Selections with values Bernstein polynomials associated to the extension operator for Lipschitz functions

Costica Mustata
(original), Approximation Theory, paper

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

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C. Mustăţa, Selections with values Bernstein polynomials associated to the extension operator for Lipschitz functions, Rev. Anal. Numér. Théor. Approx. 22 (1993) 2, 207-216 (MR # 96k: 41047).

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Revue d’Analyse Numer Theor. Approximation

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Romanian Academy

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https://ictp.acad.ro/jnaat/journal/article/view/1993-vol22-no2-art11/

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MR # 96k: 41047

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[3] Czipser, J. and L. Geher, Extension of Functions satisfying a Lipschitz Condition, Acta Math. Acad. Sci. Hungar 6(1955), 213-220.
[4] Deutsch, F., Li W. and S.H. Park, Tietze Extensions and Continuous Selections for Metric Projections, J. Approx. Theory 63 (1991), 55-88.
[5] Mc Shane, E.J., Extension of Range of Functions, Bull. Amer. Math. Soc. 40 (1934), 837-842.
[6] Mustat, C., Norm Preserving Extension of Starhaped Lipschitz Funcitons, Mathematica 19(42)2 (1977), 183-187.
[7]  Mustata C., Best Approximaiton and Uniquje Extension of Lipschitz Functions, J. Approx. Theory 19(1977), 222-230.
[8] Mustata C., M-ideals in Metric Spaces, “Babes-Bolyai” University, Fac. of Math. and Physics, Research Seminars, Seminar on Math. Analysis, Preprint nr.7 (1988), 65-74.
[9] Mustata, C., Selections Associated to Mc Shane’s Extension Theorem for Lipschitz Functions, Revue d’Analyse Numerique et de Theorie de l’Approximation, 21, 2 (1992), 135-145.

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1993-Mustata-Selections with values Bernstein polynomials-Jnaat

SELECTIONS WITH VALUES BERNSTEIN POLYNOMIALS ASSOCIATED TO THE EXTENSION OPERATOR FOR LIPSCHITZ FUNCTIONS

COSTICA MUSTATTA(Cluj-Napoca)

  1. Let ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a metric space and Y Y YYY a subset of X X XXX, containing at least two points. A function f : Y → R f : Y → R f:Y rarr Rf: Y \rightarrow Rf:Y→R is called Lipschitz if there exists a constant K Y ( f ) ⩾ 0 K Y ( f ) ⩾ 0 K_(Y)(f) >= 0K_{Y}(f) \geqslant 0KY(f)⩾0 such that
    for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X.
(1) | f ( x ) − f ( y ) | ⩽ K Y ( f ) ⋅ d ( x , y ) (1) | f ( x ) − f ( y ) | ⩽ K Y ( f ) ⋅ d ( x , y ) {:(1)|f(x)-f(y)| <= K_(Y)(f)*d(x","y):}\begin{equation*} |f(x)-f(y)| \leqslant K_{Y}(f) \cdot d(x, y) \tag{1} \end{equation*}(1)|f(x)−f(y)|⩽KY(f)⋅d(x,y)
Let
(2) Lip Y = { f ∣ f : Y → R , f is Lipschitz on Y } . (2)  Lip  Y = { f ∣ f : Y → R , f  is Lipschitz on  Y } . {:(2)" Lip "Y={f∣f:Y rarr R","f" is Lipschitz on "Y}.:}\begin{equation*} \text { Lip } Y=\{f \mid f: Y \rightarrow R, f \text { is Lipschitz on } Y\} . \tag{2} \end{equation*}(2) Lip Y={f∣f:Y→R,f is Lipschitz on Y}.
Equipped with the pointwise operations of addition and multiplication by real scalars, Lip Y Y YYY is a real linear space. The smallest constant K Y ( f ) K Y ( f ) K_(Y)(f)K_{Y}(f)KY(f) verifying (1) is denoted by ‖ f ‖ Y ‖ f ‖ Y ||f||_(Y)\|f\|_{Y}‖f‖Y and is called the Lipschitz norm of f f fff.
For Y = X Y = X Y=XY=XY=X one obtains the linear space Lip X X XXX and for F ∈ F ∈ F inF \inF∈ Lip X X XXX, ‖ F ′ ‖ X F ′ X ||F^(')||_(X)\left\|F^{\prime}\right\|_{X}‖F′‖X is the smallest Lipsehitz constant for F ′ F ′ F^(')F^{\prime}F′ (on X X XXX ).
Obviously that for F ∈ Lip X F ∈ Lip X F in Lip XF \in \operatorname{Lip} XF∈LipX and Y ⊂ X , F | Y ∈ Lip Y Y ⊂ X , F Y ∈ Lip Y Y sub X,F|_(Y)in Lip YY \subset X,\left.F\right|_{Y} \in \operatorname{Lip} YY⊂X,F|Y∈LipY and ‖ F | Y ‖ Y ⩽⩽ ‖ F ′ ‖ X F Y Y ⩽⩽ F ′ X ||F|_(Y)||_(Y)⩽⩽||F^(')||_(X)\left\|\left.F\right|_{Y}\right\|_{Y} \leqslant \leqslant\left\|F^{\prime}\right\|_{X}‖F|Y‖Y⩽⩽‖F′‖X.
Let Y ⊆ X Y ⊆ X Y sube XY \subseteq XY⊆X and f ∈ f ∈ f inf \inf∈ Lip Y Y YYY. A function F ∈ F ∈ F inF \inF∈ Lip X X XXX is called norm preserving Lipschitz extension of f f fff if
Let:
(3) F | Y = f and ‖ F ‖ X = ‖ f ‖ Y (3) F Y = f  and  ‖ F ‖ X = ‖ f ‖ Y {:(3)F|_(Y)=f" and "||F||_(X)=||f||_(Y):}\begin{equation*} \left.F\right|_{Y}=f \text { and }\|F\|_{X}=\|f\|_{Y} \tag{3} \end{equation*}(3)F|Y=f and ‖F‖X=‖f‖Y
(4) E ( f ) = { F ∈ Lip X : F | Y = f and ‖ F ‖ X = ‖ f ‖ Y } (4) E ( f ) = F ∈ Lip X : F Y = f  and  ‖ F ‖ X = ‖ f ‖ Y {:(4)E(f)={F in Lip X:F|_(Y)=f" and "||F||_(X)=||f||_(Y)}:}\begin{equation*} E(f)=\left\{F \in \operatorname{Lip} X:\left.F\right|_{Y}=f \text { and }\|F\|_{X}=\|f\|_{Y}\right\} \tag{4} \end{equation*}(4)E(f)={F∈LipX:F|Y=f and ‖F‖X=‖f‖Y}
be the set of all norm preserving Lipschitz extensions of f ∈ f ∈ f inf \inf∈ Lip Y Y YYY.
By a result of Mc Shane [5], every f ∈ f ∈ f inf \inf∈ Lip Y Y YYY has a least one norm preserving extension F ∈ Lip X F ∈ Lip X F in Lip XF \in \operatorname{Lip} XF∈LipX, i.e. E ( f ) ≠ ∅ E ( f ) ≠ ∅ E(f)!=O/E(f) \neq \emptysetE(f)≠∅ for every f ∈ Lip Y f ∈ Lip Y f in Lip Yf \in \operatorname{Lip} Yf∈LipY.
Since for every constant function c ∈ Lip Y , ‖ c ‖ Y = 0 c ∈ Lip Y , ‖ c ‖ Y = 0 c in Lip Y,||c||_(Y)=0c \in \operatorname{Lip} Y,\|c\|_{Y}=0c∈LipY,‖c‖Y=0, it follows that the "norm" ‖ ‖ Y ‖ ‖ Y ||quad||_(Y)\|\quad\|_{Y}‖‖Y is only a seminorm on Lip Y Y YYY. In order to obtain a genuine norm, fix a point x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y and let
(5) Lip 0 Y = { f ∈ Lip Y : f ( x 0 ) = 0 } (5) Lip 0 Y = f ∈ Lip Y : f x 0 = 0 {:(5)Lip_(0)Y={f in Lip Y:f(x_(0))=0}:}\begin{equation*} \operatorname{Lip}_{0} Y=\left\{f \in \operatorname{Lip} Y: f\left(x_{0}\right)=0\right\} \tag{5} \end{equation*}(5)Lip0Y={f∈LipY:f(x0)=0}
Then ‖ ‖ Y ‖ Y ||||_(Y):}\left\|\|_{Y}\right.‖‖Y is a norm on Lip 0 Y Lip 0 Y Lip_(0)Y\operatorname{Lip}_{0} YLip0Y and Lip 0 Y Lip 0 Y Lip_(0)Y\operatorname{Lip}_{0} YLip0Y is a Banach space with respect to this norm.
Then, by the above quoted result of Me Shane, one obtains :
Theorem 1. Let ( X , d X , d X,dX, dX,d ) be a metric space, x 0 x 0 x_(0)x_{0}x0 a fixed point in X X XXX and Y Y YYY a subset of X X XXX containing x 0 x 0 x_(0)x_{0}x0. Then for every f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y there exists F ∈ Lip 0 0 X F ∈ Lip 0 0 X F in Lip0_(0)XF \in \operatorname{Lip} 0_{0} XF∈Lip00X such that H ′ | Y = f H ′ Y = f H^(')|_(Y)=f\left.H^{\prime}\right|_{Y}=fH′|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 easily seen (see [3], [5]) that the following functions:
F 1 ( x ) = sup { f ( y ) − ‖ f ‖ Y ⋅ d ( x , y ) : y ∈ Y } , x ∈ X (6) and F 2 ( x ) = inf { f ( y ) − ‖ f ‖ Y ⋅ d ( x , y ) : y ∈ Y } , x ∈ X F 1 ( x ) = sup f ( y ) − ‖ f ‖ Y ⋅ d ( x , y ) : y ∈ Y , x ∈ X (6)  and  F 2 ( x ) = inf f ( y ) − ‖ f ‖ Y ⋅ d ( x , y ) : y ∈ Y , x ∈ X {:[F_(1)(x)=s u p{f(y)-||f||_(Y)*d(x,y):y in Y}","quad x in X],[(6)" and "],[F_(2)(x)=i n f{f(y)-||f||_(Y)*d(x,y):y in Y}","quad x in X]:}\begin{align*} & F_{1}(x)=\sup \left\{f(y)-\|f\|_{Y} \cdot d(x, y): y \in Y\right\}, \quad x \in X \\ & \text { and } \tag{6}\\ & F_{2}(x)=\inf \left\{f(y)-\|f\|_{Y} \cdot d(x, y): y \in Y\right\}, \quad x \in X \end{align*}F1(x)=sup{f(y)−‖f‖Y⋅d(x,y):y∈Y},x∈X(6) and F2(x)=inf{f(y)−‖f‖Y⋅d(x,y):y∈Y},x∈X
are two norm preserving Lipschitz extension of f ∈ Lip 0 I f ∈ Lip 0 I f inLip_(0)If \in \operatorname{Lip}_{0} If∈Lip0I. The set W ( f ) ⊂ Lip 0 X W ( f ) ⊂ Lip 0 X W(f)subLip_(0)XW(f) \subset \operatorname{Lip}_{0} XW(f)⊂Lip0X is nonempty, convex and bounded, such that the extension operator
(7) D : Lip 0 Y → 2 Lip 0 X (7) D : Lip 0 Y → 2 Lip 0 X {:(7)D:Lip_(0)Y rarr2^(Lip_(0)X):}\begin{equation*} D: \operatorname{Lip}_{0} Y \rightarrow 2^{\operatorname{Lip}_{0} X} \tag{7} \end{equation*}(7)D:Lip0Y→2Lip0X
is well defined and multivalued.
The problem of the existence of a selection of the extension operator E E EEE (i.e. a function e : Lip 0 Y → Lip 0 X e : Lip 0 Y → Lip 0 X e:Lip_(0)Y rarrLip_(0)Xe: \operatorname{Lip}_{0} Y \rightarrow \operatorname{Lip}_{0} Xe:Lip0Y→Lip0X such that e ( f ) ∈ E ( f ) e ( f ) ∈ E ( f ) e(f)in E(f)e(f) \in E(f)e(f)∈E(f), for all f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y ) which is linear and continuous was considered in [9].
In the particular case X = R X = R X=RX=RX=R and Y = [ a , b ] , x 0 ∈ Y Y = [ a , b ] , x 0 ∈ Y Y=[a,b],x_(0)in YY=[a, b], x_{0} \in YY=[a,b],x0∈Y, a linear and continuous selection e e eee of E E EEE can be given explicitly (see [9]).
2. Suppose X X XXX is a normed space and Y ⊂ X , x 0 ∈ X Y ⊂ X , x 0 ∈ X Y sub X,x_(0)in XY \subset X, x_{0} \in XY⊂X,x0∈X. In this case there exist functions in E ( f ) E ( f ) E(f)E(f)E(f) which preserve some properties of f f fff such as starshapedness or convexity (in these case Y Y YYY is supposed to be a starshaped, respectively a convex subset of X X XXX and x 0 = θ x 0 = θ x_(0)=thetax_{0}=\thetax0=θ ) (see [2], [6]). A natural question is to give explicit selections, with the values preserving some properties of the function f f fff and to study their linearity and continuity.
In the following we shall present an example of a homogeneous and continuous selection having values Bernstein polynomials.
