Norm preserving extension of convex Lipschitz functions

Costica Mustata
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Stefan Cobzas

Ştefan Cobzaş


Institutul de Matematica Cluj-Napoca, Romania 

Costica Mustata

Costică Mustăţa


Institutul de Matematica, Cluj-Napoca, Romania (ICTP)

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S. Cobzas, C. Mustata, Norm-preserving extension of convex Lipschitz functions, J. Approx. Theory, 24 (1978) 236-244, doi: 10.1016/0021-9045(78)90028-X

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10.1016/0021-9045(78)90028-X

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MR # 82j: 41034

[1] S. BANACH, “Wstep do teorii funkji rzeczwistych,” Warszawa/Wroclav, 1951.
[2] J. CZIPSER AND L. GEHER, Extension of function satisfying a Lipschitz condition, Acta Math. Acad. Ski. Hungar. 6 (1953, 213-220.
[3] R. B. HOLMES, “A Course on Optimisation and Best Approximation,” Lecture Notes in Mathematics No. 257, Springer-Verlag, Berlin/Heidelberg/New York, 1972.
[4] I. KOLUMBAN, Ob edinstvennosti prodolienija lineinyh funkcionalov, Mathematics (Cluj) 4 (1962), 267-270.
[5] P. J. LAURENT, “Approximation et optimisation,” Hermann, Paris, 1972.
[6] C. MUSTQA, Asupra unor subspatii cebiseviene din spafiul normat al funciiilor lipschitziene, Rea. Anal. Num. Teoria Aproximaiiei 2 (1973), 81-87.
[7] C. MUSTATA, 0 proprietate de monotonie a operatorului de tea mai buni aproximarie in spatiul functiilor lipschitziene. Rev. Anal. Num. Troria Aproximatiei 3 (1974). 153-160.
[8] C. MUSTATA, Asupra unicitritii preiungirii g-seminormelor continue, Reu. Anal. :Lwn. Teoria Apvoximafiei 2 (1973), 173-l 77.
[9] C. MUSTATA, Best approximation and unique extension of Lipschitz functions, J. Approximation Theory 19 (1977), 222-230.
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[11] R. R. PHELPS, Uniqueness of Hahn-Banach extension and unique best approximation, Trans. Au7er. Math. Sot. 95 (1960), 238-255.

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1978-Mustata-J.-Approx.-Th.-Norm-preserving-extension-of-convex-Lipschitz-functions

