\right)-F\left(x_{0}\right)}{t}, t>0
is nondecreasing (see [2] p. 17), it follows that
(F(x_(0)+t(x_(1)-x_(0)))-F(x_(0)))/(t) >= F(x_(1))-F(x_(0)) > 0\frac{F\left(x_{0}+t\left(x_{1}-x_{0}\right)\right)-F\left(x_{0}\right)}{t} \geq F\left(x_{1}\right)-F\left(x_{0}\right)>0
so that F(x_(0)+t(x_(1)-x_(0))) >= F(x_(0))+t[F(x_(1))-F(x_(0))]F\left(x_{0}+t\left(x_{1}-x_{0}\right)\right) \geq F\left(x_{0}\right)+t\left[F\left(x_{1}\right)-F\left(x_{0}\right)\right], for all t >= 1t \geq 1, which shows that the function FF is unbounded.
2a. The existence of extension. In Theorem 1 below will be shown that under some suplementary hypotheses on the function f inBS_(Y)f \in \mathrm{BS}_{Y} there exists an extension F inBS_(X)F \in \mathrm{BS}_{X} of ff.
Firstly, we prove two lemmas.
lemma 1. Let XX be a real normed space and ff a starshaped function on XX. Then, for every x in X,x!=thetax \in X, x \neq \theta, the function Psi:(0,oo)rarr R\Psi:(0, \infty) \rightarrow R, defined by
(2.1)
lemma 1. Let XX be a real normed space and ff a starshaped function on XX. Then, for every x in X,x!=thetax \in X, x \neq \theta, the function Psi:(0,oo)rarr R\Psi:(0, \infty) \rightarrow R, defined by
(2.1)
Psi(t)=f(tx)//t,quad t > 0\Psi(t)=f(t x) / t, \quad t>0
is nondecreasing.
Proof. For 0 < t_(1) < t_(2)0<t_{1}<t_{2} and a fixed x in X,x!=thetax \in X, x \neq \theta, we have
Proof. For 0 < t_(1) < t_(2)0<t_{1}<t_{2} and a fixed x in X,x!=thetax \in X, x \neq \theta, we have
(f(t_(1)x))/(t_(1))=(f((t_(1)//t_(2))t_(2)x))/(t_(1)) <= ((t_(1)//t_(2))f(t_(2)x))/(t_(1))=(f(t_(2)x))/(t_(2)).\frac{f\left(t_{1} x\right)}{t_{1}}=\frac{f\left(\left(t_{1} / t_{2}\right) t_{2} x\right)}{t_{1}} \leq \frac{\left(t_{1} / t_{2}\right) f\left(t_{2} x\right)}{t_{1}}=\frac{f\left(t_{2} x\right)}{t_{2}} .
Now, for a function f:X rarr Rf: X \rightarrow R define, as usually, the epigraph of ff, by
{:(2.2)" epi "f={(x","alpha)in X xx R:f(x) <= alpha}:}\begin{equation*}
\text { epi } f=\{(x, \alpha) \in X \times R: f(x) \leq \alpha\} \tag{2.2}
\end{equation*}
LEMMA 2. A function f:X rarr R,f(theta)=0f: X \rightarrow R, f(\theta)=0, is starshaped if and only if its epigraph is starshaped.
Proof. If ff is starshaped, f(theta)=0f(\theta)=0, and (x,alpha)in(x, \alpha) \in epi ff, then for every lambda in[0,1],f(lambda x) <= lambda f(x) <= lambda alpha\lambda \in[0,1], f(\lambda x) \leq \lambda f(x) \leq \lambda \alpha, so that lambda(x,alpha)=(lambda x,lambda alpha)in epi f\lambda(x, \alpha)=(\lambda x, \lambda \alpha) \in \operatorname{epi} f. Conversely, if epi ff is starshaped then (x,f(x))in epi f(x, f(x)) \in \operatorname{epi} f implies (lambda x,lambda f(x))∈∈(\lambda x, \lambda f(x)) \in \in epi ff, i.e. f(lambda x) <= lambda f(x)f(\lambda x) \leq \lambda f(x), for all lambda in[0,1]\lambda \in[0,1].