Let X = [ 0 , 1 ] , Y = { 0 , 1 } , x 0 = 0 X = [ 0 , 1 ] , Y = { 0 , 1 } , x 0 = 0 X=[0,1],Y={0,1},x_(0)=0X=[0,1], Y=\{0,1\}, x_{0}=0X=[0,1],Y={0,1},x0=0 and d ( x , y ) = | x − y | d ( x , y ) = | x − y | d(x,y)=|x-y|d(x, y)=|x-y|d(x,y)=|x−y|. Tri this case
Lip 0 I = Lip 0 { 0 , 1 } = { f : { 0 , 1 } → R , f ( 0 ) = 0 } Lip 0 I = Lip 0 { 0 , 1 } = { f : { 0 , 1 } → R , f ( 0 ) = 0 } Lip_(0)I=Lip_(0){0,1}={f:{0,1}rarr R,f(0)=0}\operatorname{Lip}_{0} I=\operatorname{Lip}_{0}\{0,1\}=\{f:\{0,1\} \rightarrow R, f(0)=0\}Lip0I=Lip0{0,1}={f:{0,1}→R,f(0)=0}
and
(8) Lip 0 X = Lip 0 [ 0 , 1 ] = { F : [ 0 , 1 ] → R , F ( 0 ) = 0 (8) Lip 0 X = Lip 0 [ 0 , 1 ] = { F : [ 0 , 1 ] → R , F ( 0 ) = 0 {:(8)Lip_(0)X=Lip_(0)[0","1]={F:[0","1]rarr R","F(0)=0:}\begin{equation*} \operatorname{Lip}_{0} X=\operatorname{Lip}_{0}[0,1]=\{F:[0,1] \rightarrow R, F(0)=0 \tag{8} \end{equation*}(8)Lip0X=Lip0[0,1]={F:[0,1]→R,F(0)=0
F F FFF is Lipschitz on [ 0 , 1 ] } [ 0 , 1 ] } [0,1]}[0,1]\}[0,1]}
For f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y we have ‖ f ‖ Y = | f ( 1 ) | ‖ f ‖ Y = | f ( 1 ) | ||f||_(Y)=|f(1)|\|f\|_{Y}=|f(1)|‖f‖Y=|f(1)| and
(9) ‖ T ‖ X = sup { | F ( x ) − F ( y ) | / | x − y | : x , y ∈ [ 0 , 1 ] , x ≠ y } . (9) ‖ T ‖ X = sup { | F ( x ) − F ( y ) | / | x − y | : x , y ∈ [ 0 , 1 ] , x ≠ y } . {:(9)||T||_(X)=s u p{|F(x)-F(y)|//|x-y|:x","y in[0","1]","x!=y}.:}\begin{equation*} \|T\|_{X}=\sup \{|F(x)-F(y)| /|x-y|: x, y \in[0,1], x \neq y\} . \tag{9} \end{equation*}(9)‖T‖X=sup{|F(x)−F(y)|/|x−y|:x,y∈[0,1],x≠y}.
for F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X.
In this case E E EEE is single-valued, namely E ( f ) = { F } E ( f ) = { F } E(f)={F}E(f)=\{F\}E(f)={F} where F ( x ) == f ( 1 ) x , x ∈ [ 0 , 1 ] F ( x ) == f ( 1 ) x , x ∈ [ 0 , 1 ] F(x)==f(1)x,x in[0,1]F(x)= =f(1) x, x \in[0,1]F(x)==f(1)x,x∈[0,1] and the following result hold:
Theorem 2. The application E : Lip 0 Y → Lip 0 X E : Lip 0 Y → Lip 0 X E:Lip_(0)Y rarrLip_(0)XE: \operatorname{Lip}_{0} Y \rightarrow \operatorname{Lip}_{0} XE:Lip0Y→Lip0X, where E ( f ) = { F } E ( f ) = { F } E(f)={F}E(f)=\{F\}E(f)={F} with F ( x ) = f ( 1 ) x , x ∈ [ 0 , 1 ] F ( x ) = f ( 1 ) x , x ∈ [ 0 , 1 ] F(x)=f(1)x,x in[0,1]F(x)=f(1) x, x \in[0,1]F(x)=f(1)x,x∈[0,1] is linear and continuous.
Proof. The functions F 1 F 1 F_(1)F_{1}F1 and F 2 F 2 F_(2)F_{2}F2 given by (6) are equals and F 1 ( x ) == F 2 ( x ) = f ( 1 ) x , x ∈ [ 0 , 1 ] F 1 ( x ) == F 2 ( x ) = f ( 1 ) x , x ∈ [ 0 , 1 ] F_(1)(x)==F_(2)(x)=f(1)x,x in[0,1]F_{1}(x)= =F_{2}(x)=f(1) x, x \in[0,1]F1(x)==F2(x)=f(1)x,x∈[0,1]. In [9, Th. 4 and Corollary 5] it was proved that e ( f ) = ( 1 / 2 ) ( F 1 + F 2 ) e ( f ) = ( 1 / 2 ) F 1 + F 2 e(f)=(1//2)(F_(1)+F_(2))e(f)=(1 / 2)\left(F_{1}+F_{2}\right)e(f)=(1/2)(F1+F2) is a linear and continuous selection for E E EEE, so that e ( f ) = E ( f ) e ( f ) = E ( f ) e(f)=E(f)e(f)=E(f)e(f)=E(f) is linear and continuous.
Remark 1. It is well known (see [1]) that for 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] the Bernstein polynomial of degree n ( n ⩾ 1 ) n ( n ⩾ 1 ) n(n >= 1)n(n \geqslant 1)n(n⩾1) given by
(10) B n ( F ; x ) = ∑ k = 0 n ( n k ) F ( k n ) x k ( 1 − x ) n − k , x ∈ [ 0 , 1 ] (10) B n ( F ; x ) = ∑ k = 0 n   ( n k ) F k n x k ( 1 − x ) n − k , x ∈ [ 0 , 1 ] {:(10)B_(n)(F;x)=sum_(k=0)^(n)((n)/(k))F((k)/(n))x^(k)(1-x)^(n-k)","quad x in[0","1]:}\begin{equation*} B_{n}(F ; x)=\sum_{k=0}^{n}\binom{n}{k} F\left(\frac{k}{n}\right) x^{k}(1-x)^{n-k}, \quad x \in[0,1] \tag{10} \end{equation*}(10)Bn(F;x)=∑k=0n(nk)F(kn)xk(1−x)n−k,x∈[0,1]
is Lipschitz and ‖ B n ( F ; ) ‖ X ⩽ ‖ F ‖ X B n ( F ; ) X ⩽ ‖ F ‖ X ||B_(n)(F;)||_(X) <= ||F||_(X)\left\|B_{n}(F ;)\right\|_{X} \leqslant\|F\|_{X}‖Bn(F;)‖X⩽‖F‖X. Because B n ( F ; 0 ) = F ( 0 ) == f ( 0 ) B n ( F ; 0 ) = F ( 0 ) == f ( 0 ) B_(n)(F;0)=F(0)==f(0)B_{n}(F ; 0)=F(0)= =f(0)Bn(F;0)=F(0)==f(0) and B n ( l ′ ; 1 ) = F ( 1 ) = f ( 1 ) B n l ′ ; 1 = F ( 1 ) = f ( 1 ) B_(n)(l^(');1)=F(1)=f(1)B_{n}\left(l^{\prime} ; 1\right)=F(1)=f(1)Bn(l′;1)=F(1)=f(1) for every F ∈ E ( f ) F ∈ E ( f ) F in E(f)F \in E(f)F∈E(f), it follows that B n ( F ; ) ∈ M ( f ) B n ( F ; ) ∈ M ( f ) B_(n)(F;)inM(f)B_{n}(F 😉 \in \mathcal{M}(f)Bn(F;)∈M(f), for every f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y. In this case B n ( F ; x ) = f ( 1 ) x B n ( F ; x ) = f ( 1 ) x B_(n)(F;x)=f(1)xB_{n}(F ; x)=f(1) xBn(F;x)=f(1)x, for all n ∈ N , n ⩾ 1 n ∈ N , n ⩾ 1 n in N,n >= 1n \in N, n \geqslant 1n∈N,n⩾1, so that the application
(11) f ↦ B n ( F ; ⋅ ) = E ( f ) = B 1 ( F ; ⋅ ) (11) f ↦ B n ( F ; ⋅ ) = E ( f ) = B 1 ( F ; ⋅ ) {:(11)f|->B_(n)(F;*)=E(f)=B_(1)(F;*):}\begin{equation*} f \mapsto B_{n}(F ; \cdot)=E(f)=B_{1}(F ; \cdot) \tag{11} \end{equation*}(11)f↦Bn(F;⋅)=E(f)=B1(F;⋅)
is linear and continuous. Therefore, in this case, the extension operator E : Lip 0 Y → 2 Lip 0 x E : Lip 0 Y → 2 Lip  0 x E:Lip_(0)Y rarr2^("Lip "_(0)x)E: \operatorname{Lip}_{0} Y \rightarrow 2^{\text {Lip }_{0} x}E:Lip0Y→2Lip 0x, admits a linear and continuous selection with values Bernstein polynomials (of a fixed, but arbitrary, degree n n nnn ).
It we are looking for a Lipschitz extension with a greater Lipschitz constant α ‖ ‖ y α ‖ ‖ y alpha||quad||_(y)\alpha\|\quad\|_{y}α‖‖y, where α > 1 α > 1 alpha > 1\alpha>1α>1 is fixed, then the extension operator denoted by E α E α E_(alpha)E_{\alpha}Eα, will be multivalued.
From Theorem 2 one obtains the following corollary :
Corollary 1. For every f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y there exists F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X such that
(12) F | Y = f and ‖ F ‖ X = α , | f ( 1 ) | = α ‖ f ‖ Y (12) F Y = f  and  ‖ F ‖ X = α , | f ( 1 ) | = α ‖ f ‖ Y {:(12)F|_(Y)=f" and "||F||_(X)=alpha","|f(1)|=alpha||f||_(Y):}\begin{equation*} \left.F\right|_{Y}=f \text { and }\|F\|_{X}=\alpha,|f(1)|=\alpha\|f\|_{Y} \tag{12} \end{equation*}(12)F|Y=f and ‖F‖X=α,|f(1)|=α‖f‖Y
Proof. It is easy to verify that the functions
x ¯ 1 ( x ) = max { f ( y ) − α | f ( 1 ) | | x − y | : y ∈ { 0 , 1 } } (13) and I ¯ 2 ( x ) = min { f ( y ) + α | f ( 1 ) | | x − y | : y ∈ { 0 , 1 } } x ¯ 1 ( x ) = max { f ( y ) − α | f ( 1 ) | | x − y | : y ∈ { 0 , 1 } } (13)  and  I ¯ 2 ( x ) = min { f ( y ) + α | f ( 1 ) | | x − y | : y ∈ { 0 , 1 } } {:[ bar(x)_(1)(x)=max{f(y)-alpha|f(1)||x-y|:y in{0","1}}],[(13)" and "],[ bar(I)_(2)(x)=min{f(y)+alpha|f(1)||x-y|:y in{0","1}}]:}\begin{align*} & \bar{x}_{1}(x)=\max \{f(y)-\alpha|f(1)||x-y|: y \in\{0,1\}\} \\ & \text { and } \tag{13}\\ & \bar{I}_{2}(x)=\min \{f(y)+\alpha|f(1)||x-y|: y \in\{0,1\}\} \end{align*}x¯1(x)=max{f(y)−α|f(1)||x−y|:y∈{0,1}}(13) and I¯2(x)=min{f(y)+α|f(1)||x−y|:y∈{0,1}}
x ∈ [ 0 , 1 ] x ∈ [ 0 , 1 ] x in[0,1]x \in[0,1]x∈[0,1], have the properties
and
F ¯ 1 ( 0 ) = F ¯ 2 ( 0 ) = f ( 0 ) = 0 , F ¯ 1 ( 1 ) = F ¯ 2 ( 1 ) = f ( 1 ) F ¯ 1 ( 0 ) = F ¯ 2 ( 0 ) = f ( 0 ) = 0 , F ¯ 1 ( 1 ) = F ¯ 2 ( 1 ) = f ( 1 ) bar(F)_(1)(0)= bar(F)_(2)(0)=f(0)=0,quad bar(F)_(1)(1)= bar(F)_(2)(1)=f(1)\bar{F}_{1}(0)=\bar{F}_{2}(0)=f(0)=0, \quad \bar{F}_{1}(1)=\bar{F}_{2}(1)=f(1)F¯1(0)=F¯2(0)=f(0)=0,F¯1(1)=F¯2(1)=f(1)
Let
‖ F ¯ 1 ‖ X = α | f ( 1 ) | = ‖ H ¯ 2 ‖ X F ¯ 1 X = α | f ( 1 ) | = H ¯ 2 X || bar(F)_(1)||_(X)=alpha|f(1)|=|| bar(H)_(2)||_(X)\left\|\bar{F}_{1}\right\|_{X}=\alpha|f(1)|=\left\|\bar{H}_{2}\right\|_{X}‖F¯1‖X=α|f(1)|=‖H¯2‖X
(14) E ( f ) = { F ¯ ∈ Lip p 0 X : H ′ ― | Y = f , ‖ F ¯ | X ⩽ α | f ( 1 ) | } (14) E ( f ) = F ¯ ∈ Lip p 0 X : H ′ ¯ Y = f , ‖ F ¯ X ⩽ α | f ( 1 ) | {:(14)E(f)={( bar(F))in Lipp_(0)X: bar(H^('))|_(Y)=f,||( bar(F))|_(X) <= alpha|f(1)|}:}\begin{equation*} E(f)=\left\{\bar{F} \in \operatorname{Lip} p_{0} X:\left.\overline{H^{\prime}}\right|_{Y}=f, \|\left.\bar{F}\right|_{X} \leqslant \alpha|f(1)|\right\} \tag{14} \end{equation*}(14)E(f)={F¯∈Lipp0X:H′―|Y=f,‖F¯|X⩽α|f(1)|}
denote the set of the Lipschitz extensions of the function f f fff which preserve the norm α ‖ f ‖ Y α ‖ f ‖ Y alpha||f||_(Y)\alpha\|f\|_{Y}α‖f‖Y. Then F ¯ 1 , F ¯ 2 ∈ E α ( f ) F ¯ 1 , F ¯ 2 ∈ E α ( f ) bar(F)_(1), bar(F)_(2)inE_(alpha)(f)\bar{F}_{1}, \bar{F}_{2} \in E_{\alpha}(f)F¯1,F¯2∈Eα(f) and F ¯ 1 ( x ) ≠ F ¯ 2 ( x ) F ¯ 1 ( x ) ≠ F ¯ 2 ( x ) bar(F)_(1)(x)!= bar(F)_(2)(x)\bar{F}_{1}(x) \neq \bar{F}_{2}(x)F¯1(x)≠F¯2(x) for all x ∈ ( 0 , 1 ) x ∈ ( 0 , 1 ) x in(0,1)x \in(0,1)x∈(0,1), so that the extension operator
H α : Lip 0 Y → 2 Lip 0 X H α : Lip 0 Y → 2 Lip 0 X H_(alpha):Lip_(0)Y rarr2^(Lip_(0)X)H_{\alpha}: \operatorname{Lip}_{0} Y \rightarrow 2^{\operatorname{Lip}_{0} X}Hα:Lip0Y→2Lip0X
is well defined and multivalued.
