Selections associated to McShane’s extension theorem for Lipschitz functions

Costica Mustata
(original), Approximation Theory, paper

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

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C. Mustăţa, Selections associated to McShane’s extension theorem for Lipschitz functions, Rev. Anal. Numér. Théor. Approx. 21 (1992) 2, 135-145 (MR # 94h: 46008).

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MR # 94h: 46008

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[1] Alfsen, E.M., Effros, E., Structure in real Banach spaces, Ann. of Math. 96 (1972), 98-173.
[2] Azipser, J., Géher, L., Extension of Functions satisfying a Lipschitz conditions Acta Math. Acad. Sci. Hungar 6 (1955), 213-220.
[3] Deutsch, F.., Wu Li, Sung-Ho Park, Tietze Extensions and Conditions Selections for Metric Projecitons J.A.T. 64 (1991), 55-68.
[4] Fakhoury, H., SéIections linéaires associées au théorème de Hahn-Banach, J. Funct. Analysis  11 (7972),436-452.
[5] Mc shane; E, J., Extensions of range of functions, Bull, Amer. Math. Soc. 40 (1934), 837-842.
[6] Mustata C., Best Approximation and Unique Extension of Lipschilz Functions’J.A.T. 19 (1977), 222-230.
[7] Mustata, C., M- ideals in metric spaccs, “Babes-Bolyai,, University, Fac. of Math. and Physics, Research Seminars, Seminar on Math. Anal., preprint Nr. 7 (1988), 65-74.
8. Rudin W., Functional Analysis, McGraw-Hill 1973.

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1992-Mustata-Selections associated to McShane’s extension-Jnaat

SELECTIONS ASSOCIATED TO McSHANE'S EXTENSION THEOREM FOR LIPSCHITZ FUNCTIONS

COSTICA MUSTATA(Cluj-Napocal)

  1. Let ( X , d X , d X,dX, dX,d ) be a metric space and Y Y YYY a nonvoid subset of X X XXX. A fung tion f : Y → R f : Y → R f:Y rarr Rf: Y \rightarrow Rf:Y→R is called Lipschitz on Y ¯ Y ¯ bar(Y)\bar{Y}Y¯ if there exists K ⩾ 0 K ⩾ 0 K >= 0K \geqslant 0K⩾0. such that
    ( 1 , 1 1 ) 1 , ∵ | f ( x ) − f ( y ) | ⩽ I d ( x , y ) 1 , 1 1 1 , ∵ | f ( x ) − f ( y ) | ⩽ I d ( x , y ) (1,(1)/(1))_(1),quad:'|f(x)-f(y)| <= Id(x,y)\left(1, \frac{1}{1}\right)_{1}, \quad \because|f(x)-f(y)| \leqslant I d(x, y)(1,11)1,∵|f(x)−f(y)|⩽Id(x,y),
    for all x , y ∈ Y x , y ∈ Y x,y in Yx, y \in Yx,y∈Y. A number K ⩾ 0 K ⩾ 0 K >= 0K \geqslant 0K⩾0 verifying (1.1) is called a Lipschiti constant for f f fff.
It is easy to show that the quantity ‖ f ‖ X ‖ f ‖ X ||f||_(X)\|f\|_{X}‖f‖X defined by
(1.2) ‖ f ‖ X = sup { | f ( x ) − f ( y ) | / d ( x , y ) : x , y ∈ Y , x ≠ y } (1.2) ‖ f ‖ X = sup { | f ( x ) − f ( y ) | / d ( x , y ) : x , y ∈ Y , x ≠ y } {:(1.2)||f||_(X)=s u p{|f(x)-f(y)|//d(x","y):x","y in Y","x!=y}:}\begin{equation*} \|f\|_{X}=\sup \{|f(x)-f(y)| / \mathrm{d}(x, y): x, y \in Y, x \neq y\} \tag{1.2} \end{equation*}(1.2)‖f‖X=sup{|f(x)−f(y)|/d(x,y):x,y∈Y,x≠y}
is the smallest Lipschitz constant for f f fff and we shall call it the Lipschitz norm of f f fff.
Denote by Lip Y Y YYY the set of all real-valued Lipsehitz functions on Y . Y .  Y_(". ")Y_{\text {. }}Y.  The set Lip X X XXX and the quantity ‖ f ‖ x ‖ f ‖ x ||f||_(x)\|f\|_{x}‖f‖x are defined similarly.
Obviously, with respect to the usual operations of addition and multiplication by scalars for functions, Lip Y Y YYY and Lip X X XXX are linear spaces and the functionals ‖ ‖ Y ‖ Y ||||_(Y):}\left\|\|_{Y}\right.‖‖Y and ‖ ‖ X ‖ ‖ X ||||_(X)\| \|_{X}‖‖X are seminorms on these spaces. The functionals ‖ ‖ r ‖ r ||||_(r):}\left\|\|_{r}\right.‖‖r and ‖ ‖ x ‖ ‖ x ||||_(x)\| \|_{x}‖‖x are not norms because they vanish on constant functions.
Mc Shane [5] proved the following extension theorem for Lipschitz functions:
Theorem 1. Let ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a metric space and ∅ ≠ Y ⊂ X ∅ ≠ Y ⊂ X O/_(!=Y sub X)\varnothing_{\neq Y \subset X}∅≠Y⊂X. Then every Lipschitz function on Y Y YYY has a norm preserving extensions to X X XXX, i.e. for every f ∈ f ∈ f inf \inf∈ Lip Y Y YYY there exists F ∈ F ∈ F inF \inF∈ Lip X X XXX such that
F | Y = f and ‖ F ‖ x = ‖ f ‖ Y . F Y = f  and  ‖ F ‖ x = ‖ f ‖ Y . F|_(Y)=f" and "||F||_(x)=||f||_(Y).\left.F\right|_{Y}=f \text { and }\|F\|_{x}=\|f\|_{Y} .F|Y=f and ‖F‖x=‖f‖Y.
Now let x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y be fixed and denote by Lip 0 Y 0 Y _(0)Y{ }_{0} Y0Y the subspace of Lip Y Y YYY formed of all functions in Lip Y Y YYY vanishing at x 0 x 0 x_(0)x_{0}x0, i.e.
(1.3) Lip θ Y = { f ∈ Lip Y : f ( x θ ) = 0 } . (1.3) Lip θ Y = f ∈ Lip Y : f x θ = 0 . {:(1.3)Lip_(theta)Y={f in Lip Y:f(x_(theta))=0}.:}\begin{equation*} \operatorname{Lip}_{\theta} Y=\left\{f \in \operatorname{Lip} Y: f\left(x_{\theta}\right)=0\right\} . \tag{1.3} \end{equation*}(1.3)LipθY={f∈LipY:f(xθ)=0}.
s. As ‖ F ‖ x = ‖ f ‖ y ‖ F ‖ x = ‖ f ‖ y ||F||_(x)=||f||_(y)\|F\|_{x}=\|f\|_{y}‖F‖x=‖f‖y, for all F ∈ E ( f ) F ∈ E ( f ) F in E(f)F \in E(f)F∈E(f), it follows that E ( f ) E ( f ) E(f)E(f)E(f) is bounded.
To prove the closedness of E ( f ) E ( f ) E(f)E(f)E(f), let ( F n ) F n (F_(n))\left(F_{n}\right)(Fn) be a sequence in E ( f ) E ( f ) E(f)E(f)E(f) converging to a function F ′ ∈ Lip p 0 X F ′ ∈ Lip p 0 X F^(')in Lipp_(0)XF^{\prime} \in \operatorname{Lip} p_{0} XF′∈Lipp0X. As l ′ ( x 0 ) = 0 = F n ( x 0 ) l ′ x 0 = 0 = F n x 0 l^(')(x_(0))=0=F_(n)(x_(0))l^{\prime}\left(x_{0}\right)=0=F_{n}\left(x_{0}\right)l′(x0)=0=Fn(x0), for all n ∈ N n ∈ N n in Nn \in Nn∈N, it follows that
| F n ( x ) : − F ( x ) | = | F n ( x ) + F ( x ) − ( F n ( x 0 ) − F ( x 0 ) ) | ⋯ ⩽ ⩽ ‖ F n − F ‖ x d ( x ; x 0 ) → 0 F n ( x ) : − F ( x ) = F n ( x ) + F ( x ) − F n x 0 − F x 0 ⋯ ⩽ ⩽ F n − F x d x ; x 0 → 0 {:[|F_(n)(x):-F(x)|=|F_(n)(x)+F(x)-(F_(n)(x_(0))-F(x_(0)))|cdots <= ],[ <= ||F_(n)-F||_(x)d(x;x_(0))rarr0]:}\begin{aligned} \left|F_{n}(x):-F(x)\right| & =\left|F_{n}(x)+F(x)-\left(F_{n}\left(x_{0}\right)-F\left(x_{0}\right)\right)\right| \cdots \leqslant \\ & \leqslant\left\|F_{n}-F\right\|_{x} \mathrm{~d}\left(x ; x_{0}\right) \rightarrow 0 \end{aligned}|Fn(x):−F(x)|=|Fn(x)+F(x)−(Fn(x0)−F(x0))|⋯⩽⩽‖Fn−F‖x d(x;x0)→0
for every x ∈ X x ∈ X x in Xx \in Xx∈X, implying the pointwise convergence of the sequence ( F n ( x ) ) F n ( x ) (F_(n)(x))\left(F_{n}(x)\right)(Fn(x)) to F ′ ( x ) F ′ ( x ) F^(')(x)F^{\prime}(x)F′(x), for all x ∈ X x ∈ X x in Xx \in Xx∈X. Since F n ( x ) = f ( x ) F n ( x ) = f ( x ) F_(n)(x)=f(x)F_{n}(x)=f(x)Fn(x)=f(x) for all x ∈ T ⊤ x ∈ T ⊤ x inT^(TT)x \in T^{\top}x∈T⊤ and all n ∈ N n ∈ N n in Nn \in Nn∈N it. follows F ( x ) = f ( x ) F ( x ) = f ( x ) F(x)=f(x)F(x)=f(x)F(x)=f(x), for all x ∈ Y x ∈ Y x in Yx \in Yx∈Y. Also F n → F 2 F n → F 2 F_(n)rarrF^(2)F_{n} \rightarrow F^{2}Fn→F2 in Lip 0 X 0 X _(0)X{ }_{0} X0X and ‖ F n ‖ x = ‖ f ‖ x , n ∈ N F n x = ‖ f ‖ x , n ∈ N ||F_(n)||_(x)=||f||_(x),n in N\left\|F_{n}\right\|_{x}=\|f\|_{x}, n \in N‖Fn‖x=‖f‖x,n∈N, imply ‖ F ′ ‖ x = lim n → ∞ ‖ F n ‖ x = ‖ f ‖ y F ′ x = lim n → ∞   F n x = ‖ f ‖ y ||F^(')||_(x)=lim_(n rarr oo)||F_(n)||_(x)=||f||_(y)\left\|F^{\prime}\right\|_{x}=\lim _{n \rightarrow \infty}\left\|F_{n}\right\|_{x}=\|f\|_{y}‖F′‖x=limn→∞‖Fn‖x=‖f‖y showing that F ′ ∈ F ( f ) F ′ ∈ F ( f ) F^(')inF(f)F^{\prime} \in \mathbb{F}(f)F′∈F(f);
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘. To prove the second inequality in (1.8) suppose, on the contrary, that there exists x 1 ∈ X x 1 ∈ X x_(1)in Xx_{1} \in Xx1∈X such that F ′ ( x 1 ) > F 2 ( x 1 ) F ′ x 1 > F 2 x 1 F^(')(x_(1)) > F_(2)(x_(1))F^{\prime}\left(x_{1}\right)>F_{2}\left(x_{1}\right)F′(x1)>F2(x1). Since the functions F F FFF, F 2 F 2 F_(2)F_{2}F2 are continuous on X X XXX and F ′ | X = f = F 2 | Y F ′ X = f = F 2 Y F^(')|_(X)=f=F_(2)|_(Y)\left.F^{\prime}\right|_{X}=f=\left.F_{2}\right|_{Y}F′|X=f=F2|Y, it follows that they agree on the closure Y ¯ Y ¯ bar(Y)\bar{Y}Y¯ of Y Y YYY. Therefore x 1 ∈ X ∖ Y ¯ x 1 ∈ X ∖ Y ¯ x_(1)in X\\ bar(Y)x_{1} \in X \backslash \bar{Y}x1∈X∖Y¯, implying d ( x 1 , y ) > 0 d x 1 , y > 0 d(x_(1),y) > 0\mathrm{d}\left(x_{1}, y\right)>0d(x1,y)>0 for all y ∈ Y y ∈ Y y in Yy \in Yy∈Y. Taking into account definition (1.6) of F 2 F 2 F_(2)F_{2}F2 and the inequality F 2 ( x 1 ) < F ( x 1 ) F 2 x 1 < F x 1 F_(2)(x_(1)) < F(x_(1))F_{2}\left(x_{1}\right)< F\left(x_{1}\right)F2(x1)<F(x1), there exists an element y 1 ∈ Y y 1 ∈ Y y_(1)in Yy_{1} \in Yy1∈Y such that
(1.9) f ( y 1 ) + ‖ f ‖ Y d ( x 1 , y 1 ) < F ( x 1 ) . (1.9) f y 1 + ‖ f ‖ Y d x 1 , y 1 < F x 1 . {:(1.9)f(y_(1))+||f||_(Y)d(x_(1),y_(1)) < F(x_(1)).:}\begin{equation*} f\left(y_{1}\right)+\|f\|_{Y} \mathrm{~d}\left(x_{1}, y_{1}\right)<F\left(x_{1}\right) . \tag{1.9} \end{equation*}(1.9)f(y1)+‖f‖Y d(x1,y1)<F(x1).
