Best approximation and unique extension of Lipschitz functions

Costica Mustata
(original), Approximation Theory, best approximation, ISI/JCR, paper

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Costica Mustata
Institutul de Calcul, Romania (ICTP)

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C. Mustata, Best approximation and unique extension of Lipschitz functions,  J. Approx. Theory 19 (1977), 222-230, doi: 10.1016/0021-9045(77)90053-3

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Journal of Approximation Theory

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10.1016/0021-9045(77)90053-3

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[1] D.E BOOR, “On “Best” Interpolation,” Universky of Wisconsin, MRC 7% No. 1426, 1971.
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[5] C. MUSTATA, Asupra unor subspalii cebiseviene din spatiul normat al functiilor lipschitziene, Reo. Anal Mumer. Teoria Aproximatiei 2 (1973), 81-87.
[6] R. R. PHELPS, Uniqueness of Hahn-Banach extension and unique best approximation, Trans. Amer. Math. Sot. 95 (1960), 238-255.
[7] A. K. ROY, Extreme points and linear isometries of Banach space of Lipschitz functions, Canad. J. IWZ~A. 20 (1968), 1150-1164.
[8] I. SINGER, Cea mai buna aproximare in spalii vectoriale normate prin elemente din subspalii vectoriale, Edit. Acad. R. S. Romania, Bucuresti, (1967).
[9] D. R. SHERBERT, The structure of ideals and point derivations in Banach algebras of Lipschitz functions, Trans. Amer. Math. Sot. 111 (1964), 240-272.

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1977-Mustata-J.-Approx.-Th.-Best-approximation-and-unique-extension-of-Lipschitz-functions

Best Approximation and Unique Extension of Lipschitz Functions

Costică MustățaInstitutul de Calcul, Cluj-Napoca, R. S. RontaniaCommunicated by Lothar Collatz

Received July 21, 1975

1. Introduction

In the sequel, X X XXX will always denote a metric space with the metric d , x 0 d , x 0 d,x_(0)d, x_{0}d,x0 a fixed point from X X XXX, and Y Y YYY a subset of X X XXX such that x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y. If f f fff is a real-valued function defined on X X XXX, denote
(1.1) ‖ f ‖ Y = sup { | f ( x ) − f ( y ) | / d ( x , y ) : x , y ∈ Y , x ≠ y } . (1.1) ‖ f ‖ Y = sup { | f ( x ) − f ( y ) | / d ( x , y ) : x , y ∈ Y , x ≠ y } . {:(1.1)||f||_(Y)=s u p{|f(x)-f(y)|//d(x","y):x","y in Y","x!=y}.:}\begin{equation*} \|f\|_{Y}=\sup \{|f(x)-f(y)| / d(x, y): x, y \in Y, x \neq y\} . \tag{1.1} \end{equation*}(1.1)‖f‖Y=sup{|f(x)−f(y)|/d(x,y):x,y∈Y,x≠y}.
A Lipschitz function on X X XXX is a function f : X → R f : X → R f:X rarr Rf: X \rightarrow Rf:X→R such that ‖ f ‖ X < ∞ ‖ f ‖ X < ∞ ||f||_(X) < oo\|f\|_{X}<\infty‖f‖X<∞. Denote by Lip 0 X 0 X _(0)X{ }_{0} X0X the Banach space of all Lipschitz functions on X X XXX which vanish at x 0 x 0 x_(0)x_{0}x0, with the norm ‖ f ‖ = ‖ f ‖ X ‖ f ‖ = ‖ f ‖ X ||f||=||f||_(X)\|f\|=\|f\|_{X}‖f‖=‖f‖X. Put also
(1.2) Y ⊥ = { f : f ∈ Lip 0 X , f | Y = 0 } . (1.2) Y ⊥ = f : f ∈ Lip 0 X , f Y = 0 . {:(1.2)Y^(_|_)={f:f inLip_(0)X,f|_(Y)=0}.:}\begin{equation*} Y^{\perp}=\left\{f: f \in \operatorname{Lip}_{0} X,\left.f\right|_{Y}=0\right\} . \tag{1.2} \end{equation*}(1.2)Y⊥={f:f∈Lip0X,f|Y=0}.
A Lipschitz extension of a function f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y is a function F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X such that F | Y = f F Y = f F|_(Y)=f\left.F\right|_{Y}=fF|Y=f and ‖ F ‖ X = ‖ f ‖ Y ‖ F ‖ X = ‖ f ‖ Y ||F||_(X)=||f||_(Y)\|F\|_{X}=\|f\|_{Y}‖F‖X=‖f‖Y. It is known (see, e.g., [2]) that every f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y has a Lipschitz extension in Lip Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X.
For a subset Y Y YYY of X X XXX and x ∈ X x ∈ X x in Xx \in Xx∈X we put
(1.3) d ( x , Y ) = inf { d ( x , y ) : y ∈ Y } . (1.3) d ( x , Y ) = inf { d ( x , y ) : y ∈ Y } . {:(1.3)d(x","Y)=i n f{d(x","y):y in Y}.:}\begin{equation*} d(x, Y)=\inf \{d(x, y): y \in Y\} . \tag{1.3} \end{equation*}(1.3)d(x,Y)=inf{d(x,y):y∈Y}.
Now, let E E EEE be a normed linear space, G G GGG a nonempty subset of E , x E , x E,xE, xE,x an element from E E EEE, and
(1.4) P G ( x ) = { y ∈ G : ‖ x − y ‖ = d ( x , G ) } . (1.4) P G ( x ) = { y ∈ G : ‖ x − y ‖ = d ( x , G ) } . {:(1.4)P_(G)(x)={y in G:||x-y||=d(x","G)}.:}\begin{equation*} P_{G}(x)=\{y \in G:\|x-y\|=d(x, G)\} . \tag{1.4} \end{equation*}(1.4)PG(x)={y∈G:‖x−y‖=d(x,G)}.
An element from P G ( x ) P G ( x ) P_(G)(x)P_{G}(x)PG(x) is called a best approximation to x x xxx from G G GGG. If M M MMM is a subset of E E EEE we say that G G GGG is M M MMM-proximinal if P G ( x ) ≠ ∅ P G ( x ) ≠ ∅ P_(G)(x)!=O/P_{G}(x) \neq \varnothingPG(x)≠∅, for all x ∈ M x ∈ M x in Mx \in Mx∈M. If P G ( x ) P G ( x ) P_(G)(x)P_{G}(x)PG(x) contains exactly one element for every x ∈ M x ∈ M x in Mx \in Mx∈M, then G G GGG is called M M MMM-chebyshevian. If the set G G GGG is E E EEE-proximinal (respectively E E EEE-chebyshevian) then we say, simply, that G G GGG is proximinal (respectively chebyshevian).
We say that a linear subspace Z Z ZZZ of E E EEE has the property ( U U UUU ) if every continuous linear functional on Z Z ZZZ has a unique Hahn-Banach extension to E E EEE (i.e., linear and norm preserving) [6]. Let us denote by E ∗ E ∗ E^(**)E^{*}E∗ the conjugate spacs of E E EEE and by Z l Z l  Z^("l ")Z^{\text {l }}Zl  the annihilator of the subspace Z Z ZZZ in E ∗ E ∗ E^(**)E^{*}E∗, i.e.,
(1.5) Z Ω = { φ ∈ E ∗ ; φ | Z = 0 } (1.5) Z Ω = φ ∈ E ∗ ; φ Z = 0 {:(1.5)Z^(Omega)={varphi inE^(**); varphi|_(Z)=0}:}\begin{equation*} Z^{\Omega}=\left\{\varphi \in E^{*} ;\left.\varphi\right|_{Z}=0\right\} \tag{1.5} \end{equation*}(1.5)ZΩ={φ∈E∗;φ|Z=0}
Phelps [6] showed that the subspace Z Z ZZZ of E E EEE has property ( U U UUU ) if and only if its annihilator Z 1 Z 1 Z^(1)Z^{\mathbb{1}}Z1 is chebyshevian. This result can be extended to Lipschitz functions:
Theorem 1 ([5, Lemma 2]). Let X X XXX be a metric space, x 0 x 0 x_(0)x_{0}x0 in X X XXX, and Y ⊆ X Y ⊆ X Y sube XY \subseteq XY⊆X such that x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y. The space Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is chebyshevian for f ∈ Lip 0 X f ∈ Lip 0 X f inLip_(0)Xf \in \operatorname{Lip}_{0} Xf∈Lip0X if and only if f | Y ∈ Lip 0 Y f Y ∈ Lip 0 Y f|_(Y)inLip_(0)Y\left.f\right|_{Y} \in \operatorname{Lip}_{0} Yf|Y∈Lip0Y has a unique Lipschitz extension in Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X.
We also need the following lemma.
Lemma 1. Every best approximation to f ∈ Lip 0 X f ∈ Lip 0 X f inLip_(0)Xf \in \operatorname{Lip}_{0} Xf∈Lip0X from Y − 1 Y − 1 Y^(-1)Y^{-1}Y−1 is of the form f − F f − F f-Ff-Ff−F, where F F FFF is a Lipschitz extension of f | Y f Y f_(|_(Y))f_{\left.\right|_{Y}}f|Y to X X XXX.
Proof. Suppose F F FFF is a Lipschit extension of f | y f y f|_(y)\left.f\right|_{y}f|y to X X XXX. Then, by [ 5 [ 5 [5[5[5, Theorem 2 and Lemma 1], we get
| f − ( f − I ) | X − ‖ F | X − ‖ f ‖ X − d ( f , Y ⊥ ) | f − ( f − I ) | X − F X − f ‖ X − d f , Y ⊥ |f-(f-I)|_(X)-||F|_(X)-||f||_(X)-d(f,Y^(_|_))|f-(f-I)|_{X}-\left\|\left.F\right|_{X}-\right\| f \|_{X}-d\left(f, Y^{\perp}\right)|f−(f−I)|X−‖F|X−‖f‖X−d(f,Y⊥)
Conversely, if g ∈ Y ⊥ g ∈ Y ⊥ g inY^(_|_)g \in Y^{\perp}g∈Y⊥ is a best approximation to f f fff, then ‖ f − g ‖ X = d ( f , Y ⊥ ) = | i f | Y ‖ f − g ‖ X = d f , Y ⊥ = | i f | Y ||f-g||_(X)=d(f,Y^(_|_))=|if|_(Y)\|f-g\|_{X}= d\left(f, Y^{\perp}\right)=|i f|_{Y}‖f−g‖X=d(f,Y⊥)=|if|Y and ( f − g ) | Y = f | Y ( f − g ) Y = f Y (f-g)|_(Y)=f|_(Y)\left.(f-g)\right|_{Y}=\left.f\right|_{Y}(f−g)|Y=f|Y. Therefore F = f − g F = f − g F=f-gF=f-gF=f−g is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y,