Concerning this operator E α E α E_(alpha)E_{\alpha}Eα one can prove the following theorem:
Theorem 3. a) The operator E α E α E_(alpha)E_{\alpha}Eα admits a homogeneous and continuous selection;
b) For every n ∈ N , n ⩾ 1 n ∈ N , n ⩾ 1 n in N,n >= 1n \in N, n \geqslant 1n∈N,n⩾1, the operator E α E α E_(alpha)E_{\alpha}Eα admits a homogeneous and continuous selection with values Bernstein polynomials of degree n n nnn.
Proof. a) Consider the following two selections e 1 , e 2 e 1 , e 2 e_(1),e_(2)e_{1}, e_{2}e1,e2 defined by
(15) e 1 ( f ) = F ¯ 1 and e 2 ( f ) = F ¯ 2 , f ∈ Lip p 0 I , (15) e 1 ( f ) = F ¯ 1  and  e 2 ( f ) = F ¯ 2 , f ∈ Lip p 0 I , {:(15)e_(1)(f)= bar(F)_(1)" and "e_(2)(f)= bar(F)_(2)","f in Lipp_(0)I",":}\begin{equation*} e_{1}(f)=\bar{F}_{1} \text { and } e_{2}(f)=\bar{F}_{2}, f \in \operatorname{Lip} p_{0} I, \tag{15} \end{equation*}(15)e1(f)=F¯1 and e2(f)=F¯2,f∈Lipp0I,
where
B ― → 1 ( x ) = max { − α | f ( 1 ) | x ; f ( 1 ) − α | f ( 1 ) | ( 1 − x } , a ∈ [ 0 , 1 ] B _ → 1 ( x ) = max { − α | f ( 1 ) | x ; f ( 1 ) − α | f ( 1 ) | ( 1 − x } , a ∈ [ 0 , 1 ] vec(B_)_(1)(x)=max{-alpha|f(1)|x;f(1)-alpha|f(1)|(1-x},a in[0,1]\overrightarrow{\underline{B}}_{1}(x)=\max \{-\alpha|f(1)| x ; f(1)-\alpha|f(1)|(1-x\}, a \in[0,1]B―→1(x)=max{−α|f(1)|x;f(1)−α|f(1)|(1−x},a∈[0,1]
and
(16) E ¯ 2 ( x ) = min { a | f ( 1 ) | x ; f ( 1 ) + α | f ( 1 ) | ( 1 − x ) } , x ∈ [ 0 , 1 ] . (16) E ¯ 2 ( x ) = min { a | f ( 1 ) | x ; f ( 1 ) + α | f ( 1 ) | ( 1 − x ) } , x ∈ [ 0 , 1 ] . {:(16) bar(E)_(2)(x)=min{a|f(1)|x;f(1)+alpha|f(1)|(1-x)}","x in[0","1].:}\begin{equation*} \bar{E}_{2}(x)=\min \{a|f(1)| x ; f(1)+\alpha|f(1)|(1-x)\}, x \in[0,1] . \tag{16} \end{equation*}(16)E¯2(x)=min{a|f(1)|x;f(1)+α|f(1)|(1−x)},x∈[0,1].
Then, for λ ⩾ 0 , e 1 ( λ f ) = λ ⋅ e 1 ( f ) λ ⩾ 0 , e 1 ( λ f ) = λ ⋅ e 1 ( f ) lambda >= 0,e_(1)(lambda f)=lambda*e_(1)(f)\lambda \geqslant 0, e_{1}(\lambda f)=\lambda \cdot e_{1}(f)λ⩾0,e1(λf)=λ⋅e1(f) and e 2 ( λ f ) = λ ⋅ e 2 ( f ) e 2 ( λ f ) = λ ⋅ e 2 ( f ) e_(2)(lambda f)=lambda*e_(2)(f)e_{2}(\lambda f)=\lambda \cdot e_{2}(f)e2(λf)=λ⋅e2(f). By the definition of e 1 e 1 e_(1)e_{1}e1 and e 2 , e 1 ( f ) = − e 2 ( − f ) e 2 , e 1 ( f ) = − e 2 ( − f ) e_(2),e_(1)(f)=-e_(2)(-f)e_{2}, e_{1}(f)=-e_{2}(-f)e2,e1(f)=−e2(−f), implying that the selection
(17)
e ( f ) = ( 1 / 2 ) ( e 1 ( f ) + e 2 ( f ) . ) e ( f ) = ( 1 / 2 ) e 1 ( f ) + e 2 ( f ) . e(f)=(1//2)(e_(1)(f)+e_(2)(f).)e(f)=(1 / 2)\left(e_{1}(f)+e_{2}(f) .\right)e(f)=(1/2)(e1(f)+e2(f).)
is homogeneous, i.e.
e ( λ f ) = λ e ( f ) , λ ∈ R , f ∈ Lip 0 Y e ( λ f ) = λ e ( f ) , λ ∈ R , f ∈ Lip 0 Y e(lambda f)=lambda e(f),quad lambda in R,quad f inLip_(0)Ye(\lambda f)=\lambda e(f), \quad \lambda \in R, \quad f \in \operatorname{Lip}_{0} Ye(λf)=λe(f),λ∈R,f∈Lip0Y
Now, we show that e 1 , e 2 e 1 , e 2 e_(1),e_(2)e_{1}, e_{2}e1,e2 are continuous selections which will imply the continuity of e e eee, too.
Let ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0 and 0 < δ < ε 0 < δ < ε 0 < delta < epsi0<\delta<\varepsilon0<δ<ε. We shall show that for f , g ∈ Lip p 0 Y f , g ∈ Lip p 0 Y f,g in Lipp_(0)Yf, g \in \operatorname{Lip} p_{0} Yf,g∈Lipp0Y, a | f ( 1 ) − g ( 1 ) | < δ a | f ( 1 ) − g ( 1 ) | < δ a|f(1)-g(1)| < deltaa|f(1)-g(1)|<\deltaa|f(1)−g(1)|<δ implies ‖ F ¯ 1 − G ¯ 1 ‖ X < ε F ¯ 1 − G ¯ 1 X < ε || bar(F)_(1)- bar(G)_(1)||_(X) < epsi\left\|\bar{F}_{1}-\bar{G}_{1}\right\|_{X}<\varepsilon‖F¯1−G¯1‖X<ε where F ¯ 1 F ¯ 1 bar(F)_(1)\bar{F}_{1}F¯1 is defined by (16) and
G ¯ 1 ( x ) = max { − α | g ( 1 ) | x ; g ( 1 ) − α | g ( 1 ) | ( 1 − x ) } , x ∈ [ 0 , 1 ] . G ¯ 1 ( x ) = max { − α | g ( 1 ) | x ; g ( 1 ) − α | g ( 1 ) | ( 1 − x ) } , x ∈ [ 0 , 1 ] . bar(G)_(1)(x)=max{-alpha|g(1)|x;g(1)-alpha|g(1)|(1-x)},x in[0,1].\bar{G}_{1}(x)=\max \{-\alpha|g(1)| x ; g(1)-\alpha|g(1)|(1-x)\}, x \in[0,1] .G¯1(x)=max{−α|g(1)|x;g(1)−α|g(1)|(1−x)},x∈[0,1].
We have to consider the following cases:
1 ∘ f ( 1 ) > r , g ( 1 ) > 0 1 ∘ f ( 1 ) > r , g ( 1 ) > 0 1^(@)f(1) > r,g(1) > 01^{\circ} f(1)>r, g(1)>01∘f(1)>r,g(1)>0.
In this case
F ¯ 1 ( x ) − G ¯ 1 ( x ) = α [ g ( 1 ) − f ( 1 ) ] x , for x ∈ [ 0 , α − 1 2 α ] = f ( 1 ) − g ( 1 ) − α [ f ( 1 ) − g ( 1 ) ] ( 1 − x ) , for x ∈ ( α − 1 2 α , 1 ] F ¯ 1 ( x ) − G ¯ 1 ( x ) = α [ g ( 1 ) − f ( 1 ) ] x ,  for  x ∈ 0 , α − 1 2 α = f ( 1 ) − g ( 1 ) − α [ f ( 1 ) − g ( 1 ) ] ( 1 − x ) ,  for  x ∈ α − 1 2 α , 1 {:[ bar(F)_(1)(x)- bar(G)_(1)(x)=alpha[g(1)-f(1)]x","" for "x in[0,(alpha-1)/(2alpha)]],[=f(1)-g(1)-alpha[f(1)-g(1)](1-x)","" for "x in((alpha-1)/(2alpha),1]]:}\begin{aligned} \bar{F}_{1}(x)-\bar{G}_{1}(x) & =\alpha[g(1)-f(1)] x, \text { for } x \in\left[0, \frac{\alpha-1}{2 \alpha}\right] \\ & =f(1)-g(1)-\alpha[f(1)-g(1)](1-x), \text { for } x \in\left(\frac{\alpha-1}{2 \alpha}, 1\right] \end{aligned}F¯1(x)−G¯1(x)=α[g(1)−f(1)]x, for x∈[0,α−12α]=f(1)−g(1)−α[f(1)−g(1)](1−x), for x∈(α−12α,1]
implying ‖ F ¯ 1 − G ¯ 1 ‖ x = α | f ( 1 ) − g ( 1 ) | < δ < ε F ¯ 1 − G ¯ 1 x = α | f ( 1 ) − g ( 1 ) | < δ < ε || bar(F)_(1)- bar(G)_(1)||_(x)=alpha|f(1)-g(1)| < delta < epsi\left\|\bar{F}_{1}-\bar{G}_{1}\right\|_{x}=\alpha|f(1)-g(1)|<\delta<\varepsilon‖F¯1−G¯1‖x=α|f(1)−g(1)|<δ<ε.
2 ∘ f ( 1 ) < 0 , g ( 1 ) < 0 2 ∘ f ( 1 ) < 0 , g ( 1 ) < 0 2^(@)f(1) < 0,g(1) < 02^{\circ} f(1)<0, g(1)<02∘f(1)<0,g(1)<0.
In this case
F → 1 ( x ) − G ¯ 1 ( x ) = α [ | f ( 1 ) | − | g ( 1 ) | ] x , for x ∈ [ 0 , α + 1 2 α ] = = f ( 1 ) − g ( 1 ) − α [ | f ( 1 ) | − | g ( 1 ) | ] ( 1 − x ) , for x ∈ ( α + 1 2 α , 1 ] F → 1 ( x ) − G ¯ 1 ( x ) = α [ | f ( 1 ) | − | g ( 1 ) | ] x ,  for  x ∈ 0 , α + 1 2 α = = f ( 1 ) − g ( 1 ) − α [ | f ( 1 ) | − | g ( 1 ) | ] ( 1 − x ) ,  for  x ∈ α + 1 2 α , 1 {:[ vec(F)_(1)(x)- bar(G)_(1)(x)=alpha[|f(1)|-|g(1)|]x","" for "x in[0,(alpha+1)/(2alpha)]=],[=f(1)-g(1)-alpha[|f(1)|-|g(1)|](1-x)","" for "x in((alpha+1)/(2alpha),1]]:}\begin{aligned} \vec{F}_{1}(x)-\bar{G}_{1}(x) & =\alpha[|f(1)|-|g(1)|] x, \text { for } x \in\left[0, \frac{\alpha+1}{2 \alpha}\right]= \\ & =f(1)-g(1)-\alpha[|f(1)|-|g(1)|](1-x), \text { for } x \in\left(\frac{\alpha+1}{2 \alpha}, 1\right] \end{aligned}F→1(x)−G¯1(x)=α[|f(1)|−|g(1)|]x, for x∈[0,α+12α]==f(1)−g(1)−α[|f(1)|−|g(1)|](1−x), for x∈(α+12α,1]
implying ‖ P ¯ 1 − G ¯ 1 ‖ X = α | | f ( 1 ) | − | g ( 1 ) | | ⩽ α | f ( 1 ) − g ( 1 ) | < δ < ε P ¯ 1 − G ¯ 1 X = α | | f ( 1 ) | − | g ( 1 ) | | ⩽ α | f ( 1 ) − g ( 1 ) | < δ < ε || bar(P)_(1)- bar(G)_(1)||_(X)=alpha||f(1)|-|g(1)|| <= alpha|f(1)-g(1)| < delta < epsi\left\|\bar{P}_{1}-\bar{G}_{1}\right\|_{X}=\alpha| | f(1)|-|g(1)|| \leqslant \alpha|f(1)-g(1)|<\delta<\varepsilon‖P¯1−G¯1‖X=α||f(1)|−|g(1)||⩽α|f(1)−g(1)|<δ<ε. 3 ∘ f ( 1 ) > 0 , g ( 1 ) < 0 ( 3 ∘ f ( 1 ) > 0 , g ( 1 ) < 0 ( 3^(@)f(1) > 0,g(1) < 0(3^{\circ} f(1)>0, g(1)<0(3∘f(1)>0,g(1)<0( or f ( 1 ) < 0 f ( 1 ) < 0 f(1) < 0f(1)<0f(1)<0 and g ( 1 ) > 0 ) g ( 1 ) > 0 ) g(1) > 0)g(1)>0)g(1)>0).
