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Institutul de Matematica Cluj-Napoca, Romania
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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MR # 82j: 41034
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[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.
[10] C. MUSTATA, A characterisation of Chebyshevian subspace of Y–type, Rec. Atlal. Num. ThPor. Approximation 6 (1977), 51-56.
[11] R. R. PHELPS, Uniqueness of Hahn-Banach extension and unique best approximation, Trans. Au7er. Math. Sot. 95 (1960), 238-255.
Paper (preprint) in HTML form
Norm-Preserving Extension of Convex Lipschitz Functions
Received March 15, 1977
Let ( X,dX, d ) be a metric space. A function f:X rarr Rf: X \rightarrow R is called Lipschitz if there exists a number M >= 0M \geqslant 0 such that
{:(1)f(x)-f(y)∣⩽Md(x","y):}\begin{equation*}
f(x)-f(y) \mid \leqslant M d(x, y) \tag{1}
\end{equation*}
for all x,y in Xx, y \in X. The smallest constant MM verifying (1) is called the norm of ff and is denoted by ||f_(X)\| f_{X}.
We have
{:(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*}
Denote by Lip XX the linear space of all Lipschitz functions on XX. Actually, il *|_(X)\left.\cdot\right|_{X} is not a norm on the space Lip XX, since ||f||_(X)=0\|f\|_{X}=0 if ff is constant.
Now let YY be a nonvoid subset of XX. A norm-preserving extension of a function f in Lip Yf \in \operatorname{Lip} Y to XX is a function F in Lip XF \in \operatorname{Lip} X such that F|_(Y)=f\left.F\right|_{Y}=f and ||f||_(Y)=||F||_(X)\|f\|_{Y}=\|F\|_{X}. By a result of Banach [1] (see also Czipser and Geher [2]) every f in Lip Yf \in \operatorname{Lip} Y has a norm-preserving extension FF in Lip XX. Two of these extensions are given by
{:(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*}
and
{:(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*}
Every norm-preserving extension FF of ff satisfies
{:(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*}
for all x in Xx \in X (see [7]).
Now, let XX be a normed linear space and let YY be a nonvoid convex subset of XX. Concerning the convex norm-preserving extension to XX of the convex functions in Lip YY, we can prove the following theorem:
Now, let XX be a normed linear space and let YY be a nonvoid convex subset of XX. Concerning the convex norm-preserving extension to XX of the convex functions in Lip YY, we can prove the following theorem:
Theorem 1. If XX is a normed linear space and YY a nonvoid convex subset of XX, then every convex function ff in Lip YY has a convex norm preserving extension FF in Lip XX.
Proof. We show that the maximal norm-preserving extension (4) of ff is also convex. Let 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}, y_(2)in Yy_{2} \in Y, and alpha in[0,1]\alpha \in[0,1]. Then
{:[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}
Taking the infimum with respect to y_(1),y_(2)in Yy_{1}, y_{2} \in Y, we obtain
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),
which shows that the function FF is convex.
In general, this extension is not unique. Indeed, let X=RX=R, with the usual absolute value norm, Y=[-1,1]Y=[-1,1], and f:Y rarr Rf: Y \rightarrow R be given by f(x)=-xf(x)=-x for x in[-1,0]x \in[-1,0] and f(x)=2xf(x)=2 x for x in]0,1]x \in] 0,1]. Then the maximal normpreserving extension (4) of ff is given by F(x)=-2xF(x)=-2 x for x in]-oo,-1[x \in]-\infty,-1[, F(x)=-2xF(x)=-2 x for x in[-1,0[x \in[-1,0[, and F(x)=2xF(x)=2 x for x in[0,+oo[x \in[0,+\infty[. But the function G(x)=-xG(x)=-x for x in]-oo,0[x \in]-\infty, 0[ and G(x)=2xG(x)=2 x for x in[0,+oo[x \in[0,+\infty[ is also a convex norm-preserving extension of ff, and so is every convex combination alpha F+(1-alpha)G,alpha in[0,1]\alpha F+ (1-\alpha) G, \alpha \in[0,1], of the functions FF and GG.
