The extension of starshaped bounded Lipschitz functions

Costica Mustata
(original), Approximation Theory, paper

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

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C. Mustăţa, The extension of starshaped bounded Lipschitz functions, Anal. Numer. Théor. Approx. 9 (1980) 1, 93-99

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Revue d’Analyse Numer. Theor.Approximation

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Publishing Romanian Academy

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https://ictp.acad.ro/jnaat/journal/article/view/1980-vol9-no1-art12/

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2457-6794

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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.

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1980-Mustata-The extension of starshaped bounded Lipschitz functions-Jnaat

THE EXTENSION OF STARSHAPED BOUNDED LIPSCHITZ FUNCTIONS

byc. MUSTATA

  1. Let X X XXX be a normed space and Y Y YYY a nonvoid subset of X X XXX. The set Y Y YYY is called starshaped (with respect to θ ∈ X θ ∈ X theta in X\theta \in Xθ∈X ) if α y ∈ Y α y ∈ Y alpha y in Y\alpha y \in Yαy∈Y for all α ∈ [ 0 , 1 ] α ∈ [ 0 , 1 ] alpha in[0,1]\alpha \in[0,1]α∈[0,1] and y ∈ Y y ∈ Y y in Yy \in Yy∈Y. A real function f f fff defined on Y Y YYY is called starshaped if
(1.1) f ( α y ) ≤ α f ( y ) , (1.1) f ( α y ) ≤ α f ( y ) , {:(1.1)f(alpha y) <= alpha f(y)",":}\begin{equation*} f(\alpha y) \leq \alpha f(y), \tag{1.1} \end{equation*}(1.1)f(αy)≤αf(y),
for all α ∈ [ 0 , 1 ] α ∈ [ 0 , 1 ] alpha in[0,1]\alpha \in[0,1]α∈[0,1] and y ∈ Y y ∈ Y y in Yy \in Yy∈Y. From (1.1) follows f ( θ ) ≤ 0 f ( θ ) ≤ 0 f(theta) <= 0f(\theta) \leq 0f(θ)≤0. In the following we consider only starshaped functions vanishing at θ . Y θ . Y theta.Y\theta . Yθ.Y will denote always a starshaped set.
A function f : Y → R f : Y → R f:Y rarr Rf: Y \rightarrow Rf:Y→R is called Lipschitz on Y Y YYY if there exists K ≥ 0 K ≥ 0 K >= 0K \geq 0K≥0 such that
(1.2) | f ( x ) − f ( y ) | ≤ K ⋅ ‖ x − y ‖ (1.2) | f ( x ) − f ( y ) | ≤ K ⋅ ‖ x − y ‖ {:(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*}(1.2)|f(x)−f(y)|≤K⋅‖x−y‖
for all x , y ∈ Y x , y ∈ Y x,y in Yx, y \in Yx,y∈Y. Denote by Lip 0 Y Lip 0 Y Lip_(0)Y\operatorname{Lip}_{0} YLip0Y ( Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X ) the space of all Lipschitz functions on Y Y YYY (respectively on X X XXX ) vanishing at θ ∈ Y ⊂ X θ ∈ Y ⊂ X theta in Y sub X\theta \in Y \subset Xθ∈Y⊂X ([5]) and by BLip 0 Y BLip 0 Y BLip_(0)Y\mathrm{BLip}_{0} YBLip0Y ( BLip 0 X BLip 0 X BLip_(0)X\mathrm{BLip}_{0} XBLip0X ) their subspaces formed of bounded functions on Y Y YYY (respectively on X X XXX ). Denote also
(1.3) BS Y = { f : f ∈ BLip 0 Y , f , is starshaped } (1.4) BS X = { F : F ∈ BLip 0 X , F is starshaped } (1.3) BS Y = f : f ∈ BLip 0 Y , f ,  is starshaped  (1.4) BS X = F : F ∈ BLip 0 X , F  is starshaped  {:[(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*}(1.3)BSY={f:f∈BLip0Y,f, is starshaped }(1.4)BSX={F:F∈BLip0X,F is starshaped }
The sets BS Y BS Y BS_(Y)\mathrm{BS}_{Y}BSY and BS X BS X BS_(X)\mathrm{BS}_{X}BSX are convex cones in BLip 0 Y BLip 0 Y BLip_(0)Y\mathrm{BLip}_{0} YBLip0Y and in BLip 0 X BLip 0 X BLip_(0)X\mathrm{BLip}_{0} XBLip0X, respectively, i.e. f + g f + g f+gf+gf+g and λ f λ f lambda f\lambda fλf are in BS Y BS Y BS_(Y)\mathrm{BS}_{Y}BSY (in BS X BS X BS_(X)\mathrm{BS}_{X}BSX ) for all f , g f , g f,gf, gf,g in BS Y BS Y BS_(Y)\mathrm{BS}_{Y}BSY (in BS X BS X BS_(X)\mathrm{BS}_{X}BSX ) and λ ≥ 0 λ ≥ 0 lambda >= 0\lambda \geq 0λ≥0.