2. Main Theorem

A metric space X X XXX is called uniformly discrete if there exists a number δ > 0 δ > 0 delta > 0\delta>0δ>0, such that d ( x , y ) ⩾ δ d ( x , y ) ⩾ δ d(x,y) >= deltad(x, y) \geqslant \deltad(x,y)⩾δ for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X with x ≠ y x ≠ y x!=yx \neq yx≠y. The following theorem appears in [5], in the hypothesis that Y Y YYY has an accumulation point in X X XXX. The main result is:
Theorem 2. Let X , x 0 X , x 0 X,x_(0)X, x_{0}X,x0, and Y Y YYY be as in Theorem 1. Suppose, further, that Y Y YYY is nonuniformly discrete. If every f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y has a unique Lipschitz extension. then Y → = X Y → = X vec(Y)=X\vec{Y}=XY→=X (or equivalently Y L = { 0 } Y L = { 0 } Y^(L)={0}Y^{\mathrm{L}}=\{0\}YL={0} ).
Proof. Since Y Y YYY is nonuniformly discrete, for every n ∈ N n ∈ N n in Nn \in Nn∈N, there exist x n , y n ∈ Y , x n ≠ y n x n , y n ∈ Y , x n ≠ y n x_(n),y_(n)in Y,x_(n)!=y_(n)x_{n}, y_{n} \in Y, x_{n} \neq y_{n}xn,yn∈Y,xn≠yn such that d ( x n , y n ) < 1 / n d x n , y n < 1 / n d(x_(n),y_(n)) < 1//nd\left(x_{n}, y_{n}\right)<1 / nd(xn,yn)<1/n. Defining f n : X → R f n : X → R f_(n):X rarr Rf_{n}: X \rightarrow Rfn:X→R by
f n ( x ) = d ( x , x n ) − d ( x , y n ) − d ( x 0 , x n ) + d ( x 0 , y n ) , n = 1.2 , 3 , … . f n ( x ) = d x , x n − d x , y n − d x 0 , x n + d x 0 , y n , n = 1.2 , 3 , … . f_(n)(x)=d(x,x_(n))-d(x,y_(n))-d(x_(0),x_(n))+d(x_(0),y_(n)),quad n=1.2,3,dots.f_{n}(x)=d\left(x, x_{n}\right)-d\left(x, y_{n}\right)-d\left(x_{0}, x_{n}\right)+d\left(x_{0}, y_{n}\right), \quad n=1.2,3, \ldots .fn(x)=d(x,xn)−d(x,yn)−d(x0,xn)+d(x0,yn),n=1.2,3,….
we have
f n ( x 0 ) = 0 , n = 1 , 2 , 3 , … , − 2 d ( x n , y n ) ⩽ f n ( x n ) = − d ( x n , y n ) − d ( x 0 , x n ) + d ( x 0 , y n ) ⩽ 0 , n = 1 , 2 , 3 , … , 0 ⩽ f n ( y n ) = d ( x n , y n ) − d ( x 0 , x n ) + d ( x 0 , y n ) ⩽ 2 d ( x n , y n ) , n = 1 , 2 , 3 , … , ‖ f n ‖ X = sup { | d ( x , x n ) − d ( x , y n ) − d ( y , x n ) + d ( y , y n ) | d ( x , y ) : x , y ∈ Y , x ≠ y } ⩽ 2 , n = 1 , 2 , 3 , … , f n x 0 = 0 , n = 1 , 2 , 3 , … , − 2 d x n , y n ⩽ f n x n = − d x n , y n − d x 0 , x n + d x 0 , y n ⩽ 0 , n = 1 , 2 , 3 , … , 0 ⩽ f n y n = d x n , y n − d x 0 , x n + d x 0 , y n ⩽ 2 d x n , y n , n = 1 , 2 , 3 , … , f n X = sup d x , x n − d x , y n − d y , x n + d y , y n d ( x , y ) : x , y ∈ Y , x ≠ y } ⩽ 2 , n = 1 , 2 , 3 , … , {:[f_(n)(x_(0))=0","quad n=1","2","3","dots","],[-2d(x_(n),y_(n)) <= f_(n)(x_(n))=-d(x_(n),y_(n))-d(x_(0),x_(n))+d(x_(0),y_(n))],[ <= 0","quad n=1","2","3","dots","],[0 <= f_(n)(y_(n))=d(x_(n),y_(n))-d(x_(0),x_(n))+d(x_(0),y_(n))],[ <= 2d(x_(n),y_(n))","quad n=1","2","3","dots","],[||f_(n)||_(X)=s u p{|d(x,x_(n))-d(x,y_(n))-d(y,x_(n))+d(y,y_(n))|d(x,y)::}],[x","y in Y","x!=y} <= 2","quad n=1","2","3","dots","]:}\begin{aligned} & f_{n}\left(x_{0}\right)= 0, \quad n=1,2,3, \ldots, \\ &-2 d\left(x_{n}, y_{n}\right) \leqslant f_{n}\left(x_{n}\right)=-d\left(x_{n}, y_{n}\right)-d\left(x_{0}, x_{n}\right)+d\left(x_{0}, y_{n}\right) \\ & \leqslant 0, \quad n=1,2,3, \ldots, \\ & 0 \leqslant f_{n}\left(y_{n}\right)=d\left(x_{n}, y_{n}\right)-d\left(x_{0}, x_{n}\right)+d\left(x_{0}, y_{n}\right) \\ & \leqslant 2 d\left(x_{n}, y_{n}\right), \quad n=1,2,3, \ldots, \\ &\left\|f_{n}\right\|_{X}= \sup \left\{\left|d\left(x, x_{n}\right)-d\left(x, y_{n}\right)-d\left(y, x_{n}\right)+d\left(y, y_{n}\right)\right| d(x, y):\right. \\ &x, y \in Y, x \neq y\} \leqslant 2, \quad n=1,2,3, \ldots, \end{aligned}fn(x0)=0,n=1,2,3,…,−2d(xn,yn)⩽fn(xn)=−d(xn,yn)−d(x0,xn)+d(x0,yn)⩽0,n=1,2,3,…,0⩽fn(yn)=d(xn,yn)−d(x0,xn)+d(x0,yn)⩽2d(xn,yn),n=1,2,3,…,‖fn‖X=sup{|d(x,xn)−d(x,yn)−d(y,xn)+d(y,yn)|d(x,y):x,y∈Y,x≠y}⩽2,n=1,2,3,…,
so that f n ∈ Lip 0 X f n ∈ Lip 0 X f_(n)inLip_(0)Xf_{n} \in \operatorname{Lip}_{0} Xfn∈Lip0X for n = 1 , 2 , 3 , … n = 1 , 2 , 3 , … n=1,2,3,dotsn=1,2,3, \ldotsn=1,2,3,….