THEOREM 1. Let XX be a normed space, YY a starshaped subset of XX, theta in Y\theta \in Y, and f inBS_(Y)f \in \mathrm{BS}_{Y}. Then, there exists F inBS_(X)F \in \mathrm{BS}_{X} such that
(i) F|_(Y)=f\left.F\right|_{Y}=f,
(ii) ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d},
(iii) ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty},
if and only if f(y) <= 0f(y) \leq 0, for all y in Yy \in Y.
Proof. Let f inBS_(Y)f \in \mathrm{BS}_{Y} and suppose f <= 0f \leq 0 on YY. Define G:X rarr RG: X \rightarrow R by
(i) F|_(Y)=f\left.F\right|_{Y}=f,
(ii) ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d},
(iii) ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty},
if and only if f(y) <= 0f(y) \leq 0, for all y in Yy \in Y.
Proof. Let f inBS_(Y)f \in \mathrm{BS}_{Y} and suppose f <= 0f \leq 0 on YY. Define G:X rarr RG: X \rightarrow R by
{:(2.3)G(x)=i n f_(y in Y)[f(y)+||f||_(d)||x-y||]:}\begin{equation*}
G(x)=\inf _{y \in Y}\left[f(y)+\|f\|_{d}\|x-y\|\right] \tag{2.3}
\end{equation*}
The function GG defined by (2.3) is starshaped and satisfies G|_(Y)=f\left.G\right|_{Y}=f, ||G||_(d)=||f||_(d)\|G\|_{d}=\|f\|_{d} (see [7]).
Let
Let
{:(2.4)F(x)={[0," if "G(x) > 0],[G(x)," if "G(x) <= 0]:}:}F(x)= \begin{cases}0 & \text { if } G(x)>0 \tag{2.4}\\ G(x) & \text { if } G(x) \leq 0\end{cases}
Since G|_(Y)=f <= 0\left.G\right|_{Y}=f \leq 0, it follows F|_(Y)=f\left.F\right|_{Y}=f and ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d}. Obviously ||F||_(oo) >= ||f||_(oo)\|F\|_{\infty} \geq\|f\|_{\infty}. If x in Xx \in X is such that G(x) <= 0G(x) \leq 0, then
0 >= F(x)=G(x) >= f(y)+||f||_(d)*||x-y|| >= f(y),0 \geq F(x)=G(x) \geq f(y)+\|f\|_{d} \cdot\|x-y\| \geq f(y),
so that 0 <= -F(x) <= -f(y)0 \leq-F(x) \leq-f(y), for all y in Yy \in Y. Therefore ||F||_(oo)=i n f_(x in X)(-F(x))≤≤||f||_(oo)\|F\|_{\infty}=\inf _{x \in X}(-F(x)) \leq \leq\|f\|_{\infty} and ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty}. Since the epigraph of FF is starshaped, by Lemma 2,F2, F is starshaped. Consequently, FF is the required extension of ff.
Suppose now, that there exists y_(0)in Yy_{0} \in Y such that f(y_(0)) > 0f\left(y_{0}\right)>0, and let FF be a starshaped extension of ff. By Lemma 1
0 < f(y_(0))=F(y_(0)) <= (F(ty_(0))//t),0<f\left(y_{0}\right)=F\left(y_{0}\right) \leq\left(F\left(t y_{0}\right) / t\right),
so that F(ty_(0)) >= tf(y_(0))F\left(t y_{0}\right) \geq t f\left(y_{0}\right), for all t >= 1t \geq 1, which shows that the function FF is unbounded. Therefore ff has no bounded strashaped extension, which ends the proof of Theorem 1.