For f ∈ BS Y ( f ∈ BLip 0 Y ) f ∈ BS Y f ∈ BLip 0 Y f inBS_(Y)(f inBLip_(0)Y)f \in \mathrm{BS}_{Y}\left(f \in \mathrm{BLip}_{0} Y\right)f∈BSY(f∈BLip0Y) put
(1.5)) ‖ f ‖ d = sup { | f ( y 1 ) − f ( y 2 ) | / ‖ y 1 − y 2 ‖ : y 1 , y 2 ∈ Y , y 1 ≠ y 2 } (1.5)) ‖ f ‖ d = sup f y 1 − f y 2 / y 1 − y 2 : y 1 , y 2 ∈ Y , y 1 ≠ y 2 {:(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*}(1.5))‖f‖d=sup{|f(y1)−f(y2)|/‖y1−y2‖:y1,y2∈Y,y1≠y2}
the Lipschitz norm of f f fff, and
(1.6) ‖ f ‖ ∞ = sup { | f ( y ) | : y ∈ Y } , (1.6) ‖ f ‖ ∞ = sup { | f ( y ) | : y ∈ Y } , {:(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*}(1.6)‖f‖∞=sup{|f(y)|:y∈Y},
the uniform norm of f f fff.
For F ∈ BLip 0 X F ∈ BLip 0 X F inBLip_(0)XF \in \mathrm{BLip}_{0} XF∈BLip0X, the Lipschitz and the uniform norms are defined similarly.
2. It is well known (see e.g. [4]) that a Lipschitz function f f fff defined on a nonvoid subset Y Y YYY of a metric space X X XXX has a norm preserving Lipschitz extension F F FFF on X X XXX, i.e. F | Y = f F Y = f F|_(Y)=f\left.F\right|_{Y}=fF|Y=f and ‖ F ‖ d = ‖ f ‖ d ‖ F ‖ d = ‖ f ‖ d ||F||_(d)=||f||_(d)\|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 X X XXX, then every convex Lipschitz function on Y Y YYY has a convex norm preserving Lipschitz extension F F FFF on X X XXX. 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 ∈ BS Y f ∈ BS Y f inBS_(Y)f \in \mathrm{BS}_{Y}f∈BSY find F ∈ BS X F ∈ BS X F inBS_(X)F \in \mathrm{BS}_{X}F∈BSX such that F | Y = f , ‖ F ‖ d = ‖ f ‖ d F Y = f , ‖ F ‖ d = ‖ f ‖ d F|_(Y)=f,||F||_(d)=||f||_(d)\left.F\right|_{Y}=f,\|F\|_{d}=\|f\|_{d}F|Y=f,‖F‖d=‖f‖d and ‖ F ‖ ∞ = ‖ f ‖ ∞ ‖ F ‖ ∞ = ‖ f ‖ ∞ ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty}‖F‖∞=‖f‖∞. The function F F FFF is called briefly an extension of f f fff. 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 Y Y YYY is a convex subset of X X XXX, such that θ ∈ Y θ ∈ Y theta in Y\theta \in Yθ∈Y, then the nul function is the only bounded convex (Lipschitz) function on Y Y YYY which has a bounded convex Lipschitz extension on X X XXX. This follows from the fact that the constant functions are only bounded convex functions on X X XXX. Indeed, if F : X → R F : X → R F:X rarr RF: X \rightarrow RF:X→R is a nonconstant there X X XXX. X X XXX unction, then there exist two points x 0 , x 1 ∈ X x 0 , x 1 ∈ X x_(0),x_(1)in Xx_{0}, x_{1} \in Xx0,x1∈X sich that F ( x 0 ) ≠ F ( x 1 ) F x 0 ≠ F x 1 F(x_(0))!=F(x_(1))F\left(x_{0}\right) \neq F\left(x_{1}\right)F(x0)≠F(x1), say F ( x 0 ) << F ( x 1 ) F x 0 << F x 1 F(x_(0))<<F(x_(1))F\left(x_{0}\right)< <F\left(x_{1}\right)F(x0)<<F(x1). But then, since the function φ : ( 0 , ∞ ) → R φ : ( 0 , ∞ ) → R varphi:(0,oo)rarr R\varphi:(0, \infty) \rightarrow Rφ:(0,∞)→R, defined by
φ ( t ) = F ( x 0 + t ( x 1 − x 0 ) ) − F ( x 0 ) t , t > 0 φ ( t ) = F x 0 + t x 1 − x 0 − F x 0 t , t > 0 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>0φ(t)=F(x0+t(x1−x0))−F(x0)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 F x 0 + t x 1 − x 0 − F x 0 t ≥ F x 1 − F x 0 > 0 (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)>0F(x0+t(x1−x0))−F(x0)t≥F(x1)−F(x0)>0
so that F ( x 0 + t ( x 1 − x 0 ) ) ≥ F ( x 0 ) + t [ F ( x 1 ) − F ( x 0 ) ] F x 0 + t x 1 − x 0 ≥ F x 0 + t F x 1 − F x 0 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]F(x0+t(x1−x0))≥F(x0)+t[F(x1)−F(x0)], for all t ≥ 1 t ≥ 1 t >= 1t \geq 1t≥1, which shows that the function F F FFF is unbounded.
2a. The existence of extension. In Theorem 1 below will be shown that under some suplementary hypotheses on the function f ∈ BS Y f ∈ BS Y f inBS_(Y)f \in \mathrm{BS}_{Y}f∈BSY there exists an extension F ∈ BS X F ∈ BS X F inBS_(X)F \in \mathrm{BS}_{X}F∈BSX of f f fff.