Let a n = d ( x 0 , y n ) − d ( x 0 , x n ) a n = d x 0 , y n − d x 0 , x n a_(n)=d(x_(0),y_(n))-d(x_(0),x_(n))a_{n}=d\left(x_{0}, y_{n}\right)-d\left(x_{0}, x_{n}\right)an=d(x0,yn)−d(x0,xn), and suppose that the set I = { n ∈ N I = { n ∈ N I={n in NI=\{n \in NI={n∈N : a n ⩽ 0 } a n ⩽ 0 {:a_(n) <= 0}\left.a_{n} \leqslant 0\right\}an⩽0} is infinite, say I = { n j : j ∈ N } I = n j : j ∈ N I={n_(j):j in N}I=\left\{n_{j}: j \in N\right\}I={nj:j∈N}. Then, we have f n j ( x 0 ) = 0 , f n j ( x n j ) < 0 f n j x 0 = 0 , f n j x n j < 0 f_(n_(j))(x_(0))=0,f_(n_(j))(x_(n_(j))) < 0f_{n_{j}}\left(x_{0}\right)=0, f_{n_{j}}\left(x_{n_{j}}\right)<0fnj(x0)=0,fnj(xnj)<0, f n j ( y n j ) ⩾ 0 , j = 1 , 2 , 3 , … f n j y n j ⩾ 0 , j = 1 , 2 , 3 , … f_(n_(j))(y_(n_(j))) >= 0,j=1,2,3,dotsf_{n_{j}}\left(y_{n_{j}}\right) \geqslant 0, j=1,2,3, \ldotsfnj(ynj)⩾0,j=1,2,3,…. Now, we consider the sequence { ψ j } ψ j {psi_(j)}\left\{\psi_{j}\right\}{ψj} of functions ψ i : f n j ( X ) → [ 0 , 1 ] ψ i : f n j ( X ) → [ 0 , 1 ] psi_(i):f_(n_(j))(X)rarr[0,1]\psi_{i}: f_{n_{j}}(X) \rightarrow[0,1]ψi:fnj(X)→[0,1] defined by
ψ j ( t ) = 1 , t < f n j ( x n j ) , = t / f n j ( x n j ) , f n j ( x n j ) ⩽ t < 0 = f n j ( x 0 ) , = 0 , t ⩾ 0 , ψ j ( t ) = 1 , t < f n j x n j , = t / f n j x n j , f n j x n j ⩽ t < 0 = f n j x 0 , = 0 , t ⩾ 0 , {:[psi_(j)(t)=1","t < f_(n_(j))(x_(n_(j)))","],[=t//f_(n_(j))(x_(n_(j)))","f_(n_(j))(x_(n_(j))) <= t < 0=f_(n_(j))(x_(0))","],[=0","t >= 0","]:}\begin{aligned} \psi_{j}(t) & =1, & & t<f_{n_{j}}\left(x_{n_{j}}\right), \\ & =t / f_{n_{j}}\left(x_{n_{j}}\right), & & f_{n_{j}}\left(x_{n_{j}}\right) \leqslant t<0=f_{n_{j}}\left(x_{0}\right), \\ & =0, & & t \geqslant 0, \end{aligned}ψj(t)=1,t<fnj(xnj),=t/fnj(xnj),fnj(xnj)⩽t<0=fnj(x0),=0,t⩾0,
for j = 1 , 2 , 3 , … j = 1 , 2 , 3 , … j=1,2,3,dotsj=1,2,3, \ldotsj=1,2,3,…. Putting q j = ψ j ∘ f n j q j = ψ j ∘ f n j q_(j)=psi_(j)@f_(n_(j))q_{j}=\psi_{j} \circ f_{n_{j}}qj=ψj∘fnj, we have
‖ q j ‖ Y ⩾ | ψ j ( f n j ( x n j ) ) − ψ j ( f n j ( y n j ) ) | / d ( x n j , y n j ) ⩾ n j . q j Y ⩾ ψ j f n j x n j − ψ j f n j y n j / d x n j , y n j ⩾ n j . ||q_(j)||_(Y) >= |psi_(j)(f_(n_(j))(x_(n_(j))))-psi_(j)(f_(n_(j))(y_(n_(j))))|//d(x_(n_(j)),y_(n_(j))) >= n_(j).\left\|q_{j}\right\|_{Y} \geqslant\left|\psi_{j}\left(f_{n_{j}}\left(x_{n_{j}}\right)\right)-\psi_{j}\left(f_{n_{j}}\left(y_{n_{j}}\right)\right)\right| / d\left(x_{n_{j}}, y_{n_{j}}\right) \geqslant n_{j} .‖qj‖Y⩾|ψj(fnj(xnj))−ψj(fnj(ynj))|/d(xnj,ynj)⩾nj.
By [5, Corollary 2] it follows that
d ( x , Y ) ⩽ ( sup { ψ j ( f n j ( y ) ) : y ∈ Y } − inf { ψ j ( f n j ( y ) ) : y ∈ Y } ) / ( 2 ‖ q j ‖ Y ) = 1 / ( 2 ‖ q j ‖ Y ) ⩽ 1 / n j → 0 d ( x , Y ) ⩽ sup ψ j f n j ( y ) : y ∈ Y − inf ψ j f n j ( y ) : y ∈ Y / 2 q j Y = 1 / 2 q j Y ⩽ 1 / n j → 0 {:[d(x","Y) <= (s u p{psi_(j)(f_(n_(j))(y)):y in Y}-i n f{psi_(j)(f_(n_(j))(y)):y in Y})//(2||q_(j)||_(Y))],[=1//(2||q_(j)||_(Y)) <= 1//n_(j)rarr0]:}\begin{aligned} d(x, Y) & \leqslant\left(\sup \left\{\psi_{j}\left(f_{n_{j}}(y)\right): y \in Y\right\}-\inf \left\{\psi_{j}\left(f_{n_{j}}(y)\right): y \in Y\right\}\right) /\left(2\left\|q_{j}\right\|_{Y}\right) \\ & =1 /\left(2\left\|q_{j}\right\|_{Y}\right) \leqslant 1 / n_{j} \rightarrow 0 \end{aligned}d(x,Y)⩽(sup{ψj(fnj(y)):y∈Y}−inf{ψj(fnj(y)):y∈Y})/(2‖qj‖Y)=1/(2‖qj‖Y)⩽1/nj→0
so that x ∈ Y ¯ x ∈ Y ¯ x in bar(Y)x \in \bar{Y}x∈Y¯, for all x ∈ X x ∈ X x in Xx \in Xx∈X, that is Y ¯ = X Y ¯ = X bar(Y)=X\bar{Y}=XY¯=X.
By Theorems 1 and 2, we have
Corollary 1. Suppose that Y Y YYY is nonuniformly discrete. Then Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is chebyshevian in Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X if and only if Y ⊥ = { 0 } Y ⊥ = { 0 } Y^(_|_)={0}Y^{\perp}=\{0\}Y⊥={0}.
We can also prove the following result.
Theorem 3. Let X , x 0 X , x 0 X,x_(0)X, x_{0}X,x0, and Y Y YYY be as in Theorem 1. If ( Y ⊥ ) 1 Y ⊥ 1 (Y^(_|_))^(1)\left(Y^{\perp}\right)^{1}(Y⊥)1 has the property ( U U UUU ) then every f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y has a unique Lipschitz extension F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X.
Proof. Follows from [8, Corollary 3.1.b)] and the above Theorem 1.
Corollary 2. Let X , x 0 X , x 0 X,x_(0)X, x_{0}X,x0, and Y Y YYY be as in Theorem 1. Suppose that Y Y YYY is nonuniformly discrete. If ( Y ⊥ ) 1 Y ⊥ 1 (Y^(_|_))^(1)\left(Y^{\perp}\right)^{\mathbb{1}}(Y⊥)1 has the property ( U U UUU ), then Y ¯ = X Y ¯ = X bar(Y)=X\bar{Y}=XY¯=X (or equivalently Y ⊥ = { 0 } Y ⊥ = { 0 } Y^(_|_)={0}Y^{\perp}=\{0\}Y⊥={0} ).