As f ( y 1 ) = F ( y 1 ) f y 1 = F y 1 f(y_(1))=F(y_(1))f\left(y_{1}\right)=F\left(y_{1}\right)f(y1)=F(y1), the inequality (1.9), gives the contradiction ‖ f ‖ r = ‖ f ‖ r = ||f||_(r)=\|f\|_{r}=‖f‖r= : = ‖ F ‖ X ⩾ ( F ′ ( x 1 ) − F ′ ( y 1 ) ) / d ( x 1 , y 1 ) > ‖ f ‖ Y = ‖ F ‖ X ⩾ F ′ x 1 − F ′ y 1 / d x 1 , y 1 > ‖ f ‖ Y =||F||_(X) >= (F^(')(x_(1))-F^(')(y_(1)))//d(x_(1),y_(1)) > ||f||_(Y)=\|F\|_{X} \geqslant\left(F^{\prime}\left(x_{1}\right)-F^{\prime}\left(y_{1}\right)\right) / \mathrm{d}\left(x_{1}, y_{1}\right)>\|f\|_{Y}=‖F‖X⩾(F′(x1)−F′(y1))/d(x1,y1)>‖f‖Y.
The first inequality in (1,8) can be proved in a similar way.
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘. Let B Lip 0 Y , B Lip 0 x B Lip 0 Y , B Lip 0 x B_(Lip_(0)Y),B_(Lip_(0)x)B_{\mathrm{Lip}_{0} \mathrm{Y}}, B_{\mathrm{Lip}_{0} \mathrm{x}}BLip0Y,BLip0x be the closed unit balls of the spaces Lip 0 Y 0 Y _(0)Y{ }_{0} Y0Y respectively Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X. Let f f fff be an extremal point of B Lipy B Lipy  B_("Lipy ")B_{\text {Lipy }}BLipy  and stippose that H 1 , H 2 ∈ B Lip , x H 1 , H 2 ∈ B Lip  , x H_(1),H_(2)inB_("Lip ",x)H_{1}, H_{2} \in B_{\text {Lip }, \mathrm{x}}H1,H2∈BLip ,x and λ ∈ ( 0 , 1 ) λ ∈ ( 0 , 1 ) lambda in(0,1)\lambda \in(0,1)λ∈(0,1) are such that λ H 1 + ( 1 − λ ) H 2 ∈ E ( f ) λ H 1 + ( 1 − λ ) H 2 ∈ E ( f ) lambdaH_(1)+(1-lambda)H_(2)in E(f)\lambda H_{1}+(1-\lambda) H_{2} \in E(f)λH1+(1−λ)H2∈E(f). Since f f fff is extremal the equality λ H 1 | Y + ( 1 − λ ) H 2 | Y = f λ H 1 Y + ( 1 − λ ) H 2 Y = f lambdaH_(1)|_(Y)+(1-lambda)H_(2)|_(Y)=f\left.\lambda H_{1}\right|_{Y}+\left.(1-\lambda) H_{2}\right|_{Y}=fλH1|Y+(1−λ)H2|Y=f implies H 1 | Y == f = H 2 | x H 1 Y == f = H 2 x H_(1)|_(Y)==f=H_(2)|_(x)\left.H_{1}\right|_{Y}= =f=\left.H_{2}\right|_{x}H1|Y==f=H2|x. Also 1 = ‖ f ‖ Y ⩽ ‖ λ H 1 + ( 1 − λ ) H 2 ‖ x ⩽ λ ‖ H x ‖ x + ( 1 λ ) ‖ H 2 ‖ x ⩽ 1 1 = ‖ f ‖ Y ⩽ λ H 1 + ( 1 − λ ) H 2 x ⩽ λ H x x + ( 1 λ ) H 2 x ⩽ 1 1=||f||_(Y) <= ||lambdaH_(1)+(1-lambda)H_(2)||_(x) <= lambda||H_(x)||_(x)+(1quad lambda)||H_(2)||_(x) <= 11=\|f\|_{Y} \leqslant\left\|\lambda H_{1}+(1-\lambda) H_{2}\right\|_{x} \leqslant \lambda\left\|H_{x}\right\|_{x}+(1 \quad \lambda) \left\|H_{2}\right\|_{x} \leqslant 11=‖f‖Y⩽‖λH1+(1−λ)H2‖x⩽λ‖Hx‖x+(1λ)‖H2‖x⩽1 implies ‖ H 1 ‖ x = ‖ H 2 ‖ x = 1 = ‖ f ‖ x H 1 x = H 2 x = 1 = ‖ f ‖ x ||H_(1)||_(x)=||H_(2)||_(x)=1=||f||_(x)\left\|H_{1}\right\|_{x}=\left\|H_{2}\right\|_{x}=1=\|f\|_{x}‖H1‖x=‖H2‖x=1=‖f‖x, showing that H 1 , H 2 ∈ E ( f ) H 1 , H 2 ∈ E ( f ) H_(1),H_(2)in E(f)H_{1}, H_{2} \in E(f)H1,H2∈E(f).
Now, suppose that f 2 ‖ f ‖ Y = 1 f 2 ‖ f ‖ Y = 1 f_(2)||f||_(Y)=1f_{2}\|f\|_{Y}=1f2‖f‖Y=1, is not an extremal point of B Lip 0 Y B Lip 0 Y B_(Lip_(0)Y)B_{\mathrm{Lip}_{0} Y}BLip0Y. Then there exist two distinct elements f 1 , f 2 f 1 , f 2 f_(1),f_(2)f_{1}, f_{2}f1,f2 in B Lip 0 Y B Lip  0 Y B_("Lip "_(0))YB_{\text {Lip }_{0}} YBLip 0Y and λ ∈ ( 0 , 1 ) λ ∈ ( 0 , 1 ) lambda in(0,1)\lambda \in(0,1)λ∈(0,1) such that f = λ f 1 + ( 1 − λ ) f 2 f = λ f 1 + ( 1 − λ ) f 2 f=lambdaf_(1)+(1-lambda)f_(2)f=\lambda f_{1}+(1-\lambda) f_{2}f=λf1+(1−λ)f2. If H i ∈ E ( f i ) , i = 1 , 2 H i ∈ E f i , i = 1 , 2 H_(i)in E(f_(i)),i=1,2H_{i} \in E\left(f_{i}\right), i=1,2Hi∈E(fi),i=1,2, then H 1 | x + ( 1 − λ ) H 2 | x == λ f 1 + ( 1 − λ ) f 2 = f H 1 x + ( 1 − λ ) H 2 x == λ f 1 + ( 1 − λ ) f 2 = f H_(1)|_(x)+(1-lambda)H_(2)|_(x)==lambdaf_(1)+(1-lambda)f_(2)=f\left.H_{1}\right|_{x}+\left.(1-\lambda) H_{2}\right|_{x}= =\lambda f_{1}+(1-\lambda) f_{2}=fH1|x+(1−λ)H2|x==λf1+(1−λ)f2=f and 1 = ‖ f ‖ x = ‖ λ H 1 ‖ x + ( 1 − λ ) H 2 | x ‖ x ⩽⩽ ‖ λ H 1 + ( 1 − λ ) H 2 ‖ x ⩽ 1 1 = ‖ f ‖ x = λ H 1 x + ( 1 − λ ) H 2 x ‖ x ⩽⩽ λ H 1 + ( 1 − λ ) H 2 x ⩽ 1 1=||f||_(x)=||lambdaH_(1)||_(x)+(1-lambda)H_(2)|_(x)||_(x)⩽⩽||lambdaH_(1)+(1-lambda)H_(2)||_(x) <= 11=\|f\|_{x}=\left\|\lambda H_{1}\right\|_{x}+\left.(1-\lambda) H_{2}\right|_{x} \|_{x} \leqslant \leqslant\left\|\lambda H_{1}+(1-\lambda) H_{2}\right\|_{x} \leqslant 11=‖f‖x=‖λH1‖x+(1−λ)H2|x‖x⩽⩽‖λH1+(1−λ)H2‖x⩽1, showing that λ H 1 + ( 1 − λ ) H 2 ∈ E ( f ) λ H 1 + ( 1 − λ ) H 2 ∈ E ( f ) lambdaH_(1)+(1-lambda)H_(2)in E(f)\lambda H_{1}+(1-\lambda) H_{2} \in E(f)λH1+(1−λ)H2∈E(f). Since H i | r = f i ≠ f , i = 1 , 2 H i r = f i ≠ f , i = 1 , 2 H_(i)|_(r)=f_(i)!=f,i=1,2\left.H_{i}\right|_{r}=f_{i} \neq f, i=1,2Hi|r=fi≠f,i=1,2, it follows that H i ∈ E ( f ) , i = 1 , 2 H i ∈ E ( f ) , i = 1 , 2 H_(i)in E(f),i=1,2H_{i} \in E(f), i=1,2Hi∈E(f),i=1,2 and E ( f ) E ( f ) E(f)E(f)E(f) is not a face of B △ p 0 x x B △ p 0 x x B_(/_\p_(0)x)xB_{\triangle \mathrm{p}_{0} x} xB△p0xx.
Remark 1. The extensions F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2 of x x xxx function f ∈ Lip p 0 Y f ∈ Lip p 0 Y f in Lipp_(0)Yf \in \operatorname{Lip} p_{0} Yf∈Lipp0Y, given by ( 1.5 ) ( 1.5 ) (1.5)(1.5)(1.5) and ( 1.6 ) ( 1.6 ) (1.6)(1.6)(1.6), are extremal points of the face E ( f ) E ( f ) E(f)E(f)E(f) and consequently they are extremal elements of the unit ball B 1 + D 0 x B 1 + D 0 x B_(1+D_(0)x)B_{1+D_{0} x}B1+D0x, too. Indeed, if F , G ∈∈ B ( f ) F , G ∈∈ B ( f ) F,G∈∈B(f)F, G \in \in B(f)F,G∈∈B(f) and λ ∈ ( 0 , 1 ) λ ∈ ( 0 , 1 ) lambda in(0,1)\lambda \in(0,1)λ∈(0,1) are such that F 1 = λ F + ( 1 − λ ) G F 1 = λ F + ( 1 − λ ) G F_(1)=lambda F+(1-lambda)GF_{1}=\lambda F+(1-\lambda) GF1=λF+(1−λ)G then, by (1.8), F 1 ⩽ F F 1 ⩽ F F_(1) <= FF_{1} \leqslant FF1⩽F and F 1 ⩽ G F 1 ⩽ G F_(1) <= GF_{1} \leqslant GF1⩽G implying F = F 1 = G F = F 1 = G F=F_(1)=GF=F_{1}=GF=F1=G. The extremality of F 2 F 2 F_(2)F_{2}F2 is proved similarly.

