Costica Mustata
“Tiberiu Popoviciu” Institute of Numerical Analysis, Romanian Academy, Romania
Keywords
Paper coordinates
C. Mustăţa, Extension of Hölder functions and some related problems of best approximation, ”Babeş-Bolyai” Univ., Research Seminars, Seminar on Mathematical Analysis, Preprint nr. 7 (1991), 71-86.
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[4] McShane, E.J., Extension of range of functions, Bull. Amer. Soc. 40 (1934), 837-842.
[5] Mustăța, C., Best approximation and unique extension of Lipschitz funcitons, Journal Approx. Theory 19 (1977), 222-230.
[6] Singer, I., Best approximation in normed linear spaces by elements of linear subspaces, Springer-Verlag, 171, 1970.
[7] Shubert, B., A sequential method seeking the global maximum of a function, SIAM J.Num. Anal. 9(1972), 379-388.
[8] Shubert, B., The Otto Dunkel Memorial Problem Book, New York, 1956
EXTENSION OF HÖLDER FUNCTIONS AND SOME RELATED
PROBLEMS OF BEST APPROXIMATION
Costică Mustata
Let (X,d)(X, d) be a metric space and alpha in(0,1]\alpha \in(0,1]. A function f:XrarrR\mathrm{f}: \mathrm{X} \rightarrow \mathrm{R} is called HÖlder of class alpha\alpha on X if there exists K >= 0K \geq 0 such that
(1.1) quad|f(x)-f(y)| <= Kd^(alpha)(x,y)\quad|f(x)-f(y)| \leq K d^{\alpha}(x, y),
for all x,y in Xx, y \in X.
Put
(1.2) quad||f||_(alpha,x)=s u p{|f(x)-f(y)|//d^(alpha)(x,y):x,y in x,xvy}\quad\|f\|_{\alpha, x}=\sup \left\{|f(x)-f(y)| / d^{\alpha}(x, y): x, y \in x, x v y\right\}. Then ||f||_(alpha,X)\|f\|_{\alpha, X} is the smallest constant K >= 0K \geq 0 for which the inequality (1.1) holds and is called Hölder norm of ff.
Denote by Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d) the set of all Hölder functions of class alpha\alpha on X[3]X[3]. Ther Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d) is a vector lattice, that is, it is closed under the operations of addition, multiplication by scalars and formation of supremum and infimum of two of its elements.
For a nonvoid subset YY of XX, the Hölder norm ||f||_(alpha,Y)\|f\|_{\alpha, Y} and the space Lambda_(alpha)(Y,d)\Lambda_{\alpha}(Y, d) are defined similarly.
THEOREM 1. Let ( X,d\mathrm{X}, \mathrm{d} ) be a metric space, YsubX\mathrm{Y} \subset \mathrm{X} and
{:[zinquad" If "finLambda_(alpha)(y","d)" then the functions "],[(1.3)quadF_(1)(x)=i n f{f(y)+||f||_(alpha,y)d^(alpha)(x,y):yiny}","xinX],[(1.3)F_(2)(x)=s u p{f(y)-||f||_(alpha,y)(d)(x,y):yiny}","xinX]:}\begin{align*}
& \mathrm{z} \in \quad \text { If } \mathrm{f} \in \Lambda_{\alpha}(\mathrm{y}, \mathrm{~d}) \text { then the functions } \\
& \mathrm{(1.3)} \quad \mathrm{~F}_{1}(\mathrm{x})=\inf \left\{\mathrm{f}(\mathrm{y})+\|\mathrm{f}\|_{\alpha, \mathrm{y}} \mathrm{~d}^{\alpha}(\mathrm{x}, \mathrm{y}): \mathrm{y} \in \mathrm{y}\right\}, \mathrm{x} \in \mathrm{X} \\
& \mathrm{~F}_{2}(\mathrm{x})=\sup \left\{\mathrm{f}(\mathrm{y})-\|\mathrm{f}\|_{\alpha, \mathrm{y}} \mathrm{~d}(\mathrm{x}, \mathrm{y}): \mathrm{y} \in \mathrm{y}\right\}, \mathrm{x} \in \mathrm{X} \tag{1.3}
\end{align*}
are extension of ff, i.e.