3. Examples

(a) Let X = [ 0 , 1 ] , d ( x , y ) = | x − y | , x 0 = 0 X = [ 0 , 1 ] , d ( x , y ) = | x − y | , x 0 = 0 X=[0,1],d(x,y)=|x-y|,x_(0)=0X=[0,1], d(x, y)=|x-y|, x_{0}=0X=[0,1],d(x,y)=|x−y|,x0=0, and Y = { 0 , 1 } Y = { 0 , 1 } Y={0,1}Y=\{0,1\}Y={0,1}. Then every f ∈ Lip 0 { 0 , 1 } f ∈ Lip 0 { 0 , 1 } f inLip_(0){0,1}f \in \operatorname{Lip}_{0}\{0,1\}f∈Lip0{0,1} has a unique Lipschitz extension F ∈ Lip 0 [ 0 , 1 ] F ∈ Lip 0 [ 0 , 1 ] F inLip_(0)[0,1]F \in \operatorname{Lip}_{0}[0,1]F∈Lip0[0,1], namely, F ( x ) = f ( 1 ) x F ( x ) = f ( 1 ) x F(x)=f(1)xF(x)=f(1) xF(x)=f(1)x. This example shows that the supposition that Y Y YYY is nonuniformly discrete is essential in Theorem 2.
(b) Let f ∈ Lip 0 [ 0 , 1 ] f ∈ Lip 0 [ 0 , 1 ] f inLip_(0)[0,1]f \in \operatorname{Lip}_{0}[0,1]f∈Lip0[0,1] and let Y Y YYY be the set of points 0 = x 0 < x 1 < ⋯ < x n + 1 = 1 0 = x 0 < x 1 < ⋯ < x n + 1 = 1 0=x_(0) < x_(1) < cdots < x_(n+1)=10=x_{0}<x_{1}<\cdots <x_{n+1}=10=x0<x1<⋯<xn+1=1. Then, we have:
Theorem 4. The following conditions are equivalent:
( 1 ∘ ) Y ⊥ 1 ∘ Y ⊥ (1^(@))quadY^(_|_)\left(1^{\circ}\right) \quad Y^{\perp}(1∘)Y⊥ is f-chebyshevian,
(2) ‖ f ‖ Y = ‖ [ x k , x i + 1 ; f ] ‖ , k = 0 , 1 , 2 , … , n ‖ f ‖ Y = x k , x i + 1 ; f , k = 0 , 1 , 2 , … , n quad||f||_(Y)=||[x_(k),x_(i+1);f]||,k=0,1,2,dots,n\quad\|f\|_{Y}=\left\|\left[x_{k}, x_{i+1} ; f\right]\right\|, k=0,1,2, \ldots, n‖f‖Y=‖[xk,xi+1;f]‖,k=0,1,2,…,n,
where [ x k , x k + 1 ; f ] = ( f ( x k + 1 ) − f ( x k ) ) / ( x k + 1 − x k ) x k , x k + 1 ; f = f x k + 1 − f x k / x k + 1 − x k [x_(k),x_(k+1);f]=(f(x_(k+1))-f(x_(k)))//(x_(k+1)-x_(k))\left[x_{k}, x_{k+1} ; f\right]=\left(f\left(x_{k+1}\right)-f\left(x_{k}\right)\right) /\left(x_{k+1}-x_{k}\right)[xk,xk+1;f]=(f(xk+1)−f(xk))/(xk+1−xk).
Proof. ( 1 ∘ ) ⇒ ( 2 ∘ ) 1 ∘ ⇒ 2 ∘ quad(1^(@))=>(2^(@))\quad\left(1^{\circ}\right) \Rightarrow\left(2^{\circ}\right)(1∘)⇒(2∘) Obviously,
L ( x ) = [ x k , x k + 1 ; f ] ( x − x k ) + f ( x k ) , x ∈ ( x k , x k + 1 ) , k = 0 , 1 , … , n L ( x ) = x k , x k + 1 ; f x − x k + f x k , x ∈ x k , x k + 1 , k = 0 , 1 , … , n L(x)=[x_(k),x_(k+1);f](x-x_(k))+f(x_(k)),quad x in(x_(k),x_(k+1)),k=0,1,dots,nL(x)=\left[x_{k}, x_{k+1} ; f\right]\left(x-x_{k}\right)+f\left(x_{k}\right), \quad x \in\left(x_{k}, x_{k+1}\right), k=0,1, \ldots, nL(x)=[xk,xk+1;f](x−xk)+f(xk),x∈(xk,xk+1),k=0,1,…,n,
is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y and ‖ f | Y ⩾ [ x k , x k + 1 ; f ] | i k = 0 , 1 5 , … , n ‖ f Y ⩾ x k , x k + 1 ; f i k = 0 , 1 5 , … , n ||f|_(Y) >= [x_(k),x_(k+1);f]|_(i)k=0,1_(5),dots,n\|\left. f\right|_{Y} \geqslant\left.\left[x_{k}, x_{k+1} ; f\right]\right|_{i} k=0,1_{5}, \ldots, n‖f|Y⩾[xk,xk+1;f]|ik=0,15,…,n. Suppose that k 0 , 0 ⩽ k 0 < n k 0 , 0 ⩽ k 0 < n k_(0),0 <= k_(0) < nk_{0}, 0 \leqslant k_{0}<nk0,0⩽k0<n is such that ‖ f ‖ Y >∣ [ x h 0 , x I 0 + 1 ; f ] ‖ f ‖ Y >∣ x h 0 , x I 0 + 1 ; f ||f||_(Y)>∣[x_(h_(0)),x_(I_(0)+1);f]\|f\|_{Y}>\mid\left[x_{h_{0}}, x_{I_{0}+1} ; f\right]‖f‖Y>∣[xh0,xI0+1;f]. We have to consider the following cases:
(i) f ( x k 0 ) < f ( x k 0 + 1 ) f x k 0 < f x k 0 + 1 f(x_(k_(0))) < f(x_(k_(0)+1))f\left(x_{k_{0}}\right)<f\left(x_{k_{0}+1}\right)f(xk0)<f(xk0+1),
(ii) f ( x k 0 ) > f ( x k 0 + 1 ) f x k 0 > f x k 0 + 1 f(x_(k_(0))) > f(x_(k_(0)+1))f\left(x_{k_{0}}\right)>f\left(x_{k_{0}+1}\right)f(xk0)>f(xk0+1),
(iii) f ( x k 0 ) = f ( x k 0 + 1 ) f x k 0 = f x k 0 + 1 f(x_(k_(0)))=f(x_(k_(0)+1))f\left(x_{k_{0}}\right)=f\left(x_{k_{0}+1}\right)f(xk0)=f(xk0+1).
If condition (i) holds, put z 1 = x k 0 + ( f ( x k 0 + 1 ) − f ( x k 0 ) ) / ‖ f ‖ r z 1 = x k 0 + f x k 0 + 1 − f x k 0 / ‖ f ‖ r z_(1)=x_(k_(0))+(f(x_(k_(0)+1))-f(x_(k_(0))))//||f||_(r)z_{1}=x_{k_{0}}+\left(f\left(x_{k_{0}+1}\right)-f\left(x_{k_{0}}\right)\right) /\|f\|_{r}z1=xk0+(f(xk0+1)−f(xk0))/‖f‖r and define the function F 1 : [ 0 , 1 ] → R F 1 : [ 0 , 1 ] → R F_(1):[0,1]rarr RF_{1}:[0,1] \rightarrow RF1:[0,1]→R by
F 1 ( x ) = L ( x ) , x ∈ [ 0 , 1 ] − [ x k 0 , x k 0 − 1 ] , (3.2) = f ( x k 0 ) − ∣ f ‖ Y ( x − x k 0 ) , x ∈ ( x k 1 , z 1 ) , = f ( x k 0 ) , x ∈ [ z 1 , x k 0 + 1 ) . F 1 ( x ) = L ( x ) , x ∈ [ 0 , 1 ] − x k 0 , x k 0 − 1 , (3.2) = f x k 0 − ∣ f ‖ Y x − x k 0 , x ∈ x k 1 , z 1 , = f x k 0 , x ∈ z 1 , x k 0 + 1 . {:[F_(1)(x)=L(x)","x in[0","1]-[x_(k_(0)),x_(k_(0)-1)]","],[(3.2)=f(x_(k_(0)))-∣f||_(Y)(x-x_(k_(0)))","x in(x_(k_(1)),z_(1))","],[=f(x_(k_(0)))","x in[z_(1),x_(k_(0)+1)).]:}\begin{align*} F_{1}(x) & =L(x), & & x \in[0,1]-\left[x_{k_{0}}, x_{k_{0}-1}\right], \\ & =f\left(x_{k_{0}}\right)-\mid f \|_{Y}\left(x-x_{k_{0}}\right), & & x \in\left(x_{k_{1}}, z_{1}\right), \tag{3.2}\\ & =f\left(x_{k_{0}}\right), & & x \in\left[z_{1}, x_{k_{0}+1}\right) . \end{align*}F1(x)=L(x),x∈[0,1]−[xk0,xk0−1],(3.2)=f(xk0)−∣f‖Y(x−xk0),x∈(xk1,z1),=f(xk0),x∈[z1,xk0+1).
It is easy to see that F 1 F 1 F_(1)F_{1}F1 is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y, distinct from L , L , L_(,)L_{,}L, and then, by Theorem 1 , Y ⊥ 1 , Y ⊥ 1,Y^(_|_)1, Y^{\perp}1,Y⊥ is not f f fff-chebyshevian.
In case (ii) the proof proceeds similarly. If condition (iii) holds, put z 2.2 = ( 2 x k 11 + x k 0 + 1 ) / 3 z 2.2 = 2 x k 11 + x k 0 + 1 / 3 z_(2.2)=(2x_(k_(11))+x_(k_(0)+1))//3z_{2.2}=\left(2 x_{k_{11}}+x_{k_{0}+1}\right) / 3z2.2=(2xk11+xk0+1)/3 and define