In general, this extension is not unique. Indeed, let X=RX=R, with the usual absolute value norm, Y=[-1,1]Y=[-1,1], and f:Y rarr Rf: Y \rightarrow R be given by f(x)=-xf(x)=-x for x in[-1,0]x \in[-1,0] and f(x)=2xf(x)=2 x for x in]0,1]x \in] 0,1]. Then the maximal normpreserving extension (4) of ff is given by F(x)=-2xF(x)=-2 x for x in]-oo,-1[x \in]-\infty,-1[, F(x)=-2xF(x)=-2 x for x in[-1,0[x \in[-1,0[, and F(x)=2xF(x)=2 x for x in[0,+oo[x \in[0,+\infty[. But the function G(x)=-xG(x)=-x for x in]-oo,0[x \in]-\infty, 0[ and G(x)=2xG(x)=2 x for x in[0,+oo[x \in[0,+\infty[ is also a convex norm-preserving extension of ff, and so is every convex combination alpha F+(1-alpha)G,alpha in[0,1]\alpha F+ (1-\alpha) G, \alpha \in[0,1], of the functions FF and GG.
Let, as above, XX be a normed linear space and ZZ a convex subset of XX such that 0in Z0 \in Z. Denote by Lip_(0)Z\operatorname{Lip}_{0} Z the space
{:(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*}
Then (2) is a norm on Lip_(0)Z\operatorname{Lip}_{0} Z and Lip_(0)Z\operatorname{Lip}_{0} Z is a Banach space with respect to this norm.
We use also the following notations:
{:(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*}
-the convex cone of convex functions in Lip_(0)Z\operatorname{Lip}_{0} Z;
{:(8)X_(c)=K_(X)-K_(X)",":}\begin{equation*}
X_{c}=K_{X}-K_{X}, \tag{8}
\end{equation*}
-the linear space generated by the cone K_(X)K_{X};
{:(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*}
-the null space of the set ZZ in X_(c)X_{c}.
If EE is a normed linear space, MM a nonvoid subset of EE and x in Ex \in E, we denote by d(x,M)d(x, M) the distance from xx to MM, i.e.,
d(x,M)=i n f{||x-y:y in M}d(x, M)=\inf \{\| x-y: y \in M\}
and by P_(M)P_{M} the metric projection of XX onto MM, i.e.,
P_(M)(x)={y in M:|x-y|=d(x,M)}.P_{M}(x)=\{y \in M:|x-y|=d(x, M)\} .
If KK is a subset of XX, then the set MM is called KK-proximinal ( KK-Chebyshevian) if P_(M)(x)!=O/P_{M}(x) \neq \varnothing (respectively card (P_(M)(x))=1\left(P_{M}(x)\right)=1 ), for all x in Kx \in K.
In the sequel XX denotes a normed linear space and YY a convex subset of XX such that 0in Y0 \in Y. It follows that K_(Y)K_{Y} is a PP-cone in the sense of [10], and as a particular case of the results proved there, one obtains:
Theorem 2. (a) If f inK_(X)f \in K_{X} then
|f|_(Y)|_(Y)=d(f,Y_(c)^(_|_))\left.|f|_{Y}\right|_{Y}=d\left(f, Y_{c}^{\perp}\right)
(b) The space Y_(c)^(_|_)Y_{c}{ }^{\perp} is K_(X)K_{X}-proximinal. For f inK_(X)f \in K_{X}, the function gg is in P_(Y_(e)^(_|_))(f)P_{Y_{e}^{\perp}}(f) if and only if g=f-Fg=f-F, where FF is a convex norm-preserving extension off f|_(Y)\left.f\right|_{Y}.