Let
(2.5)
(2.5)
{:[BSS_(Y)={f inBS_(Y):f <= 0}],[BS_(X)^(-)={F inBS_(X):F <= 0}]:}\begin{gathered}
\mathrm{BS} S_{Y}=\left\{f \in \mathrm{BS}_{Y}: f \leq 0\right\} \\
\mathrm{BS}_{X}^{-}=\left\{F \in \mathrm{BS}_{X}: F \leq 0\right\}
\end{gathered}
By Theorem 1 follows:
corolyary 1. Every function f inBS_( bar(Y))f \in \mathrm{BS}_{\bar{Y}} has an extension F inBS_( bar(X))F \in \mathrm{BS}_{\bar{X}}.
2b. The unicity of extension. By Theorem 1 and Corollary 1, every nonpositive bounded starshaped function, defined on a starshaped subset YY of a normed space XX, has a nonpositive bounded starshaped extension to whole XX. Furthemore, these are the only bounded starshaped function on YY admitting bounded starshaped Lipschitz extension on XX.
corolyary 1. Every function f inBS_( bar(Y))f \in \mathrm{BS}_{\bar{Y}} has an extension F inBS_( bar(X))F \in \mathrm{BS}_{\bar{X}}.
2b. The unicity of extension. By Theorem 1 and Corollary 1, every nonpositive bounded starshaped function, defined on a starshaped subset YY of a normed space XX, has a nonpositive bounded starshaped extension to whole XX. Furthemore, these are the only bounded starshaped function on YY admitting bounded starshaped Lipschitz extension on XX.
Equiped with the norms
{:[||f||_(Y)=max(||f||_(d),:}{:||f||_(oo))","quad f inBLip_(0)Y],[(2.6)||F||_(X)=max(||F||_(d),:}{:||F||_(oo))","quad F inBLip_(0)X]:}\begin{align*}
\|f\|_{Y}=\max \left(\|f\|_{d},\right. & \left.\|f\|_{\infty}\right), \quad f \in \mathrm{BLip}_{0} Y \\
\|F\|_{X}=\max \left(\|F\|_{d},\right. & \left.\|F\|_{\infty}\right), \quad F \in \mathrm{BLip}_{0} X \tag{2.6}
\end{align*}
BLip_(0)Y\mathrm{BLip}_{0} Y and BLip_(0)X\mathrm{BLip}_{0} X become Banach spaces (see [3]). Let H=BS_( bar(X))H=\mathrm{BS}_{\bar{X}} --BS_(X)-\mathrm{BS}_{X} the subspace of BLip_(0)X\mathrm{BLip}_{0} X generated by the convex cone BS_(X)\mathrm{BS}_{X} and
Y⊥={g in H:g|_(Y)=0}Y \perp=\left\{g \in H:\left.g\right|_{Y}=0\right\}
the anihilator of the set YY in HH. Obviously, Y _|_Y \perp is a subspace of HH. A subset ZZ of a normed space XX is called proximinal for W sub XW \subset X if for every f in Wf \in W there exists g_(0)in Zg_{0} \in Z, such that
{:(2.7)||f-g_(0)||=d(f","Z)=i n f{||f-g||:g in Z}.:}\begin{equation*}
\left\|f-g_{0}\right\|=d(f, Z)=\inf \{\|f-g\|: g \in Z\} . \tag{2.7}
\end{equation*}
If for every f in Wf \in W the element g_(0)in Zg_{0} \in Z satisfying (2.7) is unique then the set ZZ is called Chebyshevian for WW. An element g_(0)in Zg_{0} \in Z satisfying (2.7) is called an element of best approximation of ff by elements of ZZ.
THEOREM 2. Y _|_Y \perp is a Chebyshevian subsapce for BS_( bar(X))\mathrm{BS}_{\bar{X}} if and only if, every f inBSf \in \mathrm{BS} has a unique (preserving the uniform and Lipschitz norms) extension FF in BS_( bar(x))\mathrm{BS}_{\bar{x}}.
Proof. Follows from Theorem 1 in [6].