a) quadF_(1)|_(Y)=F_(2)|_(Y)=f\left.\quad F_{1}\right|_{Y}=\left.F_{2}\right|_{Y}=f,
b) quad||F_(1)||_(alpha,X)=||F_(2)||_(alpha,X)=||f||_(alpha,Y)\quad\left\|F_{1}\right\|_{\alpha, X}=\left\|F_{2}\right\|_{\alpha, X}=\|f\|_{\alpha, Y}.
Theorem 1 follows from Corollary 1.2 in [3].
For f inLambda_(alpha)(Y,d)f \in \Lambda_{\alpha}(Y, d) denote by E_(Y)(f)E_{Y}(f) the set of all extensions of ff in Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d), i.e.
(1.4) quadE_(Y)(E)={F inLambda_(alpha)(X,d):F|_(Y)=f,||F||_(alpha,X)=||f||_(a,Y)}\quad E_{Y}(E)=\left\{F \in \Lambda_{\alpha}(X, d):\left.F\right|_{Y}=f,\|F\|_{\alpha, X}=\|f\|_{a, Y}\right\}.
Let CC be a convex subset of a vector space VV. AA subset: HH of CC is called a face of CC if lambda x+(1-lambda)y in H\lambda x+(1-\lambda) y \in H for some lambda in(0,1)\lambda \in(0,1) and some x,y in Cx, y \in C, implies x,y in Hx, y \in H. A one-point face of CC is called an extremal element of CC.
In Theorem 2 below we present some properties of the set E_(y)(f)E_{y}(\mathrm{f}).
THEOREM 2. Let ( X,d\mathrm{X}, \mathrm{d} ) be a metric space, Y a nonvoid subset of X,alpha in(0,1]\mathrm{X}, \alpha \in(0,1] and finLambda_(alpha)(Y,d)\mathrm{f} \in \Lambda_{\alpha}(\mathrm{Y}, \mathrm{d}). Then
a) E_(Y)\mathrm{E}_{Y} (f) is a convex subset of Lambda_(alpha)(X,d)\Lambda_{\alpha}(\mathrm{X}, \mathrm{d});
b) For every F inE_(1y)(f),F_(1)(x) >= F(x) >= F_(2)(x),quad x in XF \in E_{1 y}(f), F_{1}(x) \geq F(x) \geq F_{2}(x), \quad x \in X, where the functions F_(1)\mathrm{F}_{1} and F_(2)\mathrm{F}_{2} are defined by (1.3);
c) The functions F_(1)\mathrm{F}_{1} and F_(2)\mathrm{F}_{2} defined by (1,3)(1,3) are extremal elements of E_(Y)(f)\mathrm{E}_{Y}(f).
Proof. a) For F,G inE_(Y)(f)F, G \in E_{Y}(f) and lambda in[0,1]\lambda \in[0,1] we have (lambdaF+(1-lambda)G)|_(Y)=lambdaF_(Y)+(1-lambda)G_(Y)=lambdaf+(1-lambda)f=f\left.(\lambda \mathrm{F}+(1-\lambda) \mathrm{G})\right|_{\mathrm{Y}}=\lambda \mathrm{F}_{\mathrm{Y}}+(1-\lambda) \mathrm{G}_{\mathrm{Y}}=\lambda \mathrm{f}+(1-\lambda) \mathrm{f}=\mathrm{f}.