Firstly, we prove two lemmas.
lemma 1. Let X X XXX be a real normed space and f f fff a starshaped function on X X XXX. Then, for every x ∈ X , x ≠ θ x ∈ X , x ≠ θ x in X,x!=thetax \in X, x \neq \thetax∈X,x≠θ, the function Ψ : ( 0 , ∞ ) → R Ψ : ( 0 , ∞ ) → R Psi:(0,oo)rarr R\Psi:(0, \infty) \rightarrow RΨ:(0,∞)→R, defined by
(2.1)
Ψ ( t ) = f ( t x ) / t , t > 0 Ψ ( t ) = f ( t x ) / t , t > 0 Psi(t)=f(tx)//t,quad t > 0\Psi(t)=f(t x) / t, \quad t>0Ψ(t)=f(tx)/t,t>0
is nondecreasing.
Proof. For 0 < t 1 < t 2 0 < t 1 < t 2 0 < t_(1) < t_(2)0<t_{1}<t_{2}0<t1<t2 and a fixed x ∈ X , x ≠ θ x ∈ X , x ≠ θ x in X,x!=thetax \in X, x \neq \thetax∈X,x≠θ, 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 . 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 . (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}} .f(t1x)t1=f((t1/t2)t2x)t1≤(t1/t2)f(t2x)t1=f(t2x)t2.
Now, for a function f : X → R f : X → R f:X rarr Rf: X \rightarrow Rf:X→R define, as usually, the epigraph of f f fff, by
(2.2) epi f = { ( x , α ) ∈ X × R : f ( x ) ≤ α } (2.2)  epi  f = { ( x , α ) ∈ X × R : f ( x ) ≤ α } {:(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*}(2.2) epi f={(x,α)∈X×R:f(x)≤α}
LEMMA 2. A function f : X → R , f ( θ ) = 0 f : X → R , f ( θ ) = 0 f:X rarr R,f(theta)=0f: X \rightarrow R, f(\theta)=0f:X→R,f(θ)=0, is starshaped if and only if its epigraph is starshaped.
Proof. If f f fff is starshaped, f ( θ ) = 0 f ( θ ) = 0 f(theta)=0f(\theta)=0f(θ)=0, and ( x , α ) ∈ ( x , α ) ∈ (x,alpha)in(x, \alpha) \in(x,α)∈ epi f f fff, then for every λ ∈ [ 0 , 1 ] , f ( λ x ) ≤ λ f ( x ) ≤ λ α λ ∈ [ 0 , 1 ] , f ( λ x ) ≤ λ f ( x ) ≤ λ α 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 \alphaλ∈[0,1],f(λx)≤λf(x)≤λα, so that λ ( x , α ) = ( λ x , λ α ) ∈ epi f λ ( x , α ) = ( λ x , λ α ) ∈ epi f lambda(x,alpha)=(lambda x,lambda alpha)in epi f\lambda(x, \alpha)=(\lambda x, \lambda \alpha) \in \operatorname{epi} fλ(x,α)=(λx,λα)∈epif. Conversely, if epi f f fff is starshaped then ( x , f ( x ) ) ∈ epi f ( x , f ( x ) ) ∈ epi f (x,f(x))in epi f(x, f(x)) \in \operatorname{epi} f(x,f(x))∈epif implies ( λ x , λ f ( x ) ) ∈∈ ( λ x , λ f ( x ) ) ∈∈ (lambda x,lambda f(x))∈∈(\lambda x, \lambda f(x)) \in \in(λx,λf(x))∈∈ epi f f fff, i.e. f ( λ x ) ≤ λ f ( x ) f ( λ x ) ≤ λ f ( x ) f(lambda x) <= lambda f(x)f(\lambda x) \leq \lambda f(x)f(λx)≤λf(x), for all λ ∈ [ 0 , 1 ] λ ∈ [ 0 , 1 ] lambda in[0,1]\lambda \in[0,1]λ∈[0,1].
THEOREM 1. Let X X XXX be a normed space, Y Y YYY a starshaped subset of X X XXX, θ ∈ Y θ ∈ Y theta in Y\theta \in Yθ∈Y, and f ∈ BS Y f ∈ BS Y f inBS_(Y)f \in \mathrm{BS}_{Y}f∈BSY. Then, there exists F ∈ BS X F ∈ BS X F inBS_(X)F \in \mathrm{BS}_{X}F∈BSX such that
(i) F | Y = f F Y = f F|_(Y)=f\left.F\right|_{Y}=fF|Y=f,
(ii) ‖ F ‖ d = ‖ f ‖ d ‖ F ‖ d = ‖ f ‖ d ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d}‖F‖d=‖f‖d,
(iii) ‖ F ‖ ∞ = ‖ f ‖ ∞ ‖ F ‖ ∞ = ‖ f ‖ ∞ ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty}‖F‖∞=‖f‖∞,
if and only if f ( y ) ≤ 0 f ( y ) ≤ 0 f(y) <= 0f(y) \leq 0f(y)≤0, for all y ∈ Y y ∈ Y y in Yy \in Yy∈Y.
Proof. Let f ∈ BS Y f ∈ BS Y f inBS_(Y)f \in \mathrm{BS}_{Y}f∈BSY and suppose f ≤ 0 f ≤ 0 f <= 0f \leq 0f≤0 on Y Y YYY. Define G : X → R G : X → R G:X rarr RG: X \rightarrow RG:X→R by
(2.3) G ( x ) = inf y ∈ Y [ f ( y ) + ‖ f ‖ d ‖ x − y ‖ ] (2.3) G ( x ) = inf y ∈ Y   f ( y ) + ‖ f ‖ d ‖ x − y ‖ {:(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*}(2.3)G(x)=infy∈Y[f(y)+‖f‖d‖x−y‖]
The function G G GGG defined by (2.3) is starshaped and satisfies G | Y = f G Y = f G|_(Y)=f\left.G\right|_{Y}=fG|Y=f, ‖ G ‖ d = ‖ f ‖ d ‖ G ‖ d = ‖ f ‖ d ||G||_(d)=||f||_(d)\|G\|_{d}=\|f\|_{d}‖G‖d=‖f‖d (see [7]).