Remark. Observe that Theorem 2 remains true if the spaces BLip_(0)Y\mathrm{BLip}_{0} Y and BLip_(0)X\mathrm{BLip}_{0} X are equiped with the norms
Remark. Observe that Theorem 2 remains true if the spaces BLip_(0)Y\mathrm{BLip}_{0} Y and BLip_(0)X\mathrm{BLip}_{0} X are equiped with the norms
{:||f||_(1)=||f||_(a)+||f||_(oo)," for "f inBLip_(0)Y" (respectively "BLip_(0)X)\left.\|f\|_{1}=\|f\|_{a}+\|f\|_{\infty}, \text { for } f \in \mathrm{BLip}_{0} Y \text { (respectively } \mathrm{BLip}_{0} X\right)
Theorem 2 is analogous with a theorem of PHELPS [8], in the linear case.
3. Now, we try to find conditions on the function ff ensuring the unicity of the extension.
3. Now, we try to find conditions on the function ff ensuring the unicity of the extension.
Consider, firstly, the case X=RX=R with the usual norm |*||\cdot| (the absolute value).
THEOREM 3. Let Y=[a,b]sub R,a < 0 < bY=[a, b] \subset R, a<0<b. A function f inBS_(Y)f \in \mathrm{BS}_{Y} has a unique extension F inBS_(R)^(-)F \in \mathrm{BS}_{R}^{-}if and only if f(a)=f(b)=0f(a)=f(b)=0.
Proof. Suppose that f inBS_(Y)^(-)f \in \mathrm{BS}_{Y}^{-}has two distinct extensions F_(1),F_(2)F_{1}, F_{2} in BS_( bar(R))^(-)\mathrm{BS}_{\bar{R}}^{-}. Let x in R\\[a,b]x \in R \backslash[a, b] be such that F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x), say F_(1)(x) < F_(2)(x) <= 0F_{1}(x)<F_{2}(x) \leq 0. Suppose x > bx>b. The function F_(1)F_{1} being starshaped it follows
F_(1)(lambda x) <= lambdaF_(1)(x) < 0,F_{1}(\lambda x) \leq \lambda F_{1}(x)<0,
for all lambda in(0,1]\lambda \in(0,1]. In particular, since b=lambda_(b)xb=\lambda_{b} x for lambda_(b)in(0,1)\lambda_{b} \in(0,1), it follows f(b)=F_(1)(b) < 0f(b)=F_{1}(b)<0. If x < ax<a, then a=lambda_(a)xa=\lambda_{a} x, for a lambda_(a)in(0,1)\lambda_{a} \in(0,1), and similarly, f(a)=F_(1)(a) < 0f(a)=F_{1}(a)<0.
Conversely, we shall show that if f inBS_(Y)^(-)f \in \mathrm{BS}_{Y}^{-}is such that f(a) < 0f(a)<0 of f(b) < 0f(b)<0, then ff has at least two distinct extensions F_(1)F_{1} and F_(2)F_{2} in BS_( bar(R))\mathrm{BS}_{\bar{R}}. If f(b) < 0f(b)<0, then
F_(1)(x)={[f(x)",",x in[0","b]],[f(b)+||f||_(d)(x-b)",",x in(b,b-f(b)(||f||_(d))^(-1)]],[0",",x in(-oo","a)uu(b-f(b)*(||f||_(d))^(-1),-oo)]:}F_{1}(x)= \begin{cases}f(x), & x \in[0, b] \\ f(b)+\|f\|_{d}(x-b), & x \in\left(b, b-f(b)\left(\|f\|_{d}\right)^{-1}\right] \\ 0, & x \in(-\infty, a) \cup\left(b-f(b) \cdot\left(\|f\|_{d}\right)^{-1},-\infty\right)\end{cases}
and
F_(2)(x)={[f(x),","x in[0","b]],[(f(b)//b)x,","x in(b,(-||f||_(oo)//f(b))b)],[-||f||_(oo),","x in[-(||f||_(oo)//f(b))b,+oo)],[0,","x in(-oo","a)]:}F_{2}(x)= \begin{cases}f(x) & , x \in[0, b] \\ (f(b) / b) x & , x \in\left(b,\left(-\|f\|_{\infty} / f(b)\right) b\right) \\ -\|f\|_{\infty} & , x \in\left[-\left(\|f\|_{\infty} / f(b)\right) b,+\infty\right) \\ 0 & , x \in(-\infty, a)\end{cases}
are two distinct extensions of ff, i.e. F_(1)!=F_(2),F_(1)|_(Y)=F_(2)|_(Y)=f,||F^(')_(1)||_(a)==||F_(2)||_(d)=||f||_(d),||F_(1)||_(oo)=||F_(2)||_(oo)=||f||_(oo)F_{1} \neq F_{2},\left.F_{1}\right|_{Y}=\left.F_{2}\right|_{Y}=f,\left\|F^{\prime}{ }_{1}\right\|_{a}= =\left\|F_{2}\right\|_{d}=\|f\|_{d},\left\|F_{1}\right\|_{\infty}=\left\|F_{2}\right\|_{\infty}=\|f\|_{\infty}.