(c) The space Y_(c)^(_|_)Y_{c}{ }^{\perp} is K_(X)K_{X}-Chebyshevian if and only if every f inK_(Y)f \in K_{Y} has a unique convex norm-preserving extension to XX.
(c) The space Y_(c)^(_|_)Y_{c}{ }^{\perp} is K_(X)K_{X}-Chebyshevian if and only if every f inK_(Y)f \in K_{Y} has a unique convex norm-preserving extension to XX.
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 inK_(Y)f \in K_{Y} let us denote by E_(Y)^(c)(f)E_{Y}{ }^{c}(f) the set of all convex norm preserving extensions of ff. We denote the norm *∣x\cdot \mid x by **∣\cdot \cdot \mid.
Theorem 3. If f inK_(Y)f \in K_{Y} then there exist two functions F_(1),F_(2)F_{1}, F_{2} in E_(Y)^(c)(f)E_{Y}{ }^{c}(f) such that
{:(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*}
for all x in Xx \in X and F inE_(Y)^(c)(f)F \in E_{Y}{ }^{c}(f).
For the proof we need the following lemma:
For the proof we need the following lemma:
Lemma 4. The set E_(Y)^(c)(f)E_{Y}{ }^{c}(f) is downward directed (with respect to the pointwise ordering).
Proof of Lemma 4. We have to show that for G_(1),G_(2)inE_(Y)^(c)(f)G_{1}, G_{2} \in E_{Y}{ }^{c}(f) there exists G inE_(Y)^(c)(f)G \in E_{Y}{ }^{c}(f) such that
{:(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*}
for all x in Xx \in X.
If EE is a linear space and varphi:E rarr R uu{+-oo}\varphi: E \rightarrow R \cup\{ \pm \infty\} is a function, then the strict epigraph of varphi\varphi is defined by
If EE is a linear space and varphi:E rarr R uu{+-oo}\varphi: E \rightarrow R \cup\{ \pm \infty\} is a function, then the strict epigraph of varphi\varphi is defined by
epi^(')varphi={(x,a)in E xx R:varphi(x) < a}.\operatorname{epi}^{\prime} \varphi=\{(x, a) \in E \times R: \varphi(x)<a\} .
The function varphi\varphi is convex if and only if its strict epigraph is a convex subset of E xx RE \times R (see Laurent [5, Theorem 6.1.5, Remark 6.1.6]).
For G_(1),G_(2)inE_(Y)^(c)(f)G_{1}, G_{2} \in E_{Y}{ }^{c}(f) put
{:(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*}
where co(A)\operatorname{co}(A) denotes the convex hull of the set AA.
Define G:X rarr R uu{+-oo}G: X \rightarrow R \cup\{ \pm \infty\} by
Define G:X rarr R uu{+-oo}G: X \rightarrow R \cup\{ \pm \infty\} by
{:(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*}
We show that G inE_(Y)^(c)(f)G \in E_{Y}{ }^{c}(f) and that GG verifies the inequality (11). The proof is divided into several steps.
(i) The set Gamma\Gamma is open. Since the functions G_(1)G_{1} and G_(2)G_{2} are continuous, the sets epi' G_(1)G_{1} and epi' G_(2)G_{2} are open, and so is their convex hull Gamma\Gamma.
(ii) If (z,c)in Gamma(z, c) \in \Gamma and d >= cd \geqslant c then (z,d)in Gamma(z, d) \in \Gamma. Let z=alpha x+(1-alpha)yz=\alpha x+(1-\alpha) y, c=alpha a+(1-alpha)hc=\alpha a+(1-\alpha) h, for alpha in[0,1],(x,a)in\alpha \in[0,1],(x, a) \in epi ^(')G_(1),(y,b)in^{\prime} G_{1},(y, b) \in epi ^(')G_(2)^{\prime} G_{2} and let epsilon > 0\epsilon>0 be an arbitrary number. Then (x,a+epsilon)inepi^(')G_(1)(x, a+\epsilon) \in \operatorname{epi}^{\prime} G_{1} and (y,b+epsilon)in(y, b+\epsilon) \in epi' G_(2)G_{2}, so that (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.