2. Let

Y ⊥ = { F n i ∈ Lip 0 X n ′ : F n i M ∣ Y j = 0 } , Y ⊥ = F n i ∈ Lip 0 X n ′ : F n i M ∣ Y j = 0 , Y^(_|_)={F_(n_(i))inLip_(0)X_(n^(')):F_(n_(i))^(M)∣Y_(j)=0},Y^{\perp}=\left\{F_{n_{i}} \in \operatorname{Lip}_{0} X_{n^{\prime}}: F_{n_{i}}^{M} \mid Y_{j}=0\right\},Y⊥={Fni∈Lip0Xn′:FniM∣Yj=0},
be the annihilator subspace of Y Y YYY in and X X XXX.
(1.8) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) , x ∈ X , (1.8) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) , x ∈ X , {:(1.8)F_(1)(x) <= F(x) <= F_(2)(x)","x in X",":}\begin{equation*} F_{1}(x) \leqslant F(x) \leqslant F_{2}(x), x \in X, \tag{1.8} \end{equation*}(1.8)F1(x)⩽F(x)⩽F2(x),x∈X,
where F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2 are the extensions of f f fff given by (1.5) and (1.6) ;
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘ For ‖ f ‖ Y = 1 ‖ f ‖ Y = 1 ||f||_(Y)=1\|f\|_{Y}=1‖f‖Y=1, the set E ( f ) E ( f ) E(f)E(f)E(f) is a face of the unit ball B Lip X B Lip  X B_("Lip "_(X))B_{\text {Lip }{ }_{X}}BLip X if and only if f f fff is an extremal point of the unit ball of Lip 0 Y 0 Y _(0)Y{ }_{0} \mathrm{Y}0Y.
Proof. 1 ∘ 1 ∘ 1^(@)1^{\circ}1∘. Let F , G ∈ E ( f ) F , G ∈ E ( f ) F,G in E(f)F, G \in E(f)F,G∈E(f) and λ ∈ [ 0 , 1 ] λ ∈ [ 0 , 1 ] lambda in[0,1]\lambda \in[0,1]λ∈[0,1]. Obviously that ( λ F + ( 1 − − λ ) G ) | X = f ( λ F + ( 1 − − λ ) G ) X = f (lambda F+(1--lambda)G)|_(X)=f(\lambda F+(1- -\lambda) G)\left.\right|_{X}=f(λF+(1−−λ)G)|X=f and ‖ λ F + ( 1 − λ ) G ‖ X ⩽ λ ‖ F ‖ X + ( 1 − λ ) ‖ G ‖ X == λ ‖ f ′ ‖ X + ( 1 − λ ) ‖ f ‖ Y = ‖ f ‖ Y ‖ λ F + ( 1 − λ ) G ‖ X ⩽ λ ‖ F ‖ X + ( 1 − λ ) ‖ G ‖ X == λ f ′ X + ( 1 − λ ) ‖ f ‖ Y = ‖ f ‖ Y ||lambda F+(1-lambda)G||_(X) <= lambda||F||_(X)+(1-lambda)||G||_(X)==lambda||f^(')||_(X)+(1-lambda)||f||_(Y)=||f||_(Y)\|\lambda F+(1-\lambda) G\|_{X} \leqslant \lambda\|F\|_{X}+(1-\lambda)\|G\|_{X}= =\lambda\left\|f^{\prime}\right\|_{X}+(1-\lambda)\|f\|_{Y}=\|f\|_{Y}‖λF+(1−λ)G‖X⩽λ‖F‖X+(1−λ)‖G‖X==λ‖f′‖X+(1−λ)‖f‖Y=‖f‖Y. Since ‖ λ F + ( 1 − λ ) G ‖ X ⩾ ‖ ( λ F + + ( 1 − λ ) G | Y ‖ Y = ‖ f ‖ Y ‖ λ F + ( 1 − λ ) G ‖ X ⩾ ‖ ( λ F + + ( 1 − λ ) G Y Y = f ‖ Y ||lambda F+(1-lambda)G||_(X) >= ||(lambda F++(1-lambda)G|_(Y)||_(Y)=||f||_(Y)\|\lambda F+(1-\lambda) G\|_{X} \geqslant \|(\lambda F+ +\left.(1-\lambda) G\right|_{Y}\left\|_{Y}=\right\| f \|_{Y}‖λF+(1−λ)G‖X⩾‖(λF++(1−λ)G|Y‖Y=‖f‖Y, it follows that λ F ′ + ( 1 − λ ) G ∈ E ( f ) λ F ′ + ( 1 − λ ) G ∈ E ( f ) lambdaF^(')+(1-lambda)G in E(f)\lambda F^{\prime}+(1-\lambda) G \in E(f)λF′+(1−λ)G∈E(f), proving the convexity of E ( f ) E ( f ) E(f)E(f)E(f).
A subset Λ Λ Lambda\LambdaΛ of Lip 0 X 0 X _(0)X{ }_{0} X0X is called proximinal in Lip 0 X 0 X _(0)X{ }_{0} X0X if every F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X has a nearest point in Λ Λ Lambda\LambdaΛ, i.e. there exists G ∈ A G ∈ A G in AG \in AG∈A such that ‖ F − G ‖ x ¯ =⇋ d ( F , Λ ) ‖ F − G ‖ x ¯ =⇋ d ( F , Λ ) ||F-G||_( bar(x))= ⇋d(F,Lambda)\|F-G\|_{\bar{x}}= \leftrightharpoons \mathrm{d}(F, \Lambda)‖F−G‖x¯=⇋d(F,Λ), where
d ( F , Λ ) = inf ′ { ‖ F − H ‖ X : H ∈ Λ } . d ( F , Λ ) = inf ′   ‖ F − H ‖ X : H ∈ Λ . d(F,Lambda)=i n f'{||F-H||_(X):H in Lambda}.\mathrm{d}(F, \Lambda)=\inf ^{\prime}\left\{\|F-H\|_{X}: H \in \Lambda\right\} .d(F,Λ)=inf′{‖F−H‖X:H∈Λ}.
The metric projection P Λ : P Λ : P_(Lambda):P_{\Lambda}:PΛ: Lip 0 X → 2 Λ 0 X → 2 Λ _(0)X rarr2^(Lambda){ }_{0} X \rightarrow 2^{\Lambda}0X→2Λ is defined by
P Λ ( F ) = { G ∈ Λ : ‖ F − G ‖ x = d ( F , A ) } . P Λ ( F ) = G ∈ Λ : ‖ F − G ‖ x = d ( F , A ) . P_(Lambda)(F)={G in Lambda:||F-G||_(x)=d(F,(A))}.P_{\Lambda}(F)=\left\{G \in \Lambda:\|F-G\|_{x}=\mathrm{d}(F, \mathrm{~A})\right\} .PΛ(F)={G∈Λ:‖F−G‖x=d(F, A)}.
If P Λ ( F ) P Λ ( F ) P_(Lambda)(F)P_{\Lambda}(F)PΛ(F) is a singleton for every F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X then Λ Λ Lambda\LambdaΛ is called a Ohebyshevian subset of: Lip 0 X 0 X _(0)X_{0} X0X.
There is a closed relation between the extension operator E E EEE and the projection pperator P Y ⊥ P Y ⊥ P_(Y^(_|_))P_{Y^{\perp}}PY⊥, ilustrated in the following theorem:
Theorem 3. The following assertions hold:
1 ∘ 1 ∘ 1^(@)1^{\circ}1∘ The subspace Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is proximinal in Lip 0 X 0 X _(0)X_{0} X0X;
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘ The equality
(2.1) d ( F ′ , Y ⊥ ) ≐ ‖ F ′ ) = ‖ F ′ H ′ (2.1) d F ′ , Y ⊥ ≐ ‖ F ′ = ‖ F ′ H ′ {:(2.1){:d(F^('),Y^(_|_))≐||F^('))=||F^(')H^('):}\begin{equation*} \left.\mathrm{d}\left(F^{\prime}, \mathrm{Y}^{\perp}\right) \doteq \| F^{\prime}\right)=\| F^{\prime} H^{\prime} \tag{2.1} \end{equation*}(2.1)d(F′,Y⊥)≐‖F′)=‖F′H′
is true for all F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X;
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘ A function G ∈ Y ⊥ G ∈ Y ⊥ G in Y _|_G \in Y \perpG∈Y⊥ is a best approximation element for F F FFF if and only if there exists H ∈ E ( F | Y ) H ∈ E F Y H in E(F|_(Y))H \in E\left(\left.F\right|_{Y}\right)H∈E(F|Y) such that G = F − H G = F − H G=F-HG=F-HG=F−H, or equivalently
(2.2) P Y ⊥ ( F ) = F − E ( F | Y ) . (2.2) P Y ⊥ ( F ) = F − E F Y . {:(2.2)P_(Y _|_)(F)=F-E(F|_(Y)).:}\begin{equation*} P_{Y \perp}(F)=F-E\left(\left.F\right|_{Y}\right) . \tag{2.2} \end{equation*}(2.2)PY⊥(F)=F−E(F|Y).
Proof. First we prove formula (2.1). If F ∈ Lip 0 X then for any G ∈ Y ⊥ F ∈ Lip 0 X then for any  G ∈ Y ⊥ F inLip_(0)X_("then for any ")G inY^(_|_)F \in \operatorname{Lip}_{0} X_{\text {then for any }} G \in \mathbf{Y}^{\perp}F∈Lip0Xthen for any G∈Y⊥,
‖ F | x ‖ Y = sup { | F ′ ( y ) − F ( y ′ ) | / d ( y , y ′ ) : y , y ′ ∈ Y , y ≠ y ′ } = = sup { | ( F − G ) ( y ) − ( F − G ) ( y ′ ) | / d ( y , y ′ ) : y , y ′ ∈ Y , y ≠ y ′ } ⩽ sup { | ( F − G ) ( x ) − ( F − G ) ( x ′ ) | / d ( x , x ′ ) : x , x ′ ∈ X , x ≠ x ′ } = F x Y = sup F ′ ( y ) − F y ′ / d y , y ′ : y , y ′ ∈ Y , y ≠ y ′ = = sup ( F − G ) ( y ) − ( F − G ) y ′ / d y , y ′ : y , y ′ ∈ Y , y ≠ y ′ ⩽ sup ( F − G ) ( x ) − ( F − G ) x ′ / d x , x ′ : x , x ′ ∈ X , x ≠ x ′ = {:[||F|_(x)||_(Y)=s u p{|F^(')(y)-F(y^('))|//d(y,y^(')):y,y^(')in Y,y!=y^(')}=],[=s u p{|(F-G)(y)-(F-G)(y^('))|//d(y,y^(')):y,y^(')in Y,y!=y^(')}],[ <= s u p{|(F-G)(x)-(F-G)(x^('))|//d(x,x^(')):x,x^(')in X,x!=x^(')}],[=]:}\begin{aligned} \left\|\left.F\right|_{x}\right\|_{Y} & =\sup \left\{\left|F^{\prime}(y)-F\left(y^{\prime}\right)\right| / \mathrm{d}\left(y, y^{\prime}\right): y, y^{\prime} \in Y, y \neq y^{\prime}\right\}= \\ & =\sup \left\{\left|(F-G)(y)-(F-G)\left(y^{\prime}\right)\right| / \mathrm{d}\left(y, y^{\prime}\right): y, y^{\prime} \in Y, y \neq y^{\prime}\right\} \\ & \leqslant \sup \left\{\left|(F-G)(x)-(F-G)\left(x^{\prime}\right)\right| / \mathrm{d}\left(x, x^{\prime}\right): x, x^{\prime} \in X, x \neq x^{\prime}\right\} \\ & = \end{aligned}‖F|x‖Y=sup{|F′(y)−F(y′)|/d(y,y′):y,y′∈Y,y≠y′}==sup{|(F−G)(y)−(F−G)(y′)|/d(y,y′):y,y′∈Y,y≠y′}⩽sup{|(F−G)(x)−(F−G)(x′)|/d(x,x′):x,x′∈X,x≠x′}=
implying that ‖ F | Y ‖ Y ⩽ d ( F , Y ⊥ ) F Y Y ⩽ d F , Y ⊥ ||F|_(Y)||_(Y) <= d(F,Y^(_|_))\left\|\left.F\right|_{Y}\right\|_{Y} \leqslant \mathrm{~d}\left(F, Y^{\perp}\right)‖F|Y‖Y⩽ d(F,Y⊥).