Norm-Preserving Extension of Convex Lipschitz Functions

S. Cobzaş and C. MustạțaInstitutul de Matematică Cluj-Napoca, RomaniaCommunicated by R. Bojanic

Received March 15, 1977
Let ( X , d X , d X,dX, dX,d ) be a metric space. A function f : X → R f : X → R f:X rarr Rf: X \rightarrow Rf:X→R is called Lipschitz if there exists a number M ⩾ 0 M ⩾ 0 M >= 0M \geqslant 0M⩾0 such that
(1) f ( x ) − f ( y ) ∣⩽ M d ( x , y ) (1) f ( x ) − f ( y ) ∣⩽ M d ( x , y ) {:(1)f(x)-f(y)∣⩽Md(x","y):}\begin{equation*} f(x)-f(y) \mid \leqslant M d(x, y) \tag{1} \end{equation*}(1)f(x)−f(y)∣⩽Md(x,y)
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X. The smallest constant M M MMM verifying (1) is called the norm of f f fff and is denoted by ‖ f X ‖ f X ||f_(X)\| f_{X}‖fX.
We have
(2) | f | X = sup { | f ( x ) − f ( y ) | / d ( x , y ) : x , y ∈ X , x ≠ y } . (2) | f | X = sup { | f ( x ) − f ( y ) | / d ( x , y ) : x , y ∈ X , x ≠ y } . {:(2)|f|_(X)=s u p{|f(x)-f(y)|//d(x","y):x","y in X","x!=y}.:}\begin{equation*} |f|_{X}=\sup \{|f(x)-f(y)| / d(x, y): x, y \in X, x \neq y\} . \tag{2} \end{equation*}(2)|f|X=sup{|f(x)−f(y)|/d(x,y):x,y∈X,x≠y}.
Denote by Lip X X XXX the linear space of all Lipschitz functions on X X XXX. Actually, il ⋅ | X ⋅ X *|_(X)\left.\cdot\right|_{X}⋅|X is not a norm on the space Lip X X XXX, since ‖ f ‖ X = 0 ‖ f ‖ X = 0 ||f||_(X)=0\|f\|_{X}=0‖f‖X=0 if f f fff is constant.
Now let Y Y YYY be a nonvoid subset of X X XXX. A norm-preserving extension of a function f ∈ Lip Y f ∈ Lip Y f in Lip Yf \in \operatorname{Lip} Yf∈LipY to X X XXX is a function F ∈ Lip X F ∈ Lip X F in Lip XF \in \operatorname{Lip} XF∈LipX such that F | Y = f F Y = f F|_(Y)=f\left.F\right|_{Y}=fF|Y=f and ‖ f ‖ Y = ‖ F ‖ X ‖ f ‖ Y = ‖ F ‖ X ||f||_(Y)=||F||_(X)\|f\|_{Y}=\|F\|_{X}‖f‖Y=‖F‖X. By a result of Banach [1] (see also Czipser and Geher [2]) every f ∈ Lip Y f ∈ Lip Y f in Lip Yf \in \operatorname{Lip} Yf∈LipY has a norm-preserving extension F F FFF in Lip X X XXX. Two of these extensions are given by
(3) F 1 ( x ) = sup { f ( y ) − | f | Y d ( x , y ) : y ∈ Y } (3) F 1 ( x ) = sup f ( y ) − | f | Y d ( x , y ) : y ∈ Y {:(3)F_(1)(x)=s u p{f(y)-|f|_(Y)d(x,y):y in Y}:}\begin{equation*} F_{1}(x)=\sup \left\{f(y)-|f|_{Y} d(x, y): y \in Y\right\} \tag{3} \end{equation*}(3)F1(x)=sup{f(y)−|f|Yd(x,y):y∈Y}
and
(4) F 2 ( x ) = inf { f ( y ) + f | Y d ( x , y ) : y ∈ Y } . (4) F 2 ( x ) = inf f ( y ) + f Y d ( x , y ) : y ∈ Y . {:(4)F_(2)(x)=i n f{f(y)+f|_(Y)d(x,y):y in Y}.:}\begin{equation*} F_{2}(x)=\inf \left\{f(y)+\left.f\right|_{Y} d(x, y): y \in Y\right\} . \tag{4} \end{equation*}(4)F2(x)=inf{f(y)+f|Yd(x,y):y∈Y}.
Every norm-preserving extension F F FFF of f f fff satisfies
(5) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) (5) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) {:(5)F_(1)(x) <= F(x) <= F_(2)(x):}\begin{equation*} F_{1}(x) \leqslant F(x) \leqslant F_{2}(x) \tag{5} \end{equation*}(5)F1(x)⩽F(x)⩽F2(x)
for all x ∈ X x ∈ X x in Xx \in Xx∈X (see [7]).
Now, let X X XXX be a normed linear space and let Y Y YYY be a nonvoid convex subset of X X XXX. Concerning the convex norm-preserving extension to X X XXX of the convex functions in Lip Y Y YYY, we can prove the following theorem:
Theorem 1. If X X XXX is a normed linear space and Y Y YYY a nonvoid convex subset of X X XXX, then every convex function f f fff in Lip Y Y YYY has a convex norm preserving extension F F FFF in Lip X X XXX.
Proof. We show that the maximal norm-preserving extension (4) of f f fff is also convex. Let F ( x ) = inf { f ( y ) + ‖ f ‖ Y ‖ x − y ‖ : y ∈ Y } , x 1 , x 2 ∈ X , y 1 F ( x ) = inf f ( y ) + ‖ f ‖ Y ‖ x − y ‖ : y ∈ Y , x 1 , x 2 ∈ X , y 1 F(x)=i n f{f(y)+||f||_(Y)||x-y||:y in Y},x_(1),x_(2)in X,y_(1)F(x)=\inf \left\{f(y)+\|f\|_{Y}\|x-y\|: y \in Y\right\}, x_{1}, x_{2} \in X, y_{1}F(x)=inf{f(y)+‖f‖Y‖x−y‖:y∈Y},x1,x2∈X,y1, y 2 ∈ Y y 2 ∈ Y y_(2)in Yy_{2} \in Yy2∈Y, and α ∈ [ 0 , 1 ] α ∈ [ 0 , 1 ] alpha in[0,1]\alpha \in[0,1]α∈[0,1]. Then
F ( α x 1 + ( 1 − α ) x 2 ) ⩽ f ( α y 1 + ( 1 − α ) y 2 ) + ‖ f ‖ Y ‖ α x 1 + ( 1 − α ) x 2 − α y 1 − ( 1 − α ) y 2 ‖ ⩽ α f ( y 1 ) + ( 1 − α ) f ( y 2 ) + ‖ f ‖ Y ( α ‖ x 1 − y 1 ‖ + ( 1 − α ) ‖ x 2 − y 2 ‖ ) = α ( f ( y 1 ) + ‖ f ‖ Y ‖ x 1 − y 1 ‖ ) + ( 1 − α ) ( f ( y 2 ) + ‖ f ‖ Y ‖ x 2 − y 2 ‖ ) F α x 1 + ( 1 − α ) x 2 ⩽ f α y 1 + ( 1 − α ) y 2 + ‖ f ‖ Y α x 1 + ( 1 − α ) x 2 − α y 1 − ( 1 − α ) y 2 ⩽ α f y 1 + ( 1 − α ) f y 2 + ‖ f ‖ Y α x 1 − y 1 + ( 1 − α ) x 2 − y 2 = α f y 1 + ‖ f ‖ Y x 1 − y 1 + ( 1 − α ) f y 2 + ‖ f ‖ Y x 2 − y 2 {:[F(alphax_(1)+(1-alpha)x_(2))],[quad <= f(alphay_(1)+(1-alpha)y_(2))+||f||_(Y)||alphax_(1)+(1-alpha)x_(2)-alphay_(1)-(1-alpha)y_(2)||],[quad <= alpha f(y_(1))+(1-alpha)f(y_(2))+||f||_(Y)(alpha||x_(1)-y_(1)||+(1-alpha)||x_(2)-y_(2)||)],[quad=alpha(f(y_(1))+||f||_(Y)||x_(1)-y_(1)||)+(1-alpha)(f(y_(2))+||f||_(Y)||x_(2)-y_(2)||)]:}\begin{aligned} & F\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \\ & \quad \leqslant f\left(\alpha y_{1}+(1-\alpha) y_{2}\right)+\|f\|_{Y}\left\|\alpha x_{1}+(1-\alpha) x_{2}-\alpha y_{1}-(1-\alpha) y_{2}\right\| \\ & \quad \leqslant \alpha f\left(y_{1}\right)+(1-\alpha) f\left(y_{2}\right)+\|f\|_{Y}\left(\alpha\left\|x_{1}-y_{1}\right\|+(1-\alpha)\left\|x_{2}-y_{2}\right\|\right) \\ & \quad=\alpha\left(f\left(y_{1}\right)+\|f\|_{Y}\left\|x_{1}-y_{1}\right\|\right)+(1-\alpha)\left(f\left(y_{2}\right)+\|f\|_{Y}\left\|x_{2}-y_{2}\right\|\right) \end{aligned}F(αx1+(1−α)x2)⩽f(αy1+(1−α)y2)+‖f‖Y‖αx1+(1−α)x2−αy1−(1−α)y2‖⩽αf(y1)+(1−α)f(y2)+‖f‖Y(α‖x1−y1‖+(1−α)‖x2−y2‖)=α(f(y1)+‖f‖Y‖x1−y1‖)+(1−α)(f(y2)+‖f‖Y‖x2−y2‖)
Taking the infimum with respect to y 1 , y 2 ∈ Y y 1 , y 2 ∈ Y y_(1),y_(2)in Yy_{1}, y_{2} \in Yy1,y2∈Y, we obtain