(iii) epi' G=GammaG=\Gamma and GG is a convex function. Let (x,a)in(x, a) \in epi' GG, i.e., G(x) < aG(x)<a. By (13) there exists b in Rb \in R such that (x,b)in Gamma(x, b) \in \Gamma and b < ab<a. By (ii), (x,a)in Gamma(x, a) \in \Gamma, proving the inclusion epi' G sub GammaG \subset \Gamma.
(i) The set Gamma\Gamma is open. Since the functions G_(1)G_{1} and G_(2)G_{2} are continuous, the sets epi' G_(1)G_{1} and epi' G_(2)G_{2} are open, and so is their convex hull Gamma\Gamma.
(ii) If (z,c)in Gamma(z, c) \in \Gamma and d >= cd \geqslant c then (z,d)in Gamma(z, d) \in \Gamma. Let z=alpha x+(1-alpha)yz=\alpha x+(1-\alpha) y, c=alpha a+(1-alpha)hc=\alpha a+(1-\alpha) h, for alpha in[0,1],(x,a)in\alpha \in[0,1],(x, a) \in epi ^(')G_(1),(y,b)in^{\prime} G_{1},(y, b) \in epi ^(')G_(2)^{\prime} G_{2} and let epsilon > 0\epsilon>0 be an arbitrary number. Then (x,a+epsilon)inepi^(')G_(1)(x, a+\epsilon) \in \operatorname{epi}^{\prime} G_{1} and (y,b+epsilon)in(y, b+\epsilon) \in epi' G_(2)G_{2}, so that (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.
(iii) epi' G=GammaG=\Gamma and GG is a convex function. Let (x,a)in(x, a) \in epi' GG, i.e., G(x) < aG(x)<a. By (13) there exists b in Rb \in R such that (x,b)in Gamma(x, b) \in \Gamma and b < ab<a. By (ii), (x,a)in Gamma(x, a) \in \Gamma, proving the inclusion epi' G sub GammaG \subset \Gamma.
Conversely, let (x,a)in Gamma(x, a) \in \Gamma. By (i) Gamma\Gamma is open, so that there exist a neighborhood UU of xx and epsilon > 0\epsilon>0 such that U xx]a-epsilon,a+epsilon[sub GammaU \times] a-\epsilon, a+\epsilon[\subset \Gamma. Therefore {x}xx]a-epsilon,a+epsilon[sub Gamma\{x\} \times] a-\epsilon, a+\epsilon[\subset \Gamma and, by (13), G(x) <= a-epsilon < aG(x) \leqslant a-\epsilon<a, which shows that (x,a)in(x, a) \in epi ^(')G^{\prime} G and Gamma sub\Gamma \subset epi' GG.
The convexity of GG follows from the above quoted result in Laurent [5].
(iv) We have G(x) <= min(G_(1)(x),G_(2)(x))G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right) for all x in Xx \in X and G(z)=G_(1)(z)=G_(2)(z)G(z)= G_{1}(z)=G_{2}(z) for all z in Yz \in Y. Let x in Xx \in X. Then for all a > G_(1)(x)a>G_{1}(x) and b > G_(2)(x)b>G_{2}(x) we have (x,a)in(x, a) \in epi ^(')G_(1)sub Gamma^{\prime} G_{1} \subset \Gamma and (y,b)in(y, b) \in epi ^(')G_(2)sub Gamma^{\prime} G_{2} \subset \Gamma, so that, by (13), G(x) <= min(G_(1)(x),G_(2)(x))G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right).