Since
b) Let F inE_(Y)(f)F \in E_{Y}(f) and x in Xx \in X. Then for y in Yy \in Y we have F(x)-f(y)=F(x)-F(y) >= -||F||_(alpha,x)*d^(alpha)(x,y)=-||f||_(alpha,x)d^(alpha)(x,y)F(x)-f(y)=F(x)-F(y) \geq-\|F\|_{\alpha, x} \cdot d^{\alpha}(x, y)=-\|f\|_{\alpha, x} d^{\alpha}(x, y) so that
F(x) >= f(y)-||f||_(alpha,y)d^(alpha)(x,y)", for all "y in Y". "F(x) \geq f(y)-\|f\|_{\alpha, y} d^{\alpha}(x, y) \text {, for all } y \in Y \text {. }
Therefore
F(x) >= s u p{f(y)-||f||_(alpha,y)d^(alpha)(x,y):y in y}=F_(2)(x)F(x) \geq \sup \left\{f(y)-\|f\|_{\alpha, y} d^{\alpha}(x, y): y \in y\right\}=F_{2}(x)
F(x) <= f(y)+||f||_(alpha,y)d^(alpha)(x,y)," for all "y in YF(x) \leq f(y)+\|f\|_{\alpha, y} d^{\alpha}(x, y), \text { for all } y \in Y
so that
F(x) <= i n f|f(y)+||f||_(alpha,y)d^(alpha)(x,y):y in Y|=F_(1)(x).F(x) \leq \inf \left|f(y)+\|f\|_{\alpha, y} d^{\alpha}(x, y): y \in Y\right|=F_{1}(x) .
c) If F,G inE_(Y)(f)F, G \in E_{Y}(f) and lambda in(0,1)\lambda \in(0,1) are such that lambda F+(1-lambda)G=F_(1)=lambdaF_(1)+(1-lambda)F_(1)\lambda F+ (1-\lambda) G=F_{1}=\lambda F_{1}+(1-\lambda) F_{1}, then lambda(F_(1)-F)=(1-lambda)(G-F_(1))\lambda\left(F_{1}-F\right)=(1-\lambda)\left(G-F_{1}\right) and since by b), G-F_(1) <= 0G-F_{1} \leq 0 it follows that F_(1) <= FF_{1} \leq F. But F_(1) >= FF_{1} \geq F and hence F=F_(1)F=F_{1}. Then the relation lambda F+(1-lambda)G=F_(1)\lambda F+(1-\lambda) G=F_{1} yields also G=F_(1)\mathrm{G}=\mathrm{F}_{1}.
The case of the function F_(2)F_{2} can be treated similarly.
2. For a nónvoid subset YY of metric space (X,d)(X, d) denote (2.1) quadY^(_|_)={f inLambda_(alpha)(X,d):4_(Y)=0}\quad Y^{\perp}=\left\{f \in \Lambda_{\alpha}(X, d): 4_{Y}=0\right\}.
Obviously, Y^(_|_)Y^{\perp} is a closed subspace of A_(alpha)(X,d)A_{\alpha}(X, d).
A subset SS of a normed space (V,||||)(V,\| \|) is called proximinal if for every x in Vx \in V there exists y_(0)in Sy_{0} \in S such that (2.2) quad||x-y_(0)||=d(x,S)=i n f{||x-y||:y in S}\quad\left\|x-y_{0}\right\|=d(x, S)=\inf \{\|x-y\|: y \in S\}.
An element Y_(0)in SY_{0} \in S for which the infimum in (2.2) is attained is called an element of best approximation of x by elements in s. If for every x in Vx \in V there exists a unique element of best approximation of xx in SS, then the set SS is called Chebyshevian [6]. Denote by P_(S)(x)P_{S}(x) the set of all best approximation elements of xx in SS.