F ( α x 1 + ( 1 − α ) x 2 ) ⩽ α F ( x 1 ) + ( 1 − α ) F ( x 2 ) , F α x 1 + ( 1 − α ) x 2 ⩽ α F x 1 + ( 1 − α ) F x 2 , F(alphax_(1)+(1-alpha)x_(2)) <= alpha F(x_(1))+(1-alpha)F(x_(2)),F\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \leqslant \alpha F\left(x_{1}\right)+(1-\alpha) F\left(x_{2}\right),F(αx1+(1−α)x2)⩽αF(x1)+(1−α)F(x2),
which shows that the function F F FFF is convex.
In general, this extension is not unique. Indeed, let X = R X = R X=RX=RX=R, with the usual absolute value norm, Y = [ − 1 , 1 ] Y = [ − 1 , 1 ] Y=[-1,1]Y=[-1,1]Y=[−1,1], and f : Y → R f : Y → R f:Y rarr Rf: Y \rightarrow Rf:Y→R be given by f ( x ) = − x f ( x ) = − x f(x)=-xf(x)=-xf(x)=−x for x ∈ [ − 1 , 0 ] x ∈ [ − 1 , 0 ] x in[-1,0]x \in[-1,0]x∈[−1,0] and f ( x ) = 2 x f ( x ) = 2 x f(x)=2xf(x)=2 xf(x)=2x for x ∈ ] 0 , 1 ] x ∈ ] 0 , 1 ] x in]0,1]x \in] 0,1]x∈]0,1]. Then the maximal normpreserving extension (4) of f f fff is given by F ( x ) = − 2 x F ( x ) = − 2 x F(x)=-2xF(x)=-2 xF(x)=−2x for x ∈ ] − ∞ , − 1 [ x ∈ ] − ∞ , − 1 [ x in]-oo,-1[x \in]-\infty,-1[x∈]−∞,−1[, F ( x ) = − 2 x F ( x ) = − 2 x F(x)=-2xF(x)=-2 xF(x)=−2x for x ∈ [ − 1 , 0 [ x ∈ [ − 1 , 0 [ x in[-1,0[x \in[-1,0[x∈[−1,0[, and F ( x ) = 2 x F ( x ) = 2 x F(x)=2xF(x)=2 xF(x)=2x for x ∈ [ 0 , + ∞ [ x ∈ [ 0 , + ∞ [ x in[0,+oo[x \in[0,+\infty[x∈[0,+∞[. But the function G ( x ) = − x G ( x ) = − x G(x)=-xG(x)=-xG(x)=−x for x ∈ ] − ∞ , 0 [ x ∈ ] − ∞ , 0 [ x in]-oo,0[x \in]-\infty, 0[x∈]−∞,0[ and G ( x ) = 2 x G ( x ) = 2 x G(x)=2xG(x)=2 xG(x)=2x for x ∈ [ 0 , + ∞ [ x ∈ [ 0 , + ∞ [ x in[0,+oo[x \in[0,+\infty[x∈[0,+∞[ is also a convex norm-preserving extension of f f fff, and so is every convex combination α F + ( 1 − α ) G , α ∈ [ 0 , 1 ] α F + ( 1 − α ) G , α ∈ [ 0 , 1 ] alpha F+(1-alpha)G,alpha in[0,1]\alpha F+ (1-\alpha) G, \alpha \in[0,1]αF+(1−α)G,α∈[0,1], of the functions F F FFF and G G GGG.
Let, as above, X X XXX be a normed linear space and Z Z ZZZ a convex subset of X X XXX such that 0 ∈ Z 0 ∈ Z 0in Z0 \in Z0∈Z. Denote by Lip 0 Z Lip 0 Z Lip_(0)Z\operatorname{Lip}_{0} ZLip0Z the space
(6) Lip 0 Z = { f ∈ Lip Z : f ( 0 ) = 0 } . (6) Lip 0 Z = { f ∈ Lip Z : f ( 0 ) = 0 } . {:(6)Lip_(0)Z={f in Lip Z:f(0)=0}.:}\begin{equation*} \operatorname{Lip}_{0} Z=\{f \in \operatorname{Lip} Z: f(0)=0\} . \tag{6} \end{equation*}(6)Lip0Z={f∈LipZ:f(0)=0}.
Then (2) is a norm on Lip 0 Z Lip 0 Z Lip_(0)Z\operatorname{Lip}_{0} ZLip0Z and Lip 0 Z Lip 0 Z Lip_(0)Z\operatorname{Lip}_{0} ZLip0Z is a Banach space with respect to this norm.
We use also the following notations:
(7) K Z = { f ∈ Lip 0 Z : f is convex on Z } (7) K Z = f ∈ Lip 0 Z : f  is convex on  Z {:(7)K_(Z)={f inLip_(0)Z:f" is convex on "Z}:}\begin{equation*} K_{Z}=\left\{f \in \operatorname{Lip}_{0} Z: f \text { is convex on } Z\right\} \tag{7} \end{equation*}(7)KZ={f∈Lip0Z:f is convex on Z}
-the convex cone of convex functions in Lip 0 Z Lip 0 Z Lip_(0)Z\operatorname{Lip}_{0} ZLip0Z;
(8) X c = K X − K X , (8) X c = K X − K X , {:(8)X_(c)=K_(X)-K_(X)",":}\begin{equation*} X_{c}=K_{X}-K_{X}, \tag{8} \end{equation*}(8)Xc=KX−KX,
-the linear space generated by the cone K X K X K_(X)K_{X}KX;
(9) Z c ⊥ = { f ∈ X c : f | Z = 0 } (9) Z c ⊥ = f ∈ X c : f Z = 0 {:(9)Z_(c)^(_|_)={f inX_(c):f|_(Z)=0}:}\begin{equation*} Z_{c}{ }^{\perp}=\left\{f \in X_{c}:\left.f\right|_{Z}=0\right\} \tag{9} \end{equation*}(9)Zc⊥={f∈Xc:f|Z=0}
-the null space of the set Z Z ZZZ in X c X c X_(c)X_{c}Xc.
If E E EEE is a normed linear space, M M MMM a nonvoid subset of E E EEE and x ∈ E x ∈ E x in Ex \in Ex∈E, we denote by d ( x , M ) d ( x , M ) d(x,M)d(x, M)d(x,M) the distance from x x xxx to M M MMM, i.e.,
d ( x , M ) = inf { ‖ x − y : y ∈ M } d ( x , M ) = inf { ‖ x − y : y ∈ M } d(x,M)=i n f{||x-y:y in M}d(x, M)=\inf \{\| x-y: y \in M\}d(x,M)=inf{‖x−y:y∈M}
and by P M P M P_(M)P_{M}PM the metric projection of X X XXX onto M M MMM, i.e.,
P M ( x ) = { y ∈ M : | x − y | = d ( x , M ) } . P M ( x ) = { y ∈ M : | x − y | = d ( x , M ) } . P_(M)(x)={y in M:|x-y|=d(x,M)}.P_{M}(x)=\{y \in M:|x-y|=d(x, M)\} .PM(x)={y∈M:|x−y|=d(x,M)}.
If K K KKK is a subset of X X XXX, then the set M M MMM is called K K KKK-proximinal ( K K KKK-Chebyshevian) if P M ( x ) ≠ ∅ P M ( x ) ≠ ∅ P_(M)(x)!=O/P_{M}(x) \neq \varnothingPM(x)≠∅ (respectively card ( P M ( x ) ) = 1 P M ( x ) = 1 (P_(M)(x))=1\left(P_{M}(x)\right)=1(PM(x))=1 ), for all x ∈ K x ∈ K x in Kx \in Kx∈K.
In the sequel X X XXX denotes a normed linear space and Y Y YYY a convex subset of X X XXX such that 0 ∈ Y 0 ∈ Y 0in Y0 \in Y0∈Y. It follows that K Y K Y K_(Y)K_{Y}KY is a P P PPP-cone in the sense of [10], and as a particular case of the results proved there, one obtains:
Theorem 2. (a) If f ∈ K X f ∈ K X f inK_(X)f \in K_{X}f∈KX then
| f | Y | Y = d ( f , Y c ⊥ ) | f | Y Y = d f , Y c ⊥ |f|_(Y)|_(Y)=d(f,Y_(c)^(_|_))\left.|f|_{Y}\right|_{Y}=d\left(f, Y_{c}^{\perp}\right)|f|Y|Y=d(f,Yc⊥)
(b) The space Y c ⊥ Y c ⊥ Y_(c)^(_|_)Y_{c}{ }^{\perp}Yc⊥ is K X K X K_(X)K_{X}KX-proximinal. For f ∈ K X f ∈ K X f inK_(X)f \in K_{X}f∈KX, the function g g ggg is in P Y e ⊥ ( f ) P Y e ⊥ ( f ) P_(Y_(e)^(_|_))(f)P_{Y_{e}^{\perp}}(f)PYe⊥(f) if and only if g = f − F g = f − F g=f-Fg=f-Fg=f−F, where F F FFF is a convex norm-preserving extension off f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y.
(c) The space Y c ⊥ Y c ⊥ Y_(c)^(_|_)Y_{c}{ }^{\perp}Yc⊥ is K X K X K_(X)K_{X}KX-Chebyshevian if and only if every f ∈ K Y f ∈ K Y f inK_(Y)f \in K_{Y}f∈KY has a unique convex norm-preserving extension to X X XXX.
Remark. Similar duality results appear in [4, 11] for linear functionals and in [6-10] for Lipschitz functions.