Let
(2.4) F ( x ) = { 0 if G ( x ) > 0 G ( x ) if G ( x ) ≤ 0 (2.4) F ( x ) = 0  if  G ( x ) > 0 G ( x )  if  G ( x ) ≤ 0 {:(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}(2.4)F(x)={0 if G(x)>0G(x) if G(x)≤0
Since G | Y = f ≤ 0 G Y = f ≤ 0 G|_(Y)=f <= 0\left.G\right|_{Y}=f \leq 0G|Y=f≤0, it follows F | Y = f F Y = f F|_(Y)=f\left.F\right|_{Y}=fF|Y=f and ‖ F ‖ d = ‖ f ‖ d ‖ F ‖ d = ‖ f ‖ d ||F||_(d)=||f||_(d)\|F\|_{d}=\|f\|_{d}‖F‖d=‖f‖d. Obviously ‖ F ‖ ∞ ≥ ‖ f ‖ ∞ ‖ F ‖ ∞ ≥ ‖ f ‖ ∞ ||F||_(oo) >= ||f||_(oo)\|F\|_{\infty} \geq\|f\|_{\infty}‖F‖∞≥‖f‖∞. If x ∈ X x ∈ X x in Xx \in Xx∈X is such that G ( x ) ≤ 0 G ( x ) ≤ 0 G(x) <= 0G(x) \leq 0G(x)≤0, then
0 ≥ F ( x ) = G ( x ) ≥ f ( y ) + ‖ f ‖ d ⋅ ‖ x − y ‖ ≥ f ( y ) , 0 ≥ F ( x ) = G ( x ) ≥ f ( y ) + ‖ f ‖ d ⋅ ‖ x − y ‖ ≥ f ( y ) , 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),0≥F(x)=G(x)≥f(y)+‖f‖d⋅‖x−y‖≥f(y),
so that 0 ≤ − F ( x ) ≤ − f ( y ) 0 ≤ − F ( x ) ≤ − f ( y ) 0 <= -F(x) <= -f(y)0 \leq-F(x) \leq-f(y)0≤−F(x)≤−f(y), for all y ∈ Y y ∈ Y y in Yy \in Yy∈Y. Therefore ‖ F ‖ ∞ = inf x ∈ X ( − F ( x ) ) ≤≤ ‖ f ‖ ∞ ‖ F ‖ ∞ = inf x ∈ X   ( − F ( x ) ) ≤≤ ‖ f ‖ ∞ ||F||_(oo)=i n f_(x in X)(-F(x))≤≤||f||_(oo)\|F\|_{\infty}=\inf _{x \in X}(-F(x)) \leq \leq\|f\|_{\infty}‖F‖∞=infx∈X(−F(x))≤≤‖f‖∞ and ‖ F ‖ ∞ = ‖ f ‖ ∞ ‖ F ‖ ∞ = ‖ f ‖ ∞ ||F||_(oo)=||f||_(oo)\|F\|_{\infty}=\|f\|_{\infty}‖F‖∞=‖f‖∞. Since the epigraph of F F FFF is starshaped, by Lemma 2 , F 2 , F 2,F2, F2,F is starshaped. Consequently, F F FFF is the required extension of f f fff.
Suppose now, that there exists y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y such that f ( y 0 ) > 0 f y 0 > 0 f(y_(0)) > 0f\left(y_{0}\right)>0f(y0)>0, and let F F FFF be a starshaped extension of f f fff. By Lemma 1
0 < f ( y 0 ) = F ( y 0 ) ≤ ( F ( t y 0 ) / t ) , 0 < f y 0 = F y 0 ≤ F t y 0 / t , 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),0<f(y0)=F(y0)≤(F(ty0)/t),
so that F ( t y 0 ) ≥ t f ( y 0 ) F t y 0 ≥ t f y 0 F(ty_(0)) >= tf(y_(0))F\left(t y_{0}\right) \geq t f\left(y_{0}\right)F(ty0)≥tf(y0), for all t ≥ 1 t ≥ 1 t >= 1t \geq 1t≥1, which shows that the function F F FFF is unbounded. Therefore f f fff has no bounded strashaped extension, which ends the proof of Theorem 1.
Let
(2.5)
BS S Y = { f ∈ BS Y : f ≤ 0 } BS X − = { F ∈ BS X : F ≤ 0 } BS S Y = f ∈ BS Y : f ≤ 0 BS X − = F ∈ BS X : F ≤ 0 {:[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}BSSY={f∈BSY:f≤0}BSX−={F∈BSX:F≤0}
By Theorem 1 follows:
corolyary 1. Every function f ∈ BS Y ¯ f ∈ BS Y ¯ f inBS_( bar(Y))f \in \mathrm{BS}_{\bar{Y}}f∈BSY¯ has an extension F ∈ BS X ¯ F ∈ BS X ¯ F inBS_( bar(X))F \in \mathrm{BS}_{\bar{X}}F∈BSX¯.