On the other hand, by Theorem 1, there exists H ∈ H ∈ H inH \inH∈ Lip 0 X 0 X _(0)X{ }_{0} X0X such that H | Y = F | Y H Y = F Y H|_(Y)=F|_(Y)\left.H\right|_{Y}=\left.F\right|_{Y}H|Y=F|Y and ‖ H ‖ X = ‖ F Y | Y ‖ Y ‖ H ‖ X = F Y Y Y ||H||_(X)=||F^(Y)|_(Y)||_(Y)\|H\|_{X}=\left\|\left.F^{Y}\right|_{Y}\right\|_{Y}‖H‖X=‖FY|Y‖Y. It follows that F F FFF. H ∈ Y ⊥ H ∈ Y ⊥ H inY^(_|_)H \in Y^{\perp}H∈Y⊥ and : ‖ F Y | Y ‖ Y = ‖ F − ( F − H ) x ‖ ⩾ inf { ‖ F − G ‖ x : G ∈ Y ⊥ } = d ( F ′ , Y ⊥ ) F Y Y Y = F − ( F − H ) x ⩾ inf ‖ F − G ‖ x : G ∈ Y ⊥ = d F ′ , Y ⊥ ||F^(Y)|_(Y)||_(Y)=||F-(F-H)_(x)|| >= i n f{||F-G||_(x):G inY^(_|_)}=d(F^('),Y _|_)\left\|\left.F^{Y}\right|_{Y}\right\|_{Y} =\left\|F-(F-H)_{x}\right\| \geqslant \inf \left\{\|F-G\|_{x}: G \in Y^{\perp}\right\}=\mathrm{d}\left(F^{\prime}, Y \perp\right)‖FY|Y‖Y=‖F−(F−H)x‖⩾inf{‖F−G‖x:G∈Y⊥}=d(F′,Y⊥), showing that ‖ F | Y ‖ Y = d ( F , Y ⊥ ) F Y Y = d F , Y ⊥ ||F|_(Y)||_(Y)=d(F,Y^(_|_))\left\|\left.F\right|_{Y}\right\|_{Y}=\mathrm{d}\left(F, Y^{\perp}\right)‖F|Y‖Y=d(F,Y⊥).
Assertion 3 ∘ 3 ∘ 3^(@)3^{\circ}3∘ and formula (2.2) follow from [6], Lemma 1, p. 223 and assertion 1 ∘ 1 ∘ 1^(@)1^{\circ}1∘ follows from 3 ∘ 3 ∘ 3^(@)3^{\circ}3∘.
Now, by Theorem 2, 1 ∘ 1 ∘ 1^(@)1^{\circ}1∘, the set P Y ( T ) = F − E ( F | Y ) P Y ( T ) = F − E F Y P_(Y)(T)=F-E(F|_(Y))P_{Y}(T)=F-E\left(\left.F\right|_{Y}\right)PY(T)=F−E(F|Y) is bounded, convex and closed, for any F ∈ Lip 0 X F ∈ Lip 0 X F inLip_(0)XF \in \operatorname{Lip}_{0} XF∈Lip0X.
We shall say that the set Y ⊂ X Y ⊂ X Y sub XY \subset XY⊂X has the property ( U ) ( U ) (U)(U)(U) if every f ∈∈ Lip 0 Y f ∈∈ Lip 0 Y f∈∈Lip_(0)Yf \in \in \operatorname{Lip}_{0} Yf∈∈Lip0Y has a unique Lipschitz extension F ′ ∈ Lip 0 X F ′ ∈ Lip 0 X F^(')inLip_(0)XF^{\prime} \in \operatorname{Lip}_{0} XF′∈Lip0X, i.e. E ( f ) E ( f ) E(f)E(f)E(f) is a singleton for every f ∈ Lip p 0 Y f ∈ Lip p 0 Y f in Lipp_(0)Yf \in \operatorname{Lip} p_{0} Yf∈Lipp0Y. By Theorem 3, the set Y Y YYY has property ( U ) ( U ) (U)(U)(U) if and only if Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is a Chebyshevian subspace of Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X.
2. A natural question is when have the set valued operators E E EEE and P Y ⊥ P Y ⊥ P_(Y _|_)P_{Y \perp}PY⊥ continuous selections. If S : A → 2 B S : A → 2 B S:A rarr2^(B)S: A \rightarrow 2^{\mathrm{B}}S:A→2B is a set-valued application, a function s : A → B s : A → B s:A rarr Bs: A \rightarrow Bs:A→B is called a selection for S S SSS if s ( x ) ∈ S ( x ) s ( x ) ∈ S ( x ) s(x)in S(x)s(x) \in S(x)s(x)∈S(x), for all x ∈ A x ∈ A x in Ax \in Ax∈A.
In the following theorems, we shall prove the existence of continuous selections for the operators E E EEE and P Y . L P Y . L P_(Y.L)P_{Y . L}PY.L in the particular case X = R X = R X=RX=RX=R, with the usual distance d ( x , y ) = | x − y | d ( x , y ) = | x − y | d(x,y)=|x-y|\mathrm{d}(x, y)=|x-y|d(x,y)=|x−y|.
Theorem 4. Let X = R , Y = [ a , b ] ∈ R X = R , Y = [ a , b ] ∈ R X=R,Y=[a,b]in RX=R, Y=[a, b] \in RX=R,Y=[a,b]∈R and x 0 ∈ [ a , b ] x 0 ∈ [ a , b ] x_(0)in[a,b]x_{0} \in[a, b]x0∈[a,b] fixed. Then the extension operator E : Lip 0 Y → 2 Lip 0 R E : Lip 0 Y → 2 Lip 0 R E:Lip_(0)Y rarr2^(Lip_(0)R)E: \operatorname{Lip}_{0} Y \rightarrow 2^{\operatorname{Lip}_{0} R}E:Lip0Y→2Lip0R has a continuous and positively homogeneous selection e e eee.
Proof. Define ε 2 : Lip 0 Y → Lip 0 R ˙ ε 2 : Lip 0 Y → Lip 0 R ˙ epsi_(2):Lip_(0)Y rarrLip_(0)R^(˙)\varepsilon_{2}: \operatorname{Lip}_{0} Y \rightarrow \operatorname{Lip}_{0} \dot{R}ε2:Lip0Y→Lip0R˙ by
e 2 ( f ) = F 2 , f ∈ Lip 0 Y e 2 ( f ) = F 2 , f ∈ Lip 0 Y e_(2)(f)=F_(2),quad f inLip_(0)Ye_{2}(f)=F_{2}, \quad f \in \operatorname{Lip}_{0} Ye2(f)=F2,f∈Lip0Y
where F 2 F 2 F_(2)F_{2}F2 is the maximal extension of f f fff given by (1.6).
If α > 0 α > 0 alpha > 0\alpha>0α>0 then
e 2 ( a f ) ( x ) = in { a f ( y ) + ‖ a f ‖ Y | x − y | : y ∈ [ a , b ] } = a : inf { f ( y ) + ‖ f ‖ Y | x − y | 1 : y ∈ [ a ⋅ b ] } = = a ⋅ F 2 ( x ) e 2 ( a f ) ( x ) = in a f ( y ) + ‖ a f ‖ Y | x − y | : y ∈ [ a , b ] = a : inf f ( y ) + ‖ f ‖ Y | x − y | 1 : y ∈ [ a ⋅ b ] = = a ⋅ F 2 ( x ) {:[e_(2)(af)(x)=in{af(y)+||af||_(Y)|x-y|:y in[a,b]}],[=a:i n f{f(y)+||f||_(Y)|x-y|_(1):y in[a*b]}=],[=a*F_(2)(x)]:}\begin{aligned} e_{2}(a f)(x) & =\operatorname{in}\left\{a f(y)+\|a f\|_{Y}|x-y|: y \in[a, b]\right\} \\ & =a: \inf \left\{f(y)+\|f\|_{Y}|x-y|_{1}: y \in[a \cdot b]\right\}= \\ & =a \cdot F_{2}(x) \end{aligned}e2(af)(x)=in{af(y)+‖af‖Y|x−y|:y∈[a,b]}=a:inf{f(y)+‖f‖Y|x−y|1:y∈[a⋅b]}==a⋅F2(x)
showing that e 2 e 2 e_(2)e_{2}e2 is positively homogeneous.
To prove the continuity of e 2 e 2 e_(2)e_{2}e2 for ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0, take δ = ε / 3 δ = ε / 3 delta=epsi//3\delta=\varepsilon / 3δ=ε/3 and show that
(2.3) | e 2 ( f ) − e 2 ( g ) | ⩽ ε , (2.3) e 2 ( f ) − e 2 ( g ) ⩽ ε , {:(2.3)|e_(2)(f)-e_(2)(g)| <= epsi",":}\begin{equation*} \left|e_{2}(f)-e_{2}(g)\right| \leqslant \varepsilon, \tag{2.3} \end{equation*}(2.3)|e2(f)−e2(g)|⩽ε,
for all f , g ∈ Lip 0 Y f , g ∈ Lip 0 Y f,g inLip_(0)Yf, g \in \operatorname{Lip}_{0} Yf,g∈Lip0Y such that ‖ f − g ‖ Y < δ ‖ f − g ‖ Y < δ ||f-g||_(Y) < delta\|f-g\|_{Y}<\delta‖f−g‖Y<δ.
It is easy to check that the maximal extension F 2 F 2 F_(2)F_{2}F2 of f f fff has the form
F 2 ( x ) = f ( a ) − ‖ f ‖ Y ( x − a ) , for x < a = f ( x ) , for x ∈ [ a , b ] = f ( b ) + ‖ f ‖ Y ( x − b ) , for x > b . F 2 ( x ) = f ( a ) − ‖ f ‖ Y ( x − a ) ,  for  x < a = f ( x ) ,  for  x ∈ [ a , b ] = f ( b ) + ‖ f ‖ Y ( x − b ) ,  for  x > b . {:[F_(2)(x)=f(a)-||f||_(Y)(x-a)","" for "x < a],[=f(x)","quad" for "x in[a","b]],[=f(b)+||f||_(Y)(x-b)","" for "x > b.]:}\begin{aligned} F_{2}(x) & =f(a)-\|f\|_{Y}(x-a), \text { for } x<a \\ & =f(x), \quad \text { for } x \in[a, b] \\ & =f(b)+\|f\|_{Y}(x-b), \text { for } x>b . \end{aligned}F2(x)=f(a)−‖f‖Y(x−a), for x<a=f(x), for x∈[a,b]=f(b)+‖f‖Y(x−b), for x>b.
If g ∈ Lip 0 Y g ∈ Lip 0 Y g inLip_(0)Yg \in \operatorname{Lip}_{0} Yg∈Lip0Y is such that ‖ f − g ‖ Y < δ = ε / 3 ‖ f − g ‖ Y < δ = ε / 3 ||f-g||_(Y) < delta=epsi//3\|f-g\|_{Y}<\delta=\varepsilon / 3‖f−g‖Y<δ=ε/3, then similarly the maximal extension G 2 G 2 G_(2)G_{2}G2 of g g ggg has the expression:
θ 2 ( x ) = g ( a ) − ‖ g ‖ Y ( x − a ) , for x < a = g ( x ) , for x ∈ [ a , b ] = g ( b ) + ‖ g ‖ Y ( x − b ) , for x > b . θ 2 ( x ) = g ( a ) − ‖ g ‖ Y ( x − a ) ,  for  x < a = g ( x ) ,  for  x ∈ [ a , b ] = g ( b ) + ‖ g ‖ Y ( x − b ) ,  for  x > b . {:[theta_(2)(x)=g(a)-||g||_(Y)(x-a)","" for "x < a],[=g(x)","quad" for "x in[a","b]],[=g(b)+||g||_(Y)(x-b)","" for "x > b.]:}\begin{aligned} \theta_{2}(x) & =g(a)-\|g\|_{Y}(x-a), \text { for } x<a \\ & =g(x), \quad \text { for } x \in[a, b] \\ & =g(b)+\|g\|_{Y}(x-b), \text { for } x>b . \end{aligned}θ2(x)=g(a)−‖g‖Y(x−a), for x<a=g(x), for x∈[a,b]=g(b)+‖g‖Y(x−b), for x>b.