(iv) We have G(x) <= min(G_(1)(x),G_(2)(x))G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right) for all x in Xx \in X and G(z)=G_(1)(z)=G_(2)(z)G(z)= G_{1}(z)=G_{2}(z) for all z in Yz \in Y. Let x in Xx \in X. Then for all a > G_(1)(x)a>G_{1}(x) and b > G_(2)(x)b>G_{2}(x) we have (x,a)in(x, a) \in epi ^(')G_(1)sub Gamma^{\prime} G_{1} \subset \Gamma and (y,b)in(y, b) \in epi ^(')G_(2)sub Gamma^{\prime} G_{2} \subset \Gamma, so that, by (13), G(x) <= min(G_(1)(x),G_(2)(x))G(x) \leqslant \min \left(G_{1}(x), G_{2}(x)\right).
Let zz be in YY and cc in RR such that (z,c)in Gamma(z, c) \in \Gamma. Then (z,c)=-a(x,a)(1-alpha)(y,b)(z, c)=-a(x, a) (1-\alpha)(y, b), for a number alpha in[0,1],(x,a)in\alpha \in[0,1],(x, a) \in epi ^(')G_(1)^{\prime} G_{1}, and (y,b)in(y, b) \in epi ^(')G_(2)^{\prime} G_{2}. But, by the convexity of G_(1)G_{1} and 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=c, for i==1i==1, 2. Taking the infimum with respect to all c in Rc \in R such that (z,c)in Gamma(z, c) \in \Gamma we obtain G(z) >= G_(1)(z)=G_(2)(z)G(z) \geqslant G_{1}(z)= G_{2}(z). Since the converse inequality holds for all x in Xx \in X, it follows G(z)G(z) G_(1)(z)=G_(2)(z)G_{1}(z)=G_{2}(z), for all z in Yz \in Y.
(v) -oo < G(x) < +oo-\infty<G(x)<+\infty for all x in Xx \in X. The relations (x,G_(1)(x)-1)in\left(x, G_{1}(x)-1\right) \in epi' G_(1)sub GammaG_{1} \subset \Gamma and (13) imply G(x) <= G_(1)(x)+1 < ooG(x) \leqslant G_{1}(x)+1<\infty. Suppose there exists x in Xx \in X such that G(x)=-ooG(x)=-\infty. Choose an element y in Yy \in Y and put z=2y-xz=2 y-x. Then, by (iv) and the convexity of GG we get
(v) -oo < G(x) < +oo-\infty<G(x)<+\infty for all x in Xx \in X. The relations (x,G_(1)(x)-1)in\left(x, G_{1}(x)-1\right) \in epi' G_(1)sub GammaG_{1} \subset \Gamma and (13) imply G(x) <= G_(1)(x)+1 < ooG(x) \leqslant G_{1}(x)+1<\infty. Suppose there exists x in Xx \in X such that G(x)=-ooG(x)=-\infty. Choose an element y in Yy \in Y and put z=2y-xz=2 y-x. Then, by (iv) and the convexity of GG we get
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,
implying G_(1)(y)=-ooG_{1}(y)=-\infty, which is impossible.