2b. The unicity of extension. By Theorem 1 and Corollary 1, every nonpositive bounded starshaped function, defined on a starshaped subset Y Y YYY of a normed space X X XXX, has a nonpositive bounded starshaped extension to whole X X XXX. Furthemore, these are the only bounded starshaped function on Y Y YYY admitting bounded starshaped Lipschitz extension on X X XXX.
Equiped with the norms
‖ f ‖ Y = max ( ‖ f ‖ d , ‖ f ‖ ∞ ) , f ∈ BLip 0 Y (2.6) ‖ F ‖ X = max ( ‖ F ‖ d , ‖ F ‖ ∞ ) , F ∈ BLip 0 X ‖ f ‖ Y = max ‖ f ‖ d , ‖ f ‖ ∞ , f ∈ BLip 0 Y (2.6) ‖ F ‖ X = max ‖ F ‖ d , ‖ F ‖ ∞ , F ∈ BLip 0 X {:[||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*}‖f‖Y=max(‖f‖d,‖f‖∞),f∈BLip0Y(2.6)‖F‖X=max(‖F‖d,‖F‖∞),F∈BLip0X
BLip 0 Y BLip 0 Y BLip_(0)Y\mathrm{BLip}_{0} YBLip0Y and BLip 0 X BLip 0 X BLip_(0)X\mathrm{BLip}_{0} XBLip0X become Banach spaces (see [3]). Let H = BS X ¯ H = BS X ¯ H=BS_( bar(X))H=\mathrm{BS}_{\bar{X}}H=BSX¯ - − BS X − BS X -BS_(X)-\mathrm{BS}_{X}−BSX the subspace of BLip 0 X BLip 0 X BLip_(0)X\mathrm{BLip}_{0} XBLip0X generated by the convex cone BS X BS X BS_(X)\mathrm{BS}_{X}BSX and
Y ⊥= { g ∈ H : g | Y = 0 } Y ⊥= g ∈ H : g Y = 0 Y⊥={g in H:g|_(Y)=0}Y \perp=\left\{g \in H:\left.g\right|_{Y}=0\right\}Y⊥={g∈H:g|Y=0}
the anihilator of the set Y Y YYY in H H HHH. Obviously, Y ⊥ Y ⊥ Y _|_Y \perpY⊥ is a subspace of H H HHH. A subset Z Z ZZZ of a normed space X X XXX is called proximinal for W ⊂ X W ⊂ X W sub XW \subset XW⊂X if for every f ∈ W f ∈ W f in Wf \in Wf∈W there exists g 0 ∈ Z g 0 ∈ Z g_(0)in Zg_{0} \in Zg0∈Z, such that
(2.7) ‖ f − g 0 ‖ = d ( f , Z ) = inf { ‖ f − g ‖ : g ∈ Z } . (2.7) f − g 0 = d ( f , Z ) = inf { ‖ f − g ‖ : g ∈ Z } . {:(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*}(2.7)‖f−g0‖=d(f,Z)=inf{‖f−g‖:g∈Z}.
If for every f ∈ W f ∈ W f in Wf \in Wf∈W the element g 0 ∈ Z g 0 ∈ Z g_(0)in Zg_{0} \in Zg0∈Z satisfying (2.7) is unique then the set Z Z ZZZ is called Chebyshevian for W W WWW. An element g 0 ∈ Z g 0 ∈ Z g_(0)in Zg_{0} \in Zg0∈Z satisfying (2.7) is called an element of best approximation of f f fff by elements of Z Z ZZZ.
THEOREM 2. Y ⊥ Y ⊥ Y _|_Y \perpY⊥ is a Chebyshevian subsapce for BS X ¯ BS X ¯ BS_( bar(X))\mathrm{BS}_{\bar{X}}BSX¯ if and only if, every f ∈ BS f ∈ BS f inBSf \in \mathrm{BS}f∈BS has a unique (preserving the uniform and Lipschitz norms) extension F F FFF in BS x ¯ BS x ¯ BS_( bar(x))\mathrm{BS}_{\bar{x}}BSx¯.
Proof. Follows from Theorem 1 in [6].
Remark. Observe that Theorem 2 remains true if the spaces BLip 0 Y BLip 0 Y BLip_(0)Y\mathrm{BLip}_{0} YBLip0Y and BLip 0 X BLip 0 X BLip_(0)X\mathrm{BLip}_{0} XBLip0X are equiped with the norms
‖ f ‖ 1 = ‖ f ‖ a + ‖ f ‖ ∞ , for f ∈ BLip 0 Y (respectively BLip 0 X ) ‖ f ‖ 1 = ‖ f ‖ a + ‖ f ‖ ∞ ,  for  f ∈ BLip 0 Y  (respectively  BLip 0 X {:||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)‖f‖1=‖f‖a+‖f‖∞, for f∈BLip0Y (respectively BLip0X)
Theorem 2 is analogous with a theorem of PHELPS [8], in the linear case.
3. Now, we try to find conditions on the function f f fff ensuring the unicity of the extension.
Consider, firstly, the case X = R X = R X=RX=RX=R with the usual norm | ⋅ | | ⋅ | |*||\cdot||⋅| (the absolute value).