F 2 ( x ) = L ( x ) , x ∈ [ 0 , 1 ] − [ x k 0 , x k 0 + 1 ] , (3.3) = f ( x k 0 ) + :∣ f ‖ Y ( x − x k 0 ) , x ∈ ( x k 0 , z 2 ] , = f ( x k 0 + 1 ) − ‖ f Y ‖ Y ( x − x k 0 + 1 ) , x ∈ ( z 2 , x k 0 − 1 ) , F 2 ( x ) = L ( x ) , x ∈ [ 0 , 1 ] − x k 0 , x k 0 + 1 , (3.3) = f x k 0 + :∣ f ‖ Y x − x k 0 , x ∈ x k 0 , z 2 , = f x k 0 + 1 − f Y Y x − x k 0 + 1 , x ∈ z 2 , x k 0 − 1 , {:[F_(2)(x)=L(x)","x in[0","1]-[x_(k_(0)),x_(k_(0)+1)]","],[(3.3)=f(x_(k_(0)))+:∣f||_(Y)(x-x_(k_(0)))","x in(x_(k_(0)),z_(2)]","],[=f(x_(k_(0)+1))-||f_(Y)||_(Y)(x-x_(k_(0)+1))","x in(z_(2),x_(k_(0)-1))","]:}\begin{align*} F_{2}(x) & =L(x), & & x \in[0,1]-\left[x_{k_{0}}, x_{k_{0}+1}\right], \\ & =f\left(x_{k_{0}}\right)+: \mid f \|_{Y}\left(x-x_{k_{0}}\right), & & x \in\left(x_{k_{0}}, z_{2}\right], \tag{3.3}\\ & =f\left(x_{k_{0}+1}\right)-\left\|f_{Y}\right\|_{Y}\left(x-x_{k_{0}+1}\right), & & x \in\left(z_{2}, x_{k_{0}-1}\right), \end{align*}F2(x)=L(x),x∈[0,1]−[xk0,xk0+1],(3.3)=f(xk0)+:∣f‖Y(x−xk0),x∈(xk0,z2],=f(xk0+1)−‖fY‖Y(x−xk0+1),x∈(z2,xk0−1),
Then F 2 F 2 F_(2)F_{2}F2 is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y, different from L L LLL. By Theorem 1, Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is not f f fff-chebyshevian.
( 2 ∘ ) ⇒ ( 1 ∘ ) 2 ∘ ⇒ 1 ∘ (2^(@))=>(1^(@))\left(2^{\circ}\right) \Rightarrow\left(1^{\circ}\right)(2∘)⇒(1∘) If | [ x k , x k + 1 ; f ] | = ‖ f ‖ Y x k , x k + 1 ; f = ‖ f ‖ Y |[x_(k),x_(k+1);f]|=||f||_(Y)\left|\left[x_{k}, x_{k+1} ; f\right]\right|=\|f\|_{Y}|[xk,xk+1;f]|=‖f‖Y for k = 0 , 1 , 2 , … , n k = 0 , 1 , 2 , … , n k=0,1,2,dots,nk=0,1,2, \ldots, nk=0,1,2,…,n, then the function L L LLL defined by (3.1) is the only Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y.
A consequence of Theorem 4 is:
Corollary 3. Let Y Y YYY be the set of points 0 = x 0 < x 1 < ⋯ < x n + 1 = 1 0 = x 0 < x 1 < ⋯ < x n + 1 = 1 0=x_(0) < x_(1) < cdots < x_(n+1)=10=x_{0}<x_{1}<\cdots<x_{n+1}=10=x0<x1<⋯<xn+1=1, f ∈ Lip 0 [ 0 , 1 ] f ∈ Lip 0 [ 0 , 1 ] f inLip_(0)[0,1]f \in \operatorname{Lip}_{0}[0,1]f∈Lip0[0,1] and
(3.4) K = { h : h ∈ Lip 0 [ 0 , 1 ] , h ( x k ) = f ( x k ) , k = 0 , 1 , 2 , … , n + 1 } . (3.4) K = h : h ∈ Lip 0 [ 0 , 1 ] , h x k = f x k , k = 0 , 1 , 2 , … , n + 1 . {:(3.4)K={h:h inLip_(0)[0,1],h(x_(k))=f(x_(k)),k=0,1,2,dots,n+1}.:}\begin{equation*} K=\left\{h: h \in \operatorname{Lip}_{0}[0,1], h\left(x_{k}\right)=f\left(x_{k}\right), k=0,1,2, \ldots, n+1\right\} . \tag{3.4} \end{equation*}(3.4)K={h:h∈Lip0[0,1],h(xk)=f(xk),k=0,1,2,…,n+1}.
Then Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is K K KKK-chebyshevian if and only if Y − Y − Y^(-)Y^{-}Y−is f f fff-chebyshevian.
(c) Let C 1 [ 0 , 1 ] C 1 [ 0 , 1 ] C^(1)[0,1]C^{1}[0,1]C1[0,1] be the space of all continuously differentiable functions on [ 0,1 ] and let Y Y YYY be the set of points 0 = x 0 < x 1 < ⋯ < x n + 1 = 1 0 = x 0 < x 1 < ⋯ < x n + 1 = 1 0=x_(0) < x_(1) < cdots < x_(n+1)=10=x_{0}<x_{1}<\cdots<x_{n+1}=10=x0<x1<⋯<xn+1=1. Put
(3.5) Z = C 1 [ 0 , 1 ] ∩ Lip 0 [ 0 , 1 ] , W = C 1 [ 0 , 1 ] ∩ Y ⊥ . (3.5) Z = C 1 [ 0 , 1 ] ∩ Lip 0 [ 0 , 1 ] , W = C 1 [ 0 , 1 ] ∩ Y ⊥ . {:(3.5)Z=C^(1)[0","1]nnLip_(0)[0","1]","quad W=C^(1)[0","1]nnY^(_|_).:}\begin{equation*} Z=C^{1}[0,1] \cap \operatorname{Lip}_{0}[0,1], \quad W=C^{1}[0,1] \cap Y^{\perp} . \tag{3.5} \end{equation*}(3.5)Z=C1[0,1]∩Lip0[0,1],W=C1[0,1]∩Y⊥.
For f ∈ Z f ∈ Z f in Zf \in Zf∈Z, we have
(3.6) ‖ f ‖ [ 0 , 1 ] = max { | f ′ ( x ) | : x ∈ [ 0 , 1 ] } . (3.6) ‖ f ‖ [ 0 , 1 ] = max f ′ ( x ) : x ∈ [ 0 , 1 ] . {:(3.6)||f||_([0,1])=max{|f^(')(x)|:x in[0,1]}.:}\begin{equation*} \|f\|_{[0,1]}=\max \left\{\left|f^{\prime}(x)\right|: x \in[0,1]\right\} . \tag{3.6} \end{equation*}(3.6)‖f‖[0,1]=max{|f′(x)|:x∈[0,1]}.
Let us define the function set S S SSS by
S = { h : h ∈ Z , [ x k , x k + 1 ; h ] [ x k + 1 , x k + 2 ; h ] (3.7) ≠ − ‖ h ‖ Y 2 , k = 0 , 1 , 2 , … , n − 1 } . S = h : h ∈ Z , x k , x k + 1 ; h x k + 1 , x k + 2 ; h (3.7) ≠ − ‖ h ‖ Y 2 , k = 0 , 1 , 2 , … , n − 1 . {:[S={h:h in Z,[x_(k),x_(k+1);h][x_(k+1),x_(k+2);h]:}],[(3.7){:!=-||h||_(Y)^(2),k=0,1,2,dots,n-1}.]:}\begin{align*} S & =\left\{h: h \in Z,\left[x_{k}, x_{k+1} ; h\right]\left[x_{k+1}, x_{k+2} ; h\right]\right. \\ & \left.\neq-\|h\|_{Y}^{2}, k=0,1,2, \ldots, n-1\right\} . \tag{3.7} \end{align*}S={h:h∈Z,[xk,xk+1;h][xk+1,xk+2;h](3.7)≠−‖h‖Y2,k=0,1,2,…,n−1}.
We need the following two lemmas:
Lemma 2. Let [ p , q ] ⊂ R , f ( x ) = a x + b , a , b ∈ R , a > 0 [ p , q ] ⊂ R , f ( x ) = a x + b , a , b ∈ R , a > 0 [p,q]sub R,f(x)=ax+b,a,b in R,a > 0[p, q] \subset R, f(x)=a x+b, a, b \in R, a>0[p,q]⊂R,f(x)=ax+b,a,b∈R,a>0, and M > a M > a M > aM>aM>a. Then there exists a function g ∈ C 1 [ p , q ] g ∈ C 1 [ p , q ] g inC^(1)[p,q]g \in C^{1}[p, q]g∈C1[p,q] such that f ( p ) = g ( p ) , f ( q ) = g ( q ) f ( p ) = g ( p ) , f ( q ) = g ( q ) f(p)=g(p),f(q)=g(q)f(p)=g(p), f(q)=g(q)f(p)=g(p),f(q)=g(q), f ′ ( p ) = M ( f ′ ( q ) = M ) , f ′ ( q ) = g ′ ( q ) ( f ′ ( p ) = g ′ ( p ) ) f ′ ( p ) = M f ′ ( q ) = M , f ′ ( q ) = g ′ ( q ) f ′ ( p ) = g ′ ( p ) f^(')(p)=M(f^(')(q)=M),f^(')(q)=g^(')(q)(f^(')(p)=g^(')(p))f^{\prime}(p)=M\left(f^{\prime}(q)=M\right), f^{\prime}(q)=g^{\prime}(q)\left(f^{\prime}(p)=g^{\prime}(p)\right)f′(p)=M(f′(q)=M),f′(q)=g′(q)(f′(p)=g′(p)) and max { | g ′ ( x ) | : x ∈ [ p , q ] } = M max g ′ ( x ) : x ∈ [ p , q ] } = M max{|g^(')(x)|::}x in[p,q]}=M\max \left\{\left|g^{\prime}(x)\right|:\right. x \in[p, q]\}=Mmax{|g′(x)|:x∈[p,q]}=M.
Proof. The proof of the lemma is obvious from Fig. 1:
Figure 1