(vi) Equality of the norms: ||G||=||f|_(Y)=||G_(1)||=||G_(2)||\|G\|=\left\|\left.f\right|_{Y}=\right\| G_{1}\|=\| G_{2} \|. Since G|_(Y)=G_(1)|_(Y)-f\left.G\right|_{Y}=\left.G_{1}\right|_{Y}-f, it follows ||G|| >= ||G_(1)|_(1)\|G\| \geqslant \|\left. G_{1}\right|_{1}. Suppose ||G|| > ||G_(1)||\|G\|>\left\|G_{1}\right\|. By the definition (2) of the norm in Lip XX, there exist x,y in X,x!=yx, y \in X, x \neq y such that |G(x)-G(y)|//||x-y|| > ||G_(1)|||G(x)-G(y)| /||x-y||>\left|\left|G_{1}\right|\right|, say
(vi) Equality of the norms: ||G||=||f|_(Y)=||G_(1)||=||G_(2)||\|G\|=\left\|\left.f\right|_{Y}=\right\| G_{1}\|=\| G_{2} \|. Since G|_(Y)=G_(1)|_(Y)-f\left.G\right|_{Y}=\left.G_{1}\right|_{Y}-f, it follows ||G|| >= ||G_(1)|_(1)\|G\| \geqslant \|\left. G_{1}\right|_{1}. Suppose ||G|| > ||G_(1)||\|G\|>\left\|G_{1}\right\|. By the definition (2) of the norm in Lip XX, there exist x,y in X,x!=yx, y \in X, x \neq y such that |G(x)-G(y)|//||x-y|| > ||G_(1)|||G(x)-G(y)| /||x-y||>\left|\left|G_{1}\right|\right|, say
|G(x)-G(y)|//||x-y||=|G_(1)|∣+epsilon,|G(x)-G(y)| /\|x-y\|=\left|G_{1}\right| \mid+\epsilon,
for an epsilon≫0\epsilon \gg 0. Without loss of generality we can suppose
{:(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*}
Let vec(xy)={x+t(y-x):t >= 0}\overrightarrow{x y}=\{x+t(y-x): t \geqslant 0\} be the half-line determined by xx and yy. Define varphi:]0,oo[rarr R:}\varphi:] 0, \infty\left[\rightarrow R\right. by varphi(t)=t^(-1)(G(x+t(y-x))-G(x))\varphi(t)=t^{-1}(G(x+t(y-x))-G(x)). By Holmes [3, p. 17], the function varphi\varphi is nondecreasing, so that
{:[(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}
for all t >= 1t \geqslant 1.
Therefore
Therefore
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)
for all t >= 1t \geqslant 1. But for tt sufficiently large, G(x)-G_(1)(x)+t epsilon||y-x|| > 0G(x)-G_{1}(x)+t \epsilon\|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)), contradicting the inequality G <= G_(1)(iv)G \leqslant G_{1}(\mathrm{iv}).
that G_(1)(x+t(y-x)) < G(x+t(y-x))G_{1}(x+t(y-x))<G(x+t(y-x)), contradicting the inequality G <= G_(1)(iv)G \leqslant G_{1}(\mathrm{iv}).
Lemma 4 is completely proved.
Proof of Theorem 3. Let F_(2)F_{2} be the maximal norm-preserving extension (4) of ff. By the proof of Theorem 1, F_(2)F_{2} is convex and since F_(2)(x) >= F(x)F_{2}(x) \geqslant F(x) for every norm-preserving extension FF of ff, this is a fortiori true for the convex norm-preserving extensions of ff.
Proof of Theorem 3. Let F_(2)F_{2} be the maximal norm-preserving extension (4) of ff. By the proof of Theorem 1, F_(2)F_{2} is convex and since F_(2)(x) >= F(x)F_{2}(x) \geqslant F(x) for every norm-preserving extension FF of ff, this is a fortiori true for the convex norm-preserving extensions of ff.
Put
{:(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*}
To end the proof we have to show that F_(1)F_{1} is a convex norm-preserving extension of ff.