THEOREM 3. Let Y = [ a , b ] ⊂ R , a < 0 < b Y = [ a , b ] ⊂ R , a < 0 < b Y=[a,b]sub R,a < 0 < bY=[a, b] \subset R, a<0<bY=[a,b]⊂R,a<0<b. A function f ∈ BS Y f ∈ BS Y f inBS_(Y)f \in \mathrm{BS}_{Y}f∈BSY has a unique extension F ∈ BS R − F ∈ BS R − F inBS_(R)^(-)F \in \mathrm{BS}_{R}^{-}F∈BSR−if and only if f ( a ) = f ( b ) = 0 f ( a ) = f ( b ) = 0 f(a)=f(b)=0f(a)=f(b)=0f(a)=f(b)=0.
Proof. Suppose that f ∈ BS Y − f ∈ BS Y − f inBS_(Y)^(-)f \in \mathrm{BS}_{Y}^{-}f∈BSY−has two distinct extensions F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2 in BS R ¯ − BS R ¯ − BS_( bar(R))^(-)\mathrm{BS}_{\bar{R}}^{-}BSR¯−. Let x ∈ R ∖ [ a , b ] x ∈ R ∖ [ a , b ] x in R\\[a,b]x \in R \backslash[a, b]x∈R∖[a,b] be such that F 1 ( x ) ≠ F 2 ( x ) F 1 ( x ) ≠ F 2 ( x ) F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x)F1(x)≠F2(x), say F 1 ( x ) < F 2 ( x ) ≤ 0 F 1 ( x ) < F 2 ( x ) ≤ 0 F_(1)(x) < F_(2)(x) <= 0F_{1}(x)<F_{2}(x) \leq 0F1(x)<F2(x)≤0. Suppose x > b x > b x > bx>bx>b. The function F 1 F 1 F_(1)F_{1}F1 being starshaped it follows
F 1 ( λ x ) ≤ λ F 1 ( x ) < 0 , F 1 ( λ x ) ≤ λ F 1 ( x ) < 0 , F_(1)(lambda x) <= lambdaF_(1)(x) < 0,F_{1}(\lambda x) \leq \lambda F_{1}(x)<0,F1(λx)≤λF1(x)<0,
for all λ ∈ ( 0 , 1 ] λ ∈ ( 0 , 1 ] lambda in(0,1]\lambda \in(0,1]λ∈(0,1]. In particular, since b = λ b x b = λ b x b=lambda_(b)xb=\lambda_{b} xb=λbx for λ b ∈ ( 0 , 1 ) λ b ∈ ( 0 , 1 ) lambda_(b)in(0,1)\lambda_{b} \in(0,1)λb∈(0,1), it follows f ( b ) = F 1 ( b ) < 0 f ( b ) = F 1 ( b ) < 0 f(b)=F_(1)(b) < 0f(b)=F_{1}(b)<0f(b)=F1(b)<0. If x < a x < a x < ax<ax<a, then a = λ a x a = λ a x a=lambda_(a)xa=\lambda_{a} xa=λax, for a λ a ∈ ( 0 , 1 ) λ a ∈ ( 0 , 1 ) lambda_(a)in(0,1)\lambda_{a} \in(0,1)λa∈(0,1), and similarly, f ( a ) = F 1 ( a ) < 0 f ( a ) = F 1 ( a ) < 0 f(a)=F_(1)(a) < 0f(a)=F_{1}(a)<0f(a)=F1(a)<0.
Conversely, we shall show that if f ∈ BS Y − f ∈ BS Y − f inBS_(Y)^(-)f \in \mathrm{BS}_{Y}^{-}f∈BSY−is such that f ( a ) < 0 f ( a ) < 0 f(a) < 0f(a)<0f(a)<0 of f ( b ) < 0 f ( b ) < 0 f(b) < 0f(b)<0f(b)<0, then f f fff has at least two distinct extensions F 1 F 1 F_(1)F_{1}F1 and F 2 F 2 F_(2)F_{2}F2 in BS R ¯ BS R ¯ BS_( bar(R))\mathrm{BS}_{\bar{R}}BSR¯. If f ( b ) < 0 f ( b ) < 0 f(b) < 0f(b)<0f(b)<0, then
F 1 ( x ) = { f ( x ) , x ∈ [ 0 , b ] f ( b ) + ‖ f ‖ d ( x − b ) , x ∈ ( b , b − f ( b ) ( ‖ f ‖ d ) − 1 ] 0 , x ∈ ( − ∞ , a ) ∪ ( b − f ( b ) ⋅ ( ‖ f ‖ d ) − 1 , − ∞ ) F 1 ( x ) = f ( x ) ,      x ∈ [ 0 , b ] f ( b ) + ‖ f ‖ d ( x − b ) ,      x ∈ b , b − f ( b ) ‖ f ‖ d − 1 0 ,      x ∈ ( − ∞ , a ) ∪ b − f ( b ) ⋅ ‖ f ‖ d − 1 , − ∞ 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}F1(x)={f(x),x∈[0,b]f(b)+‖f‖d(x−b),x∈(b,b−f(b)(‖f‖d)−1]0,x∈(−∞,a)∪(b−f(b)⋅(‖f‖d)−1,−∞)
and
F 2 ( x ) = { f ( x ) , x ∈ [ 0 , b ] ( f ( b ) / b ) x , x ∈ ( b , ( − ‖ f ‖ ∞ / f ( b ) ) b ) − ‖ f ‖ ∞ , x ∈ [ − ( ‖ f ‖ ∞ / f ( b ) ) b , + ∞ ) 0 , x ∈ ( − ∞ , a ) F 2 ( x ) = f ( x )      , x ∈ [ 0 , b ] ( f ( b ) / b ) x      , x ∈ b , − ‖ f ‖ ∞ / f ( b ) b − ‖ f ‖ ∞      , x ∈ − ‖ f ‖ ∞ / f ( b ) b , + ∞ 0      , x ∈ ( − ∞ , a ) 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}F2(x)={f(x),x∈[0,b](f(b)/b)x,x∈(b,(−‖f‖∞/f(b))b)−‖f‖∞,x∈[−(‖f‖∞/f(b))b,+∞)0,x∈(−∞,a)
are two distinct extensions of f f fff, i.e. F 1 ≠ F 2 , F 1 | Y = F 2 | Y = f , ‖ F ′ 1 ‖ a == ‖ F 2 ‖ d = ‖ f ‖ d , ‖ F 1 ‖ ∞ = ‖ F 2 ‖ ∞ = ‖ f ‖ ∞ F 1 ≠ F 2 , F 1 Y = F 2 Y = f , F ′ 1 a == F 2 d = ‖ f ‖ d , F 1 ∞ = F 2 ∞ = ‖ f ‖ ∞ 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}F1≠F2,F1|Y=F2|Y=f,‖F′1‖a==‖F2‖d=‖f‖d,‖F1‖∞=‖F2‖∞=‖f‖∞.