Now, we want to show that an inequality similar to (5) holds also for the convex norm-preserving extensions of a given convex Lipschitz function. For f ∈ K Y f ∈ K Y f inK_(Y)f \in K_{Y}f∈KY let us denote by E Y c ( f ) E Y c ( f ) E_(Y)^(c)(f)E_{Y}{ }^{c}(f)EYc(f) the set of all convex norm preserving extensions of f f fff. We denote the norm ⋅ ∣ x ⋅ ∣ x *∣x\cdot \mid x⋅∣x by ⋅ ⋅ ∣ ⋅ ⋅ ∣ **∣\cdot \cdot \mid⋅⋅∣.
Theorem 3. If f ∈ K Y f ∈ K Y f inK_(Y)f \in K_{Y}f∈KY then there exist two functions F 1 , F 2 F 1 , F 2 F_(1),F_(2)F_{1}, F_{2}F1,F2 in E Y c ( f ) E Y c ( f ) E_(Y)^(c)(f)E_{Y}{ }^{c}(f)EYc(f) such that
(10) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) (10) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) {:(10)F_(1)(x) <= F(x) <= F_(2)(x):}\begin{equation*} F_{1}(x) \leqslant F(x) \leqslant F_{2}(x) \tag{10} \end{equation*}(10)F1(x)⩽F(x)⩽F2(x)
for all x ∈ X x ∈ X x in Xx \in Xx∈X and F ∈ E Y c ( f ) F ∈ E Y c ( f ) F inE_(Y)^(c)(f)F \in E_{Y}{ }^{c}(f)F∈EYc(f).
For the proof we need the following lemma:
Lemma 4. The set E Y c ( f ) E Y c ( f ) E_(Y)^(c)(f)E_{Y}{ }^{c}(f)EYc(f) is downward directed (with respect to the pointwise ordering).
Proof of Lemma 4. We have to show that for G 1 , G 2 ∈ E Y c ( f ) G 1 , G 2 ∈ E Y c ( f ) G_(1),G_(2)inE_(Y)^(c)(f)G_{1}, G_{2} \in E_{Y}{ }^{c}(f)G1,G2∈EYc(f) there exists G ∈ E Y c ( f ) G ∈ E Y c ( f ) G inE_(Y)^(c)(f)G \in E_{Y}{ }^{c}(f)G∈EYc(f) such that
(11) G ( x ) ⩽ min ( G 1 ( x ) , G 2 ( x ) ) , (11) G ( x ) ⩽ min G 1 ( x ) , G 2 ( x ) , {:(11)G(x) <= min(G_(1)(x),G_(2)(x))",":}\begin{equation*} G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right), \tag{11} \end{equation*}(11)G(x)⩽min(G1(x),G2(x)),
for all x ∈ X x ∈ X x in Xx \in Xx∈X.
If E E EEE is a linear space and φ : E → R ∪ { ± ∞ } φ : E → R ∪ { ± ∞ } varphi:E rarr R uu{+-oo}\varphi: E \rightarrow R \cup\{ \pm \infty\}φ:E→R∪{±∞} is a function, then the strict epigraph of φ φ varphi\varphiφ is defined by
epi ′ φ = { ( x , a ) ∈ E × R : φ ( x ) < a } . epi ′ φ = { ( x , a ) ∈ E × R : φ ( x ) < a } . epi^(')varphi={(x,a)in E xx R:varphi(x) < a}.\operatorname{epi}^{\prime} \varphi=\{(x, a) \in E \times R: \varphi(x)<a\} .epi′φ={(x,a)∈E×R:φ(x)<a}.
The function φ φ varphi\varphiφ is convex if and only if its strict epigraph is a convex subset of E × R E × R E xx RE \times RE×R (see Laurent [5, Theorem 6.1.5, Remark 6.1.6]).
For G 1 , G 2 ∈ E Y c ( f ) G 1 , G 2 ∈ E Y c ( f ) G_(1),G_(2)inE_(Y)^(c)(f)G_{1}, G_{2} \in E_{Y}{ }^{c}(f)G1,G2∈EYc(f) put
(12) Γ = co ( epi ′ G 1 ∪ epi ′ G 2 ) (12) Γ = co  epi  ′ G 1 ∪  epi  ′ G 2 {:(12)Gamma=co(" epi "^(')G_(1)uu" epi "^(')G_(2)):}\begin{equation*} \Gamma=\operatorname{co}\left(\text { epi }^{\prime} G_{1} \cup \text { epi }^{\prime} G_{2}\right) \tag{12} \end{equation*}(12)Γ=co( epi ′G1∪ epi ′G2)
where co ( A ) co ( A ) co(A)\operatorname{co}(A)co(A) denotes the convex hull of the set A A AAA.
Define G : X → R ∪ { ± ∞ } G : X → R ∪ { ± ∞ } G:X rarr R uu{+-oo}G: X \rightarrow R \cup\{ \pm \infty\}G:X→R∪{±∞} by
(13) G ( x ) = inf { a ∈ R : ( x , a ) ∈ Γ } , x ∈ X . (13) G ( x ) = inf { a ∈ R : ( x , a ) ∈ Γ } , x ∈ X . {:(13)G(x)=i n f{a in R:(x","a)in Gamma}","quad x in X.:}\begin{equation*} G(x)=\inf \{a \in R:(x, a) \in \Gamma\}, \quad x \in X . \tag{13} \end{equation*}(13)G(x)=inf{a∈R:(x,a)∈Γ},x∈X.
We show that G ∈ E Y c ( f ) G ∈ E Y c ( f ) G inE_(Y)^(c)(f)G \in E_{Y}{ }^{c}(f)G∈EYc(f) and that G G GGG verifies the inequality (11). The proof is divided into several steps.
(i) The set Γ Γ Gamma\GammaΓ is open. Since the functions G 1 G 1 G_(1)G_{1}G1 and G 2 G 2 G_(2)G_{2}G2 are continuous, the sets epi' G 1 G 1 G_(1)G_{1}G1 and epi' G 2 G 2 G_(2)G_{2}G2 are open, and so is their convex hull Γ Γ Gamma\GammaΓ.
(ii) If ( z , c ) ∈ Γ ( z , c ) ∈ Γ (z,c)in Gamma(z, c) \in \Gamma(z,c)∈Γ and d ⩾ c d ⩾ c d >= cd \geqslant cd⩾c then ( z , d ) ∈ Γ ( z , d ) ∈ Γ (z,d)in Gamma(z, d) \in \Gamma(z,d)∈Γ. Let z = α x + ( 1 − α ) y z = α x + ( 1 − α ) y z=alpha x+(1-alpha)yz=\alpha x+(1-\alpha) yz=αx+(1−α)y, c = α a + ( 1 − α ) h c = α a + ( 1 − α ) h c=alpha a+(1-alpha)hc=\alpha a+(1-\alpha) hc=αa+(1−α)h, for α ∈ [ 0 , 1 ] , ( x , a ) ∈ α ∈ [ 0 , 1 ] , ( x , a ) ∈ alpha in[0,1],(x,a)in\alpha \in[0,1],(x, a) \inα∈[0,1],(x,a)∈ epi ′ G 1 , ( y , b ) ∈ ′ G 1 , ( y , b ) ∈ ^(')G_(1),(y,b)in^{\prime} G_{1},(y, b) \in′G1,(y,b)∈ epi ′ G 2 ′ G 2 ^(')G_(2)^{\prime} G_{2}′G2 and let ϵ > 0 ϵ > 0 epsilon > 0\epsilon>0ϵ>0 be an arbitrary number. Then ( x , a + ϵ ) ∈ epi ′ G 1 ( x , a + ϵ ) ∈ epi ′ G 1 (x,a+epsilon)inepi^(')G_(1)(x, a+\epsilon) \in \operatorname{epi}^{\prime} G_{1}(x,a+ϵ)∈epi′G1 and ( y , b + ϵ ) ∈ ( y , b + ϵ ) ∈ (y,b+epsilon)in(y, b+\epsilon) \in(y,b+ϵ)∈ epi' G 2 G 2 G_(2)G_{2}G2, so that ( z , c + ϵ ) = α ( x , a + ϵ ) + ( 1 − α ) ( y , b + ϵ ) ∈ Γ ( z , c + ϵ ) = α ( x , a + ϵ ) + ( 1 − α ) ( y , b + ϵ ) ∈ Γ (z,c+epsilon)=alpha(x,a+epsilon)+(1-alpha)(y,b+epsilon)in Gamma(z, c+\epsilon)=\alpha(x, a+\epsilon)+(1-\alpha)(y, b+\epsilon) \in \Gamma(z,c+ϵ)=α(x,a+ϵ)+(1−α)(y,b+ϵ)∈Γ.
(iii) epi' G = Γ G = Γ G=GammaG=\GammaG=Γ and G G GGG is a convex function. Let ( x , a ) ∈ ( x , a ) ∈ (x,a)in(x, a) \in(x,a)∈ epi' G G GGG, i.e., G ( x ) < a G ( x ) < a G(x) < aG(x)<aG(x)<a. By (13) there exists b ∈ R b ∈ R b in Rb \in Rb∈R such that ( x , b ) ∈ Γ ( x , b ) ∈ Γ (x,b)in Gamma(x, b) \in \Gamma(x,b)∈Γ and b < a b < a b < ab<ab<a. By (ii), ( x , a ) ∈ Γ ( x , a ) ∈ Γ (x,a)in Gamma(x, a) \in \Gamma(x,a)∈Γ, proving the inclusion epi' G ⊂ Γ G ⊂ Γ G sub GammaG \subset \GammaG⊂Γ.