(i) F_(1)F_{1} is a convex function. Let x,y sub X,alpha in[0,1],epsilon > 0x, y \subset X, \alpha \in[0,1], \epsilon>0 and let G_(1)G_{1}, G_(2)inE_(Y)^(c)(f)G_{2} \in E_{Y}{ }^{c}(f) be such that G_(1)(x) < F_(1)(x)+epsilonG_{1}(x)<F_{1}(x)+\epsilon and G_(2)(y) < F_(1)(y)+epsilonG_{2}(y)<F_{1}(y)+\epsilon. Since, by Lemma 4, the set E_(Y)^(c)(f)E_{Y}{ }^{c}(f) is downward directed, there exists G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f) such that G_(3) <= G_(1)G_{3} \leqslant G_{1} and G_(3) <= G_(2)G_{3} \leqslant G_{2}. Then
(i) F_(1)F_{1} is a convex function. Let x,y sub X,alpha in[0,1],epsilon > 0x, y \subset X, \alpha \in[0,1], \epsilon>0 and let G_(1)G_{1}, G_(2)inE_(Y)^(c)(f)G_{2} \in E_{Y}{ }^{c}(f) be such that G_(1)(x) < F_(1)(x)+epsilonG_{1}(x)<F_{1}(x)+\epsilon and G_(2)(y) < F_(1)(y)+epsilonG_{2}(y)<F_{1}(y)+\epsilon. Since, by Lemma 4, the set E_(Y)^(c)(f)E_{Y}{ }^{c}(f) is downward directed, there exists G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f) such that G_(3) <= G_(1)G_{3} \leqslant G_{1} and G_(3) <= G_(2)G_{3} \leqslant G_{2}. Then
{:[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}
Since epsilon > 0\epsilon>0 is arbitrary, it follows that
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),
i.e., the function F_(1)F_{1} is convex.
(ii) F_(1)|_(Y)=f\left.F_{1}\right|_{Y}=f. This is obvious since F(y)=f(y)F(y)=f(y) for all y in Yy \in Y and F inE_(Y)^(c)(f)F \in E_{Y}{ }^{c}(f).
(iii) Equality of the norms: ||F_(1)||=||f||_(Y)\left\|F_{1}\right\|=\|f\|_{Y}. Obviously, ∣F_(1)|| >= ||f||_(Y)\mid F_{1}\|\geqslant\| f \|_{Y}. Let us suppose ||F_(1)|| > ||f||_(Y)\left\|F_{1}\right\|>\|f\|_{Y}. Then there exists delta > 0\delta>0 such that ||F_(1)||=||f||_(Y)+delta\left\|F_{1}\right\|= \|f\|_{Y}+\delta. By the definition of the norm in Lip XX, there exist x,y in X,x!=yx, y \in X, x \neq y such that
(ii) F_(1)|_(Y)=f\left.F_{1}\right|_{Y}=f. This is obvious since F(y)=f(y)F(y)=f(y) for all y in Yy \in Y and F inE_(Y)^(c)(f)F \in E_{Y}{ }^{c}(f).
(iii) Equality of the norms: ||F_(1)||=||f||_(Y)\left\|F_{1}\right\|=\|f\|_{Y}. Obviously, ∣F_(1)|| >= ||f||_(Y)\mid F_{1}\|\geqslant\| f \|_{Y}. Let us suppose ||F_(1)|| > ||f||_(Y)\left\|F_{1}\right\|>\|f\|_{Y}. Then there exists delta > 0\delta>0 such that ||F_(1)||=||f||_(Y)+delta\left\|F_{1}\right\|= \|f\|_{Y}+\delta. By the definition of the norm in Lip XX, there exist x,y in X,x!=yx, y \in X, x \neq y such that
{:(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*}
where 0 < epsilon < delta0<\epsilon<\delta. By definition (15) of F_(1)F_{1}, for 0 < eta < epsilon||x-y||0<\eta<\epsilon\|x-y\|, there exist G_(1),G_(2)inE_(Y)^(c)(f)G_{1}, G_{2} \in E_{Y}^{c}(f) such that G_(1)(x) < F_(1)(x)+etaG_{1}(x)<F_{1}(x)+\eta and G_(2)(y) < F_(1)(y)+etaG_{2}(y)<F_{1}(y)+\eta. The set E_(Y)^(c)(f)E_{Y}{ }^{c}(f) being downward directed (Lemma 4), there exists G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f) such that G_(3) <= G_(1)G_{3} \leqslant G_{1} and G_(3) <= G_(2)G_{3} \leqslant G_{2}. Consequently
F_(1)(x) <= G_(3)(x) < F_(1)(x)+etaF_{1}(x) \leqslant G_{3}(x)<F_{1}(x)+\eta
and
F_(1)(y) <= G_(3)(y) < F_(1)(y)+etaF_{1}(y) \leqslant G_{3}(y)<F_{1}(y)+\eta
or, equivalently,
0 <= G_(3)(x)-Gamma_(1)(x) < eta,0 \leqslant G_{3}(x)-\Gamma_{1}(x)<\eta,
and
0 <= G_(3)(y)-F_(1)(y) < eta.0 \leqslant G_{3}(y)-F_{1}(y)<\eta .