It follows that the difference H 2 = F 2 − G 2 H 2 = F 2 − G 2 H_(2)=F_(2)-G_(2)H_{2}=F_{2}-G_{2}H2=F2−G2 has the expression :
H 2 ( x ) = f ( a ) − g ( a ) − ( ‖ f ‖ Y − ‖ g ‖ Y ) ( x − a ) , for x < a = f ( x ) − g ( x ) , = f ( b ) − g ( b ) + ( ‖ f ‖ Y − ‖ g ‖ Y ) ( x − b ) , for x > b H 2 ( x ) = f ( a ) − g ( a ) − ‖ f ‖ Y − ‖ g ‖ Y ( x − a ) ,  for  x < a = f ( x ) − g ( x ) , = f ( b ) − g ( b ) + ‖ f ‖ Y − ‖ g ‖ Y ( x − b ) ,  for  x > b {:[H_(2)(x)=f(a)-g(a)-(||f||_(Y)-||g||_(Y))(x-a)","" for "x < a],[=f(x)-g(x)","],[=f(b)-g(b)+(||f||_(Y)-||g||_(Y))(x-b)","" for "x > b]:}\begin{aligned} H_{2}(x) & =f(a)-g(a)-\left(\|f\|_{Y}-\|g\|_{Y}\right)(x-a), \text { for } x<a \\ & =f(x)-g(x), \\ & =f(b)-g(b)+\left(\|f\|_{Y}-\|g\|_{Y}\right)(x-b), \text { for } x>b \end{aligned}H2(x)=f(a)−g(a)−(‖f‖Y−‖g‖Y)(x−a), for x<a=f(x)−g(x),=f(b)−g(b)+(‖f‖Y−‖g‖Y)(x−b), for x>b
We have to consider several cases:
(... Case 1 ∘ ∴ x , y > b , x ≠ y 1 ∘ ∴ x , y > b , x ≠ y 1^(@):.x,y > b,x!=y1^{\circ} \therefore x, y>b, x \neq y1∘∴x,y>b,x≠y. In this case
(2.4) | H 2 ( x ) − H 2 ( y ) | ⩽ | ‖ f ‖ X − ‖ g ‖ Y | ⋅ | x − y | ⩽ ‖ f − g ‖ Y ⋅ | x − y | < ε | x − y | (2.4) H 2 ( x ) − H 2 ( y ) ⩽ ‖ f ‖ X − ‖ g ‖ Y ⋅ | x − y | ⩽ ‖ f − g ‖ Y ⋅ | x − y | < ε | x − y | {:[(2.4)|H_(2)(x)-H_(2)(y)| <= |||f||_(X)-||g||_(Y)|*|x-y|],[ <= ||f-g||_(Y)*|x-y| < epsi|x-y|]:}\begin{align*} \left|H_{2}(x)-H_{2}(y)\right| & \leqslant\left|\|f\|_{X}-\|g\|_{Y}\right| \cdot|x-y| \tag{2.4}\\ & \leqslant\|f-g\|_{Y} \cdot|x-y|<\varepsilon|x-y| \end{align*}(2.4)|H2(x)−H2(y)|⩽|‖f‖X−‖g‖Y|⋅|x−y|⩽‖f−g‖Y⋅|x−y|<ε|x−y|
Case 2 ∘ . x , y < a , x ≠ y 2 ∘ . x , y < a , x ≠ y 2^(@).x,y < a,x!=y2^{\circ} . x, y<a, x \neq y2∘.x,y<a,x≠y. Reasoning like in Case 1 ∘ 1 ∘ 1^(@)1^{\circ}1∘ one obtains
(2.5) | H 2 ( x ) − H 2 ( y ) | < ε ⋅ | x − y | . (2.5) H 2 ( x ) − H 2 ( y ) < ε ⋅ | x − y | . {:(2.5)|H_(2)(x)-H_(2)(y)| < epsi*|x-y|.:}\begin{equation*} \left|H_{2}(x)-H_{2}(y)\right|<\varepsilon \cdot|x-y| . \tag{2.5} \end{equation*}(2.5)|H2(x)−H2(y)|<ε⋅|x−y|.
Case 3 ∘ : a ⩽ x , y ⩽ b ; x ≠ y 3 ∘ : a ⩽ x , y ⩽ b ; x ≠ y 3^(@):a <= x,y <= b;x!=y3^{\circ}: a \leqslant x, y \leqslant b ; x \neq y3∘:a⩽x,y⩽b;x≠y. In this case
(2.6) | H 2 ( x ) − H 2 ( y ) | = | ( f − g ) ( x ) − ( f − g ) ( y ) | ⩽ ⩽ | f − g | | Y ⋅ | x − y | ⩽ ( ε / 3 ) ⋅ | x − y | . (2.6) H 2 ( x ) − H 2 ( y ) = | ( f − g ) ( x ) − ( f − g ) ( y ) | ⩽ ⩽ | f − g | Y ⋅ | x − y | ⩽ ( ε / 3 ) ⋅ | x − y | . {:[(2.6)|H_(2)(x)-H_(2)(y)|=|(f-g)(x)-(f-g)(y)| <= ],[ <= |f-g||_(Y)*|x-y| <= (epsi//3)*|x-y|.]:}\begin{align*} \left|H_{2}(x)-H_{2}(y)\right| & =|(f-g)(x)-(f-g)(y)| \leqslant \tag{2.6}\\ & \leqslant\left.|f-g|\right|_{Y} \cdot|x-y| \leqslant(\varepsilon / 3) \cdot|x-y| . \end{align*}(2.6)|H2(x)−H2(y)|=|(f−g)(x)−(f−g)(y)|⩽⩽|f−g||Y⋅|x−y|⩽(ε/3)⋅|x−y|.
(2.7) | H 2 ( x ) − H 2 ( y ) | =∣ ( f − g ) ( x ) − ( f − g ) ( b ) − (2.7) H 2 ( x ) − H 2 ( y ) =∣ ( f − g ) ( x ) − ( f − g ) ( b ) − {:(2.7)|H_(2)(x)-H_(2)(y)|=∣(f-g)(x)-(f-g)(b)-:}\begin{equation*} \left|H_{2}(x)-H_{2}(y)\right|=\mid(f-g)(x)-(f-g)(b)- \tag{2.7} \end{equation*}(2.7)|H2(x)−H2(y)|=∣(f−g)(x)−(f−g)(b)−
abor
(1.5) , 11 (1.5)  ,  11 {:(1.5)" , "11:}\begin{equation*} \text { , } 11 \tag{1.5} \end{equation*}(1.5) , 11
Case 5 ∘ . y < a 1 ⩽ x ⩽ b 5 ∘ . y < a 1 ⩽ x ⩽ b 5^(@).y < a_(1) <= x <= b5^{\circ} . y<a_{1} \leqslant x \leqslant b5∘.y<a1⩽x⩽b. Beasoning like in Case 4 ∘ 4 ∘ 4^(@)4^{\circ}4∘ one obtains
(2.8) | H 2 ( x ) − H 2 ( y ) | ⩽ ( 2 ϵ / 3 ) ⋅ | x − y | . (2.8) H 2 ( x ) − H 2 ( y ) ⩽ ( 2 ϵ / 3 ) ⋅ | x − y | . {:(2.8)|H_(2)(x)-H_(2)(y)| <= (2epsilon//3)*|x-y|.:}\begin{equation*} \left|H_{2}(x)-H_{2}(y)\right| \leqslant(2 \epsilon / 3) \cdot|x-y| . \tag{2.8} \end{equation*}(2.8)|H2(x)−H2(y)|⩽(2ϵ/3)⋅|x−y|.
Case 6 ∘ ⋯ ∞ < a < b < y 6 ∘ ⋯ ∞ < a < b < y 6^(@)cdots oo < a < b < y6^{\circ} \cdots \infty<a<b<y6∘⋯∞<a<b<y. In this case
(2.9) | H 2 ( x ) − H 2 ( y ) | =∣ ( f − g ) ( a ) − ( 1 | f ‖ Y − ‖ g | Y ) ( x ) − a ) − ‖ − ( f − g ˙ ) ( b ) − ( ‖ f ‖ x − ‖ g ‖ y ) ( y − b ) ‖ 3 ⩽ ⩽ ‖ f − g ⋅ ‖ Y | a − b | + | ‖ f ‖ Y − ‖ g ‖ Y | ⋅ | x + y − a − b | < < 3 ⋅ | | f − g ‖ Y | x − y | < 3 δ : | x − y | = ε ⋅ | x − y | . (2.9) H 2 ( x ) − H 2 ( y ) =∣ ( f − g ) ( a ) − 1 f Y − g Y ( x ) − a − ‖ − ( f − g ˙ ) ( b ) − ‖ f ‖ x − ‖ g ‖ y ( y − b ) ‖ 3 ⩽ ⩽ ‖ f − g ⋅ ‖ Y | a − b | + ‖ f ‖ Y − ‖ g ‖ Y ⋅ | x + y − a − b | < < 3 ⋅ | | f − g ‖ Y | x − y | < 3 δ : | x − y | = ε ⋅ | x − y | . {:[(2.9){:|H_(2)(x)-H_(2)(y)|=∣(f-g)(a)-(1|f||_(Y)-||g|Y)(x)-a)-||],[-(f-g^(˙))(b)-(||f||_(x)-||g||_(y))(y-b)||^(3) <= ],[ <= ||f-g*||_(Y)|a-b|+|||f||_(Y)-||g||_(Y)|*|x+y-a-b| < ],[ < 3*||f-g||_(Y)|x-y| < 3delta:|x-y|=epsi*|x-y|.]:}\begin{align*} & \left.\left|H_{2}(x)-H_{2}(y)\right|=\mid(f-g)(a)-\left(1\left|f\left\|_{Y}-\right\| g\right| Y\right)(x)-a\right)-\| \tag{2.9}\\ & -(f-\dot{g})(b)-\left(\|f\|_{x}-\|g\|_{y}\right)(y-b) \|^{3} \leqslant \\ & \leqslant\|f-g \cdot\|_{Y}|a-b|+\left|\|f\|_{Y}-\|g\|_{Y}\right| \cdot|x+y-a-b|< \\ & <3 \cdot| | f-g \|_{Y}|x-y|<3 \delta:|x-y|=\varepsilon \cdot|x-y| . \end{align*}(2.9)|H2(x)−H2(y)|=∣(f−g)(a)−(1|f‖Y−‖g|Y)(x)−a)−‖−(f−g˙)(b)−(‖f‖x−‖g‖y)(y−b)‖3⩽⩽‖f−g⋅‖Y|a−b|+|‖f‖Y−‖g‖Y|⋅|x+y−a−b|<<3⋅||f−g‖Y|x−y|<3δ:|x−y|=ε⋅|x−y|.
ast Taking into account the inequalities (2.4)-(2.9), it follows that ‖ F 2 − G 2 ‖ X = ‖ H 2 ‖ X = sup { | H 2 ( x ) − H 2 ( y ) | | | x − y ∣: x , y ∈ R , x ≠ y } < ε F 2 − G 2 X = H 2 X = sup H 2 ( x ) − H 2 ( y ) | | x − y ∣: x , y ∈ R , x ≠ y < ε ||F_(2)-G_(2)||_(X)=||H_(2)||_(X)=s u p{|H_(2)(x)-H_(2)(y)|||x-y∣:x,y in R,x!=y} < epsi\left\|F_{2}-G_{2}\right\|_{X}=\left\|H_{2}\right\|_{X}=\sup \left\{\left|H_{2}(x)-H_{2}(y)\right|| | x-y \mid: x, y \in R, x \neq y\right\}<\varepsilon‖F2−G2‖X=‖H2‖X=sup{|H2(x)−H2(y)|||x−y∣:x,y∈R,x≠y}<ε i.e.
| e 2 ( f ) − e 2 ( g ) | < ε . e 2 ( f ) − e 2 ( g ) < ε . |e_(2)(f)-e_(2)(g)| < epsi.\left|e_{2}(f)-e_{2}(g)\right|<\varepsilon .|e2(f)−e2(g)|<ε.