( AC ) s 1 ( x ) = f ( p ) + M ( x − p ) , ( BC ) s 2 ( x ) = f ( q ) − M ( x − q ) , ( DE ) s 3 ( x ) = f ( r ) − M ( x − r ) , r ∈ ( p , q ) . ( AC )      s 1 ( x ) = f ( p ) + M ( x − p ) , ( BC )      s 2 ( x ) = f ( q ) − M ( x − q ) , ( DE )      s 3 ( x ) = f ( r ) − M ( x − r ) , r ∈ ( p , q ) . {:[(AC),s_(1)(x)=f(p)+M(x-p)","],[(BC),s_(2)(x)=f(q)-M(x-q)","],[(DE),s_(3)(x)=f(r)-M(x-r)","quad r in(p","q).]:}\begin{array}{ll} (\mathrm{AC}) & s_{1}(x)=f(p)+M(x-p), \\ (\mathrm{BC}) & s_{2}(x)=f(q)-M(x-q), \\ (\mathrm{DE}) & s_{3}(x)=f(r)-M(x-r), \quad r \in(p, q) . \end{array}(AC)s1(x)=f(p)+M(x−p),(BC)s2(x)=f(q)−M(x−q),(DE)s3(x)=f(r)−M(x−r),r∈(p,q).
Lemma 3. If h ∈ S h ∈ S h in Sh \in Sh∈S, then h | Y h Y h|_(Y)\left.h\right|_{Y}h|Y has at least one Lipschitz extension H ∈ Z H ∈ Z H in ZH \in ZH∈Z.
Proof. Let h ∈ S h ∈ S h in Sh \in Sh∈S and k 0 ∈ N , 0 ⩽ k 0 < n + 1 k 0 ∈ N , 0 ⩽ k 0 < n + 1 k_(0)in N,0 <= k_(0) < n+1k_{0} \in N, 0 \leqslant k_{0}<n+1k0∈N,0⩽k0<n+1, such that
(3.8) [ x k 0 , x k 0 + 1 ; h ] = ‖ 1 h ‖ Y (3.8) x k 0 , x k 0 + 1 ; h = 1 h Y {:(3.8)[x_(k_(0)),x_(k_(0)+1);h]=||_(1)h||_(Y):}\begin{equation*} \left[x_{k_{0}}, x_{k_{0}+1} ; h\right]=\left\|_{1} h\right\|_{Y} \tag{3.8} \end{equation*}(3.8)[xk0,xk0+1;h]=‖1h‖Y
By the definition of S S SSS, we have
− ; h ‖ Y < [ x k 0 − 1 , x k 0 ; h ] ⩽ | h | Y − . h ‖ Y < [ x h 0 + 1 , x k 0 + 2 ; h ] ⩽ ‖ h | Y Y . − ; h ‖ Y < x k 0 − 1 , x k 0 ; h ⩽ | h | Y − . h Y < x h 0 + 1 , x k 0 + 2 ; h ⩽ h Y Y . {:[-;h||_(Y) < [x_(k_(0)-1),x_(k_(0));h] <= |h|_(Y)],[-.h||_(Y) < [x_(h_(0)+1),x_(k_(0)+2);h] <= ||h|_(Y_(Y)).]:}\begin{aligned} & -; h \|_{\mathrm{Y}}<\left[x_{k_{0}-1}, x_{k_{0}} ; h\right] \leqslant|h|{ }_{\mathrm{Y}} \\ & -.\left.h\left\|_{\mathrm{Y}}<\left[x_{h_{0}+1}, x_{k_{0}+2} ; h\right] \leqslant\right\| h\right|_{\mathrm{Y}_{Y}} . \end{aligned}−;h‖Y<[xk0−1,xk0;h]⩽|h|Y−.h‖Y<[xh0+1,xk0+2;h]⩽‖h|YY.
Applying Lemma 2 to the intervals [ x k 0 − 1 , x k 6 ] x k 0 − 1 , x k 6 [x_(k_(0)-1),x_(k_(6))]\left[x_{k_{0}-1}, x_{k_{6}}\right][xk0−1,xk6] and [ x k 0 + 1 , x k 0 + 2 ] x k 0 + 1 , x k 0 + 2 [x_(k_(0)+1),x_(k_(0)+2)]\left[x_{k_{0}+1}, x_{k_{0}+2}\right][xk0+1,xk0+2], twice it follows that there exists a function H 1 H 1 H_(1)H_{1}H1 in C 1 [ x k 0 − 1 , x k 0 + 2 ] C 1 x k 0 − 1 , x k 0 + 2 C^(1)[x_(k_(0)-1),x_(k_(0)+2)]C^{1}\left[x_{k_{0}-1}, x_{k_{0}+2}\right]C1[xk0−1,xk0+2] such that max { | H 1 ′ ( x ) | : x ∈ [ x k 0 − 1 , x k 0 + 2 ] } = ‖ h ‖ Y max H 1 ′ ( x ) : x ∈ x k 0 − 1 , x k 0 + 2 = ‖ h ‖ Y max{|H_(1)^(')(x)|:x in[x_(k_(0)-1),x_(k_(0)+2)]}=||h||_(Y)\max \left\{\left|H_{1}{ }^{\prime}(x)\right|: x \in\left[x_{k_{0}-1}, x_{k_{0}+2}\right]\right\}=\|h\|_{Y}max{|H1′(x)|:x∈[xk0−1,xk0+2]}=‖h‖Y and which interpolates the function h h hhh at the points x k 0 − 1 , x k 0 , x k 0 + 1 , x k 0 + 2 x k 0 − 1 , x k 0 , x k 0 + 1 , x k 0 + 2 x_(k_(0)-1),x_(k_(0)),x_(k_(0)+1),x_(k_(0)+2)x_{k_{0}-1}, x_{k_{0}}, x_{k_{0}+1}, x_{k_{0}+2}xk0−1,xk0,xk0+1,xk0+2.
Appiying Lemma 2 to the intervals [ x i , x i + 1 ] , i = 0 , 1 , … , k 0 − 2 x i , x i + 1 , i = 0 , 1 , … , k 0 − 2 [x_(i),x_(i+1)],i=0,1,dots,k_(0)-2\left[x_{i}, x_{i+1}\right], i=0,1, \ldots, k_{0}-2[xi,xi+1],i=0,1,…,k0−2, k 0 + 2 , … , n k 0 + 2 , … , n k_(0)+2,dots,nk_{0}+2, \ldots, nk0+2,…,n, we get a function H ∈ Z H ∈ Z H in ZH \in ZH∈Z, which is a Lipschitz extension of h ′ v h ′ v h^(')vh^{\prime} vh′v to [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
If [ x k 0 , x k 0 + 1 ; h ] = − ‖ h ‖ Y x k 0 , x k 0 + 1 ; h = − ‖ h ‖ Y [x_(k_(0)),x_(k_(0)+1);h]=-||h||_(Y)\left[x_{k_{0}}, x_{k_{0}+1} ; h\right]=-\|h\|_{Y}[xk0,xk0+1;h]=−‖h‖Y we can proceed analogously.
Theorem 5. The subspace W W WWW is S S SSS proximinal and for each h ∈ S h ∈ S h in Sh \in Sh∈S the following equality holds:
(3.9) d ( h , W ) = d ( h , Y ⊥ ) (3.9) d ( h , W ) = d h , Y ⊥ {:(3.9)d(h","W)=d(h,Y^(_|_)):}\begin{equation*} d(h, W)=d\left(h, Y^{\perp}\right) \tag{3.9} \end{equation*}(3.9)d(h,W)=d(h,Y⊥)
Proof. Let h ∈ S h ∈ S h in Sh \in Sh∈S. By Lemma 3, h | Y h Y h|_(Y)\left.h\right|_{Y}h|Y has a Lipschitz extension H ∈ Z H ∈ Z H in ZH \in ZH∈Z. Then, h − H ∈ W h − H ∈ W h-H in Wh-H \in Wh−H∈W, and this is a best approximation to h h hhh, from Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥.
But then
d ( h , Y ⊥ ) ⩽ d ( h , W ) ⩽ ‖ h − ( h − H ) | X = d ( h , Y − ) d h , Y ⊥ ⩽ d ( h , W ) ⩽ ‖ h − ( h − H ) X = d h , Y − d(h,Y^(_|_)) <= d(h,W) <= ||h-(h-H)|_(X)=d(h,Y^(-))d\left(h, Y^{\perp}\right) \leqslant d(h, W) \leqslant \| h-\left.(h-H)\right|_{X}=d\left(h, Y^{-}\right)d(h,Y⊥)⩽d(h,W)⩽‖h−(h−H)|X=d(h,Y−)
so that
‖ h − ( h − H ) ‖ X = d ( h , W ) = d ( h , Y ⊥ ) ‖ h − ( h − H ) ‖ X = d ( h , W ) = d h , Y ⊥ ||h-(h-H)||_(X)=d(h,W)=d(h,Y^(_|_))\|h-(h-H)\|_{X}=d(h, W)=d\left(h, Y^{\perp}\right)‖h−(h−H)‖X=d(h,W)=d(h,Y⊥)
Remark 1. Let f ∈ Z − S f ∈ Z − S f in Z-Sf \in Z-Sf∈Z−S; that is, there exists 0 ⩽ k 1 < n + 1 0 ⩽ k 1 < n + 1 0 <= k_(1) < n+10 \leqslant k_{1}<n+10⩽k1<n+1 such that
[ x k 1 − 1 , x k 1 ; f ] [ x k 1 , x k 1 + 1 ; f ] = − | f | Y 2 x k 1 − 1 , x k 1 ; f x k 1 , x k 1 + 1 ; f = − | f | Y 2 [x_(k_(1)-1),x_(k_(1));f][x_(k_(1)),x_(k_(1)+1);f]=-|f|_(Y)^(2)\left[x_{k_{1}-1}, x_{k_{1}} ; f\right]\left[x_{k_{1}}, x_{k_{1}+1} ; f\right]=-|f|_{Y}^{2}[xk1−1,xk1;f][xk1,xk1+1;f]=−|f|Y2
In this case, it is possible that no Lipschitz extension to f f fff exists in Z Z ZZZ : e.g.for f ( x ) = − 4 x 2 + 4 x , Y = [ 0 , 1 2 , 1 } f ( x ) = − 4 x 2 + 4 x , Y = 0 , 1 2 , 1 f(x)=-4x^(2)+4x,Y=[0,(1)/(2),1}f(x)=-4 x^{2}+4 x, Y=\left[0, \frac{1}{2}, 1\right\}f(x)=−4x2+4x,Y=[0,12,1} we have