Conversely, let ( x , a ) ∈ Γ ( x , a ) ∈ Γ (x,a)in Gamma(x, a) \in \Gamma(x,a)∈Γ. By (i) Γ Γ Gamma\GammaΓ is open, so that there exist a neighborhood U U UUU of x x xxx and ϵ > 0 ϵ > 0 epsilon > 0\epsilon>0ϵ>0 such that U × ] a − ϵ , a + ϵ [ ⊂ Γ U × ] a − ϵ , a + ϵ [ ⊂ Γ U xx]a-epsilon,a+epsilon[sub GammaU \times] a-\epsilon, a+\epsilon[\subset \GammaU×]a−ϵ,a+ϵ[⊂Γ. Therefore { x } × ] a − ϵ , a + ϵ [ ⊂ Γ { x } × ] a − ϵ , a + ϵ [ ⊂ Γ {x}xx]a-epsilon,a+epsilon[sub Gamma\{x\} \times] a-\epsilon, a+\epsilon[\subset \Gamma{x}×]a−ϵ,a+ϵ[⊂Γ and, by (13), G ( x ) ⩽ a − ϵ < a G ( x ) ⩽ a − ϵ < a G(x) <= a-epsilon < aG(x) \leqslant a-\epsilon<aG(x)⩽a−ϵ<a, which shows that ( x , a ) ∈ ( x , a ) ∈ (x,a)in(x, a) \in(x,a)∈ epi ′ G ′ G ^(')G^{\prime} G′G and Γ ⊂ Γ ⊂ Gamma sub\Gamma \subsetΓ⊂ epi' G G GGG.
The convexity of G G GGG follows from the above quoted result in Laurent [5].
(iv) We have G ( x ) ⩽ min ( G 1 ( x ) , G 2 ( x ) ) G ( x ) ⩽ min G 1 ( x ) , G 2 ( x ) G(x) <= min(G_(1)(x),G_(2)(x))G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right)G(x)⩽min(G1(x),G2(x)) for all x ∈ X x ∈ X x in Xx \in Xx∈X and G ( z ) = G 1 ( z ) = G 2 ( z ) G ( z ) = G 1 ( z ) = G 2 ( z ) G(z)=G_(1)(z)=G_(2)(z)G(z)= G_{1}(z)=G_{2}(z)G(z)=G1(z)=G2(z) for all z ∈ Y z ∈ Y z in Yz \in Yz∈Y. Let x ∈ X x ∈ X x in Xx \in Xx∈X. Then for all a > G 1 ( x ) a > G 1 ( x ) a > G_(1)(x)a>G_{1}(x)a>G1(x) and b > G 2 ( x ) b > G 2 ( x ) b > G_(2)(x)b>G_{2}(x)b>G2(x) we have ( x , a ) ∈ ( x , a ) ∈ (x,a)in(x, a) \in(x,a)∈ epi ′ G 1 ⊂ Γ ′ G 1 ⊂ Γ ^(')G_(1)sub Gamma^{\prime} G_{1} \subset \Gamma′G1⊂Γ and ( y , b ) ∈ ( y , b ) ∈ (y,b)in(y, b) \in(y,b)∈ epi ′ G 2 ⊂ Γ ′ G 2 ⊂ Γ ^(')G_(2)sub Gamma^{\prime} G_{2} \subset \Gamma′G2⊂Γ, so that, by (13), G ( x ) ⩽ min ( G 1 ( x ) , G 2 ( x ) ) G ( x ) ⩽ min G 1 ( x ) , G 2 ( x ) G(x) <= min(G_(1)(x),G_(2)(x))G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right)G(x)⩽min(G1(x),G2(x)).
Let z z zzz be in Y Y YYY and c c ccc in R R RRR such that ( z , c ) ∈ Γ ( z , c ) ∈ Γ (z,c)in Gamma(z, c) \in \Gamma(z,c)∈Γ. Then ( z , c ) = − a ( x , a ) ( 1 − α ) ( y , b ) ( z , c ) = − a ( x , a ) ( 1 − α ) ( y , b ) (z,c)=-a(x,a)(1-alpha)(y,b)(z, c)=-a(x, a) (1-\alpha)(y, b)(z,c)=−a(x,a)(1−α)(y,b), for a number α ∈ [ 0 , 1 ] , ( x , a ) ∈ α ∈ [ 0 , 1 ] , ( x , a ) ∈ alpha in[0,1],(x,a)in\alpha \in[0,1],(x, a) \inα∈[0,1],(x,a)∈ epi ′ G 1 ′ G 1 ^(')G_(1)^{\prime} G_{1}′G1, and ( y , b ) ∈ ( y , b ) ∈ (y,b)in(y, b) \in(y,b)∈ epi ′ G 2 ′ G 2 ^(')G_(2)^{\prime} G_{2}′G2. But, by the convexity of G 1 G 1 G_(1)G_{1}G1 and G 2 , G i ( z ) = G i ( α x + ( 1 − α ) y ) ⩽ α G i ( x ) + ( 1 − α ) G i ( y ) < α a + ( 1 − α ) b = c G 2 , G i ( z ) = G i ( α x + ( 1 − α ) y ) ⩽ α G i ( x ) + ( 1 − α ) G i ( y ) < α a + ( 1 − α ) b = c G_(2),G_(i)(z)=G_(i)(alpha x+(1-alpha)y) <= alphaG_(i)(x)+(1-alpha)G_(i)(y) < alpha a+(1-alpha)b=cG_{2}, G_{i}(z)=G_{i}(\alpha x+(1-\alpha) y) \leqslant \alpha G_{i}(x) +(1-\alpha) G_{i}(y)<\alpha a+(1-\alpha) b=cG2,Gi(z)=Gi(αx+(1−α)y)⩽αGi(x)+(1−α)Gi(y)<αa+(1−α)b=c, for i == 1 i == 1 i==1i==1i==1, 2. Taking the infimum with respect to all c ∈ R c ∈ R c in Rc \in Rc∈R such that ( z , c ) ∈ Γ ( z , c ) ∈ Γ (z,c)in Gamma(z, c) \in \Gamma(z,c)∈Γ we obtain G ( z ) ⩾ G 1 ( z ) = G 2 ( z ) G ( z ) ⩾ G 1 ( z ) = G 2 ( z ) G(z) >= G_(1)(z)=G_(2)(z)G(z) \geqslant G_{1}(z)= G_{2}(z)G(z)⩾G1(z)=G2(z). Since the converse inequality holds for all x ∈ X x ∈ X x in Xx \in Xx∈X, it follows G ( z ) G ( z ) G(z)G(z)G(z) G 1 ( z ) = G 2 ( z ) G 1 ( z ) = G 2 ( z ) G_(1)(z)=G_(2)(z)G_{1}(z)=G_{2}(z)G1(z)=G2(z), for all z ∈ Y z ∈ Y z in Yz \in Yz∈Y.
(v) − ∞ < G ( x ) < + ∞ − ∞ < G ( x ) < + ∞ -oo < G(x) < +oo-\infty<G(x)<+\infty−∞<G(x)<+∞ for all x ∈ X x ∈ X x in Xx \in Xx∈X. The relations ( x , G 1 ( x ) − 1 ) ∈ x , G 1 ( x ) − 1 ∈ (x,G_(1)(x)-1)in\left(x, G_{1}(x)-1\right) \in(x,G1(x)−1)∈ epi' G 1 ⊂ Γ G 1 ⊂ Γ G_(1)sub GammaG_{1} \subset \GammaG1⊂Γ and (13) imply G ( x ) ⩽ G 1 ( x ) + 1 < ∞ G ( x ) ⩽ G 1 ( x ) + 1 < ∞ G(x) <= G_(1)(x)+1 < ooG(x) \leqslant G_{1}(x)+1<\inftyG(x)⩽G1(x)+1<∞. Suppose there exists x ∈ X x ∈ X x in Xx \in Xx∈X such that G ( x ) = − ∞ G ( x ) = − ∞ G(x)=-ooG(x)=-\inftyG(x)=−∞. Choose an element y ∈ Y y ∈ Y y in Yy \in Yy∈Y and put z = 2 y − x z = 2 y − x z=2y-xz=2 y-xz=2y−x. Then, by (iv) and the convexity of G G GGG we get
G 1 ( y ) = G ( y ) ⩽ 2 − 1 ( F ( x ) + F ( z ) ) = − ∞ , G 1 ( y ) = G ( y ) ⩽ 2 − 1 ( F ( x ) + F ( z ) ) = − ∞ , G_(1)(y)=G(y) <= 2^(-1)(F(x)+F(z))=-oo,G_{1}(y)=G(y) \leqslant 2^{-1}(F(x)+F(z))=-\infty,G1(y)=G(y)⩽2−1(F(x)+F(z))=−∞,
implying G 1 ( y ) = − ∞ G 1 ( y ) = − ∞ G_(1)(y)=-ooG_{1}(y)=-\inftyG1(y)=−∞, which is impossible.
(vi) Equality of the norms: ‖ G ‖ = ‖ f | Y = ‖ G 1 ‖ = ‖ G 2 ‖ ‖ G ‖ = f Y = G 1 ‖ = ‖ G 2 ‖ ||G||=||f|_(Y)=||G_(1)||=||G_(2)||\|G\|=\left\|\left.f\right|_{Y}=\right\| G_{1}\|=\| G_{2} \|‖G‖=‖f|Y=‖G1‖=‖G2‖. Since G | Y = G 1 | Y − f G Y = G 1 Y − f G|_(Y)=G_(1)|_(Y)-f\left.G\right|_{Y}=\left.G_{1}\right|_{Y}-fG|Y=G1|Y−f, it follows ‖ G ‖ ⩾ ‖ G 1 | 1 ‖ G ‖ ⩾ ‖ G 1 1 ||G|| >= ||G_(1)|_(1)\|G\| \geqslant \|\left. G_{1}\right|_{1}‖G‖⩾‖G1|1. Suppose ‖ G ‖ > ‖ G 1 ‖ ‖ G ‖ > G 1 ||G|| > ||G_(1)||\|G\|>\left\|G_{1}\right\|‖G‖>‖G1‖. By the definition (2) of the norm in Lip X X XXX, there exist x , y ∈ X , x ≠ y x , y ∈ X , x ≠ y x,y in X,x!=yx, y \in X, x \neq yx,y∈X,x≠y such that | G ( x ) − G ( y ) | / | | x − y | | > | | G 1 | | | G ( x ) − G ( y ) | / | | x − y | | > G 1 |G(x)-G(y)|//||x-y|| > ||G_(1)|||G(x)-G(y)| /||x-y||>\left|\left|G_{1}\right|\right||G(x)−G(y)|/||x−y||>||G1||, say
| G ( x ) − G ( y ) | / ‖ x − y ‖ = | G 1 | ∣ + ϵ , | G ( x ) − G ( y ) | / ‖ x − y ‖ = G 1 ∣ + ϵ , |G(x)-G(y)|//||x-y||=|G_(1)|∣+epsilon,|G(x)-G(y)| /\|x-y\|=\left|G_{1}\right| \mid+\epsilon,|G(x)−G(y)|/‖x−y‖=|G1|∣+ϵ,