In this case
F ¯ 1 ( x ) − G ¯ 1 ( x ) = α [ | g ( 1 ) | − | f ( 1 ) | ] x , for x ∈ [ 0 , α − 1 2 α ] = f ( 1 ) − α f ( 1 ) + α [ | g ( 1 ) | − f ( 1 ) ] ⋅ x , for x ∈ ( α − 1 2 α , α + 1 2 α ] = f ( 1 ) − g ( 1 ) + α [ | g ( 1 ) | − f ( 1 ) ] + α [ f ( 1 ) − | g ( 1 ) | ] x for x ∈ ( α + 1 2 α , 1 ] F ¯ 1 ( x ) − G ¯ 1 ( x ) = α [ | g ( 1 ) | − | f ( 1 ) | ] x ,  for  x ∈ 0 , α − 1 2 α = f ( 1 ) − α f ( 1 ) + α [ | g ( 1 ) | − f ( 1 ) ] ⋅ x ,  for  x ∈ α − 1 2 α , α + 1 2 α = f ( 1 ) − g ( 1 ) + α [ | g ( 1 ) | − f ( 1 ) ] + α [ f ( 1 ) − | g ( 1 ) | ] x  for  x ∈ α + 1 2 α , 1 {:[ bar(F)_(1)(x)- bar(G)_(1)(x)=alpha[|g(1)|-|f(1)|]x","" for "x in[0,(alpha-1)/(2alpha)]],[=f(1)-alpha f(1)+alpha[|g(1)|-f(1)]*x","" for "x in((alpha-1)/(2alpha),(alpha+1)/(2alpha)]],[=f(1)-g(1)+alpha[|g(1)|-f(1)]+alpha[f(1)-|g(1)|]x],[" for "x in((alpha+1)/(2alpha),1]]:}\begin{aligned} & \bar{F}_{1}(x)-\bar{G}_{1}(x)=\alpha[|g(1)|-|f(1)|] x, \text { for } x \in\left[0, \frac{\alpha-1}{2 \alpha}\right] \\ &=f(1)-\alpha f(1)+\alpha[|g(1)|-f(1)] \cdot x, \text { for } x \in\left(\frac{\alpha-1}{2 \alpha}, \frac{\alpha+1}{2 \alpha}\right] \\ &=f(1)-g(1)+\alpha[|g(1)|-f(1)]+\alpha[f(1)-|g(1)|] x \\ & \text { for } x \in\left(\frac{\alpha+1}{2 \alpha}, 1\right] \end{aligned}F¯1(x)−G¯1(x)=α[|g(1)|−|f(1)|]x, for x∈[0,α−12α]=f(1)−αf(1)+α[|g(1)|−f(1)]⋅x, for x∈(α−12α,α+12α]=f(1)−g(1)+α[|g(1)|−f(1)]+α[f(1)−|g(1)|]x for x∈(α+12α,1]
implying ‖ h ¯ 1 − G ¯ 1 ‖ x = α | f ′ ( 1 ) − | g ( 1 ) | | ⩽ α | f ( 1 ) − g ( 1 ) | < δ < ε 4 ∘ f ( 1 ) = 0 h ¯ 1 − G ¯ 1 x = α f ′ ( 1 ) − | g ( 1 ) | ⩽ α | f ( 1 ) − g ( 1 ) | < δ < ε 4 ∘ f ( 1 ) = 0 || bar(h)_(1)- bar(G)_(1)||_(x)=alpha|f^(')(1)-|g(1)|| <= alpha|f(1)-g(1)| < delta < epsi4^(@)f(1)=0\left\|\bar{h}_{1}-\bar{G}_{1}\right\|_{x}=\alpha\left|f^{\prime}(1)-|g(1)|\right| \leqslant \alpha|f(1)-g(1)|<\delta<\varepsilon 4^{\circ} f(1)=0‖h¯1−G¯1‖x=α|f′(1)−|g(1)||⩽α|f(1)−g(1)|<δ<ε4∘f(1)=0 and g ( 1 ) ≠ 0 ( g ( 1 ) ≠ 0 ( g(1)!=0(g(1) \neq 0(g(1)≠0( or f ( 1 ) ≠ 0 f ( 1 ) ≠ 0 f(1)!=0f(1) \neq 0f(1)≠0 and g ( 1 ) = 0 ) g ( 1 ) = 0 ) g(1)=0)g(1)=0)g(1)=0)
In this case F ¯ 1 ( x ) = 0 , x ∈ [ 0 , 1 ] F ¯ 1 ( x ) = 0 , x ∈ [ 0 , 1 ] bar(F)_(1)(x)=0,x in[0,1]\bar{F}_{1}(x)=0, x \in[0,1]F¯1(x)=0,x∈[0,1] and ‖ F ¯ 1 − G ¯ 1 ‖ X = ‖ G ¯ 1 ‖ X == a | g ( 1 ) | < δ < ε F ¯ 1 − G ¯ 1 X = G ¯ 1 X == a | g ( 1 ) | < δ < ε || bar(F)_(1)- bar(G)_(1)||_(X)=|| bar(G)_(1)||_(X)==a|g(1)| < delta < epsi\left\|\bar{F}_{1}-\bar{G}_{1}\right\|_{X}=\left\|\bar{G}_{1}\right\|_{X}= =a|g(1)|<\delta<\varepsilon‖F¯1−G¯1‖X=‖G¯1‖X==a|g(1)|<δ<ε.
It follows that e 1 e 1 e_(1)e_{1}e1 is a continuous selections. In a similar way one can show the continuty of the selection, e 2 e 2 e_(2)e_{2}e2, implying the continuity of the selection e e eee.
b) Let n ∈ N , n ⩾ 1 n ∈ N , n ⩾ 1 n in N,n >= 1n \in N, n \geqslant 1n∈N,n⩾1, be a fixed and for f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \mathrm{Lip}_{0} Yf∈Lip0Y let B n ( e ( f ) B n ( e ( f ) B_(n)(e(f)B_{n}(e(f)Bn(e(f); .) be the Bernstein operator asociated to the function e ( f ) e ( f ) e(f)e(f)e(f) :
(18) B n ( e ( f ) ; x ) = ∑ k = 0 n ( n k ) ⋅ e ( f ) ( k n ) ⋅ x k ( 1 − x ) n − k , x ∈ [ 0 , 1 ] (18) B n ( e ( f ) ; x ) = ∑ k = 0 n   ( n k ) ⋅ e ( f ) k n ⋅ x k ( 1 − x ) n − k , x ∈ [ 0 , 1 ] {:(18)B_(n)(e(f);x)=sum_(k=0)^(n)((n)/(k))*e(f)((k)/(n))*x^(k)(1-x)^(n-k)","quad x in[0","1]:}\begin{equation*} B_{n}(e(f) ; x)=\sum_{k=0}^{n}\binom{n}{k} \cdot e(f)\left(\frac{k}{n}\right) \cdot x^{k}(1-x)^{n-k}, \quad x \in[0,1] \tag{18} \end{equation*}(18)Bn(e(f);x)=∑k=0n(nk)⋅e(f)(kn)⋅xk(1−x)n−k,x∈[0,1]
By the result from [1] it follows
‖ B n ( e ( f ) ; . ) ‖ x ⩽ ‖ e ( f ) ‖ x = a | f ( 1 ) | B n ( e ( f ) ; . ) x ⩽ ‖ e ( f ) ‖ x = a | f ( 1 ) | ||B_(n)(e(f);.)||_(x) <= ||e(f)||_(x)=a|f(1)|\left\|B_{n}(e(f) ; .)\right\|_{x} \leqslant\|e(f)\|_{x}=a|f(1)|‖Bn(e(f);.)‖x⩽‖e(f)‖x=a|f(1)|
Since B n ( e ( f ) ; 0 ) = e ( f ) ( 0 ) = f ( 0 ) = 0 B n ( e ( f ) ; 0 ) = e ( f ) ( 0 ) = f ( 0 ) = 0 B_(n)(e(f);0)=e(f)(0)=f(0)=0B_{n}(e(f) ; 0)=e(f)(0)=f(0)=0Bn(e(f);0)=e(f)(0)=f(0)=0 and B n ( e ( f ) ; 1 ) = e ( f ) ( 1 ) = f ( 1 ) B n ( e ( f ) ; 1 ) = e ( f ) ( 1 ) = f ( 1 ) B_(n)(e(f);1)=e(f)(1)=f(1)B_{n}(e(f) ; 1)=e(f)(1)=f(1)Bn(e(f);1)=e(f)(1)=f(1), it follows that B n ( e ( f ) ; . ) ∈ E α ( f ) B n ( e ( f ) ; . ) ∈ E α ( f ) B_(n)(e(f);.)inE_(alpha)(f)B_{n}(e(f) ;.) \in E_{\alpha}(f)Bn(e(f);.)∈Eα(f).
Define the selection
b n : Iip 0 { 0 , 1 } → Lip 0 [ 0 , 1 ] b n : Iip 0 { 0 , 1 } → Lip 0 [ 0 , 1 ] b_(n):Iip_(0){0,1}rarrLip_(0)[0,1]b_{n}: \operatorname{Iip}_{0}\{0,1\} \rightarrow \operatorname{Lip}_{0}[0,1]bn:Iip0{0,1}→Lip0[0,1]
by
(19) b n ( f ) = B n ( e ( f ) ; . ) (19) b n ( f ) = B n ( e ( f ) ; . ) {:(19)b_(n)(f)=B_(n)(e(f);.):}\begin{equation*} b_{n}(f)=B_{n}(e(f) ; .) \tag{19} \end{equation*}(19)bn(f)=Bn(e(f);.)
As the Bernstein operator is linear it follows that for λ ∈ R λ ∈ R lambda inR\lambda \in \mathbb{R}λ∈R, b n ( λ f ) = B n ( e ( λ f ) ; ) = . B n ( λ e ( f ) ; ) = . λ B n ( e ( f ) ; ) = . λ b n ( f ) b n ( λ f ) = B n ( e ( λ f ) ; ) = . B n ( λ e ( f ) ; ) = . λ B n ( e ( f ) ; ) = . λ b n ( f ) b_(n)(lambda f)=B_(n)(e(lambda f);)=.B_(n)(lambda e(f);)=.lambdaB_(n)(e(f);)=.lambdab_(n)(f)b_{n}(\lambda f)=B_{n}(e(\lambda f) ;)=.B_{n}(\lambda e(f) ;)=.\lambda B_{n}(e(f) ;)=.\lambda b_{n}(f)bn(λf)=Bn(e(λf);)=.Bn(λe(f);)=.λBn(e(f);)=.λbn(f), showing that b n b n b_(n)b_{n}bn is a homogeneous selection.
If f , g ∈ Lip 0 { 0 , 1 } f , g ∈ Lip 0 { 0 , 1 } f,g inLip_(0){0,1}f, g \in \operatorname{Lip}_{0}\{0,1\}f,g∈Lip0{0,1} are such that α | f ( 1 ) − g ( 1 ) | < δ < ε α | f ( 1 ) − g ( 1 ) | < δ < ε alpha|f(1)-g(1)| < delta < epsi\alpha|f(1)-g(1)|<\delta<\varepsilonα|f(1)−g(1)|<δ<ε then ‖ H ¯ 1 − G ¯ 1 ‖ x < ε H ¯ 1 − G ¯ 1 x < ε || bar(H)_(1)- bar(G)_(1)||_(x) < epsi\left\|\bar{H}_{1}-\bar{G}_{1}\right\|_{x}<\varepsilon‖H¯1−G¯1‖x<ε and ‖ F ¯ 2 − G ¯ 2 ‖ x < ε F ¯ 2 − G ¯ 2 x < ε || bar(F)_(2)- bar(G)_(2)||_(x) < epsi\left\|\bar{F}_{2}-\bar{G}_{2}\right\|_{x}<\varepsilon‖F¯2−G¯2‖x<ε, so that
‖ b 18 ( f ) − b n ( g ) ‖ X = ‖ B n ( e ( f ) ; . ) − B n ( e ( g ) ; . ) ‖ X = b 18 ( f ) − b n ( g ) X = B n ( e ( f ) ; . ) − B n ( e ( g ) ; . ) X = ||b_(18)(f)-b_(n)(g)||_(X)=||B_(n)(e(f);.)-B_(n)(e(g);.)||_(X)=\left\|b_{18}(f)-b_{n}(g)\right\|_{X}=\left\|B_{n}(e(f) ; .)-B_{n}(e(g) ; .)\right\|_{X}=‖b18(f)−bn(g)‖X=‖Bn(e(f);.)−Bn(e(g);.)‖X=
= ( 1 / 2 ) ‖ B n ( F 1 − G 1 ) + B n ( F 2 − G 2 ) ‖ X ⩽ ⩽ ( 1 / 2 ) [ ‖ B n ( F 1 − G 1 ) ‖ X + ‖ B ¯ n ( F 2 − G 2 ) ‖ X ] ⩽ ⩽ ( 1 / 2 ) ‖ F ¯ 1 − G ¯ 1 ‖ X + ( 1 / 2 ) ‖ F ¯ 2 − G ¯ 2 ‖ X < ε , = ( 1 / 2 ) B n F 1 − G 1 + B n F 2 − G 2 X ⩽ ⩽ ( 1 / 2 ) B n F 1 − G 1 X + B ¯ n F 2 − G 2 X ⩽ ⩽ ( 1 / 2 ) F ¯ 1 − G ¯ 1 X + ( 1 / 2 ) F ¯ 2 − G ¯ 2 X < ε , {:[=(1//2)||B_(n)(F_(1)-G_(1))+B_(n)(F_(2)-G_(2))||_(X) <= ],[ <= (1//2)[||B_(n)(F_(1)-G_(1))||_(X)+|| bar(B)_(n)(F_(2)-G_(2))||_(X)] <= ],[ <= (1//2)|| bar(F)_(1)- bar(G)_(1)||_(X)+(1//2)|| bar(F)_(2)- bar(G)_(2)||_(X) < epsi","]:}\begin{aligned} & =(1 / 2)\left\|B_{n}\left(F_{1}-G_{1}\right)+B_{n}\left(F_{2}-G_{2}\right)\right\|_{X} \leqslant \\ & \leqslant(1 / 2)\left[\left\|B_{n}\left(F_{1}-G_{1}\right)\right\|_{X}+\left\|\bar{B}_{n}\left(F_{2}-G_{2}\right)\right\|_{X}\right] \leqslant \\ & \leqslant(1 / 2)\left\|\bar{F}_{1}-\bar{G}_{1}\right\|_{X}+(1 / 2)\left\|\bar{F}_{2}-\bar{G}_{2}\right\|_{X}<\varepsilon, \end{aligned}=(1/2)‖Bn(F1−G1)+Bn(F2−G2)‖X⩽⩽(1/2)[‖Bn(F1−G1)‖X+‖B¯n(F2−G2)‖X]⩽⩽(1/2)‖F¯1−G¯1‖X+(1/2)‖F¯2−G¯2‖X<ε,
showing that the selection b n b n b_(n)b_{n}bn is also continuous.