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 ) ≠ F 2 ( x ) F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x)F1(x)≠F2(x) for all x > max { b − f ( b ) ⋅ ( ‖ f ‖ a ) − 1 ; − ‖ f ‖ ∞ ⋅ ( f ( b ) ) − 1 } x > max b − f ( b ) ⋅ ‖ f ‖ a − 1 ; − ‖ f ‖ ∞ ⋅ ( f ( b ) ) − 1 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\}x>max{b−f(b)⋅(‖f‖a)−1;−‖f‖∞⋅(f(b))−1}.
If f ( a ) < 0 f ( a ) < 0 f(a) < 0f(a)<0f(a)<0, then two distinct extensions F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2, may be given, in a similar way.
Remark. The hypothesis a < 0 < b a < 0 < b a < 0 < ba<0<ba<0<b in Theorem 3 is essential as it is shown by the following example. Take Y = [ 0 , b ] , b > 0 Y = [ 0 , b ] , b > 0 Y=[0,b],b > 0Y=[0, b], b>0Y=[0,b],b>0 or Y == [ a , 0 ] , a < 0 Y == [ a , 0 ] , a < 0 Y==[a,0],a < 0Y= =[a, 0], a<0Y==[a,0],a<0. Then every f ∈ BSS Y − f ∈ BSS Y − f inBSS_(Y)^(-)f \in \mathrm{BSS}_{Y}^{-}f∈BSSY−has an infinite set of extensions in BS R ¯ − BS R ¯ − BS_( bar(R))^(-)\mathrm{BS}_{\bar{R}}^{-}BSR¯−. For exemple, if Y = [ 0 , b ] Y = [ 0 , b ] Y=[0,b]Y=[0, b]Y=[0,b] and f ∈ BS Y ¯ − f ∈ BS Y ¯ − f inBS_( bar(Y))^(-)f \in \mathrm{BS}_{\bar{Y}}^{-}f∈BSY¯−is such that f ( 0 ) = 0 f ( 0 ) = 0 f(0)=0f(0)=0f(0)=0, f ( b ) = 0 f ( b ) = 0 f(b)=0f(b)=0f(b)=0, then
F λ ( x ) = { f ( x ) , x ∈ [ 0 , b ] 0 , x ∈ ( 0 , + ∞ ) ‖ f ‖ d x , x ∈ ( − λ ‖ f ‖ ∞ ⋅ ( ‖ f ‖ d ) − 1 , 0 ) − λ ‖ f ‖ ∞ , x ∈ ( − ∞ , − λ ‖ f ∣ ‖ ∞ ( ‖ f ‖ d ) − 1 ) F λ ( x ) = f ( x ) ,      x ∈ [ 0 , b ] 0 ,      x ∈ ( 0 , + ∞ ) ‖ f ‖ d x ,      x ∈ − λ ‖ f ‖ ∞ ⋅ ‖ f ‖ d − 1 , 0 − λ ‖ f ‖ ∞ ,      x ∈ − ∞ , − λ f ∣ ∞ ‖ f ‖ d − 1 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}Fλ(x)={f(x),x∈[0,b]0,x∈(0,+∞)‖f‖dx,x∈(−λ‖f‖∞⋅(‖f‖d)−1,0)−λ‖f‖∞,x∈(−∞,−λ‖f∣‖∞(‖f‖d)−1)
is an extension of f f fff for every λ ∈ [ 0 , 1 ] λ ∈ [ 0 , 1 ] lambda in[0,1]\lambda \in[0,1]λ∈[0,1].