THEOREM 3. If Y is a nonvoid subset of a metric space ( x,d\mathrm{x}, \mathrm{d} ) then
a) The subspace Y^(_|_)Y^{\perp} is proximinal and (2.3) quad d(f,Y_(.)^(_|_))=||f||_(Y)||_(alpha,Y)\quad d\left(f, Y_{.}^{\perp}\right)=\|f\|_{Y} \|_{\alpha, Y}, for every f inLambda_(alpha)(x,d)f \in \Lambda_{\alpha}(\mathrm{x}, \mathrm{d});
b) Every element g_(0)inY^(_|_)g_{0} \in Y^{\perp} of best approximation for f\mathbf{f} has the form g_(0)=f-Fg_{0}=f-F, where F inE_(Y)(f|_(Y))F \in E_{Y}\left(\left.f\right|_{Y}\right), and, conversely, for every F inE_(Y)(f|_(Y)),f-FF \in E_{Y}\left(\left.f\right|_{Y}\right), f-F is an element of best approximation for ff in Y^(_|_)Y^{\perp}, i.e.
c) The subspace Y^(_|_)Y^{\perp} is Chebyshevian if and only if for every f inLambda_(alpha)(X,d)f \in \Lambda_{\alpha}(X, d) the function f_(Y)f_{Y} has a unique extension in Lambda_(alpha)(x,d)\Lambda_{\alpha}(x, d).
Proof. a) Let f inLambda_(alpha)(X,d)^("and ")F inE_(Y)(f|_(Y))f \in \Lambda_{\alpha}(X, d)^{\text {and }} F \in E_{Y}\left(\left.f\right|_{Y}\right). Then (2.4)
i n f{||f-g||_(alpha,x):g inY^(_|_)} <= ||f||_(alpha,y)". "\inf \left\{\|f-g\|_{\alpha, x}: g \in Y^{\perp}\right\} \leq\|f\|_{\alpha, y} \text {. }
on the other hand
{:[|f|x||_(alpha,y)= s u p{|(f-g)(x)-(f-g)(y)|//d^(alpha)(x,y):x,y in Y;:}],[x!=y} <= s u p{|(f-g)(x)-(f-g)(y)|//d^(alpha)(x,y):x,y in X;:}],[{:x!=y^(')}=||f-g||_(alpha,x)]:}\begin{aligned}
|f| x \|_{\alpha, y}= & \sup \left\{|(f-g)(x)-(f-g)(y)| / d^{\alpha}(x, y): x, y \in Y ;\right. \\
x \neq y\} \leq & \sup \left\{|(f-g)(x)-(f-g)(y)| / d^{\alpha}(x, y): x, y \in X ;\right. \\
& \left.x \neq y^{\prime}\right\}=\|f-g\|_{\alpha, x}
\end{aligned}
for every g inY^(_|_)g \in Y^{\perp}, so that
|f|_(y)||_(alpha,x) <= i n f{||f-(g^(˙))||_(alpha,x):g inX^(_|_)}.|f|_{y} \|_{\alpha, x} \leq \inf \left\{\|f-\dot{g}\|_{\alpha, x}: g \in X^{\perp}\right\} .
Therefore quad||f||_(X)||_(alpha,Y)=||f-(f-F)||_(alpha,X)=d(f,Y^(_|_))\quad\|f\|_{X}\left\|_{\alpha, Y}=\right\| f-(f-F) \|_{\alpha, X}=d\left(f, Y^{\perp}\right), which shows that Y^(_|_)Y^{\perp} is proximinal and that the formula (2.3) holds.
b) By (2.4), it follows that f=F inY^(_|_)f=F \in Y^{\perp} is an element of best approximation for ff in Y^(_|_)Y^{\perp}, where F inE_(Y)(f|_(Y))F \in E_{Y}\left(\left.f\right|_{Y}\right). If g_(0)inY^(i)g_{0} \in Y^{i} is an element of best approximation for ff by elements of Y^(_|_)Y^{\perp}, then
||f-g_(0)||_(alpha,X)=||f|_(Y)||_(alpha,Y)quad" and " quad(f-g_(0))|_(Y)=f|_(Y)\left\|f-g_{0}\right\|_{\alpha, X}=\left\|\left.f\right|_{Y}\right\|_{\alpha, Y} \quad \text { and }\left.\quad\left(f-g_{0}\right)\right|_{Y}=\left.f\right|_{Y}
so that f-g_(0)inE_(Y)(f_(Y))f-g_{0} \in E_{Y}\left(f_{Y}\right) and F_(0)=f-g_(0)F_{0}=f-g_{0} is an extension of f∣Yf \mid Y.