Remark 2. (a) Let C + C + C^(+)C^{+}C+be the cone of positive functions in Lip 0 { 0 , 1 } Lip 0 { 0 , 1 } Lip_(0){0,1}\operatorname{Lip}_{0}\{0,1\}Lip0{0,1} and C − C − C^(-)C^{-}C−the cone of negative functions, i.e.
C + = { f ∈ Lip 0 { 0 , 1 } : f ( 1 ) > 0 } (20) C − = { f ∈ Lip 0 { 0 , 1 } : f ( 1 ) < 0 } C + = f ∈ Lip 0 { 0 , 1 } : f ( 1 ) > 0 (20) C − = f ∈ Lip 0 { 0 , 1 } : f ( 1 ) < 0 {:[C^(+)={f inLip_(0){0,1}:f(1) > 0}],[(20)C^(-)={f inLip_(0){0,1}:f(1) < 0}]:}\begin{align*} & C^{+}=\left\{f \in \operatorname{Lip}_{0}\{0,1\}: f(1)>0\right\} \\ & C^{-}=\left\{f \in \operatorname{Lip}_{0}\{0,1\}: f(1)<0\right\} \tag{20} \end{align*}C+={f∈Lip0{0,1}:f(1)>0}(20)C−={f∈Lip0{0,1}:f(1)<0}
Then e 1 ( C − ) ⊆ K − e 1 C − ⊆ K − e_(1)(C^(-))subeK^(-)e_{1}\left(C^{-}\right) \subseteq K^{-}e1(C−)⊆K−, where K − = { F ∈ Lip p 0 [ 0 , 1 ] , P K − = F ∈ Lip p 0 [ 0 , 1 ] , P K^(-)={F in Lipp_(0)[0,1],P:}K^{-}=\left\{F \in \operatorname{Lip} p_{0}[0,1], P\right.K−={F∈Lipp0[0,1],P is negative } } }\}}, and e 2 ( C + ) ⊆ K + e 2 C + ⊆ K + e_(2)(C^(+))subeK^(+)e_{2}\left(C^{+}\right) \subseteq K^{+}e2(C+)⊆K+, where K + = { F ∈ Lip 0 [ 0 , 1 ] , F K + = F ∈ Lip 0 [ 0 , 1 ] , F K^(+)={F in Lip_(0)[0,1],F:}K^{+}=\left\{F \in \operatorname{Lip}{ }_{0}[0,1], F\right.K+={F∈Lip0[0,1],F is positive } } }\}}.
Let
(21) H α − : C − → 2 K − , E α + : C + → 2 K + (21) H α − : C − → 2 K − , E α + : C + → 2 K + {:(21)H_(alpha)^(-):C^(-)rarr2^(K-)","E_(alpha)^(+):C^(+)rarr2^(K+):}\begin{equation*} H_{\alpha}^{-}: C^{-} \rightarrow 2^{K-}, E_{\alpha}^{+}: C^{+} \rightarrow 2^{K+} \tag{21} \end{equation*}(21)Hα−:C−→2K−,Eα+:C+→2K+
be the restrictions of E α E α E_(alpha)E_{\alpha}Eα to the cones C − C − C^(-)C^{-}C−and C + C + C^(+)C^{+}C+, respectively.
Obviously that E α − ( f ) ≠ ∅ E α − ( f ) ≠ ∅ E_(alpha)^(-)(f)!=O/E_{\alpha}^{-}(f) \neq \emptysetEα−(f)≠∅, for every f ∈ C − f ∈ C − f inC^(-)f \in C^{-}f∈C−(the set E α − ( f ) E α − ( f ) E_(alpha)^(-)(f)E_{\alpha}^{-}(f)Eα−(f) contains at least the function F ¯ 1 ∈ K − F ¯ 1 ∈ K − bar(F)_(1)inK^(-)\bar{F}_{1} \in K^{-}F¯1∈K−) and E α + ( f ) ≠ 0 E α + ( f ) ≠ 0 E_(alpha)^(+)(f)!=0E_{\alpha}^{+}(f) \neq 0Eα+(f)≠0, for every f ∈ C + f ∈ C + f inC^(+)f \in C^{+}f∈C+(the set E α + ( f ) E α + ( f ) E_(alpha)^(+)(f)E_{\alpha}^{+}(f)Eα+(f) contains at least the function F ¯ 2 ∈ K + F ¯ 2 ∈ K + bar(F)_(2)inK^(+)\bar{F}_{2} \in K^{+}F¯2∈K+).
We have the following corollary
Corollary 2. a) The selection e 1 − ( f ) = F ¯ 1 , f ∈ C − e 1 − ( f ) = F ¯ 1 , f ∈ C − e_(1)^(-)(f)= bar(F)_(1),f inC^(-)e_{1}^{-}(f)=\bar{F}_{1}, f \in C^{-}e1−(f)=F¯1,f∈C−, associated to the operator E − − E − − E_(-)^(-)E_{-}^{-}E−−is continuous, positively homogeneous and additive;
b) The selection e 2 + ( f ) = F ¯ 2 , f ∈ C + e 2 + ( f ) = F ¯ 2 , f ∈ C + e_(2)^(+)(f)= bar(F)_(2),f inC^(+)e_{2}^{+}(f)=\bar{F}_{2}, f \in C^{+}e2+(f)=F¯2,f∈C+, associated to the operator D α + D α + D_(alpha)^(+)D_{\alpha}^{+}Dα+is continuous, positively homogeneous and additive;
e) The selections b n − ( f ) = B n ( e 1 − ( f ) ; . ) b n − ( f ) = B n e 1 − ( f ) ; . b_(n)^(-)(f)=B_(n)(e_(1)^(-)(f);.)b_{n}^{-}(f)=B_{n}\left(e_{1}^{-}(f) ;.\right)bn−(f)=Bn(e1−(f);.) and b n + ( f ) = B n ( e 2 + ( f ) ; . ) b n + ( f ) = B n e 2 + ( f ) ; . b_(n)^(+)(f)=B_(n)(e_(2)^(+)(f);.)b_{n}^{+}(f)=B_{n}\left(e_{2}^{+}(f) ;.\right)bn+(f)=Bn(e2+(f);.) are continuous, positively homogeneous and additive.
Proof. The continuity and the positive homogeneity of the selections e 1 − e 1 − e_(1)^(-)e_{1}^{-}e1−and e 2 + e 2 + e_(2)^(+)e_{2}^{+}e2+follow from the proofs of Cases 1 ∘ 1 ∘ 1^(@)1^{\circ}1∘ and 2 ∘ 2 ∘ 2^(@)2^{\circ}2∘ of Theorem 3.
If f ( 1 ) < 0 f ( 1 ) < 0 f(1) < 0f(1)<0f(1)<0 and g ( 1 ) < 0 g ( 1 ) < 0 g(1) < 0g(1)<0g(1)<0 then
F ¯ 1 ( x ) = − α | f ( 1 ) | x , for a ∈ [ 0 , α + 1 2 α ] = f ( 1 ) − α | f ( 1 ) | ( 1 − x ) , for a ∈ ( α + 1 2 α , 1 ] F ¯ 1 ( x ) = − α | f ( 1 ) | x ,  for  a ∈ 0 , α + 1 2 α = f ( 1 ) − α | f ( 1 ) | ( 1 − x ) ,  for  a ∈ α + 1 2 α , 1 {:[ bar(F)_(1)(x)=-alpha|f(1)|x","quad" for "quad a in[0,(alpha+1)/(2alpha)]],[=f(1)-alpha|f(1)|(1-x)","quad" for "quad a in((alpha+1)/(2alpha),1]]:}\begin{aligned} \bar{F}_{1}(x) & =-\alpha|f(1)| x, \quad \text { for } \quad a \in\left[0, \frac{\alpha+1}{2 \alpha}\right] \\ & =f(1)-\alpha|f(1)|(1-x), \quad \text { for } \quad a \in\left(\frac{\alpha+1}{2 \alpha}, 1\right] \end{aligned}F¯1(x)=−α|f(1)|x, for a∈[0,α+12α]=f(1)−α|f(1)|(1−x), for a∈(α+12α,1]
and
G ¯ 1 ( x ) = − α | g ( 1 ) | x , for x ∈ [ 0 , α + 1 2 α ] = g ( 1 ) − α | g ( 1 ) | ( 1 − x ) , for x ∈ ( α + 1 2 α , 1 ] G ¯ 1 ( x ) = − α | g ( 1 ) | x ,  for  x ∈ 0 , α + 1 2 α = g ( 1 ) − α | g ( 1 ) | ( 1 − x ) ,  for  x ∈ α + 1 2 α , 1 {:[ bar(G)_(1)(x)=-alpha|g(1)|x","" for "quad x in[0,(alpha+1)/(2alpha)]],[=g(1)-alpha|g(1)|(1-x)","" for "x in((alpha+1)/(2alpha),1]]:}\begin{aligned} \bar{G}_{1}(x) & =-\alpha|g(1)| x, \text { for } \quad x \in\left[0, \frac{\alpha+1}{2 \alpha}\right] \\ & =g(1)-\alpha|g(1)|(1-x), \text { for } x \in\left(\frac{\alpha+1}{2 \alpha}, 1\right] \end{aligned}G¯1(x)=−α|g(1)|x, for x∈[0,α+12α]=g(1)−α|g(1)|(1−x), for x∈(α+12α,1]
implying e 1 − ( f + g ) = e 1 − ( f ) + e 1 − ( g ) e 1 − ( f + g ) = e 1 − ( f ) + e 1 − ( g ) e_(1)^(-)(f+g)=e_(1)^(-)(f)+e_(1)^(-)(g)e_{1}^{-}(f+g)=e_{1}^{-}(f)+e_{1}^{-}(g)e1−(f+g)=e1−(f)+e1−(g)
Similarly for f ( 1 ) > 0 f ( 1 ) > 0 f(1) > 0f(1)>0f(1)>0 and g ( 1 ) > 0 g ( 1 ) > 0 g(1) > 0g(1)>0g(1)>0 one obtains e 2 + ( f + g ) === e 2 + ( f ) + e 2 + ( g ) e 2 + ( f + g ) === e 2 + ( f ) + e 2 + ( g ) e_(2)^(+)(f+g)===e_(2)^(+)(f)+e_(2)^(+)(g)e_{2}^{+}(f+g)== =e_{2}^{+}(f)+e_{2}^{+}(g)e2+(f+g)===e2+(f)+e2+(g).
Assertion c) follows from the fact that the Bernstein operator is linear and positive.
(b) Remark that the selections e 1 − e 1 − e_(1)^(-)e_{1}^{-}e1−and e 1 + e 1 + e_(1)^(+)e_{1}^{+}e1+are monotonically increasing with respect to the pointwise order, i.e. 0 < f ( 1 ) < g ( 1 ) 0 < f ( 1 ) < g ( 1 ) 0 < f(1) < g(1)0<f(1)<g(1)0<f(1)<g(1) implies F ¯ 2 ( a ) ⩽⩽ G ¯ 2 ( x ) , x ∈ [ 0 , 1 ] F ¯ 2 ( a ) ⩽⩽ G ¯ 2 ( x ) , x ∈ [ 0 , 1 ] bar(F)_(2)(a)⩽⩽ bar(G)_(2)(x),x in[0,1]\bar{F}_{2}(a) \leqslant \leqslant \bar{G}_{2}(x), x \in[0,1]F¯2(a)⩽⩽G¯2(x),x∈[0,1] and 0 > f ( 1 ) > g ( 1 ) 0 > f ( 1 ) > g ( 1 ) 0 > f(1) > g(1)0>f(1)>g(1)0>f(1)>g(1) implies F 1 ( x ) ⩾ G 1 ( x ) , x ∈ [ 0 , 1 ] F 1 ( x ) ⩾ G 1 ( x ) , x ∈ [ 0 , 1 ] F_(1)(x) >= G_(1)(x),x in[0,1]F_{1}(x) \geqslant G_{1}(x), x \in[0,1]F1(x)⩾G1(x),x∈[0,1]
Furthermore, e 1 − ( f ) e 1 − ( f ) e_(1)^(-)(f)e_{1}^{-}(f)e1−(f) is a convex lunction for f ∈ C − f ∈ C − f inC^(-)f \in C^{-}f∈C−and e 2 + ( f ) e 2 + ( f ) e_(2)^(+)(f)e_{2}^{+}(f)e2+(f) is a concave function for f ∈ C + f ∈ C + f inC^(+)f \in C^{+}f∈C+.