for an ϵ ≫ 0 ϵ ≫ 0 epsilon≫0\epsilon \gg 0ϵ≫0. Without loss of generality we can suppose
(14) G ( y ) − G ( x ) ‖ x − y ‖ = ‖ G 1 ‖ + ϵ . (14) G ( y ) − G ( x ) ‖ x − y ‖ = G 1 + ϵ . {:(14)(G(y)-G(x))/(||x-y||)=||G_(1)||+epsilon.:}\begin{equation*} \frac{G(y)-G(x)}{\|x-y\|}=\left\|G_{1}\right\|+\epsilon . \tag{14} \end{equation*}(14)G(y)−G(x)‖x−y‖=‖G1‖+ϵ.
Let x y → = { x + t ( y − x ) : t ⩾ 0 } x y → = { x + t ( y − x ) : t ⩾ 0 } vec(xy)={x+t(y-x):t >= 0}\overrightarrow{x y}=\{x+t(y-x): t \geqslant 0\}xy→={x+t(y−x):t⩾0} be the half-line determined by x x xxx and y y yyy. Define φ : ] 0 , ∞ [ → R φ : ] 0 , ∞ → R varphi:]0,oo[rarr R:}\varphi:] 0, \infty\left[\rightarrow R\right.φ:]0,∞[→R by φ ( t ) = t − 1 ( G ( x + t ( y − x ) ) − G ( x ) ) φ ( t ) = t − 1 ( G ( x + t ( y − x ) ) − G ( x ) ) varphi(t)=t^(-1)(G(x+t(y-x))-G(x))\varphi(t)=t^{-1}(G(x+t(y-x))-G(x))φ(t)=t−1(G(x+t(y−x))−G(x)). By Holmes [3, p. 17], the function φ φ varphi\varphiφ is nondecreasing, so that
G ( x + t ( y − x ) ) − G ( x ) ‖ t ( y − x ) ‖ = 1 ‖ y − x ‖ ⋅ φ ( t ) ⩾ 1 ‖ y − x ‖ ⋅ φ ( 1 ) = G ( y ) − G ( x ) ‖ y − x ‖ = ‖ G 1 ‖ + ϵ ⩾ G 1 ( x + t ( y − x ) ) − G 1 ( x ) ‖ t ( y − x ) ‖ + ϵ G ( x + t ( y − x ) ) − G ( x ) ‖ t ( y − x ) ‖ = 1 ‖ y − x ‖ ⋅ φ ( t ) ⩾ 1 ‖ y − x ‖ ⋅ φ ( 1 ) = G ( y ) − G ( x ) ‖ y − x ‖ = G 1 + ϵ ⩾ G 1 ( x + t ( y − x ) ) − G 1 ( x ) ‖ t ( y − x ) ‖ + ϵ {:[(G(x+t(y-x))-G(x))/(||t(y-x)||)=(1)/(||y-x||)*varphi(t) >= (1)/(||y-x||)*varphi(1)],[=(G(y)-G(x))/(||y-x||)=||G_(1)||+epsilon],[ >= (G_(1)(x+t(y-x))-G_(1)(x))/(||t(y-x)||)+epsilon]:}\begin{aligned} \frac{G(x+t(y-x))-G(x)}{\|t(y-x)\|} & =\frac{1}{\|y-x\|} \cdot \varphi(t) \geqslant \frac{1}{\|y-x\|} \cdot \varphi(1) \\ & =\frac{G(y)-G(x)}{\|y-x\|}=\left\|G_{1}\right\|+\epsilon \\ & \geqslant \frac{G_{1}(x+t(y-x))-G_{1}(x)}{\|t(y-x)\|}+\epsilon \end{aligned}G(x+t(y−x))−G(x)‖t(y−x)‖=1‖y−x‖⋅φ(t)⩾1‖y−x‖⋅φ(1)=G(y)−G(x)‖y−x‖=‖G1‖+ϵ⩾G1(x+t(y−x))−G1(x)‖t(y−x)‖+ϵ
for all t ⩾ 1 t ⩾ 1 t >= 1t \geqslant 1t⩾1.
Therefore
G 1 ( x + t ( y − x ) ) ⩽ G ( x + t ( y − x ) ) − ( G ( x ) − G 1 ( x ) + t ϵ ‖ y − x ‖ ) G 1 ( x + t ( y − x ) ) ⩽ G ( x + t ( y − x ) ) − G ( x ) − G 1 ( x ) + t ϵ ‖ y − x ‖ G_(1)(x+t(y-x)) <= G(x+t(y-x))-(G(x)-G_(1)(x)+t epsilon||y-x||)G_{1}(x+t(y-x)) \leqslant G(x+t(y-x))-\left(G(x)-G_{1}(x)+t \epsilon\|y-x\|\right)G1(x+t(y−x))⩽G(x+t(y−x))−(G(x)−G1(x)+tϵ‖y−x‖)
for all t ⩾ 1 t ⩾ 1 t >= 1t \geqslant 1t⩾1. But for t t ttt sufficiently large, G ( x ) − G 1 ( x ) + t ϵ ‖ y − x ‖ > 0 G ( x ) − G 1 ( x ) + t ϵ ‖ y − x ‖ > 0 G(x)-G_(1)(x)+t epsilon||y-x|| > 0G(x)-G_{1}(x)+t \epsilon\|y-x\|>0G(x)−G1(x)+tϵ‖y−x‖>0, so
that G 1 ( x + t ( y − x ) ) < G ( x + t ( y − x ) ) G 1 ( x + t ( y − x ) ) < G ( x + t ( y − x ) ) G_(1)(x+t(y-x)) < G(x+t(y-x))G_{1}(x+t(y-x))<G(x+t(y-x))G1(x+t(y−x))<G(x+t(y−x)), contradicting the inequality G ⩽ G 1 ( iv ) G ⩽ G 1 ( iv ) G <= G_(1)(iv)G \leqslant G_{1}(\mathrm{iv})G⩽G1(iv).
Lemma 4 is completely proved.
Proof of Theorem 3. Let F 2 F 2 F_(2)F_{2}F2 be the maximal norm-preserving extension (4) of f f fff. By the proof of Theorem 1, F 2 F 2 F_(2)F_{2}F2 is convex and since F 2 ( x ) ⩾ F ( x ) F 2 ( x ) ⩾ F ( x ) F_(2)(x) >= F(x)F_{2}(x) \geqslant F(x)F2(x)⩾F(x) for every norm-preserving extension F F FFF of f f fff, this is a fortiori true for the convex norm-preserving extensions of f f fff.
Put
(15) F 1 ( x ) = inf { F ( x ) : F ∈ E Y c ( f ) } . (15) F 1 ( x ) = inf F ( x ) : F ∈ E Y c ( f ) . {:(15)F_(1)(x)=i n f{F(x):F inE_(Y)^(c)(f)}.:}\begin{equation*} F_{1}(x)=\inf \left\{F(x): F \in E_{Y}^{c}(f)\right\} . \tag{15} \end{equation*}(15)F1(x)=inf{F(x):F∈EYc(f)}.
To end the proof we have to show that F 1 F 1 F_(1)F_{1}F1 is a convex norm-preserving extension of f f fff.
(i) F 1 F 1 F_(1)F_{1}F1 is a convex function. Let x , y ⊂ X , α ∈ [ 0 , 1 ] , ϵ > 0 x , y ⊂ X , α ∈ [ 0 , 1 ] , ϵ > 0 x,y sub X,alpha in[0,1],epsilon > 0x, y \subset X, \alpha \in[0,1], \epsilon>0x,y⊂X,α∈[0,1],ϵ>0 and let G 1 G 1 G_(1)G_{1}G1, G 2 ∈ E Y c ( f ) G 2 ∈ E Y c ( f ) G_(2)inE_(Y)^(c)(f)G_{2} \in E_{Y}{ }^{c}(f)G2∈EYc(f) be such that G 1 ( x ) < F 1 ( x ) + ϵ G 1 ( x ) < F 1 ( x ) + ϵ G_(1)(x) < F_(1)(x)+epsilonG_{1}(x)<F_{1}(x)+\epsilonG1(x)<F1(x)+ϵ and G 2 ( y ) < F 1 ( y ) + ϵ G 2 ( y ) < F 1 ( y ) + ϵ G_(2)(y) < F_(1)(y)+epsilonG_{2}(y)<F_{1}(y)+\epsilonG2(y)<F1(y)+ϵ. Since, by Lemma 4, the set E Y c ( f ) E Y c ( f ) E_(Y)^(c)(f)E_{Y}{ }^{c}(f)EYc(f) is downward directed, there exists G 3 ∈ E Y c ( f ) G 3 ∈ E Y c ( f ) G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f)G3∈EYc(f) such that G 3 ⩽ G 1 G 3 ⩽ G 1 G_(3) <= G_(1)G_{3} \leqslant G_{1}G3⩽G1 and G 3 ⩽ G 2 G 3 ⩽ G 2 G_(3) <= G_(2)G_{3} \leqslant G_{2}G3⩽G2. Then
F 1 ( α x + ( 1 − α ) y ) ⩽ G 3 ( α x + ( 1 − α ) y ) ⩽ α G 3 ( x ) + ( 1 − α ) G 3 ( y ) ⩽ α G 1 ( x ) + ( 1 − α ) G 2 ( y ) < α F 1 ( x ) + ( 1 − α ) F 2 ( y ) + ϵ F 1 ( α x + ( 1 − α ) y ) ⩽ G 3 ( α x + ( 1 − α ) y ) ⩽ α G 3 ( x ) + ( 1 − α ) G 3 ( y ) ⩽ α G 1 ( x ) + ( 1 − α ) G 2 ( y ) < α F 1 ( x ) + ( 1 − α ) F 2 ( y ) + ϵ {:[F_(1)(alpha x+(1-alpha)y)],[quad <= G_(3)(alpha x+(1-alpha)y) <= alphaG_(3)(x)+(1-alpha)G_(3)(y)],[quad <= alphaG_(1)(x)+(1-alpha)G_(2)(y) < alphaF_(1)(x)+(1-alpha)F_(2)(y)+epsilon]:}\begin{aligned} & F_{1}(\alpha x+(1-\alpha) y) \\ & \quad \leqslant G_{3}(\alpha x+(1-\alpha) y) \leqslant \alpha G_{3}(x)+(1-\alpha) G_{3}(y) \\ & \quad \leqslant \alpha G_{1}(x)+(1-\alpha) G_{2}(y)<\alpha F_{1}(x)+(1-\alpha) F_{2}(y)+\epsilon \end{aligned}F1(αx+(1−α)y)⩽G3(αx+(1−α)y)⩽αG3(x)+(1−α)G3(y)⩽αG1(x)+(1−α)G2(y)<αF1(x)+(1−α)F2(y)+ϵ
Since ϵ > 0 ϵ > 0 epsilon > 0\epsilon>0ϵ>0 is arbitrary, it follows that