c) If chi^(_|_)\chi^{\perp} is Chebyshevian, then every f inLambda_(alpha)(X,d)f \in \Lambda_{\alpha}(X, d) has a unique element of best approximation in Y^(_|_)Y^{\perp} and by bb ), the set E_(Y)(f_(Y))E_{Y}\left(f_{Y}\right) contains only one element.
3. Let now ( x,d\mathrm{x}, \mathrm{d} ) be a metric space of finite diameter, i.e. d(X)=s u p{d(x,y):x_(p)y in X} < ood(X)=\sup \left\{d(x, y): x_{p} y \in X\right\}<\infty. Then every function f inLambda_(alpha)(X,d)f \in \Lambda_{\alpha}(X, d) is bounded, for if x_(0)in Xx_{0} \in X is fixed, then |f(x)| <= |f(x)-f(x_(0))|+|f(x_(0))| <= |f(x_(0))|+||f||_(alpha,x)d^(alpha)(x,x_(0))≤≤|f(x_(0))|+||f||_(alpha,x)[d(X)]^(alpha),quad|f(x)| \leq\left|f(x)-f\left(x_{0}\right)\right|+\left|f\left(x_{0}\right)\right| \leq\left|f\left(x_{0}\right)\right|+\|f\|_{\alpha, x} d^{\alpha}\left(x, x_{0}\right) \leq \leq\left|f\left(x_{0}\right)\right|+\|f\|_{\alpha, x}[d(X)]^{\alpha}, \quad for every x in Xx \in X.
In this case, we can define the uniform norm on Lambda_(alpha)(x,d)\Lambda_{\alpha}(x, d), (3.1)
||f||_(u,x)=s u p{|f(x)|:x in X}.\|f\|_{u, x}=\sup \{|f(x)|: x \in X\} .
Let XX be a subset of metric space of finite diameter ( X,dX, d ) and let X^(_|_)X^{\perp} be defined by (2.1). For f inLambda_(alpha)(X,d)f \in \Lambda_{\alpha}(X, d) let G(f)G(f) denote
the set of all best approximation elements, with respect to the Hölder norm, of ff by elements in Y^(-)Y^{-}. Consider the following problem:
Find gg, and g^(**)g^{*} in G(f)G(f) such that
{:[||f-g||_(u,x)=i n f{||f-g||_(u,x):g in G(f)}","],[(3.2)||f-g^(**)||_(u,x)=s u p{||f-g||_(u,x):g in G(f)}.]:}\begin{align*}
& \|f-g\|_{u, x}=\inf \left\{\|f-g\|_{u, x}: g \in G(f)\right\}, \\
& \left\|f-g^{*}\right\|_{u, x}=\sup \left\{\|f-g\|_{u, x}: g \in G(f)\right\} . \tag{3.2}
\end{align*}
By Theorem 3 b), the problem (3.2) is equivalent to the following problem:
Find two extensions F_(s)F_{s} and F^(')F^{\prime} in E_(Y)(f_(Y))E_{Y}\left(f_{Y}\right) such that
{:[{F_(a)||_(u,x)=i n f{||F||_(u,x):F inE_(Y)(f||_(Y))},:}],[(3.3)||F^(**)||_(u,x)=s u p{||F||_(u,x):F inE_(Y)(f||_(X))}.]:}\begin{align*}
& \left\{F_{\mathrm{a}} \|_{\mathrm{u}, \mathrm{x}}=\inf \left\{\|F\|_{\mathrm{u}, \mathrm{x}}: F \in E_{\mathrm{Y}}\left(f \|_{\mathrm{Y}}\right)\right\},\right. \\
& \left\|F^{*}\right\|_{\mathrm{u}, \mathrm{x}}=\sup \left\{\|F\|_{\mathrm{u}, \mathrm{x}}: F \in E_{\mathrm{Y}}\left(f \|_{\mathrm{X}}\right)\right\} . \tag{3.3}
\end{align*}
The next theorem shows that the problem considered above has always a solution.