3. Selections associated to the operator of metric projection
Let Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ be the anihilator of the set Y = { 0 , 1 } Y = { 0 , 1 } Y={0,1}Y=\{0,1\}Y={0,1} in Lip 0 [ 0 , 1 ] Lip 0 [ 0 , 1 ] Lip_(0)[0,1]\operatorname{Lip}_{0}[0,1]Lip0[0,1], i.e.
(22) Y ⊥= { G ∈ Lip 0 [ 0 , 1 ] : G ( 0 ) = G ( 1 ) = 0 } (22) Y ⊥= G ∈ Lip 0 [ 0 , 1 ] : G ( 0 ) = G ( 1 ) = 0 {:(22)Y⊥={G inLip_(0)[0,1]:G(0)=G(1)=0}:}\begin{equation*} Y \perp=\left\{G \in \operatorname{Lip}_{0}[0,1]: G(0)=G(1)=0\right\} \tag{22} \end{equation*}(22)Y⊥={G∈Lip0[0,1]:G(0)=G(1)=0}
Then X ⊥ X ⊥ X^(_|_)X^{\perp}X⊥ is a closed ideal in Lip p 0 [ 0 , 1 ] Lip p 0 [ 0 , 1 ] Lipp_(0)[0,1]\operatorname{Lip} p_{0}[0,1]Lipp0[0,1]. For X 1 ∈ Lip p 0 [ 0 , 1 ] X 1 ∈ Lip p 0 [ 0 , 1 ] X^(1)in Lipp_(0)[0,1]X^{1} \in \operatorname{Lip} p_{0}[0,1]X1∈Lipp0[0,1] let
(23) d ( F , Y ⊥ ) = inf { ‖ I Y − G ‖ X : G ∈ Y ⊥ } (23) d F , Y ⊥ = inf I Y − G X : G ∈ Y ⊥ {:(23)d(F,Y^(_|_))=i n f{||I^(Y)-G||_(X):quad G inY^(_|_)}:}\begin{equation*} d\left(F, Y^{\perp}\right)=\inf \left\{\left\|I^{Y}-G\right\|_{X}: \quad G \in Y^{\perp}\right\} \tag{23} \end{equation*}(23)d(F,Y⊥)=inf{‖IY−G‖X:G∈Y⊥}
An element G 0 ∈ Y ⊥ G 0 ∈ Y ⊥ G_(0)inY^(_|_)G_{0} \in Y^{\perp}G0∈Y⊥ for which the infimum in (23) is attained is called the nearest point to F 1 F 1 F^(1)F^{1}F1 in Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥.
Let
(24) P Y ⊥ : lip 0 [ 0 , 1 ] → 2 Y ⊥ (24) P Y ⊥ : lip 0 [ 0 , 1 ] → 2 Y ⊥ {:(24)P_(Y^(_|_)):lip_(0)[0","1]rarr2^(Y^(_|_)):}\begin{equation*} P_{Y^{\perp}}: \operatorname{lip}_{0}[0,1] \rightarrow 2^{Y^{\perp}} \tag{24} \end{equation*}(24)PY⊥:lip0[0,1]→2Y⊥
be the operator of metric projection on Y ⊥ Y ⊥ Y _|_Y \perpY⊥, defined by
P Y ⊥ ( F ⊤ ) = { G ∈ Y ⊥ : ‖ F ⊤ − G ‖ X = d ( F ⊤ , Y ⊥ ) } P Y ⊥ F ⊤ = G ∈ Y ⊥ : F ⊤ − G X = d F ⊤ , Y ⊥ P_(Y^(_|_))(F^(TT))={G inY^(_|_):||F^(TT)-G||_(X)=d(F^(TT),Y^(_|_))}P_{Y^{\perp}}\left(F^{\top}\right)=\left\{G \in Y^{\perp}:\left\|F^{\top}-G\right\|_{X}=d\left(F^{\top}, Y^{\perp}\right)\right\}PY⊥(F⊤)={G∈Y⊥:‖F⊤−G‖X=d(F⊤,Y⊥)}
for all 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].
X ⊥ X ⊥ X^(_|_)X^{\perp}X⊥ is called proximinal (resp. Chebyshev) if for each F ∈ Jip 0 [ 0 , 1 ] F ∈ Jip 0 [ 0 , 1 ] F inJip_(0)[0,1]F \in \mathrm{Jip}_{0}[0,1]F∈Jip0[0,1] the set P Y ⊥ ( F r ) P Y ⊥ F r P_(Y^(_|_))(F^(r))P_{Y^{\perp}}\left(F^{r}\right)PY⊥(Fr) is nonempty (resp. a singleton).
The following proposition holds :
Proposition 1. a) The formula
(25)
d ( F , Y ⊥ ) = | B → ( 1 ) | d F , Y ⊥ = | B → ( 1 ) | d(F,Y^(_|_))=| vec(B)(1)|d\left(F, Y^{\perp}\right)=|\vec{B}(1)|d(F,Y⊥)=|B→(1)|
is valid for every H ′ ∈ Lip 0 H ′ ∈ Lip 0 H^(')inLip_(0)H^{\prime} \in \operatorname{Lip}_{0}H′∈Lip0 [0,1]. In particular Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is a proximinal subspace of Lip Lip 0 [ 0 , 1 ] Lip 0 [ 0 , 1 ] Lip_(0)[0,1]\operatorname{Lip}_{0}[0,1]Lip0[0,1];
b) If G ∈ P x ⊥ ( F ) G ∈ P x ⊥ ( F ) G inP_(x^(_|_))(F)G \in P_{x^{\perp}}(F)G∈Px⊥(F) then G = H ⊤ − H G = H ⊤ − H G=H^(TT)-HG=H^{\top}-HG=H⊤−H, where H ∈ E α ( F ⊤ | Y ) ∣ H ∈ E α F ⊤ Y ∣ H inE_(alpha)(F^(TT)|_(Y))∣H \in E_{\alpha}\left(\left.F^{\top}\right|_{Y}\right) \midH∈Eα(F⊤|Y)∣ is such that ‖ H ‖ X = | F ( 1 ) | : ‖ H ‖ X = | F ( 1 ) | : ||H||_(X)=|F(1)|:\|H\|_{X}=|F(1)|:‖H‖X=|F(1)|:
c) There holds the equality:
(26) d ( F , X ⊥ ) = d ( F , F − E x ( F | Y ) ) (26) d ( F , X ⊥ ) = d F , F − E x F Y {:(26)d(F","X _|_)=d(F,F-E_(x)(F|_(Y))):}\begin{equation*} d(F, X \perp)=d\left(F, F-E_{x}\left(\left.F\right|_{Y}\right)\right) \tag{26} \end{equation*}(26)d(F,X⊥)=d(F,F−Ex(F|Y))
where F − D α ( F | X ) = { F − Π : H ∈ E α ( F | Y ) } , F ∈ Lip 0 [ 0 , 1 ] ; F − D α F X = F − Π : H ∈ E α F Y , F ∈ Lip 0 [ 0 , 1 ] ; F-D_(alpha)(F|_(X))={F-Pi:H inE_(alpha)(F|_(Y))},quad F in Lip_(0)[0,1];F-D_{\alpha}\left(\left.F\right|_{X}\right)=\left\{F-\Pi: H \in E_{\alpha}\left(\left.F\right|_{Y}\right)\right\}, \quad F \in \operatorname{Lip}{ }_{0}[0,1] ;F−Dα(F|X)={F−Π:H∈Eα(F|Y)},F∈Lip0[0,1];
d) The equality
(27) sup { ‖ F − G ‖ X : G ∈ F − W α ( F | Y ) } = α | F ( 1 ) | (27) sup ‖ F − G ‖ X : G ∈ F − W α F Y = α | F ( 1 ) | {:(27)s u p{||F-G||_(X):G in F-W_(alpha)(F|_(Y))}=alpha|F(1)|:}\begin{equation*} \sup \left\{\|F-G\|_{X}: G \in F-W_{\alpha}\left(\left.F\right|_{Y}\right)\right\}=\alpha|F(1)| \tag{27} \end{equation*}(27)sup{‖F−G‖X:G∈F−Wα(F|Y)}=α|F(1)|
holds for every W ∈ Lip 0 [ 0 , 1 ] W ∈ Lip 0 [ 0 , 1 ] W inLip_(0)[0,1]W \in \operatorname{Lip}_{0}[0,1]W∈Lip0[0,1].
Proof. a) Let F ∈ Lip p 0 [ 0 , 1 ] F ∈ Lip p 0 [ 0 , 1 ] F in Lipp_(0)[0,1]F \in \operatorname{Lip} p_{0}[0,1]F∈Lipp0[0,1]. Then for every G ∈ Y ⊥ G ∈ Y ⊥ G in Y _|_G \in Y \perpG∈Y⊥ one has | G ( 1 ) | == | F ( 1 ) − G ( 1 ) | ⩽ ‖ F − G ‖ X | G ( 1 ) | == | F ( 1 ) − G ( 1 ) | ⩽ ‖ F − G ‖ X |G(1)|==|F(1)-G(1)| <= ||F-G||_(X)|G(1)|= =|F(1)-G(1)| \leqslant\|F-G\|_{X}|G(1)|==|F(1)−G(1)|⩽‖F−G‖X. Taking the infimum with respect to G ∈ Y ⊥ G ∈ Y ⊥ G inY^(_|_)G \in Y^{\perp}G∈Y⊥ one obtains
F ′ ( 1 ) ∣⩽ d ( F , X ⊥ ) F ′ ( 1 ) ∣⩽ d F , X ⊥ F^(')(1)∣⩽d(F,X^(_|_))F^{\prime}(1) \mid \leqslant d\left(F, X^{\perp}\right)F′(1)∣⩽d(F,X⊥)
Let G 0 ( x ) = H ( x ) − H ( 1 ) x , x ∈ [ 0 , 1 ] G 0 ( x ) = H ( x ) − H ( 1 ) x , x ∈ [ 0 , 1 ] G_(0)(x)=H(x)-H(1)x,x in[0,1]G_{0}(x)=H(x)-H(1) x, x \in[0,1]G0(x)=H(x)−H(1)x,x∈[0,1]. It follows that G 0 ∈ Y ⊥ ( G 0 ( 0 ) = 0 = = G 0 ( 1 ) G 0 ∈ Y ⊥ G 0 ( 0 ) = 0 = = G 0 ( 1 ) G_(0)in Y _|_(G_(0)(0)=0=:}=G_(0)(1)G_{0} \in Y \perp\left(G_{0}(0)=0=\right. =G_{0}(1)G0∈Y⊥(G0(0)=0==G0(1) and. ‖ F − G 0 ‖ X = | F ′ ( 1 ) | F − G 0 X = F ′ ( 1 ) ||F-G_(0)||_(X)=|F^(')(1)|\left\|F-G_{0}\right\|_{X}=\left|F^{\prime}(1)\right|‖F−G0‖X=|F′(1)| so that ‖ F ′ − G 0 ‖ X = d ( F I , Y ⊥ ) F ′ − G 0 X = d F I , Y ⊥ ||F^(')-G_(0)||_(X)=d(F^(I),Y _|_)\left\|F^{\prime}-G_{0}\right\|_{X}=d\left(F^{I}, Y \perp\right)‖F′−G0‖X=d(FI,Y⊥). This shows that Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is a proximinal subspace of Lip 0 [ 0 , 1 ] Lip 0 [ 0 , 1 ] Lip_(0)[0,1]\operatorname{Lip}_{0}[0,1]Lip0[0,1].
b) If G ∈ P Y ⊥ ( F l ) G ∈ P Y ⊥ F l G inP_(Y^(_|_))(F^(l))G \in P_{Y^{\perp}}\left(F^{l}\right)G∈PY⊥(Fl), then
‖ F − G ‖ x = d ( F , Y 1 ) = | F ( 1 ) | ⩽ α | F ′ ( 1 ) | , ‖ F − G ‖ x = d F , Y 1 = | F ( 1 ) | ⩽ α F ′ ( 1 ) , ||F-G||_(x)=d(F,Y^(1))=|F(1)| <= alpha|F^(')(1)|,\|F-G\|_{x}=d\left(F, Y^{1}\right)=|F(1)| \leqslant \alpha\left|F^{\prime}(1)\right|,‖F−G‖x=d(F,Y1)=|F(1)|⩽α|F′(1)|,
and
( F − G ) | Y = F ′ | Y , ( F − G ) Y = F ′ Y , (F-G)|_(Y)=F^(')|_(Y),\left.(F-G)\right|_{Y}=\left.F^{\prime}\right|_{Y},(F−G)|Y=F′|Y,
showing that F − G ∈ E x ( P ′ | Y ) F − G ∈ E x P ′ Y F-G inE_(x)(P^(')|_(Y))F-G \in E_{x}\left(\left.P^{\prime}\right|_{Y}\right)F−G∈Ex(P′|Y). If follows that there exists U U UUU in E α ( F | Y ′ ) E α F Y ′ E_(alpha)(F|_(Y^(')))E_{\alpha}\left(\left.F\right|_{Y^{\prime}}\right)Eα(F|Y′) such that F ′ − G = H F ′ − G = H F^(')-G=HF^{\prime}-G=HF′−G=H and ‖ W ‖ x = ‖ I − G x ˙ = | H ′ ( 1 ) | ‖ W ‖ x = ‖ I − G x ˙ = H ′ ( 1 ) ||W||_(x)=||I-G_(x^(˙))=|H^(')(1)|\|W\|_{x}=\| I-G_{\dot{x}}=\left|H^{\prime}(1)\right|‖W‖x=‖I−Gx˙=|H′(1)|.
c) Follows from a) and b).
d) For every G ∈ F − H 3 ( F | Y ) G ∈ F − H 3 F Y G in F-H_(3)(F|_(Y))G \in F-H_{3}\left(\left.F\right|_{Y}\right)G∈F−H3(F|Y) we have
‖ F − G ‖ X = ‖ F − ( F − H ) ‖ X = ‖ H ‖ X ⩽ α | F ( 1 ) | , ‖ F − G ‖ X = ‖ F − ( F − H ) ‖ X = ‖ H ‖ X ⩽ α | F ( 1 ) | , ||F-G||_(X)=||F-(F-H)||_(X)=||H||_(X) <= alpha|F(1)|,\|F-G\|_{X}=\|F-(F-H)\|_{X}=\|H\|_{X} \leqslant \alpha|F(1)|,‖F−G‖X=‖F−(F−H)‖X=‖H‖X⩽α|F(1)|,
where H ∈ D α ( H ∣ J ) H ∈ D α H ∣ J H inD_(alpha)(H_(∣J))H \in D_{\alpha}\left(H_{\mid J}\right)H∈Dα(H∣J).