7 - Mathematica - Revue d'analyse numérique et de théorie de l'approximation, Tome 9. nr. 1/1980
F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x) for all x > max{b-f(b)*(||f||_(a))^(-1);-||f||_(oo)*(f(b))^(-1)}x>\max \left\{b-f(b) \cdot\left(\|f\|_{a}\right)^{-1} ;-\|f\|_{\infty} \cdot(f(b))^{-1}\right\}.
If f(a) < 0f(a)<0, then two distinct extensions F_(1),F_(2)F_{1}, F_{2}, may be given, in a similar way.
F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x) for all x > max{b-f(b)*(||f||_(a))^(-1);-||f||_(oo)*(f(b))^(-1)}x>\max \left\{b-f(b) \cdot\left(\|f\|_{a}\right)^{-1} ;-\|f\|_{\infty} \cdot(f(b))^{-1}\right\}.
If f(a) < 0f(a)<0, then two distinct extensions F_(1),F_(2)F_{1}, F_{2}, may be given, in a similar way.
Remark. The hypothesis a < 0 < ba<0<b in Theorem 3 is essential as it is shown by the following example. Take Y=[0,b],b > 0Y=[0, b], b>0 or Y==[a,0],a < 0Y= =[a, 0], a<0. Then every f inBSS_(Y)^(-)f \in \mathrm{BSS}_{Y}^{-}has an infinite set of extensions in BS_( bar(R))^(-)\mathrm{BS}_{\bar{R}}^{-}. For exemple, if Y=[0,b]Y=[0, b] and f inBS_( bar(Y))^(-)f \in \mathrm{BS}_{\bar{Y}}^{-}is such that f(0)=0f(0)=0, f(b)=0f(b)=0, then
F_(lambda)(x)={[f(x)",",x in[0","b]],[0",",x in(0","+oo)],[||f||_(d)x",",x in(-lambda||f||_(oo)*(||f||_(d))^(-1),0)],[-lambda||f||_(oo)",",x in(-oo,-lambda||_(f)∣||_(oo)(||f||_(d))^(-1))]:}F_{\lambda}(x)= \begin{cases}f(x), & x \in[0, b] \\ 0, & x \in(0,+\infty) \\ \|f\|_{d} x, & x \in\left(-\lambda\|f\|_{\infty} \cdot\left(\|f\|_{d}\right)^{-1}, 0\right) \\ -\lambda\|f\|_{\infty}, & x \in\left(-\infty,-\lambda\left\|_{f} \mid\right\|_{\infty}\left(\|f\|_{d}\right)^{-1}\right)\end{cases}
is an extension of ff for every lambda in[0,1]\lambda \in[0,1].