THEOREM 4. a) The infimum in (3.2) is attained, namely for every function g_(**)in G(f)g_{*} \in G(f) of the form g_(**)=f- bar(r)g_{*}=f-\bar{r}, where F_(q)inE_(Y)(f|_(Y))F_{q} \in E_{Y}\left(\left.f\right|_{Y}\right) is such that ||F_(**)||_(U,X)=|f|_(Y)||_(U,X)\left\|F_{*}\right\|_{U, X}=|f|_{Y} \|_{U, X};
b) The supremum in (3.2) is attained, namely for g^(**)=f-F_(1)g^{*}=f-F_{1} or g^(**)=f-F_(2)g^{*}=f-F_{2} or for both of these functions, where F_(1)F_{1} and F_(2)F_{2} are de ned by (1.3).
proof. a) First, observe that there exists F_(v)inE_(Y)(f∣x)F_{v} \in E_{Y}(f \mid x) such that ||F_(**)||_(U,X)=||f||_(Y)||_(U,X)\left\|F_{*}\right\|_{U, X}=\|f\|_{Y} \|_{U, X}. Indeed, if for F inE_(Y)(f∣Y)F \in E_{Y}(f \mid Y) put F_(a)(x)=|f|_(Y)||_(u,Y)quadF_{a}(x)=|f|_{Y} \|_{u, Y} \quad if F(x) > |f|_(Y)||_(u,Y)F(x)>|f|_{Y} \|_{u, Y}, (3.4) =F(x)quad=F(x) \quad if -||f|_(Y)||_(,Y) <= F(x) <= ||f||_(Y)||_(U,Y", ")=-|f|_(Y)||_(U,Y)quad-\left\|\left.f\right|_{Y}\right\|_{, Y} \leq F(x) \leq\|f\|_{Y} \|_{U, Y \text {, }} =-|f|_{Y} \|_{U, Y} \quad if F(x) < -|f|_(Y)||_(U,Y)F(x)<-|f|_{Y} \|_{U, Y},
then F_(q)F_{q} is in E_(Y)(f|_(Y))E_{Y}\left(\left.f\right|_{Y}\right) and ||F_(q)||_(U,X)=||f||_(Y)||_(U,Y)\left\|F_{q}\right\|_{U, X}=\|f\|_{Y} \|_{U, Y}.
Since
|F||_(u,x) >= s u p{|F(y)|:y in y}=|f|y||_(u,y),:}\left|F\left\|_{u, x} \geq \sup \{|F(y)|: y \in y\}=|f| y\right\|_{u, y},\right.
for every F inE_(Y)(f|_(Y))F \in E_{Y}\left(\left.f\right|_{Y}\right), it follows that
i n f{||F||_(u,x):F inE_(X)(f|_(Y))}=||F,||_(u,x)\inf \left\{\|F\|_{u, x}: F \in E_{X}\left(\left.f\right|_{Y}\right)\right\}=\|F,\|_{u, x}
where F_(z)F_{z} is defined by (3.4).
b) Since
F_(2)(x) <= F(x) <= F_(1)(x),x in X,F_{2}(x) \leq F(x) \leq F_{1}(x), x \in X,
for every F inE_(Y)(f|_(Y))F \in E_{Y}\left(\left.f\right|_{Y}\right) (Theorem 2. b)), it follows that
Remarks. a) The set of the functions of the form (3.4) is a convex subset of E_(Y)(f∣Y)E_{Y}(f \mid Y);
b) The functions F_(1),F_(2)F_{1}, F_{2} for which the supremum in (3.3) is attained are extremal elements of the convex set E_(Y)(f|_(Y))E_{Y}\left(\left.f\right|_{Y}\right).