Remark 2. The selection e 1 ( f ) = F 1 e 1 ( f ) = F 1 e_(1)(f)=F_(1)e_{1}(f)=F_{1}e1(f)=F1, where F 1 F 1 F_(1)F_{1}F1 is the minimal extension of f f fff defined by (1.5) is also continuous and positively homogeneous. This can be proved directly or taking into account the equality
(2.10) e 1 ( f ) = − e 2 ( − f ) , (2.10) e 1 ( f ) = − e 2 ( − f ) , {:(2.10)e_(1)(f)=-e_(2)(-f)",":}\begin{equation*} e_{1}(f)=-e_{2}(-f), \tag{2.10} \end{equation*}(2.10)e1(f)=−e2(−f),
which holds for all f ∈ Lip 0 Y f ∈ Lip 0 Y f in Lip_(0)Yf \in \operatorname{Lip}{ }_{0} Yf∈Lip0Y.
Combining these two results one obtains the following consequence:
Corollary 5. The extension operator E - Lip 0 Y → 2 Lip 0 E - Lip  0 Y → 2 Lip  0 E_("- Lip "0)Y rarr2^("Lip "0)E_{\text {- Lip } 0} Y \rightarrow 2^{\text {Lip } 0}E- Lip 0Y→2Lip 0 R has a contimious and homogeneous selection given by
e ( f ) = ( 1 / 2 ) ⋅ ( e 1 ( f ) + e 2 ( f ) ) , f ∈ Lip 0 Y . e ( f ) = ( 1 / 2 ) ⋅ e 1 ( f ) + e 2 ( f ) , f ∈ Lip 0 Y . e(f)=(1//2)*(e_(1)(f)+e_(2)(f)),f inLip_(0)Y.e(f)=(1 / 2) \cdot\left(e_{1}(f)+e_{2}(f)\right), f \in \operatorname{Lip}_{0} Y .e(f)=(1/2)⋅(e1(f)+e2(f)),f∈Lip0Y.
Proof. Obviously that e e eee is continuous and positively homogeneous. On the other hand by (2.10), it follows
e ( − f ) ≐ − e ( f ) , f ∈ Lip 0 Y , e ( − f ) ≐ − e ( f ) , f ∈ Lip 0 Y , e(-f)≐-e(f),f inLip_(0)Y,e(-f) \doteq-e(f), f \in \operatorname{Lip}_{0} Y,e(−f)≐−e(f),f∈Lip0Y,
implying the homogeneity of e : e ( α f ) ⇌ α : e ( f ) , α ∈ R , f ∈ Lip ] 0 I e : e ( α f ) ⇌ α : e ( f ) , α ∈ R , f ∈ Lip ] 0 I e:e(alpha f)⇌alpha:e(f),alpha in R,f in Lip]_(0)Ie: e(\alpha f) \rightleftharpoons \alpha: e(f), \alpha \in R, f \in \operatorname{Lip}]_{0} \mathcal{I}e:e(αf)⇌α:e(f),α∈R,f∈Lip]0I.
3. This section is concerned with the existence of selections for the metric projection P Y ⊥ P Y ⊥ P_(Y _|_)P_{Y \perp}PY⊥ in the case X ⇒ R , Y = [ a , b ] , x 0 ∈ [ a , b ] X ⇒ R , Y = [ a , b ] , x 0 ∈ [ a , b ] X=>R,Y=[a,b],x_(0)in[a,b]X \Rightarrow R, Y=[a, b], x_{0} \in[a, b]X⇒R,Y=[a,b],x0∈[a,b]. : usal selection p p ppp : Lip X → Y ⊥ X → Y ⊥ X rarrY^(_|_)X \rightarrow Y^{\perp}X→Y⊥ of P Y ⊥ P Y ⊥ P_(Y_(_|_))P_{Y_{\perp}}PY⊥ is called additive modulo Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥, provided:
(81) p ( F + G ) = p ( F ) + p ( G ) , (81) p ( F + G ) = p ( F ) + p ( G ) , {:(81)p(F+G)=p(F)+p(G)",":}\begin{equation*} p(F+G)=p(F)+p(G), \tag{81} \end{equation*}(81)p(F+G)=p(F)+p(G),
formall e − 1 p 0 X e − 1 p 0 X e^(-1)p_(0)Xe^{-1} p_{0} Xe−1p0X and G ∈ Y ⊥ G ∈ Y ⊥ G inY^(_|_)G \in Y^{\perp}G∈Y⊥.
Theorem 6. The metric projection P Y ⊥ P Y ⊥ P_(Y _|_)P_{Y \perp}PY⊥ has a homogeneous, additive modulo Y ⊥ Y ⊥ Y _|_Y \perpY⊥ and continuous setection p p ppp.
Proof. Let e e eee be the homogeneous and continuous selection of the extension operator E E EEE, given in Corollary 5. Taking into account Theorem 3 and equality (2.10), define p : Lip 0 X → Y ⊥ p : Lip 0 X → Y ⊥ p:Lip_(0)X rarr Y _|_p: \operatorname{Lip}_{0} X \rightarrow Y \perpp:Lip0X→Y⊥ by the formula
p = I − e ∘ r , p = I − e ∘ r , p=I-e@r,p=I-e \circ r,p=I−e∘r,
where r : Lip 0 X → Lip 0 Y r : Lip 0 X → Lip 0 Y r:Lip_(0)X rarrLip_(0)Yr: \operatorname{Lip}_{0} X \rightarrow \operatorname{Lip}_{0} Yr:Lip0X→Lip0Y is the restriction operator given by r ( F ) == F ′ | Y r ( F ) == F ′ Y r(F)==F^(')|_(Y)r(F)= =\left.F^{\prime}\right|_{Y}r(F)==F′|Y and I : Lip 0 X : Lip 0 X I : Lip 0 X : Lip 0 X I:Lip_(0)X:Lip_(0)XI: \operatorname{Lip}_{0} X: \operatorname{Lip}_{0} XI:Lip0X:Lip0X this the identity map. Then
p ( F ) = ( I − e ∘ r ) ( F ) = F − e ( F | Y ) ∈ P Y ⊥ ( F ) . p ( F ) = ( I − e ∘ r ) ( F ) = F − e F Y ∈ P Y ⊥ ( F ) . p(F)=(I-e@r)(F)=F-e(F|_(Y))inP_(Y _|_)(F).p(F)=(I-e \circ r)(F)=F-e\left(\left.F\right|_{Y}\right) \in P_{Y \perp}(F) .p(F)=(I−e∘r)(F)=F−e(F|Y)∈PY⊥(F).
Indeed F − e ( F | Y ) = ( 1 / 2 ) ( F − F 1 ) + ( 1 / 2 ) ( F − F 2 ) ∈ P Y ⊥ ( F ) F − e F Y = ( 1 / 2 ) F − F 1 + ( 1 / 2 ) F − F 2 ∈ P Y ⊥ ( F ) F-e(F|_(Y))=(1//2)(F-F_(1))+(1//2)(F-F_(2))inP_(Y _|_)(F)F-e\left(\left.F\right|_{Y}\right)=(1 / 2)\left(F-F_{1}\right)+(1 / 2)\left(F-F_{2}\right) \in P_{Y \perp}(F)F−e(F|Y)=(1/2)(F−F1)+(1/2)(F−F2)∈PY⊥(F); since the set P Y ⊥ ( F ′ ) P Y ⊥ F ′ P_(Y _|_)(F^('))P_{Y \perp}\left(F^{\prime}\right)PY⊥(F′) is convex.
Obviously the selection p p ppp is continuous and
p ( α F ) = α F + e ( α F | Y ) = α ( F + e ( | F ′ | Y ) ) = α ⋅ p ( F ) ′ ) p ( α F ) = α F + e α F Y = α F + e F ′ Y = α ⋅ p ( F ) ′ {:p(alpha F)=alpha F+e( alpha F|_(Y))=alpha(F+e(|F^(')|_(Y)))=alpha*p(F)^('))\left.p(\alpha F)=\alpha F+e\left(\left.\alpha F\right|_{Y}\right)=\alpha\left(F+e\left(\left|F^{\prime}\right|_{Y}\right)\right)=\alpha \cdot p(F)^{\prime}\right)p(αF)=αF+e(αF|Y)=α(F+e(|F′|Y))=α⋅p(F)′)
for all α ∈ R α ∈ R alpha inR\alpha \in \mathbb{R}α∈R, showing that p p ppp is homogeneous.
Now
(3.2) p ( F + G ) = F + G − e ( ( F + G ) | X ) = F + G − e ( F | Y ) = = F − e ( F | Y ) + G = p ( F ) + G = p ( F ) + p ( G ) (3.2) p ( F + G ) = F + G − e ( F + G ) X = F + G − e F Y = = F − e F Y + G = p ( F ) + G = p ( F ) + p ( G ) {:[(3.2)p(F+G)=F+G-e((F+G)|_(X))=F+G-e(F|_(Y))=],[=F-e(F|_(Y))+G=p(F)+G=p(F)+p(G)]:}\begin{align*} p(F+G) & =F+G-e\left(\left.(F+G)\right|_{X}\right)=F+G-e\left(\left.F\right|_{Y}\right)= \tag{3.2}\\ & =F-e\left(\left.F\right|_{Y}\right)+G=p(F)+G=p(F)+p(G) \end{align*}(3.2)p(F+G)=F+G−e((F+G)|X)=F+G−e(F|Y)==F−e(F|Y)+G=p(F)+G=p(F)+p(G)
for all T ∈ Lip p 0 X T ∈ Lip p 0 X T in Lipp_(0)XT \in \operatorname{Lip} p_{0} XT∈Lipp0X and G ∈ Y ⊥ G ∈ Y ⊥ G inY^(_|_)G \in Y^{\perp}G∈Y⊥, since for G ∈ Y ⊥ , P Y ⊥ ( G ) = { G } G ∈ Y ⊥ , P Y ⊥ ( G ) = { G } G inY^(_|_),P_(Y^(_|_))(G)={G}G \in Y^{\perp}, P_{Y^{\perp}}(G)=\{G\}G∈Y⊥,PY⊥(G)={G} and p ( G ) = G p ( G ) = G p(G)=Gp(G)=Gp(G)=G
By (3.2), the selection p p ppp is additive modulo Y ⊥ Y ⊥ Y _|_Y \perpY⊥ and Theorem 6 is. proved.