Taking the supremum with respect to G ∈ I − E α ( F | x ) G ∈ I − E α F x G in I-E_(alpha)(F|_(x))G \in I-E_{\alpha}\left(\left.F\right|_{x}\right)G∈I−Eα(F|x) we find
sup { ‖ F − G ‖ X : G ∈ F − D α ( F | Y ) } ⩽ α | F ( 1 ) | . sup ‖ F − G ‖ X : G ∈ F − D α F Y ⩽ α | F ( 1 ) | . s u p{||F-G||_(X):G in F-D_(alpha)(F|_(Y))} <= alpha|F(1)|.\sup \left\{\|F-G\|_{X}: G \in F-D_{\alpha}\left(\left.F\right|_{Y}\right)\right\} \leqslant \alpha|F(1)| .sup{‖F−G‖X:G∈F−Dα(F|Y)}⩽α|F(1)|.
Let
G 1 ( x ) = H ′ ( x ) − max { − α | H ′ ( 1 ) | x ; H ′ ( 1 ) − α | H ′ ( 1 ) | ( 1 − x ) } , G 1 ( x ) = H ′ ( x ) − max − α H ′ ( 1 ) x ; H ′ ( 1 ) − α H ′ ( 1 ) ( 1 − x ) , G_(1)(x)=H^(')(x)-max{-alpha|H^(')(1)|x;H^(')(1)-alpha|H^(')(1)|(1-x)},G_{1}(x)=H^{\prime}(x)-\max \left\{-\alpha\left|H^{\prime}(1)\right| x ; H^{\prime}(1)-\alpha\left|H^{\prime}(1)\right|(1-x)\right\},G1(x)=H′(x)−max{−α|H′(1)|x;H′(1)−α|H′(1)|(1−x)},
and
G 2 ( x ) = l ′ ( x ) − min { α | F ′ ( 1 ) | x ; l ′ ( 1 ) + α | F ′ ( 1 ) | ( 1 − x ) } , G 2 ( x ) = l ′ ( x ) − min α F ′ ( 1 ) x ; l ′ ( 1 ) + α F ′ ( 1 ) ( 1 − x ) , G_(2)(x)=l^(')(x)-min{alpha|F^(')(1)|x;l^(')(1)+alpha|F^(')(1)|(1-x)},G_{2}(x)=l^{\prime}(x)-\min \left\{\alpha\left|F^{\prime}(1)\right| x ; l^{\prime}(1)+\alpha\left|F^{\prime}(1)\right|(1-x)\right\},G2(x)=l′(x)−min{α|F′(1)|x;l′(1)+α|F′(1)|(1−x)},
x ∈ [ 0 , 1 ] x ∈ [ 0 , 1 ] x in[0,1]x \in[0,1]x∈[0,1].
Obviously that G 1 , G 2 ∈ F − H 2 ( I ′ | Y ) G 1 , G 2 ∈ F − H 2 I ′ Y G_(1),G_(2)in F-H_(2)(I^(')|_(Y))G_{1}, G_{2} \in F-H_{2}\left(\left.I^{\prime}\right|_{Y}\right)G1,G2∈F−H2(I′|Y) and,
‖ F − G ^ 1 ‖ x = ‖ F − G 2 ‖ x = α | F ( 1 ) | , F − G ^ 1 x = F − G 2 x = α | F ( 1 ) | , ||F- hat(G)_(1)||_(x)=||F-G_(2)||_(x)=alpha|F(1)|,\left\|F-\hat{G}_{1}\right\|_{x}=\left\|F-G_{2}\right\|_{x}=\alpha|F(1)|,‖F−G^1‖x=‖F−G2‖x=α|F(1)|,
proving the assertion d d ddd ).
Remark 3. By Proposition 1 it follows that the nearest points to F ∈ Lip [ 0 , 1 ] F ∈ Lip [ 0 , 1 ] F in Lip[0,1]F \in \operatorname{Lip}[0,1]F∈Lip[0,1] in Y ⊥ Y ⊥ Y _|_Y \perpY⊥ are the functions G ∈ F − E α ( F | Y ) ⊂ Y ⊥ , G = F − H G ∈ F − E α F Y ⊂ Y ⊥ , G = F − H G in F-E_(alpha)(F|_(Y))subY^(_|_),G=F-HG \in F-E_{\alpha}\left(\left.F\right|_{Y}\right) \subset Y^{\perp}, G=F-HG∈F−Eα(F|Y)⊂Y⊥,G=F−H with H ∈ E α ( F | Y ) H ∈ E α F Y H inE_(alpha)(F|_(Y))H \in \mathbb{E}_{\alpha}\left(\left.F\right|_{Y}\right)H∈Eα(F|Y) of minimal Lipsehitz norm and the farthest points for F ′ F ′ F^(')F^{\prime}F′ in F ′ − E a ( F ′ | Y ) ⊂ X ⊥ F ′ − E a F ′ Y ⊂ X ⊥ F^(')-E_(a)(F^(')|_(Y))subX^(_|_)F^{\prime}-E_{a}\left(\left.F^{\prime}\right|_{Y}\right) \subset X^{\perp}F′−Ea(F′|Y)⊂X⊥ are the functions G = F − I ′ G = F − I ′ G=F-I^(')G=F-I^{\prime}G=F−I′, with H ∈ B a ( H Y | Y ) H ∈ B a H Y Y H inB_(a)(H^(Y)|_(Y))H \in B_{a}\left(\left.H^{Y}\right|_{Y}\right)H∈Ba(HY|Y) of the maximal norm ( ‖ H ‖ X = α ∣ H ( 1 ) ‖ ‖ H ‖ X = α ∣ H ( 1 ) ‖ ||H||_(X)=alpha∣H(1)||\|H\|_{X}=\alpha \mid H(1) \|‖H‖X=α∣H(1)‖ ).
Let r : Lip 0 [ 0 , 1 ] → Lip 0 { 0 , 1 } r : Lip 0 [ 0 , 1 ] → Lip 0 { 0 , 1 } r:Lip_(0)[0,1]rarrLip_(0){0,1}r: \operatorname{Lip}_{0}[0,1] \rightarrow \operatorname{Lip}_{0}\{0,1\}r:Lip0[0,1]→Lip0{0,1} be the restriction operator
(28) r ( F ) = I ′ | { 0 , 1 } ∈ Lip 0 { 0 , 1 } , F ′ ∈ Lip 0 [ 0 , 1 ] . (28) r ( F ) = I ′ { 0 , 1 } ∈ Lip 0 { 0 , 1 } , F ′ ∈ Lip 0 [ 0 , 1 ] . {:(28)r(F)=I^(')|_({0,1})inLip_(0){0","1}","quadF^(')inLip_(0)[0","1].:}\begin{equation*} r(F)=\left.I^{\prime}\right|_{\{0,1\}} \in \operatorname{Lip}_{0}\{0,1\}, \quad F^{\prime} \in \operatorname{Lip}_{0}[0,1] . \tag{28} \end{equation*}(28)r(F)=I′|{0,1}∈Lip0{0,1},F′∈Lip0[0,1].
Then the operator Q α : Lip 0 [ 0 , 1 ] → 2 Y ⊥ Q α : Lip 0 [ 0 , 1 ] → 2 Y ⊥ Q_(alpha):Lip_(0)[0,1]rarr2^(Y^(_|_))Q_{\alpha}: \operatorname{Lip}_{0}[0,1] \rightarrow 2^{Y^{\perp}}Qα:Lip0[0,1]→2Y⊥, defined by
Q α = I − E α ∘ r , Q α = I − E α ∘ r , Q_(alpha)=I-E_(alpha)@r,Q_{\alpha}=I-E_{\alpha} \circ r,Qα=I−Eα∘r,
where I : Lip 0 [ 0 , 1 ] → Lip 0 [ 0 , 1 ] I : Lip 0 [ 0 , 1 ] → Lip 0 [ 0 , 1 ] I:Lip_(0)[0,1]rarrLip_(0)[0,1]I: \operatorname{Lip}_{0}[0,1] \rightarrow \operatorname{Lip}_{0}[0,1]I:Lip0[0,1]→Lip0[0,1] is the identity operator, i.e.
is a multivalued operator for α > 1 α > 1 alpha > 1\alpha>1α>1.
(29) Q α ( l ′ ) = F ′ − k α ( l ′ | l ′ ) , F ′ ∈ Lip 0 [ 0 , 1 ] (29) Q α l ′ = F ′ − k α l ′ l ′ , F ′ ∈ Lip 0 [ 0 , 1 ] {:(29)Q_(alpha)(l^('))=F^(')-k_(alpha)(l^(')|_(l^(')))","quadF^(')inLip_(0)[0","1]:}\begin{equation*} Q_{\alpha}\left(l^{\prime}\right)=F^{\prime}-k_{\alpha}\left(\left.l^{\prime}\right|_{l^{\prime}}\right), \quad F^{\prime} \in \operatorname{Lip}_{0}[0,1] \tag{29} \end{equation*}(29)Qα(l′)=F′−kα(l′|l′),F′∈Lip0[0,1]
Since the metric projection operator on X − ⊥ X − ⊥ X^(-_|_)X^{-\perp}X−⊥ verifies the equality
P Y ⊥ ( l ′ ) = { G ∈ Q α ( I ′ ) : ‖ G − I ′ ‖ x = d ( F , I ⊥ ) } , P Y ⊥ l ′ = G ∈ Q α I ′ : G − I ′ x = d F , I ⊥ , P_(Y^(_|_))(l^('))={G inQ_(alpha)(I^(')):||G-I^(')||_(x)=d(F,I^(_|_))},P_{Y^{\perp}}\left(l^{\prime}\right)=\left\{G \in Q_{\alpha}\left(I^{\prime}\right):\left\|G-I^{\prime}\right\|_{x}=d\left(F, I^{\perp}\right)\right\},PY⊥(l′)={G∈Qα(I′):‖G−I′‖x=d(F,I⊥)},
it follows that P Y ⊥ ( Y r ) ⊆ Q χ ( Y r ) P Y ⊥ Y r ⊆ Q χ Y r P_(Y^(_|_))(Y^(r))subeQ_(chi)(Y^(r))P_{Y^{\perp}}\left(Y^{r}\right) \subseteq Q_{\chi}\left(Y^{r}\right)PY⊥(Yr)⊆Qχ(Yr), for all F ∈ Lip P 0 [ 0 , 1 ] F ∈ Lip P 0 [ 0 , 1 ] F in LipP_(0)[0,1]F \in \operatorname{Lip} P_{0}[0,1]F∈LipP0[0,1].
Let T α : Lip 0 [ 0 , 1 ] → 2 r ⊥ T α : Lip 0 [ 0 , 1 ] → 2 r ⊥ T_(alpha):Lip_(0)[0,1]rarr2r^(_|_)T_{\alpha}: \operatorname{Lip}_{0}[0,1] \rightarrow 2 r^{\perp}Tα:Lip0[0,1]→2r⊥ be defined by
(30) T α ( H ′ ) = { I I ∈ Q α ( I ) : ‖ I I − H ′ ‖ X = sup { ‖ U − I ′ ‖ X : U ∈ Q α ( I ′ ) } } . (30) T α H ′ = I I ∈ Q α ( I ) : I I − H ′ X = sup U − I ′ X : U ∈ Q α I ′ . {:(30)T_(alpha)(H^('))={II inQ_(alpha)(I):||II-H^(')||_(X)=s u p{||U-I^(')||_(X):U inQ_(alpha)(I^('))}}.:}\begin{equation*} T_{\alpha}\left(H^{\prime}\right)=\left\{I I \in Q_{\alpha}(I):\left\|I I-H^{\prime}\right\|_{X}=\sup \left\{\left\|U-I^{\prime}\right\|_{X}: U \in Q_{\alpha}\left(I^{\prime}\right)\right\}\right\} . \tag{30} \end{equation*}(30)Tα(H′)={II∈Qα(I):‖II−H′‖X=sup{‖U−I′‖X:U∈Qα(I′)}}.