Consider now the general case. For x ∈ X , x ≠ θ x ∈ X , x ≠ θ x in X,x!=thetax \in X, x \neq \thetax∈X,x≠θ, the ray θ x → θ x → vec(theta x)\overrightarrow{\theta x}θx→ is defined by
θ x → = { α x : α ≥ 0 } θ x → = { α x : α ≥ 0 } vec(theta x)={alpha x:alpha >= 0}\overrightarrow{\theta x}=\{\alpha x: \alpha \geq 0\}θx→={αx:α≥0}
If the set Y Y YYY is starshaped and y ∈ Y , y ≠ θ y ∈ Y , y ≠ θ y in Y,y!=thetay \in Y, y \neq \thetay∈Y,y≠θ, then θ y → ⊂ Y θ y → ⊂ Y vec(theta y)sub Y\overrightarrow{\theta y} \subset Yθy→⊂Y or θ y → ∩ Y θ y → ∩ Y vec(theta y)nn Y\overrightarrow{\theta y} \cap Yθy→∩Y is a segment. In the second case put
α y = sup { α : α y ∈ Y } , α y = sup { α : α y ∈ Y } , alpha_(y)=s u p{alpha:alpha y in Y},\alpha_{y}=\sup \{\alpha: \alpha y \in Y\},αy=sup{α:αy∈Y},
and z y = α y ⋅ y z y = α y ⋅ y z_(y)=alpha_(y)*yz_{y}=\alpha_{y} \cdot yzy=αy⋅y. The set { z y : y ∈ Y } z y : y ∈ Y {z_(y):y in Y}\left\{z_{y}: y \in Y\right\}{zy:y∈Y} is called the algebric starshapes boundary of Y Y YYY and is denoted by F r s Y F r s Y F_(r)^(s)Y\mathrm{F}_{\mathrm{r}}^{\mathrm{s}} YFrsY.
Evidently, every z ∈ F r s Y z ∈ F r s Y z inF_(r)^(s)Yz \in \mathrm{~F}_{\mathrm{r}}^{\mathrm{s}} Yz∈ FrsY is a limit point of Y Y YYY, i.e. Y ∪ F r s Y ( Y ¯ Y ∪ F r s Y ( Y ¯ Y uuF_(r)^(s)Y( bar(Y)Y \cup \mathrm{~F}_{\mathrm{r}}^{\mathrm{s}} Y(\bar{Y}Y∪ FrsY(Y¯. Since every f ∈ BS Y f ∈ BS Y f inBS_(Y)f \in \mathrm{BS}_{Y}f∈BSY is uniformly continuous (as Lipschitz) it can be uniquely extended to Y ⋃ Fr r s Y Y ⋃ Fr r s Y Y uuuFr_(r)^(s)YY \bigcup \mathrm{Fr}_{r}^{s} YY⋃FrrsY. Therefore with no restrition of generality, we can suppose F r s Y ⊂ Y F r s Y ⊂ Y F_(r)^(s)Y sub Y\mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Y \subset YFrsY⊂Y.
THEOREM 4. Let Y Y YYY be an absorbing starshaped subset of the normed space X X XXX, such that. F r s Y ⊂ Y F r s Y ⊂ Y F_(r)^(s)Y sub Y\mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Y \subset YFrsY⊂Y. If f ∈ BS Y − f ∈ BS Y − f inBS_(Y)^(-)f \in \mathrm{BS}_{Y}^{-}f∈BSY−is such that f ( z ) = 0 f ( z ) = 0 f(z)=0f(z)=0f(z)=0, for all z ∈∈ F r s Y z ∈∈ F r s Y z∈∈F_(r)^(s)Yz \in \in \mathrm{F}_{\mathrm{r}}^{\mathrm{s}} Yz∈∈FrsY, then f f fff has a unique extension F ∈ BSS X − F ∈ BSS X − F inBSS_(X)^(-)F \in \mathrm{BSS}_{X}^{-}F∈BSSX−.
Proof. Suppose f ∈ BS Y − , f ( z ) = 0 f ∈ BS Y − , f ( z ) = 0 f inBS_(Y)^(-),f(z)=0f \in \mathrm{BS}_{Y}^{-}, f(z)=0f∈BSY−,f(z)=0, for all z ∈ F I s Y z ∈ F I s Y z inF_(I)^(s)Yz \in \mathrm{~F}_{\mathrm{I}}^{s} Yz∈ FIsY, and suppose that f f fff has two distinct extensions F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2 in BS X − BS X − BS_(X)^(-)\mathrm{BS}_{X}^{-}BSX−. Let x ∈ X ∖ Y x ∈ X ∖ Y x in X\\Yx \in X \backslash Yx∈X∖Y be such that F 1 ( x ) ≠ F 2 ( x ) F 1 ( x ) ≠ F 2 ( x ) F_(1)(x)!=F_(2)(x)F_{1}(x) \neq F_{2}(x)F1(x)≠F2(x), say F 1 ( x ) < F 2 ( x ) ≤ 0 F 1 ( x ) < F 2 ( x ) ≤ 0 F_(1)(x) < F_(2)(x) <= 0F_{1}(x)<F_{2}(x) \leq 0F1(x)<F2(x)≤0. The set Y Y YYY being absorbing and starshaped, there exists λ > 0 λ > 0 lambda > 0\lambda>0λ>0 such that λ x ∈ F r s Y λ x ∈ F r s Y lambda x inF_(r)^(s)Y\lambda x \in \mathrm{~F}_{\mathrm{r}}^{\mathrm{s}} Yλx∈ FrsY. But then, one obtains the contradiction
0 = f ( λ x ) = F 1 ( λ x ) ≤ λ F 1 ( x ) < 0 0 = f ( λ x ) = F 1 ( λ x ) ≤ λ F 1 ( x ) < 0 0=f(lambda x)=F_(1)(lambda x) <= lambdaF_(1)(x) < 00=f(\lambda x)=F_{1}(\lambda x) \leq \lambda F_{1}(x)<00=f(λx)=F1(λx)≤λF1(x)<0
Theorem 4 is proved.

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