[ 0 , 1 2 ; f ] [ 1 2 , 1 ; f ] = − 4 0 , 1 2 ; f 1 2 , 1 ; f = − 4 [0,(1)/(2);f][(1)/(2),1;f]=-4\left[0, \frac{1}{2} ; f\right]\left[\frac{1}{2}, 1 ; f\right]=-4[0,12;f][12,1;f]=−4
and the only Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y is
F ( x ) = 2 x , x ∈ [ 0 , 1 2 ) , = − 2 ( x − 1 ) , x ∈ [ 1 2 , 1 ] , F ( x ) = 2 x , x ∈ 0 , 1 2 , = − 2 ( x − 1 ) , x ∈ 1 2 , 1 , {:[F(x)=2x","x in[0,(1)/(2))","],[=-2(x-1)","x in[(1)/(2),1]","]:}\begin{aligned} F(x) & =2 x, & & x \in\left[0, \frac{1}{2}\right), \\ & =-2(x-1), & & x \in\left[\frac{1}{2}, 1\right], \end{aligned}F(x)=2x,x∈[0,12),=−2(x−1),x∈[12,1],
which, obviously, does not belong to Z Z ZZZ.
By Lemmas 2 and 3, every h ∈ S h ∈ S h in Sh \in Sh∈S has a best approximation in W W WWW, namely, h − H h − H h-Hh-Hh−H, where H H HHH is a Lipschitz extension of h h hhh, such that H ∈ Z H ∈ Z H in ZH \in ZH∈Z. We can show that every best approximation is of this form (Lemma 1). It follows that W W WWW is chebyshevian for h ∈ S h ∈ S h in Sh \in Sh∈S if and only if h | Y h Y h|_(Y)\left.h\right|_{Y}h|Y has a unique Lipschitz extension in Z Z ZZZ. A class of such functions is given by
(3.11) S 1 = { h : h ∈ S , h ( x k ) = h ( 1 ) x k , k = 0 , 1 , 2 , … , n + 1 } . (3.11) S 1 = h : h ∈ S , h x k = h ( 1 ) x k , k = 0 , 1 , 2 , … , n + 1 . {:(3.11)S_(1)={h:h in S,h(x_(k))=h(1)x_(k),k=0,1,2,dots,n+1}.:}\begin{equation*} S_{1}=\left\{h: h \in S, h\left(x_{k}\right)=h(1) x_{k}, k=0,1,2, \ldots, n+1\right\} . \tag{3.11} \end{equation*}(3.11)S1={h:h∈S,h(xk)=h(1)xk,k=0,1,2,…,n+1}.
Theorem 6. W W WWW is S 1 S 1 S_(1)S_{1}S1-chebyshevian.
Proof. If h ∈ S 1 h ∈ S 1 h inS_(1)h \in S_{1}h∈S1, then the unique Lipschitz extension of h h hhh in Z Z ZZZ is H ( x ) = h ( 1 ) x H ( x ) = h ( 1 ) x H(x)=h(1)xH(x)=h(1) xH(x)=h(1)x. Therefore h ( x ) − h ( 1 ) x h ( x ) − h ( 1 ) x h(x)-h(1)xh(x)-h(1) xh(x)−h(1)x is the only element of best approximation for h h hhh in W W WWW.
Remark 2. J. Favard and recently de Boor [1] considered a problem analogous to that in Example (c).
(d) Finally, let X X XXX be a metric space of finite diameter (i.e., sup { d ( x , y ) sup { d ( x , y ) s u p{d(x,y)\sup \{d(x, y)sup{d(x,y) : x , y ∈ X } < ∞ x , y ∈ X } < ∞ x,y in X} < oox, y \in X\}<\inftyx,y∈X}<∞ ), x 0 x 0 x_(0)x_{0}x0 a fixed element in X X XXX, and Y Y YYY a subset of X X XXX such that x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y. Let f ∈ Lip 0 X f ∈ Lip 0 X f inLip_(0)Xf \in \operatorname{Lip}_{0} Xf∈Lip0X and let G ( f ) G ( f ) G(f)G(f)G(f) be the set of best approximation to f f fff from Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥. We can define on Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X the uniform norm ‖ ⋅ ‖ u : Lip 0 X → R ‖ ⋅ ‖ u : Lip 0 X → R ||*||_(u):Lip_(0)X rarr R\|\cdot\|_{u}: \operatorname{Lip}_{0} X \rightarrow R‖⋅‖u:Lip0X→R by
(3.12) ‖ f ‖ u = sup { | f ( x ) | : x ∈ X } , f ∈ Lip 0 X . (3.12) ‖ f ‖ u = sup { | f ( x ) | : x ∈ X } , f ∈ Lip 0 X . {:(3.12)||f||_(u)=s u p{|f(x)|:x in X}","quad f inLip_(0)X.:}\begin{equation*} \|f\|_{u}=\sup \{|f(x)|: x \in X\}, \quad f \in \operatorname{Lip}_{0} X . \tag{3.12} \end{equation*}(3.12)‖f‖u=sup{|f(x)|:x∈X},f∈Lip0X.
Obviously, the set G ( f ) ⊂ Y ⊥ G ( f ) ⊂ Y ⊥ G(f)subY^(_|_)G(f) \subset Y^{\perp}G(f)⊂Y⊥ is closed, convex, and bounded, for every f ∈ Lip 0 X f ∈ Lip 0 X f inLip_(0)Xf \in \operatorname{Lip}_{0} Xf∈Lip0X. We consider the following problems: Find g ∗ , g ∗ ∈ G ( f ) g ∗ , g ∗ ∈ G ( f ) g_(**),g^(**)in G(f)g_{*}, g^{*} \in G(f)g∗,g∗∈G(f) such that
(3.13) ‖ f − g ∗ ‖ u = inf { ‖ f − g ‖ u : g ∈ G ( f ) } (3.13) f − g ∗ u = inf ‖ f − g ‖ u : g ∈ G ( f ) {:(3.13)||f-g_(**)||_(u)=i n f{||f-g||_(u):g in G(f)}:}\begin{equation*} \left\|f-g_{*}\right\|_{u}=\inf \left\{\|f-g\|_{u}: g \in G(f)\right\} \tag{3.13} \end{equation*}(3.13)‖f−g∗‖u=inf{‖f−g‖u:g∈G(f)}
and
(3.14) ‖ f − g ∗ ‖ u = sup { ‖ f − g ‖ u : g ∈ G ( f ) } ; (3.14) f − g ∗ u = sup ‖ f − g ‖ u : g ∈ G ( f ) ; {:(3.14)||f-g^(**)||_(u)=s u p{||f-g||_(u):g in G(f)};:}\begin{equation*} \left\|f-g^{*}\right\|_{u}=\sup \left\{\|f-g\|_{u}: g \in G(f)\right\} ; \tag{3.14} \end{equation*}(3.14)‖f−g∗‖u=sup{‖f−g‖u:g∈G(f)};
i.e., find the nearest and the farthest point to f f fff in G ( f ) G ( f ) G(f)G(f)G(f), in the uniform norm.
Since every element in G ( f ) G ( f ) G(f)G(f)G(f) is of the form f − F f − F f-Ff-Ff−F, where F F FFF is a Lipschitz extension of f | X f X f|_(X)\left.f\right|_{X}f|X it follows that the problems (3.13) and (3.14) are equivalent to the following problems: Find two Lipschitz extensions F ∗ F ∗ F_(**)F_{*}F∗ and F ∗ F ∗ F^(**)F^{*}F∗ of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y such that
( ′ ) ‖ F ∗ ‖ u = inf { ‖ F ‖ u : F is a Lipschitz extension of f | Y } ( ′ ) F ∗ u = inf ‖ F ‖ u : F  is a Lipschitz extension of  f Y {:('")"||F_(**)||_(u)=i n f{||F||_(u):F" is a Lipschitz extension of "f|_(Y)}:}\begin{equation*} \left\|F_{*}\right\|_{u}=\inf \left\{\|F\|_{u}: F \text { is a Lipschitz extension of }\left.f\right|_{Y}\right\} \tag{$\prime$} \end{equation*}(′)‖F∗‖u=inf{‖F‖u:F is a Lipschitz extension of f|Y}
and