The following theorem holds:
Teorem 1. a) The operator P Σ ⊥ P Σ ⊥ P_(Sigma^(_|_))P_{\Sigma^{\perp}}PΣ⊥ is a linear and continuous selection of the operator Q α Q α Q_(alpha)Q_{\alpha}Qα;
b) The subspace I ⊥ I ⊥ I^(_|_)I^{\perp}I⊥ is complemented in L 1 P 0 [ 0 , 1 ] L 1 P 0 [ 0 , 1 ] L_(1)P_(0)[0,1]L_{1} P_{0}[0,1]L1P0[0,1] by the subspace
(31) W = { H ∈ Lip 0 [ 0 , 1 ] : H ( x ) = a α , α ∈ [ 0 , 1 ] , a ∈ R j ; (31) W = H ∈ Lip 0 [ 0 , 1 ] : H ( x ) = a α , α ∈ [ 0 , 1 ] , a ∈ R j ; {:(31)W={H in Lip0[0,1]:H(x)=a alpha,alpha in[0,1],a inR_(j);:}:}\begin{equation*} W=\left\{H \in \operatorname{Lip} 0[0,1]: H(x)=a \alpha, \alpha \in[0,1], a \in R_{j} ;\right. \tag{31} \end{equation*}(31)W={H∈Lip0[0,1]:H(x)=aα,α∈[0,1],a∈Rj;
tions.
c) The operators T α T α T_(alpha)T_{\alpha}Tα and Q α Q α Q_(alpha)Q_{\alpha}Qα admit continuous and homogeneous selec-
Proof. a) The operator P y ⊥ P y ⊥ P_(y^(_|_))P_{y^{\perp}}Py⊥ is single-valued since for 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] there exists a unique element H ∈ U x ( F | Y ′ ) H ∈ U x F Y ′ H inU_(x)(F|_(Y^(')))H \in U_{x}\left(\left.F\right|_{Y^{\prime}}\right)H∈Ux(F|Y′) such that ‖ U ‖ X =∣ F ′ ( 1 ) ‖ U ‖ X =∣ F ′ ( 1 ) ||U||_(X)=∣F^(')(1)\|U\|_{X}= \mid F^{\prime}(1)‖U‖X=∣F′(1) and by Proposition 1. b), it tollows that I ′ I ′ I^(')I^{\prime}I′ has a unique nearest point in Y ⊥ Y ⊥ Y _|_Y \perpY⊥. Then
P Y ⊥ ( λ M ′ ) ( x ) = λ I ′ ( x ) − λ W ′ ( 1 ) x = λ P Y ⊥ ( W ⊤ ) ( x ) , P Y ⊥ λ M ′ ( x ) = λ I ′ ( x ) − λ W ′ ( 1 ) x = λ P Y ⊥ W ⊤ ( x ) , P_(Y^(_|_))(lambdaM^('))(x)=lambdaI^(')(x)-lambdaW^(')(1)x=lambdaP_(Y^(_|_))(W^(TT))(x),P_{Y^{\perp}}\left(\lambda M^{\prime}\right)(x)=\lambda I^{\prime}(x)-\lambda W^{\prime}(1) x=\lambda P_{Y^{\perp}}\left(W^{\top}\right)(x),PY⊥(λM′)(x)=λI′(x)−λW′(1)x=λPY⊥(W⊤)(x),
for x ∈ [ 0 , 1 ] x ∈ [ 0 , 1 ] x in[0,1]x \in[0,1]x∈[0,1] and λ ∈ R λ ∈ R lambda in R\lambda \in Rλ∈R.
For F 1 , F 2 ∈ Lip 0 [ 0 , 1 ] F 1 , F 2 ∈ Lip 0 [ 0 , 1 ] F_(1),F_(2)inLip_(0)[0,1]F_{1}, F_{2} \in \operatorname{Lip}_{0}[0,1]F1,F2∈Lip0[0,1] we have
P I ⊥ ( F 1 + F 2 ) ( x ) = F 1 ( x ) + F 2 ( x ) − ( H 1 ( 1 ) + F 2 ( 1 ) ) x = = F 1 ( x ) − F 1 ( 1 ) x + F 2 ( x ) − F 2 ( 1 ) x = P Y ⊥ ( F 1 ) ( x ) + P Y ⊥ ( F 2 ) ( x ) . P I ⊥ F 1 + F 2 ( x ) = F 1 ( x ) + F 2 ( x ) − H 1 ( 1 ) + F 2 ( 1 ) x = = F 1 ( x ) − F 1 ( 1 ) x + F 2 ( x ) − F 2 ( 1 ) x = P Y ⊥ F 1 ( x ) + P Y ⊥ F 2 ( x ) . {:[P_(I^(_|_))(F_(1)+F_(2))(x)=F_(1)(x)+F_(2)(x)-(H_(1)(1)+F_(2)(1))x=],[=F_(1)(x)-F_(1)(1)x+F_(2)(x)-F_(2)(1)x=P_(Y^(_|_))(F_(1))(x)+P_(Y^(_|_))(F_(2))(x).]:}\begin{gathered} P_{I^{\perp}}\left(F_{1}+F_{2}\right)(x)=F_{1}(x)+F_{2}(x)-\left(H_{1}(1)+F_{2}(1)\right) x= \\ =F_{1}(x)-F_{1}(1) x+F_{2}(x)-F_{2}(1) x=P_{Y^{\perp}}\left(F_{1}\right)(x)+P_{Y^{\perp}}\left(F_{2}\right)(x) . \end{gathered}PI⊥(F1+F2)(x)=F1(x)+F2(x)−(H1(1)+F2(1))x==F1(x)−F1(1)x+F2(x)−F2(1)x=PY⊥(F1)(x)+PY⊥(F2)(x).
Therefore P Y ⊥ P Y ⊥ P_(Y^(_|_))P_{Y^{\perp}}PY⊥ is homogeneous and additive.
Also
‖ P I ⊥ ( d r ) − P X ⊥ ( G ) ‖ ⩽ 2 ‖ M r − G ‖ X P I ⊥ d r − P X ⊥ ( G ) ⩽ 2 M r − G X ||P_(I^(_|_))(d^(r))-P_(X^(_|_))(G)|| <= 2||M^(r)-G||_(X)\left\|P_{I^{\perp}}\left(d^{r}\right)-P_{X^{\perp}}(G)\right\| \leqslant 2\left\|M^{r}-G\right\|_{X}‖PI⊥(dr)−PX⊥(G)‖⩽2‖Mr−G‖X
so that ‖ P Y ⊥ ( F ′ ) − P Y ⊥ ( G ) ‖ X < 2 P Y ⊥ F ′ − P Y ⊥ ( G ) X < 2 ||P_(Y^(_|_))(F^('))-P_(Y^(_|_))(G)||_(X) < 2\left\|P_{Y^{\perp}}\left(F^{\prime}\right)-P_{Y^{\perp}}(G)\right\|_{X}<2‖PY⊥(F′)−PY⊥(G)‖X<2 e for ‖ I ′ − G ‖ X < ε I ′ − G X < ε ||I^(')-G||_(X) < epsi\left\|I^{\prime}-G\right\|_{X}<\varepsilon‖I′−G‖X<ε, proving the continuity of the operator P γ ⊥ P γ ⊥ P_(gamma _|_)P_{\gamma \perp}Pγ⊥.
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]. Then G ( x ) = F ( x ) − W ( 1 ) x , x ∈ [ 0 , 1 ] G ( x ) = F ( x ) − W ( 1 ) x , x ∈ [ 0 , 1 ] G(x)=F(x)-W(1)x,x in[0,1]G(x)=F(x)-W(1) x, x \in[0,1]G(x)=F(x)−W(1)x,x∈[0,1] is an element of Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ and, since F ( 1 ) x F ( 1 ) x F(1)xF(1) xF(1)x is an element of W W WWW it follows that P ( x ) === G ( x ) + E ( 1 ) x , x ∈ [ 0 , 1 ] P ( x ) === G ( x ) + E ( 1 ) x , x ∈ [ 0 , 1 ] P(x)===G(x)+E(1)x,quad x in[0,1]P(x)== =G(x)+E(1) x, \quad x \in[0,1]P(x)===G(x)+E(1)x,x∈[0,1].
If F n → F ′ F n → F ′ F_(n)rarrF^(')F_{n} \rightarrow F^{\prime}Fn→F′ in Lip 0 [ 0 , 1 ] 0 [ 0 , 1 ] _(0)[0,1]{ }_{0}[0,1]0[0,1], i.e. ‖ F n − F ′ ‖ x → 0 F n − F ′ x → 0 ||F_(n)-F^(')||_(x)rarr0\left\|F_{n}-F^{\prime}\right\|_{x} \rightarrow 0‖Fn−F′‖x→0, then the inequality
| F n ( 1 ) − H ( 1 ) | ⩽ ‖ F n ′ − F ‖ x F n ( 1 ) − H ( 1 ) ⩽ F n ′ − F x |F_(n)(1)-H(1)| <= ||F_(n)^(')-F||_(x)\left|F_{n}(1)-H(1)\right| \leqslant\left\|F_{n}^{\prime}-F\right\|_{x}|Fn(1)−H(1)|⩽‖Fn′−F‖x
implies | F n ( 1 ) | → | F ( 1 ) | F n ( 1 ) → | F ( 1 ) | |F_(n)(1)|rarr|F(1)|\left|F_{n}(1)\right| \rightarrow|F(1)||Fn(1)|→|F(1)|, showing that the projection operator on W W WWW is continuous. Consequently Lip 0 [ 0 , 1 ] = Y ⊥ ⊕ W Lip 0 [ 0 , 1 ] = Y ⊥ ⊕ W Lip_(0)[0,1]=Y _|_ o+W\operatorname{Lip}_{0}[0,1]=Y \perp \oplus WLip0[0,1]=Y⊥⊕W.
c) Consider the selections of the metric projections
t α , 1 ( F ) = F − e 1 ( F | Y ) , F ∈ Lip 0 [ 0 , 1 ] , t α , 2 ( F ) = F − e 2 ( F Y ∣ ) , F ∈ Lip 0 [ 0 , 1 ] , t α , 1 ( F ) = F − e 1 F Y , F ∈ Lip 0 [ 0 , 1 ] , t α , 2 ( F ) = F − e 2 F Y ∣ , F ∈ Lip 0 [ 0 , 1 ] , {:[t_(alpha,1)(F)=F-e_(1)(F|_(Y))","quad F inLip_(0)[0","1]","],[t_(alpha,2)(F)=F-e_(2)(F_(Y)∣)","quad F inLip_(0)[0","1]","]:}\begin{aligned} & t_{\alpha, 1}(F)=F-e_{1}\left(\left.F\right|_{Y}\right), \quad F \in \operatorname{Lip}_{0}[0,1], \\ & t_{\alpha, 2}(F)=F-e_{2}\left(F_{Y} \mid\right), \quad F \in \operatorname{Lip}_{0}[0,1], \end{aligned}tα,1(F)=F−e1(F|Y),F∈Lip0[0,1],tα,2(F)=F−e2(FY∣),F∈Lip0[0,1],
where e 1 , e 2 e 1 , e 2 e_(1),e_(2)e_{1}, e_{2}e1,e2 are the selections defined by formulae (15) and (16) (with f = F ′ | Y f = F ′ Y f=F^(')|_(Y)f=\left.F^{\prime}\right|_{Y}f=F′|Y ). Then the selection
(32) t α = ( 1 / 2 ) ( t α , 1 + t α , 2 ) (32) t α = ( 1 / 2 ) t α , 1 + t α , 2 {:(32)t_(alpha)=(1//2)(t_(alpha,1)+t_(alpha,2)):}\begin{equation*} t_{\alpha}=(1 / 2)\left(t_{\alpha, 1}+t_{\alpha, 2}\right) \tag{32} \end{equation*}(32)tα=(1/2)(tα,1+tα,2)
is homogeneous and continuous (according to assertion a of Theorem 3).
Since T α ( F ) ⊆ Q α ( F ) T α ( F ) ⊆ Q α ( F ) T_(alpha)(F)subeQ_(alpha)(F)T_{\alpha}(F) \subseteq Q_{\alpha}(F)Tα(F)⊆Qα(F), for all 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], it follows that the selection t α t α t_(alpha)t_{\alpha}tα defined by (32) is a selection for Q α Q α Q_(alpha)Q_{\alpha}Qα, too.
Remark 4. For α = 1 α = 1 alpha=1\alpha=1α=1 one obtains P x ⊥ = T 1 = Q 1 P x ⊥ = T 1 = Q 1 P_(x^(_|_))=T_(1)=Q_(1)P_{x^{\perp}}=T_{1}=Q_{1}Px⊥=T1=Q1 implying that T 1 T 1 T_(1)T_{1}T1 and Q 1 Q 1 Q_(1)Q_{1}Q1 are single valued and therefore are linear and continuous applications from Lip 0 [ 0 , 1 ] Lip 0 [ 0 , 1 ] Lip_(0)[0,1]\operatorname{Lip}_{0}[0,1]Lip0[0,1] to Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥.

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  5. Mc Shane, E. J., Extension of Range of Functions, Bull. Amer. Math. Soc. 40 (1934), 837-842.
  6. Mustăta, C., Norm Preserving Extension of Starshaped Lipschitz Functions, Mathematica 19(42)2 (1977), 183-187.
  7. Mustăţa, C., Best Approximation and Unique Extension of Lipschilz Funclions, J. Approx. Theory 19(1977), 222-230.
  8. Mustăta, C., M-ideals in Metric Spaces, „Babeş-Bolyai" Universily, Fac. of Math. and Physics, Research Seminars, Seminar on Math. Analysis, Preprint № 7 (1988), 65-74.
  9. Mustăţa, C., Selections Associated to Mc Shane's Extension Theorem for Lipschitz Functions, Revue d'Analyse Numérique et de Théorie de l'Approx. 21, 2 (1992), 135-145.
Received 2.IV,1993
Institutul de Calcul
Str. Republicii Nr. 37
Oficiul Postal 1
3400 Cluj-Napoca
România
1993

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