4. Let (X,d)(X, d) be a metric space of finite diameter. Consider on A_(alpha)(X,d)A_{\alpha}(X, d) the norms
called the "sum-norm" and the "max-norm", respectively.
If YY is a subset of XX and ff is in Lambda_(alpha)(Y,d)\Lambda_{\alpha}(Y, d), then it is natural to ask iffi f f has an extension F inA_(alpha)(X,d)F \in A_{\alpha}(X, d) preserving the norms (4.1). An afirmative answer to this question and some consequences will be objects of the following theorems.
THEOREM 5. If (X,d)(\mathrm{X}, \mathrm{d}) is a metric space of finite diameter and YY is a nonvoid subset of XX, then for every f inLambda_(alpha)(Y,d)f \in \Lambda_{\alpha}(Y, d) there exists FinA_(alpha)(X,d)\mathrm{F} \in \mathrm{A}_{\alpha}(\mathrm{X}, \mathrm{d}) such that
FF_(y)=f quad" and "quad|F|_(s)=|f|_(s)". "F F_{y}=f \quad \text { and } \quad|F|_{s}=|f|_{s} \text {. }
proof. Let F_(1)F_{1} be defined by (1.3) and let
{:[ bar(F)_(1)(x)=F_(1)(x)quad" if "F_(1)(x) <= ||f||_(u,Y)],[(4.3)=||f||_(u,Y)quad" if "quadF_(1)(x) > ||F||_(u,Y)]:}\begin{align*}
\bar{F}_{1}(x) & =F_{1}(x) \quad \text { if } F_{1}(x) \leq\|f\|_{u, Y} \\
& =\|f\|_{u, Y} \quad \text { if } \quad F_{1}(x)>\|F\|_{u, Y} \tag{4.3}
\end{align*}
Then bar(F)_(1)]_(Y)=f\left.\bar{F}_{1}\right]_{Y}=f and bar(F)_(1)\bar{F}_{1} preserves both of the Holder and uniform norms, so that it preserves the "sum-norm", also.
similarly, the function
{:[ bar(F)_(2)(x)=F_(2)(x)" if "F_(2)(x) >= -||f||_(u,y)","],[(4.4)=||f||_(u,y)" if "F_(2)(x) < -||f||_(u,x)","]:}\begin{align*}
\bar{F}_{2}(x) & =F_{2}(x) \text { if } F_{2}(x) \geq-\|f\|_{u, y}, \\
& =\|f\|_{u, y} \text { if } F_{2}(x)<-\|f\|_{u, x}, \tag{4.4}
\end{align*}
where F_(2)F_{2} is defined by (1.3), is also an extension of ff, preserving both of the Hölder and uniform norms.
Denote by E_(g)(f)E_{g}(f) the set of all extensions of f inLambda_(alpha)(Y,d)f \in \Lambda_{\alpha}(Y, d) which preserve the "sum-norm", i.e.
(4.5)
E_(g)(f)={F inA_(alpha)(X,d):FX_(Y)=f,quad||F||_(g)=||f||_(g)}.E_{g}(f)=\left\{F \in A_{\alpha}(X, d): F X_{Y}=f, \quad\|F\|_{g}=\|f\|_{g}\right\} .