F 1 ( α x + ( 1 − α ) y ) ⩽ α F 1 ( x ) + ( 1 − α ) F 2 ( y ) , F 1 ( α x + ( 1 − α ) y ) ⩽ α F 1 ( x ) + ( 1 − α ) F 2 ( y ) , F_(1)(alpha x+(1-alpha)y) <= alphaF_(1)(x)+(1-alpha)F_(2)(y),F_{1}(\alpha x+(1-\alpha) y) \leqslant \alpha F_{1}(x)+(1-\alpha) F_{2}(y),F1(αx+(1−α)y)⩽αF1(x)+(1−α)F2(y),
i.e., the function F 1 F 1 F_(1)F_{1}F1 is convex.
(ii) F 1 | Y = f F 1 Y = f F_(1)|_(Y)=f\left.F_{1}\right|_{Y}=fF1|Y=f. This is obvious since F ( y ) = f ( y ) F ( y ) = f ( y ) F(y)=f(y)F(y)=f(y)F(y)=f(y) for all y ∈ Y y ∈ Y y in Yy \in Yy∈Y and F ∈ E Y c ( f ) F ∈ E Y c ( f ) F inE_(Y)^(c)(f)F \in E_{Y}{ }^{c}(f)F∈EYc(f).
(iii) Equality of the norms: ‖ F 1 ‖ = ‖ f ‖ Y F 1 = ‖ f ‖ Y ||F_(1)||=||f||_(Y)\left\|F_{1}\right\|=\|f\|_{Y}‖F1‖=‖f‖Y. Obviously, ∣ F 1 ‖ ⩾ ‖ f ‖ Y ∣ F 1 ‖ ⩾ ‖ f ‖ Y ∣F_(1)|| >= ||f||_(Y)\mid F_{1}\|\geqslant\| f \|_{Y}∣F1‖⩾‖f‖Y. Let us suppose ‖ F 1 ‖ > ‖ f ‖ Y F 1 > ‖ f ‖ Y ||F_(1)|| > ||f||_(Y)\left\|F_{1}\right\|>\|f\|_{Y}‖F1‖>‖f‖Y. Then there exists δ > 0 δ > 0 delta > 0\delta>0δ>0 such that ‖ F 1 ‖ = ‖ f ‖ Y + δ F 1 = ‖ f ‖ Y + δ ||F_(1)||=||f||_(Y)+delta\left\|F_{1}\right\|= \|f\|_{Y}+\delta‖F1‖=‖f‖Y+δ. By the definition of the norm in Lip X X XXX, there exist x , y ∈ X , x ≠ y x , y ∈ X , x ≠ y x,y in X,x!=yx, y \in X, x \neq yx,y∈X,x≠y such that
(16) ( F 1 ( y ) − F 1 ( x ) ) / ‖ y − x ‖ ⩾ ‖ f ‖ Y + ϵ , (16) F 1 ( y ) − F 1 ( x ) / ‖ y − x ‖ ⩾ ‖ f ‖ Y + ϵ , {:(16)(F_(1)(y)-F_(1)(x))//||y-x|| >= ||f||_(Y)+epsilon",":}\begin{equation*} \left(F_{1}(y)-F_{1}(x)\right) /\|y-x\| \geqslant\|f\|_{Y}+\epsilon, \tag{16} \end{equation*}(16)(F1(y)−F1(x))/‖y−x‖⩾‖f‖Y+ϵ,
where 0 < ϵ < δ 0 < ϵ < δ 0 < epsilon < delta0<\epsilon<\delta0<ϵ<δ. By definition (15) of F 1 F 1 F_(1)F_{1}F1, for 0 < η < ϵ ‖ x − y ‖ 0 < η < ϵ ‖ x − y ‖ 0 < eta < epsilon||x-y||0<\eta<\epsilon\|x-y\|0<η<ϵ‖x−y‖, there exist G 1 , G 2 ∈ E Y c ( f ) G 1 , G 2 ∈ E Y c ( f ) G_(1),G_(2)inE_(Y)^(c)(f)G_{1}, G_{2} \in E_{Y}^{c}(f)G1,G2∈EYc(f) such that G 1 ( x ) < F 1 ( x ) + η G 1 ( x ) < F 1 ( x ) + η G_(1)(x) < F_(1)(x)+etaG_{1}(x)<F_{1}(x)+\etaG1(x)<F1(x)+η and G 2 ( y ) < F 1 ( y ) + η G 2 ( y ) < F 1 ( y ) + η G_(2)(y) < F_(1)(y)+etaG_{2}(y)<F_{1}(y)+\etaG2(y)<F1(y)+η. The set E Y c ( f ) E Y c ( f ) E_(Y)^(c)(f)E_{Y}{ }^{c}(f)EYc(f) being downward directed (Lemma 4), there exists G 3 ∈ E Y c ( f ) G 3 ∈ E Y c ( f ) G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f)G3∈EYc(f) such that G 3 ⩽ G 1 G 3 ⩽ G 1 G_(3) <= G_(1)G_{3} \leqslant G_{1}G3⩽G1 and G 3 ⩽ G 2 G 3 ⩽ G 2 G_(3) <= G_(2)G_{3} \leqslant G_{2}G3⩽G2. Consequently
F 1 ( x ) ⩽ G 3 ( x ) < F 1 ( x ) + η F 1 ( x ) ⩽ G 3 ( x ) < F 1 ( x ) + η F_(1)(x) <= G_(3)(x) < F_(1)(x)+etaF_{1}(x) \leqslant G_{3}(x)<F_{1}(x)+\etaF1(x)⩽G3(x)<F1(x)+η
and
F 1 ( y ) ⩽ G 3 ( y ) < F 1 ( y ) + η F 1 ( y ) ⩽ G 3 ( y ) < F 1 ( y ) + η F_(1)(y) <= G_(3)(y) < F_(1)(y)+etaF_{1}(y) \leqslant G_{3}(y)<F_{1}(y)+\etaF1(y)⩽G3(y)<F1(y)+η
or, equivalently,
0 ⩽ G 3 ( x ) − Γ 1 ( x ) < η , 0 ⩽ G 3 ( x ) − Γ 1 ( x ) < η , 0 <= G_(3)(x)-Gamma_(1)(x) < eta,0 \leqslant G_{3}(x)-\Gamma_{1}(x)<\eta,0⩽G3(x)−Γ1(x)<η,
and
0 ⩽ G 3 ( y ) − F 1 ( y ) < η . 0 ⩽ G 3 ( y ) − F 1 ( y ) < η . 0 <= G_(3)(y)-F_(1)(y) < eta.0 \leqslant G_{3}(y)-F_{1}(y)<\eta .0⩽G3(y)−F1(y)<η.
From these inequalities one obtains
G 3 ( x ) − F 1 ( x ) − ( G 3 ( y ) − F 1 ( y ) ) ⩽ G 3 ( x ) − F 1 ( x ) < η , G 3 ( x ) − F 1 ( x ) − G 3 ( y ) − F 1 ( y ) ⩽ G 3 ( x ) − F 1 ( x ) < η , G_(3)(x)-F_(1)(x)-(G_(3)(y)-F_(1)(y)) <= G_(3)(x)-F_(1)(x) < eta,G_{3}(x)-F_{1}(x)-\left(G_{3}(y)-F_{1}(y)\right) \leqslant G_{3}(x)-F_{1}(x)<\eta,G3(x)−F1(x)−(G3(y)−F1(y))⩽G3(x)−F1(x)<η,
so that
(17) G 3 ( y ) − G 3 ( x ) > F 1 ( y ) − F 1 ( x ) − η . (17) G 3 ( y ) − G 3 ( x ) > F 1 ( y ) − F 1 ( x ) − η . {:(17)G_(3)(y)-G_(3)(x) > F_(1)(y)-F_(1)(x)-eta.:}\begin{equation*} G_{3}(y)-G_{3}(x)>F_{1}(y)-F_{1}(x)-\eta . \tag{17} \end{equation*}(17)G3(y)−G3(x)>F1(y)−F1(x)−η.
Taking into account (16) and (17)
G 3 ( y ) − G 3 ( x ) ‖ y − x ‖ > F 1 ( y ) − F 1 ( x ) ‖ y − x ‖ − η ‖ y − x ‖ > ‖ f ‖ Y + ϵ − η y − x > ‖ f ‖ Y . G 3 ( y ) − G 3 ( x ) ‖ y − x ‖ > F 1 ( y ) − F 1 ( x ) ‖ y − x ‖ − η ‖ y − x ‖ > ‖ f ‖ Y + ϵ − η y − x > ‖ f ‖ Y . {:[(G_(3)(y)-G_(3)(x))/(||y-x||) > (F_(1)(y)-F_(1)(x))/(||y-x||)-(eta)/(||y-x||)],[ > ||f||_(Y)+epsilon-(eta)/(y-x) > ||f||_(Y).]:}\begin{aligned} \frac{G_{3}(y)-G_{3}(x)}{\|y-x\|} & >\frac{F_{1}(y)-F_{1}(x)}{\|y-x\|}-\frac{\eta}{\|y-x\|} \\ & >\|f\|_{Y}+\epsilon-\frac{\eta}{y-x}>\|f\|_{Y} . \end{aligned}G3(y)−G3(x)‖y−x‖>F1(y)−F1(x)‖y−x‖−η‖y−x‖>‖f‖Y+ϵ−ηy−x>‖f‖Y.
But then ‖ G 3 ‖ > ‖ f ‖ Y G 3 > ‖ f ‖ Y ||G_(3)|| > ||f||_(Y)\left\|G_{3}\right\|>\|f\|_{Y}‖G3‖>‖f‖Y, in contradiction to G 3 ∈ E Y c ( f ) G 3 ∈ E Y c ( f ) G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f)G3∈EYc(f).
Theorem 3 is proved.
Remark. Let X = R X = R X=RX=RX=R and Y = [ a , b ] , 0 ∈ Y Y = [ a , b ] , 0 ∈ Y Y=[a,b],0in YY=[a, b], 0 \in YY=[a,b],0∈Y. For f ∈ K Y ′ f ∈ K Y ′ f inK_(Y)^(')f \in K_{Y}^{\prime}f∈KY′, let
m 1 = min ( | f ′ ( a + 0 ) | , | f ′ ( b − 0 ) | ) m 1 = min f ′ ( a + 0 ) , f ′ ( b − 0 ) m_(1)=min(|f^(')(a+0)|,|f^(')(b-0)|)m_{1}=\min \left(\left|f^{\prime}(a+0)\right|,\left|f^{\prime}(b-0)\right|\right)m1=min(|f′(a+0)|,|f′(b−0)|)
and
m 2 = max ( | f ′ ( a + 0 ) | , | f ′ ( b − 0 ) | ) . m 2 = max f ′ ( a + 0 ) , f ′ ( b − 0 ) . m_(2)=max(|f^(')(a+0)|,|f^(')(b-0)|).m_{2}=\max \left(\left|f^{\prime}(a+0)\right|,\left|f^{\prime}(b-0)\right|\right) .m2=max(|f′(a+0)|,|f′(b−0)|).
Then the minimal and maximal convex norm-preserving extensions F 1 F 1 F_(1)F_{1}F1 and F 2 F 2 F_(2)F_{2}F2, respectively, of f f fff, are given by