Consider now the general case. For x in X,x!=thetax \in X, x \neq \theta, the ray vec(theta x)\overrightarrow{\theta x} is defined by
Consider now the general case. For x in X,x!=thetax \in X, x \neq \theta, the ray vec(theta x)\overrightarrow{\theta x} is defined by
vec(theta x)={alpha x:alpha >= 0}\overrightarrow{\theta x}=\{\alpha x: \alpha \geq 0\}
If the set YY is starshaped and y in Y,y!=thetay \in Y, y \neq \theta, then vec(theta y)sub Y\overrightarrow{\theta y} \subset Y or vec(theta y)nn Y\overrightarrow{\theta y} \cap Y is a segment. In the second case put
alpha_(y)=s u p{alpha:alpha y in Y},\alpha_{y}=\sup \{\alpha: \alpha y \in Y\},
and z_(y)=alpha_(y)*yz_{y}=\alpha_{y} \cdot y. The set {z_(y):y in Y}\left\{z_{y}: y \in Y\right\} is called the algebric starshapes boundary of YY and is denoted by F_(r)^(s)Y\mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Y.
Evidently, every z inF_(r)^(s)Yz \in \mathrm{~F}_{\mathrm{r}}^{\mathrm{s}} Y is a limit point of YY, i.e. Y uuF_(r)^(s)Y( bar(Y)Y \cup \mathrm{~F}_{\mathrm{r}}^{\mathrm{s}} Y(\bar{Y}. Since every f inBS_(Y)f \in \mathrm{BS}_{Y} is uniformly continuous (as Lipschitz) it can be uniquely extended to Y uuuFr_(r)^(s)YY \bigcup \mathrm{Fr}_{r}^{s} Y. Therefore with no restrition of generality, we can suppose F_(r)^(s)Y sub Y\mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Y \subset Y.
THEOREM 4. Let YY be an absorbing starshaped subset of the normed space XX, such that. F_(r)^(s)Y sub Y\mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Y \subset Y. If f inBS_(Y)^(-)f \in \mathrm{BS}_{Y}^{-}is such that f(z)=0f(z)=0, for all z∈∈F_(r)^(s)Yz \in \in \mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Y, then ff has a unique extension F inBSS_(X)^(-)F \in \mathrm{BSS}_{X}^{-}.
Proof. Suppose f inBS_(Y)^(-),f(z)=0f \in \mathrm{BS}_{Y}^{-}, f(z)=0, for all z inF_(I)^(s)Yz \in \mathrm{~F}_{\mathrm{I}}^{s} Y, and suppose that ff has two distinct extensions F_(1),F_(2)F_{1}, F_{2} in BS_(X)^(-)\mathrm{BS}_{X}^{-}. Let x in X\\Yx \in X \backslash Y be such that F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x), say F_(1)(x) < F_(2)(x) <= 0F_{1}(x)<F_{2}(x) \leq 0. The set YY being absorbing and starshaped, there exists lambda > 0\lambda>0 such that lambda x inF_(r)^(s)Y\lambda x \in \mathrm{~F}_{\mathrm{r}}^{\mathrm{s}} Y. But then, one obtains the contradiction
0=f(lambda x)=F_(1)(lambda x) <= lambdaF_(1)(x) < 00=f(\lambda x)=F_{1}(\lambda x) \leq \lambda F_{1}(x)<0
Theorem 4 is proved.
BIBLIOGRAPHY
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[2] Holmes, R. B., A Course on Optimisation and Best Approximation, Lectures Notes in s, R. B., A Course on Poringer Verlag, Berlin-Heidelberg-New York, 1972.
[3] J. A. Banach Spaces of Lipschitz Functions and Vector-Valued Lipschitz Fi Mons, Trans. Amer. Math. Soc. 148 (1970), 147-169.
Funch. Extension of Range of Functions, Bull. Amer. Math. Soc. 40 (1934),
[4] McSh 837-842.
[5] Lipschitz Functions, J. [5] Mustăta, C.,
[5] Must Approx. The Approximation and 230.
Approx. Theory 19 (1977), 222-230.
[6] Mustăta, C., A Characterisainon of Cheox., L'Analyse Num. Teor. Approx. 6, 1 (1977), 51-56.
[7] Musiăta, C., Norm Preserving Extension of Starshoped Lipschitz Functions, Mathe-
matica 19 (12) 2 Haness of Hahn-Banach Extension and Unique Best Approximation, s, R. R., Uniqueness of Han- Amer. Math. Soc. 95 (1960), 238-255.
Received. 21.XII.1989.