From these inequalities one obtains
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,
so that
{:(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*}
Taking into account (16) and (17)
{:[(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}
But then ||G_(3)|| > ||f||_(Y)\left\|G_{3}\right\|>\|f\|_{Y}, in contradiction to G_(3)inE_(Y)^(c)(f)G_{3} \in E_{Y}{ }^{c}(f).
Theorem 3 is proved.
Remark. Let X=RX=R and Y=[a,b],0in YY=[a, b], 0 \in Y. For f inK_(Y)^(')f \in K_{Y}^{\prime}, let
Theorem 3 is proved.
Remark. Let X=RX=R and Y=[a,b],0in YY=[a, b], 0 \in Y. For f inK_(Y)^(')f \in K_{Y}^{\prime}, let
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)
and
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) .
Then the minimal and maximal convex norm-preserving extensions F_(1)F_{1} and F_(2)F_{2}, respectively, of ff, are given by
{:[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}
i=1,2i=1,2.
Let now XX be a normed linear space, YY a convex subset of XX such that 0in Y0 \in Y, and ZZ a nonvoid bounded subset of XX.
Let now XX be a normed linear space, YY a convex subset of XX such that 0in Y0 \in Y, and ZZ a nonvoid bounded subset of XX.
Consider the space
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\},
normed by the uniform norm
||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\} .
Consider the following problem:
(A) For f inK_(X)f \in K_{X}, find two elements g_(**)g_{*} and g^(**)g^{*} in P_(Y_(c)^(_|_))(f)P_{Y_{c}^{\perp}}(f) such that
(A) For f inK_(X)f \in K_{X}, find two elements g_(**)g_{*} and g^(**)g^{*} in P_(Y_(c)^(_|_))(f)P_{Y_{c}^{\perp}}(f) such that
||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\}
and
||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\}
Theorem 5. Problem (A) has a solution for all f inK_(X)f \in K_{X}.
Proof. By Theorem 2(b) every gg in P_(Y_(o)^(_|_))(f)P_{Y_{o}^{\perp}}(f) has the form g=f-Fg=f-F for a convex norm-preserving extension FF of f|_(Y)\left.f\right|_{Y}. By Theorem 3, there exist two convex norm-preserving extensions F_(1)F_{1} and F_(2)F_{2} of f|_(Y)\left.f\right|_{Y} such that
Proof. By Theorem 2(b) every gg in P_(Y_(o)^(_|_))(f)P_{Y_{o}^{\perp}}(f) has the form g=f-Fg=f-F for a convex norm-preserving extension FF of f|_(Y)\left.f\right|_{Y}. By Theorem 3, there exist two convex norm-preserving extensions F_(1)F_{1} and F_(2)F_{2} of f|_(Y)\left.f\right|_{Y} such that
F_(1)(x) <= F(x) <= F_(2)(x)F_{1}(x) \leqslant F(x) \leqslant F_{2}(x)
for all x in Xx \in X, i.e.,
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),
for all x in Xx \in X, where g_(i)=f-F_(i),i=1,2g_{i}=f-F_{i}, i=1,2. Therefore
{:[ 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}
It follows that a solution of Problem (A) is given by g_(**)=g_(i)g_{*}=g_{i} and g^(**)=g_(j)g^{*}=g_{j}, where i,j in{1,2}i, j \in\{1,2\} are such that
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)
and
|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) .
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