F i ( x ) − f ( x ) for x ∈ [ a , b ] , = f ( x ) − m i ( x − a ) for x ∈ ] − ∞ , a [ , = f ( x ) + m i ( x − b ) for x ∈ ] b , + ∞ [ ; F i ( x ) − f ( x )  for  x ∈ [ a , b ] , = f ( x ) − m i ( x − a )  for  x ∈ ] − ∞ , a [ , = f ( x ) + m i ( x − b )  for  x ∈ ] b , + ∞ [ ; {:[F_(i)(x)-f(x)" for "x in[a","b]","],[=f(x)-m_(i)(x-a)" for "x in]-oo","a[","],[=f(x)+m_(i)(x-b)" for "x in]b","+oo[;]:}\begin{aligned} F_{i}(x) & -f(x) & & \text { for } x \in[a, b], \\ & =f(x)-m_{i}(x-a) & & \text { for } x \in]-\infty, a[, \\ & =f(x)+m_{i}(x-b) & & \text { for } x \in] b,+\infty[; \end{aligned}Fi(x)−f(x) for x∈[a,b],=f(x)−mi(x−a) for x∈]−∞,a[,=f(x)+mi(x−b) for x∈]b,+∞[;
i = 1 , 2 i = 1 , 2 i=1,2i=1,2i=1,2.
Let now X X XXX be a normed linear space, Y Y YYY a convex subset of X X XXX such that 0 ∈ Y 0 ∈ Y 0in Y0 \in Y0∈Y, and Z Z ZZZ a nonvoid bounded subset of X X XXX.
Consider the space
Lip 0 ( X , Z ) = { f | Z : f ∈ Lip 0 X } , Lip 0 ( X , Z ) = f Z : f ∈ Lip 0 X , Lip_(0)(X,Z)={f|_(Z):f inLip_(0)X},\operatorname{Lip}_{0}(X, Z)=\left\{\left.f\right|_{Z}: f \in \operatorname{Lip}_{0} X\right\},Lip0(X,Z)={f|Z:f∈Lip0X},
normed by the uniform norm
‖ f | Z ‖ u = sup { | f | Z ( x ) ∣: x ∈ Z } . f Z u = sup | f | Z ( x ) ∣: x ∈ Z . ||f|_(Z)||_(u)=s u p{|f|_(Z)(x)∣:x in Z}.\left\|\left.f\right|_{Z}\right\|_{u}=\sup \left\{|f|_{Z}(x) \mid: x \in Z\right\} .‖f|Z‖u=sup{|f|Z(x)∣:x∈Z}.
Consider the following problem:
(A) For f ∈ K X f ∈ K X f inK_(X)f \in K_{X}f∈KX, find two elements g ∗ g ∗ g_(**)g_{*}g∗ and g ∗ g ∗ g^(**)g^{*}g∗ in P Y c ⊥ ( f ) P Y c ⊥ ( f ) P_(Y_(c)^(_|_))(f)P_{Y_{c}^{\perp}}(f)PYc⊥(f) such that
‖ f | Z − g ∗ | Z ‖ u = inf { ‖ f | Z − g | Z ‖ u : g ∈ P Y c ⊥ ( f ) } f Z − g ∗ Z u = inf f Z − g Z u : g ∈ P Y c ⊥ ( f ) ||f|_(Z)-g_(**)|_(Z)||_(u)=i n f{||f|_(Z)-g|_(Z)||_(u):g inP_(Y_(c)^(_|_))(f)}\left\|\left.f\right|_{Z}-\left.g_{*}\right|_{Z}\right\|_{u}=\inf \left\{\left\|\left.f\right|_{Z}-\left.g\right|_{Z}\right\|_{u}: g \in P_{Y_{c}^{\perp}}(f)\right\}‖f|Z−g∗|Z‖u=inf{‖f|Z−g|Z‖u:g∈PYc⊥(f)}
and
‖ f | z − g ∗ | z ‖ u = sup { ‖ f | z − g | z | u : g ∈ P Y c ⊥ ( f ) } f z − g ∗ z u = sup ‖ f z − g z u : g ∈ P Y c ⊥ ( f ) ||f|_(z)-g^(**)|_(z)||_(u)=s u p{||f|_(z)-g|_(z)|_(u):g inP_(Y_(c)^(_|_))(f)}\left\|\left.f\right|_{z}-\left.g^{*}\right|_{z}\right\|_{u}=\sup \left\{\|\left. f\right|_{z}-\left.\left.g\right|_{z}\right|_{u}: g \in P_{Y_{c}^{\perp}}(f)\right\}‖f|z−g∗|z‖u=sup{‖f|z−g|z|u:g∈PYc⊥(f)}
Theorem 5. Problem (A) has a solution for all f ∈ K X f ∈ K X f inK_(X)f \in K_{X}f∈KX.
Proof. By Theorem 2(b) every g g ggg in P Y o ⊥ ( f ) P Y o ⊥ ( f ) P_(Y_(o)^(_|_))(f)P_{Y_{o}^{\perp}}(f)PYo⊥(f) has the form g = f − F g = f − F g=f-Fg=f-Fg=f−F for a convex norm-preserving extension F F FFF of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y. By Theorem 3, there exist two convex norm-preserving extensions F 1 F 1 F_(1)F_{1}F1 and F 2 F 2 F_(2)F_{2}F2 of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y such that
F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) F 1 ( x ) ⩽ F ( x ) ⩽ F 2 ( x ) F_(1)(x) <= F(x) <= F_(2)(x)F_{1}(x) \leqslant F(x) \leqslant F_{2}(x)F1(x)⩽F(x)⩽F2(x)
for all x ∈ X x ∈ X x in Xx \in Xx∈X, i.e.,
f ( x ) − g 1 ( x ) ⩽ f ( x ) − g ( x ) ⩽ f ( x ) − g 2 ( x ) , f ( x ) − g 1 ( x ) ⩽ f ( x ) − g ( x ) ⩽ f ( x ) − g 2 ( x ) , f(x)-g_(1)(x) <= f(x)-g(x) <= f(x)-g_(2)(x),f(x)-g_{1}(x) \leqslant f(x)-g(x) \leqslant f(x)-g_{2}(x),f(x)−g1(x)⩽f(x)−g(x)⩽f(x)−g2(x),
for all x ∈ X x ∈ X x in Xx \in Xx∈X, where g i = f − F i , i = 1 , 2 g i = f − F i , i = 1 , 2 g_(i)=f-F_(i),i=1,2g_{i}=f-F_{i}, i=1,2gi=f−Fi,i=1,2. Therefore
min ( ‖ f | Z − g 1 | Z ‖ u , ‖ f | Z − g 2 | Z ‖ u ) ⩽ ‖ f | Z − g | Z ‖ u ⩽ max ( ‖ f | Z − g 1 | Z ‖ u , | f | Z − g 2 | Z ‖ u ) min f Z − g 1 Z u , f Z − g 2 Z u ⩽ f Z − g Z u ⩽ max f Z − g 1 Z u , | f | Z − g 2 Z ‖ u {:[ min(||f|_(Z)-g_(1)|_(Z)||_(u),||f|_(Z)-g_(2)|_(Z)||_(u)) <= ||f|_(Z)-g|_(Z)||_(u)],[quad <= max(||f|_(Z)-g_(1)|_(Z)||_(u),|f|_(Z)-g_(2)|_(Z)||_(u))]:}\begin{aligned} & \min \left(\left\|\left.f\right|_{Z}-\left.g_{1}\right|_{Z}\right\|_{u},\left\|\left.f\right|_{Z}-\left.g_{2}\right|_{Z}\right\|_{u}\right) \leqslant\left\|\left.f\right|_{Z}-\left.g\right|_{Z}\right\|_{u} \\ & \quad \leqslant \max \left(\left\|\left.f\right|_{Z}-\left.g_{1}\right|_{Z}\right\|_{u},|f|_{Z}-\left.g_{2}\right|_{Z} \|_{u}\right) \end{aligned}min(‖f|Z−g1|Z‖u,‖f|Z−g2|Z‖u)⩽‖f|Z−g|Z‖u⩽max(‖f|Z−g1|Z‖u,|f|Z−g2|Z‖u)
It follows that a solution of Problem (A) is given by g ∗ = g i g ∗ = g i g_(**)=g_(i)g_{*}=g_{i}g∗=gi and g ∗ = g j g ∗ = g j g^(**)=g_(j)g^{*}=g_{j}g∗=gj, where i , j ∈ { 1 , 2 } i , j ∈ { 1 , 2 } i,j in{1,2}i, j \in\{1,2\}i,j∈{1,2} are such that
i 1 f | Z − g i | Z ‖ u = min ( ‖ f | Z g 1 | Z ‖ u , ‖ f | Z − g 2 | Z | u ) i 1 f Z − g i Z ‖ u = min f Z g 1 Z u , ‖ f Z − g 2 Z u i_(1)f|_(Z)-g_(i)|_(Z)||_(u)=min(||f|_(Z)quadg_(1)|_(Z)||_(u),||f|_(Z)-g_(2)|_(Z)|_(u))\left.i_{1} f\right|_{Z}-\left.g_{i}\right|_{Z} \|_{u}=\min \left(\left\|\left.\left.f\right|_{Z} \quad g_{1}\right|_{Z}\right\|_{u}, \|\left. f\right|_{Z}-\left.\left.g_{2}\right|_{Z}\right|_{u}\right)i1f|Z−gi|Z‖u=min(‖f|Zg1|Z‖u,‖f|Z−g2|Z|u)
and
| f | Z − g i | Z ‖ u = max ( ‖ f | Z − g 1 | Z ‖ u , ‖ f | Z − g 9 | Z ‖ u ) . | f | Z − g i Z ‖ u = max f Z − g 1 Z u , f Z − g 9 Z u . |f|_(Z)-g_(i)|_(Z)||_(u)=max(||f|_(Z)-g_(1)|_(Z)||_(u),||f|_(Z)-g_(9)|_(Z)||_(u)).|f|_{Z}-\left.g_{i}\right|_{Z} \|_{u}=\max \left(\left\|\left.f\right|_{Z}-\left.g_{1}\right|_{Z}\right\|_{u},\left\|\left.f\right|_{Z}-\left.g_{9}\right|_{Z}\right\|_{u}\right) .|f|Z−gi|Z‖u=max(‖f|Z−g1|Z‖u,‖f|Z−g9|Z‖u).

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1978

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