It is easily seen that the kernel of P Y L P Y L P_(YL)P_{Y L}PYL,
Ker P Y ⊥ = { F ∈ Lip 0 X , 0 ∈ P Y ⊥ ( F ) } Ker P Y ⊥ = F ∈ Lip 0 X , 0 ∈ P Y ⊥ ( F ) KerP_(Y _|_)={F inLip_(0)X,0inP_(Y _|_)(F)}\operatorname{Ker} P_{Y \perp}=\left\{F \in \operatorname{Lip}_{0} X, 0 \in P_{Y \perp}(F)\right\}KerPY⊥={F∈Lip0X,0∈PY⊥(F)}
verifies the equality
(3.3) Ker P Y ′ = { F ∈ Lip p 0 X : ‖ F ‖ X = ‖ F | Y ‖ Y } (3.3) Ker P Y ′ = F ∈ Lip p 0 X : ‖ F ‖ X = F Y Y {:(3.3)KerP_(Y^('))={F in Lipp_(0)X:||F||_(X)=||F|_(Y)||_(Y)}:}\begin{equation*} \operatorname{Ker} P_{Y^{\prime}}=\left\{F \in \operatorname{Lip} p_{0} X:\|F\|_{X}=\left\|\left.F\right|_{Y}\right\|_{Y}\right\} \tag{3.3} \end{equation*}(3.3)KerPY′={F∈Lipp0X:‖F‖X=‖F|Y‖Y}
Cozollary 7. For X = R , Y = [ a ; b ] X = R , Y = [ a ; b ] X=R,Y=[a;b]X=R, \mathrm{Y}=[a ; b]X=R,Y=[a;b] and x 0 ∈ [ a ; b ] x 0 ∈ [ a ; b ] x_(0)in[a;b]x_{0} \in[a ; b]x0∈[a;b] the following. assertions are true:
(a) The extension operator Thes a linear and continuous setection;
(b) The metric projection P Y ⊥ P Y ⊥ P_(Y _|_)P_{Y \perp}PY⊥ has linear and continuous selection;
(c) There exists a subspace W W WWW of the subspace Ker P Y ⊥ P Y ⊥ P_(Y _|_)P_{Y \perp}PY⊥ such that every T ′ ∈ Lip 0 X T ′ ∈ Lip 0 X T^(')inLip_(0)XT^{\prime} \in \mathrm{Lip}_{0} XT′∈Lip0X can be uniquely represented in the form F = H + G F = H + G F=H+GF=H+GF=H+G, with H ∈ W ⊤ , G ∈ Y ⊥ H ∈ W ⊤ , G ∈ Y ⊥ H inW^(TT),G inY^(_|_)H \in W^{\top}, G \in Y^{\perp}H∈W⊤,G∈Y⊥, i.e. the subspace Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is complemented in Lip 0 X 0 X _(0)X{ }_{0} X0X.
Proof. (a) We show that the application e : Lip p 0 Ψ → e : Lip p 0 Ψ → e:Lipp_(0)Psi rarre: \operatorname{Lip} p_{0} \Psi \rightarrowe:Lipp0Ψ→ Lip p 0 X defi- p 0 X defi-  p_(0)X_("defi- ")p_{0} X_{\text {defi- }}p0Xdefi-  ned by
(3.4) e ( f ) = ( 1 / 2 ) ( F 1 + F 2 ) , (3.4) e ( f ) = ( 1 / 2 ) F 1 + F 2 , {:(3.4)e(f)=(1//2)(F_(1)+F_(2))",":}\begin{equation*} e(f)=(1 / 2)\left(F_{1}+F_{2}\right), \tag{3.4} \end{equation*}(3.4)e(f)=(1/2)(F1+F2),
where H 1 , F 2 H 1 , F 2 H_(1),F_(2)H_{1}, F_{2}H1,F2 are the extremal extensions of f f fff given by ( 1 , 5 ) ( 1 , 5 ) (1,5)(1,5)(1,5) and ( 1 , 6 ) ( 1 , 6 ) (1,6)(1,6)(1,6), is linear and continuous selection of E E EEE.
Writing explicitly e e eee we find that
e ( f ) ( x ) = f ( a ) , for x < a = f ( x ) , for x ∈ [ a , b ] = f ( b ) , for x > b e ( f ) ( x ) = f ( a ) ,  for  x < a = f ( x ) ,  for  x ∈ [ a , b ] = f ( b ) ,  for  x > b {:[e(f)(x)=f(a)","" for "x < a],[=f(x)","" for "x in[a","b]],[=f(b)","" for "x > b]:}\begin{aligned} e(f)(x) & =f(a), \text { for } x<a \\ & =f(x), \text { for } x \in[a, b] \\ & =f(b), \text { for } x>b \end{aligned}e(f)(x)=f(a), for x<a=f(x), for x∈[a,b]=f(b), for x>b
for any f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} \mathbf{Y}f∈Lip0Y. Obviously e ( α f + β g ) ( x ) = α ⋅ e ( f ) ( x ) + β ⋅ e ( g ) ( x ) e ( α f + β g ) ( x ) = α ⋅ e ( f ) ( x ) + β ⋅ e ( g ) ( x ) e(alpha f+beta g)(x)=alpha*e(f)(x)+beta*e(g)(x)e(\alpha f+\beta g)(x)=\alpha \cdot e(f)(x)+\beta \cdot e(g)(x)e(αf+βg)(x)=α⋅e(f)(x)+β⋅e(g)(x) for all x ∈ R x ∈ R x in Rx \in Rx∈R and all f , g ∈ Lip θ θ Y , α , β ∈ R f , g ∈ Lip θ θ Y , α , β ∈ R f,g in Liptheta_(theta)Y,alpha,beta in Rf, g \in \operatorname{Lip} \theta_{\theta} Y, \alpha, \beta \in Rf,g∈LipθθY,α,β∈R, showing that e e eee is linear. The continuity of eiwas proved in Corollary 5.
(b) The application p : Lip 0 Y → Y ⊥ p : Lip 0 Y → Y ⊥ p:Lip_(0)Y rarrY^(_|_)p: \operatorname{Lip}_{0} Y \rightarrow Y^{\perp}p:Lip0Y→Y⊥ defined, for F ∈ Lip p 0 X F ∈ Lip p 0 X F in Lipp_(0)XF \in \operatorname{Lip} p_{0} XF∈Lipp0X, by
(3.5) p ( F ) = F − e ( F ′ | Y ) = F − ( 1 / 2 ) ( F 1 + F 2 ) (3.5) p ( F ) = F − e F ′ Y = F − ( 1 / 2 ) F 1 + F 2 {:(3.5)p(F)=F-e(F^(')|_(Y))=F-(1//2)(F_(1)+F_(2)):}\begin{equation*} p(F)=F-e\left(\left.F^{\prime}\right|_{Y}\right)=F-(1 / 2)\left(F_{1}+F_{2}\right) \tag{3.5} \end{equation*}(3.5)p(F)=F−e(F′|Y)=F−(1/2)(F1+F2)
where F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2 are the extensions given by (1.5), (1.6), is linear and continuous.
(c) Let
(3:6) W = { e ( F | X ) : F ∈ Lip 0 X } (3:6) W = e F X : F ∈ Lip 0 X {:(3:6)W={e(F|_(X)):F inLip_(0)X}:}\begin{equation*} W=\left\{e\left(\left.F\right|_{X}\right): F \in \operatorname{Lip}_{0} X\right\} \tag{3:6} \end{equation*}(3:6)W={e(F|X):F∈Lip0X}
The linearity of e e eee implies that W W WWW is a subspace of Lip p 0 X Lip p 0 X Lipp_(0)X\operatorname{Lip} p_{0} XLipp0X. By (3.3) and the equality
‖ e ( F ∣ ) Y ‖ X = ‖ F | Y ‖ Y e ( F ∣ ) Y X = F Y Y ||e(F∣)_(Y)||_(X)=||F|_(Y)||_(Y)\left\|e(F \mid)_{Y}\right\|_{X}=\left\|\left.F\right|_{Y}\right\|_{Y}‖e(F∣)Y‖X=‖F|Y‖Y
it follows that W ⊂ Ker P Y ⊥ W ⊂ Ker P Y ⊥ W sub KerP_(Y _|_)W \subset \operatorname{Ker} P_{Y \perp}W⊂KerPY⊥.
For F ∈ Lip p 0 X F ∈ Lip p 0 X F in Lipp_(0)XF \in \operatorname{Lip} p_{0} XF∈Lipp0X define
(3.7) G ( x ) = 0 , for x ∈ [ a , b ] = F ′ ( x ) − F ( a ) , for x < a = F ′ ( x ) − F ( b ) , for x > b (3.7) G ( x ) = 0 ,  for  x ∈ [ a , b ] = F ′ ( x ) − F ( a ) ,  for  x < a = F ′ ( x ) − F ( b ) ,  for  x > b {:[(3.7)G(x)=0","" for "x in[a","b]],[=F^(')(x)-F(a)","" for "x < a],[=F^(')(x)-F(b)","" for "x > b]:}\begin{align*} G(x) & =0, \text { for } x \in[a, b] \tag{3.7}\\ & =F^{\prime}(x)-F(a), \text { for } x<a \\ & =F^{\prime}(x)-F(b), \text { for } x>b \end{align*}(3.7)G(x)=0, for x∈[a,b]=F′(x)−F(a), for x<a=F′(x)−F(b), for x>b
and
(3.8) H ( x ) = F ¯ ( x ) , for x ∈ [ a , b ] = F ( a ) , for x < a = F ( b ) , for x > b (3.8) H ( x ) = F ¯ ( x ) ,  for  x ∈ [ a , b ] = F ( a ) ,  for  x < a = F ( b ) ,  for  x > b {:[(3.8)H(x)= bar(F)(x)","" for "x in[a","b]],[=F(a)","" for "x < a],[=F(b)","" for "x > b]:}\begin{align*} H(x) & =\bar{F}(x), \text { for } x \in[a, b] \tag{3.8}\\ & =F(a), \text { for } x<a \\ & =F(b), \text { for } x>b \end{align*}(3.8)H(x)=F¯(x), for x∈[a,b]=F(a), for x<a=F(b), for x>b
Then F = H + G , G ∈ Y ⊥ F = H + G , G ∈ Y ⊥ F=H+G,G inY^(_|_)F=H+G, G \in Y^{\perp}F=H+G,G∈Y⊥ and H = e ( F | Y ) ∈ W H = e F Y ∈ W H=e(F|_(Y))in WH=e\left(\left.F\right|_{Y}\right) \in WH=e(F|Y)∈W.
To prove that Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X is the topological direct sum of the subspaces Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ and W W WWW, it is sufficient to prove that the projection operator on X ¯ ⊥ X ¯ ⊥ bar(X)^(_|_)\bar{X}^{\perp}X¯⊥ is continuous, i.e. that the application F → G F → G F rarr GF \rightarrow GF→G, where G ∈ Y ⊥ G ∈ Y ⊥ G in Y _|_G \in Y \perpG∈Y⊥ is the function defined in (3.7) is a linear and continuous operator from Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X to Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥.
寻 1 The linearity is obvious. To prove the continuity suppose that F n , T 0 , F F n , T 0 , F F_(n),T_(0),FF_{n}, T_{0}, FFn,T0,F in Lip 0 X Lip 0 X Lip_(0)X\operatorname{Lip}_{0} XLip0X, i.e. ‖ F n − F ‖ x → 0 F n − F x → 0 ||F_(n)-F||_(x)rarr0\left\|F_{n}-F\right\|_{x} \rightarrow 0‖Fn−F‖x→0. Then
G n ( x ) = 0 , x ∈ [ a , b ] = F n ( x ) − F n ( a ) , x < a = F n ( x ) − F n ( b ) , x > b G n ( x ) = 0 , x ∈ [ a , b ] = F n ( x ) − F n ( a ) , x < a = F n ( x ) − F n ( b ) , x > b {:[G_(n)(x)=0","x in[a","b]],[=F_(n)(x)-F_(n)(a)","x < a],[=F_(n)(x)-F_(n)(b)","x > b]:}\begin{aligned} G_{n}(x) & =0, x \in[a, b] \\ & =F_{n}(x)-F_{n}(a), x<a \\ & =F_{n}(x)-F_{n}(b), x>b \end{aligned}Gn(x)=0,x∈[a,b]=Fn(x)−Fn(a),x<a=Fn(x)−Fn(b),x>b
and
U n ( x ) = G n ( x ) − G ( x ) = 0 , x ∈ [ a ; b ] = F n ( x ) − F ( x ) + ( F ( a ) − F n ( a ) ) , x < a = F n ( x ) − F ( x ) + ( F ( b ) − F n ( b ) ) , x > b U n ( x ) = G n ( x ) − G ( x ) = 0 , x ∈ [ a ; b ] = F n ( x ) − F ( x ) + F ( a ) − F n ( a ) , x < a = F n ( x ) − F ( x ) + F ( b ) − F n ( b ) , x > b {:[U_(n)(x)=G_(n)(x)-G(x)=0","quad x in[a;b]],[=F_(n)(x)-F(x)+(F(a)-F_(n)(a))","quad x < a],[=F_(n)(x)-F(x)+(F(b)-F_(n)(b))","quad x > b]:}\begin{aligned} U_{n}(x) & =G_{n}(x)-G(x)=0, \quad x \in[a ; b] \\ & =F_{n}(x)-F(x)+\left(F(a)-F_{n}(a)\right), \quad x<a \\ & =F_{n}(x)-F(x)+\left(F(b)-F_{n}(b)\right), \quad x>b \end{aligned}Un(x)=Gn(x)−G(x)=0,x∈[a;b]=Fn(x)−F(x)+(F(a)−Fn(a)),x<a=Fn(x)−F(x)+(F(b)−Fn(b)),x>b
Again, we have to consider "several cases:
Case 1. x , y < a x , y < a x,y < ax, y<ax,y<a. In this case
| U n ( x ) − U n ( y ) | = | ( F n − F ) ′ ( x ) − ( F n − F ) ( y ) | ⩽ ‖ F n − F ‖ X ⋅ | x − y | U n ( x ) − U n ( y ) = F n − F ′ ( x ) − F n − F ( y ) ⩽ F n − F X ⋅ | x − y | |U_(n)(x)-U_(n)(y)|=|(F_(n)-F)^(')(x)-(F_(n)-F)(y)| <= ||F_(n)-F||_(X)*|x-y|\left|U_{n}(x)-U_{n}(y)\right|=\left|\left(F_{n}-F\right)^{\prime}(x)-\left(F_{n}-F\right)(y)\right| \leqslant\left\|F_{n}-F\right\|_{X} \cdot|x-y||Un(x)−Un(y)|=|(Fn−F)′(x)−(Fn−F)(y)|⩽‖Fn−F‖X⋅|x−y|
The same inequality is obtained for x , y > b x , y > b x,y > bx, y>bx,y>b.