(3.14') ‖ F ∗ ‖ u = sup { ‖ F | u : F is a Lipschitz extension of f | Y } . (3.14') F ∗ u = sup ‖ F u : F  is a Lipschitz extension of  f Y . {:(3.14')||F^(**)||_(u)=s u p{||F|_(u):F" is a Lipschitz extension of "f|_(Y)}.:}\begin{equation*} \left\|F^{*}\right\|_{u}=\sup \left\{\|\left. F\right|_{u}: F \text { is a Lipschitz extension of }\left.f\right|_{Y}\right\} . \tag{3.14'} \end{equation*}(3.14')‖F∗‖u=sup{‖F|u:F is a Lipschitz extension of f|Y}.
THEOREM 6. The infimum (3.13) is attained for every g ∗ = f − F ∗ g ∗ = f − F ∗ g_(**)=f-F_(**)g_{*}=f-F_{*}g∗=f−F∗ such that F ∗ F ∗ F_(**)F_{*}F∗ is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y and ‖ F ∗ ‖ ‖ = ‖ f | Y ‖ ‖ F ∗ ‖ = ‖ f Y ‖ ‖ ||F_(**)||_(||)=_(||)f|_(Y)||_(||)\left\|F_{*}\right\|_{\|}=\left.{ }_{\|} f\right|_{Y} \|_{\|}‖F∗‖‖=‖f|Y‖‖. The set of these extensions is nonempty.
Proof. If F F FFF is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y then
| F ′ | u ⩾ sup { | F ( y ) | : y ∈ Y } = sup { | f ( y ) | : y ∈ Y } = | f | Y | a F ′ u ⩾ sup { | F ( y ) | : y ∈ Y } = sup | f ( y ) | : y ∈ Y = | f | Y a |F^(')|_(u) >= s u p{|F(y)|:y in Y}=s u p{|f(y)|_(:)y in Y}=|f|_(Y)|_(a)\left|F^{\prime}\right|_{u} \geqslant \sup \{|F(y)|: y \in Y\}=\sup \left\{|f(y)|_{:} y \in Y\right\}=\left.|f|_{Y}\right|_{a}|F′|u⩾sup{|F(y)|:y∈Y}=sup{|f(y)|:y∈Y}=|f|Y|a
Therefore, if ‖ F ∗ ‖ u = ‖ f ′ Y ‖ u F ∗ u = f ′ Y u ||F_(**)||_(u)=||f^(')_(Y)||_(u)\left\|F_{*}\right\|_{u}=\left\|f^{\prime}{ }_{Y}\right\|_{u}‖F∗‖u=‖f′Y‖u then inf { ‖ F ‖ u : F inf ‖ F ‖ u : F i n f{||F||_(u):F:}\inf \left\{\|F\|_{u}: F\right.inf{‖F‖u:F is a Lipschitz extension of f | Y } = { F ∗ | u , = ‖ f | Y | u f Y = F ∗ u , = ‖ f Y u {:f|_(Y)}={F_(**)|_(u,)=||f|_(Y)|_(u):}\left.\left.f\right|_{Y}\right\}=\left\{\left.F_{*}\right|_{u,}=\|\left.\left. f\right|_{Y}\right|_{u}\right.f|Y}={F∗|u,=‖f|Y|u. Now, if F F FFF is a Lipschitz extension of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y. we define a new Lipschitz function F ∗ F ∗ F_(**)F_{*}F∗ by
( { 3.15 } ) F Y ( x ) = ‖ f | X ‖ u b if F ( x ) > { f | Y ‖ u , = F ( x ) if − ‖ f | Y | u ⩽ F ( x ) ⩽ ‖ f | Y ! u , = − ‖ f | Y ‖ u b if F ( x ) < − | f | Y | u . ( { 3.15 } ) F Y ( x ) = f X u b  if  F ( x ) > f Y ‖ u , = F ( x )  if  − f Y u ⩽ F ( x ) ⩽ f Y ! u , = − f Y u b  if  F ( x ) < − f | Y | u . {:(({3.15})")"{:[F_(Y)(x),=||f|_(X)||_(ub),," if ",F(x), > {f|_(Y)||_(u),:}],[,=F(x),," if ",-||f|_(Y)|_(u) <= F(x) <= ||f|_(Y)!_(u)","],[,=-||f|_(Y)||_(ub),," if ",F(x) < -|_(f)|Y|_(u).]:}:}\begin{array}{rlrlrl} F_{Y}(x) & =\left\|\left.f\right|_{X}\right\|_{u b} & & \text { if } & F(x) & >\left\{\left.f\right|_{Y} \|_{u},\right. \\ & =F(x) & & \text { if } & -\left.\left\|\left.\left.f\right|_{Y}\right|_{u} \leqslant F(x) \leqslant\right\| f\right|_{Y}!_{u}, \tag{$\{3.15\}$}\\ & =-\left\|\left.f\right|_{Y}\right\|_{u b} & & \text { if } & F(x)<-\left.\right|_{f}|Y|_{u} . \end{array}({3.15})FY(x)=‖f|X‖ub if F(x)>{f|Y‖u,=F(x) if −‖f|Y|u⩽F(x)⩽‖f|Y!u,=−‖f|Y‖ub if F(x)<−|f|Y|u.
It is easy to see that F ∗ F ∗ F_(**)F_{*}F∗ is a Lipschitz extension of f ∣ Y f ∣ Y f_(∣Y)f_{\mid Y}f∣Y such that | F x | ∣ u = | ′ f ∣ u | u F x ∣ u = ′ f ∣ u u |F_(x)|_(∣u)=|^(')f_(∣u)|_(u)\left|F_{x}\right|_{\mid u}= \left.\left.\right|^{\prime} f_{\mid u}\right|_{u}|Fx|∣u=|′f∣u|u.
Theorem 7. The supremum (3.14) is attained for f − F 1 ∗ f − F 1 ∗ f-F_(1)^(**)f-F_{1}^{*}f−F1∗ or f − F 2 ∗ f − F 2 ∗ f-F_(2)^(**)f-F_{2}^{*}f−F2∗ or for both of these functions, where
( { 3.16 } ) F 1 ∗ ( x ) = inf { [ f ( y ) + | f | 1 ′ d ( x , y ) ] : y ∈ Y } . ( { 3.16 } ) F 1 ∗ ( x ) = inf f ( y ) + | f | 1 ′ d ( x , y ) : y ∈ Y . {:(({3.16})")"F_(1)^(**)(x)=i n f{[f(y)+|f|_(1)^(')d(x,y)]:y in Y}.:}\begin{equation*} F_{1}^{*}(x)=\inf \left\{\left[f(y)+|f|_{1}^{\prime} d(x, y)\right]: y \in Y\right\} . \tag{$\{3.16\}$} \end{equation*}({3.16})F1∗(x)=inf{[f(y)+|f|1′d(x,y)]:y∈Y}.
and
(3.17) F 2 ∗ ( x ) = sup { [ f ( y ) − ∣ f ‖ ‖ Y d ( x , y ) ] : y ∈ Y } . (3.17) F 2 ∗ ( x ) = sup f ( y ) − ∣ f ‖ ‖ Y d ( x , y ) : y ∈ Y . {:(3.17)F_(2)^(**)(x)=s u p{[f(y)-∣f_(||)||_(Y)d(x,y)]:y in Y}.:}\begin{equation*} F_{2}^{*}(x)=\sup \left\{\left[f(y)-\mid f_{\|} \|_{Y} d(x, y)\right]: y \in Y\right\} . \tag{3.17} \end{equation*}(3.17)F2∗(x)=sup{[f(y)−∣f‖‖Yd(x,y)]:y∈Y}.
Proof. By [2], F 1 ∗ F 1 ∗ F_(1)^(**)F_{1}{ }^{*}F1∗ and F 2 ∗ F 2 ∗ F_(2)^(**)F_{2}{ }^{*}F2∗ are Lipschitz extensions of f | r f r f|_(r)\left.f\right|_{r}f|r and obviously. for every Lipschitz extension F F FFF of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y we have
F 2 ∗ ( x ) ⩽ F ( x ) ⩽ F 1 ∗ ( x ) , x ∈ X ′ F 2 ∗ ( x ) ⩽ F ( x ) ⩽ F 1 ∗ ( x ) , x ∈ X ′ F_(2)^(**)(x) <= F(x) <= F_(1)^(**)(x),quad x inX^(')F_{2}^{*}(x) \leqslant F(x) \leqslant F_{1}^{*}(x), \quad x \in X^{\prime}F2∗(x)⩽F(x)⩽F1∗(x),x∈X′
From these inequalities, it follows that
‖ ; F ‖ u ⩽ max ( ‖ F 1 ∗ ‖ u , | F e ∗ | u ) ‖ ; F ‖ u ⩽ max F 1 ∗ u , F e ∗ u ||;F||_(u) <= max(||F_(1)**||_(u),|F_(e)**|_(u))\|; F\|_{u} \leqslant \max \left(\left\|F_{1} *\right\|_{u},\left|F_{\mathrm{e}} *\right|_{u}\right)‖;F‖u⩽max(‖F1∗‖u,|Fe∗|u)
Remark 3. Dunham [3] has considered a problem similar to the problem in (d) in the case when G ( f ) G ( f ) G(f)G(f)G(f) has the betweenness property (see [3] for definition). In (d) the set G ( f ) G ( f ) G(f)G(f)G(f), being convex, has the betweenness property. We found explicitly the nearest and the farthest points of f f fff in G ( f ) G ( f ) G(f)G(f)G(f).

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1977

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