Case 2. a < x < b < y a < x < b < y a < x < b < ya<x<b<ya<x<b<y. In this case
| U n ( x ) − U n ( y ) | = | U n ( y ) | = | F n ( y ) − F ( y ) − ( F ( b ) − F n ( b ) ) | = = | ( F n − F ) ( y ) − ( F n − F ) ( b ) | ⩽ ‖ F n − F ‖ x | y − b | ⩽ | | F n − F ‖ x ⋅ | x − y | U n ( x ) − U n ( y ) = U n ( y ) = F n ( y ) − F ( y ) − F ( b ) − F n ( b ) = = F n − F ( y ) − F n − F ( b ) ⩽ F n − F x | y − b | ⩽ F n − F ‖ x ⋅ x − y {:[|U_(n)(x)-U_(n)(y)|=|U_(n)(y)|=|F_(n)(y)-F(y)-(F(b)-F_(n)(b))|=],[=|(F_(n)-F)(y)-(F_(n)-F)(b)| <= ||F_(n)-F||_(x)|y-b| <= ||F_(n)-F||_(x)*|x-y|]:}\begin{gathered} \left|U_{n}(x)-U_{n}(y)\right|=\left|U_{n}(y)\right|=\left|F_{n}(y)-F(y)-\left(F(b)-F_{n}(b)\right)\right|= \\ =\left|\left(F_{n}-F\right)(y)-\left(F_{n}-F\right)(b)\right| \leqslant\left\|F_{n}-F\right\|_{x}|y-b| \leqslant\left|\left|F_{n}-F \|_{x} \cdot\right| x-y\right| \end{gathered}|Un(x)−Un(y)|=|Un(y)|=|Fn(y)−F(y)−(F(b)−Fn(b))|==|(Fn−F)(y)−(Fn−F)(b)|⩽‖Fn−F‖x|y−b|⩽||Fn−F‖x⋅|x−y|
The same inequality holds for x < a < y ˙ < b x < a < y ˙ < b x < a < y^(˙) < bx<a<\dot{y}<bx<a<y˙<b.
Case 3. x < a < b < y x < a < b < y x < a < b < yx<a<b<yx<a<b<y. In this case
| U n ( x ) − U n ( y ) | =∣ F n ( x ) − F ( x ) − ( F n ( a ) − F ( a ) ) − F n ( y ) + F ( y ) + + F n ( b ) − F ( b ) | ⩽ | ( F n − F ) ( x ) − ( F n − F ) ( y ) | + | ( F n − F ) ( b ) − − ( F n − F ) ( a ) | ⩽ ‖ F n − F ‖ X ⋅ | x − y ∣ + ‖ F n − F ‖ X ⋅ ( b − a ) ⩽ 2 ‖ F n − F ‖ X ⋅ | x − y | U n ( x ) − U n ( y ) =∣ F n ( x ) − F ( x ) − F n ( a ) − F ( a ) − F n ( y ) + F ( y ) + + F n ( b ) − F ( b ) ⩽ F n − F ( x ) − F n − F ( y ) + F n − F ( b ) − − F n − F ( a ) ⩽ F n − F X ⋅ x − y ∣ + F n − F X ⋅ ( b − a ) ⩽ 2 F n − F X ⋅ | x − y | {:[|U_(n)(x)-U_(n)(y)|=∣F_(n)(x)-F(x)-(F_(n)(a)-F(a))-F_(n)(y)+F(y)+],[+F_(n)(b)-F(b)| <= |(F_(n)-F)(x)-(F_(n)-F)(y)|+|(F_(n)-F)(b)-],[-(F_(n)-F)(a)| <= ||F_(n)-F||_(X)*|x-y∣+||F_(n)-F||_(X)*(b-a)],[ <= 2||F_(n)-F||_(X)*|x-y|]:}\begin{gathered} \left|U_{n}(x)-U_{n}(y)\right|=\mid F_{n}(x)-F(x)-\left(F_{n}(a)-F(a)\right)-F_{n}(y)+F(y)+ \\ +F_{n}(b)-F(b)\left|\leqslant\left|\left(F_{n}-F\right)(x)-\left(F_{n}-F\right)(y)\right|+\right|\left(F_{n}-F\right)(b)- \\ -\left(F_{n}-F\right)(a)\left|\leqslant\left\|F_{n}-F\right\|_{X} \cdot\right| x-y \mid+\left\|F_{n}-F\right\|_{X} \cdot(b-a) \\ \leqslant 2\left\|F_{n}-F\right\|_{X} \cdot|x-y| \end{gathered}|Un(x)−Un(y)|=∣Fn(x)−F(x)−(Fn(a)−F(a))−Fn(y)+F(y)++Fn(b)−F(b)|⩽|(Fn−F)(x)−(Fn−F)(y)|+|(Fn−F)(b)−−(Fn−F)(a)|⩽‖Fn−F‖X⋅|x−y∣+‖Fn−F‖X⋅(b−a)⩽2‖Fn−F‖X⋅|x−y|
It follows that
| U n ( x ) − U n ( y ) | ⩽ 2 ‖ F n − F ‖ X ⋅ | x − y | , U n ( x ) − U n ( y ) ⩽ 2 F n − F X ⋅ | x − y | , |U_(n)(x)-U_(n)(y)| <= 2||F_(n)-F||_(X)*|x-y|,\left|U_{n}(x)-U_{n}(y)\right| \leqslant 2\left\|F_{n}-F\right\|_{X} \cdot|x-y|,|Un(x)−Un(y)|⩽2‖Fn−F‖X⋅|x−y|,
for all x , y ∈ R x , y ∈ R x,y in Rx, y \in Rx,y∈R, implying ‖ D n ‖ X ⩽ 2 ‖ F n − F ‖ X → 0 D n X ⩽ 2 F n − F X → 0 ||D_(n)||_(X) <= 2||F_(n)-F||_(X)rarr0\left\|D_{n}\right\|_{X} \leqslant 2\left\|F_{n}-F\right\|_{X} \rightarrow 0‖Dn‖X⩽2‖Fn−F‖X→0.
It follows that F n → F F n → F F_(n)rarr FF_{n} \rightarrow FFn→F implies G n → G G n → G G_(n)rarr GG_{n} \rightarrow GGn→G, showing that the projection operator on Y ⊥ Y ⊥ Y^(_|_)Y^{\perp}Y⊥ is continuous and consequently L 1 p 0 X L 1 p 0 X L^(1)p_(0)XL^{1} p_{0} XL1p0X is the direct sum of Y i Y i Y^(i)Y^{i}Yi and W W WWW. Corollary 7 is completely proved.
: : Remarks 3. (a) In the considered case ( X = R , Y ˙ = [ a , b ] , x 0 ∈ ∈ [ a , b ] ) X = R , Y ˙ = [ a , b ] , x 0 ∈ ∈ [ a , b ] ) (X=R,(Y^(˙))=[a,b],x_(0)in:}in[a,b])\left(X=R, \dot{Y}=[a, b], x_{0} \in\right. \in[a, b])(X=R,Y˙=[a,b],x0∈∈[a,b]), we have e 1 ( f ) ≠ e 2 ( f ) e 1 ( f ) ≠ e 2 ( f ) e_(1)(f)!=e_(2)(f)e_{1}(f) \neq e_{2}(f)e1(f)≠e2(f), for all f ∈ Lip 0 Y , f ≠ 0 f ∈ Lip 0 Y , f ≠ 0 f inLip_(0)Y,f!=0f \in \operatorname{Lip}_{0} Y, f \neq 0f∈Lip0Y,f≠0. Tn fact e 1 ( f ) < e ( f ) << e 2 ( f ) , f Lip 0 Y e 1 ( f ) < e ( f ) << e 2 ( f ) , f Lip 0 Y e_(1)(f) < e(f)<<e_(2)(f),fLip_(0)Ye_{1}(f)<e(f)< <e_{2}(f), f \operatorname{Lip}_{0} Ye1(f)<e(f)<<e2(f),fLip0Y.
(b) Let X = [ 0 , 1 ] X = [ 0 , 1 ] X=[0,1]X=[0,1]X=[0,1] with d ( x , y ) = | x − y | , Y = { 0 , 1 } d ( x , y ) = | x − y | , Y = { 0 , 1 } d(x,y)=|x-y|,Y={0,1}\mathrm{d}(x, y)=|x-y|, Y=\{0,1\}d(x,y)=|x−y|,Y={0,1} and x 0 = 0 x 0 = 0 x_(0)=0x_{0}=0x0=0 Then e 1 ( f ) = e 2 ( f ) = e ( f ) e 1 ( f ) = e 2 ( f ) = e ( f ) e_(1)(f)=e_(2)(f)=e(f)e_{1}(f)=e_{2}(f)=e(f)e1(f)=e2(f)=e(f), for all f ∈ Lip 0 Y f ∈ Lip 0 Y f inLip_(0)Yf \in \operatorname{Lip}_{0} Yf∈Lip0Y. It follows that Y ⊥ = { H ′ ∈ Lip 0 X : F ( 0 ) = F ( 1 ) = 0 } Y ⊥ = H ′ ∈ Lip 0 X : F ( 0 ) = F ( 1 ) = 0 } Y^(_|_)={H^(')inLip_(0)X:}:F(0)=F(1)=0}Y^{\perp}=\left\{H^{\prime} \in \operatorname{Lip}_{0} X\right. : F(0)=F(1)=0\}Y⊥={H′∈Lip0X:F(0)=F(1)=0} is a Chebyshevian subspace of Lip 0 X 0 X _(0)X{ }_{0} X0X. In this case Lip 0 X = Ker P Y ⊥ ⊕ Y ⊥ Lip 0 X = Ker P Y ⊥ ⊕ Y ⊥ Lip_(0)X=KerP_(Y _|_)o+Y^(_|_)\operatorname{Lip}_{0} X=\operatorname{Ker} P_{Y \perp} \oplus Y^{\perp}Lip0X=KerPY⊥⊕Y⊥ and the extension operator E E EEE and the metric projection P Y ⊥ P Y ⊥ P_(Y^(_|_))P_{Y^{\perp}}PY⊥ are linear and single valued.

REFERENCES

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Received 15.V. 1992
Institutul de Calcul
Oficiul Poștal 1 C.P. 68
3400 Cluj-Napoca Romania
1992

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