THEOREA 6. If ( x,d\mathrm{x}, \mathrm{d} ) is a metric space of finite diameter, Y a subset of X and finLambda_(a)(Y,d)\mathrm{f} \in \Lambda_{\mathrm{a}}(\mathrm{Y}, \mathrm{d}), then
(4.6) a) quad||F||_(a,X)=||f||_(a,Y)\quad\|F\|_{a, X}=\|f\|_{a, Y} and ||F||_(u,X)=||f||_(u,X)\|F\|_{u, X}=\|f\|_{u, X}
for every F inE_(g)(f)F \in \mathrm{E}_{g}(f);
b) rx quad bar(F)_(1)r x \quad \bar{F}_{1} and bar(F)_(2)\bar{F}_{2} are defined by (4.3) and (4.4) then (4.7) quad bar(F)_(1)(x) >= E(x) >= bar(F)_(2)(x)quad,x in X\quad \bar{F}_{1}(x) \geq E(x) \geq \bar{F}_{2}(x) \quad, x \in X,
for every F inE_(s)(f)F \in E_{s}(f) :
c) The set E_(s)\mathrm{E}_{s} (t) is a convex subset of the ball (with respect to "sum-norm") of center 0 and radius ||f||_(s)\|f\|_{s} in A_(alpha)(X,d)\mathrm{A}_{\alpha}(\mathrm{X}, \mathrm{d});
d) The functions bar(F)_(1)\bar{F}_{1} and bar(F)_(2)\bar{F}_{2} are extremal elements of E_(g)(f)E_{g}(f);
e) If ||f||_(g)=1\|f\|_{g}=1, then ff is an extremal element of the unit ball of A_(alpha)(Y,d)\mathrm{A}_{\alpha}(\mathrm{Y}, \mathrm{d}) (with respect to the "sum-norm") if and only if E_(s)(x)\mathrm{E}_{\mathrm{s}}(\mathrm{x}) is a face of the unit ball of Lambda_(alpha)(x,d)\Lambda_{\alpha}(\mathrm{x}, \mathrm{d}).
Pxoof. a) Let F inE_(g)(f)F \in E_{g}(f). Since F_(Y)=fF_{Y}=f, it follows that =||F||_(alpha,X) >= ||f||_(alpha,Y)=\|F\|_{\alpha, X} \geq\|f\|_{\alpha, Y} and ||F||_(U,X) >= ||f||_(U,Y)\|F\|_{U, X} \geq\|f\|_{U, Y}. If ||F||_(alpha,X) > ||f||_(alpha,Y)\|F\|_{\alpha, X}>\|f\|_{\alpha, Y} then ||f||_(alpha,Y)+||f||_(u,Y) < ||F||_(alpha,X)+||F||_(u,X)=||F||_(s)=||f||_(s)=||f||_(alpha,Y)+||f||_(u,Y)\|f\|_{\alpha, Y}+\|f\|_{u, Y}<\|F\|_{\alpha, X}+\|F\|_{u, X}=\|F\|_{s}=\|f\|_{s}=\|f\|_{\alpha, Y}+\|f\|_{u, Y}, which is impossible. Therefore ||F||_(alpha,v)=||f||_(alpha,Y)\|F\|_{\alpha, v}=\|f\|_{\alpha, Y} and ||F||_(u,X)==||F||_(s)-||F||_(alpha,X)=||f||_(s)-||f||_(alpha,Y)=||f||_(u,nu)\|F\|_{u, X}= =\|F\|_{s}-\|F\|_{\alpha, X}=\|f\|_{s}-\|f\|_{\alpha, Y}=\|f\|_{u, \nu}.
The proof of b), c), d) proceedes similarly to the proofs of assertions a), b), c) of Theorem 2.
e) Let ff be an extremal element of the unit ball of Lambda_(alpha)(Y,d)\Lambda_{\alpha}(Y, d) and let lambda epsilon(0,1)\lambda \epsilon(0,1). If F_(1),F_(2)F_{1}, F_{2} are elements of the unit ball of A_(alpha)(x,d)A_{\alpha}(x, d) such that lambdaF_(1)+(1-lambda)F_(2)inE_(s)(f)\lambda F_{1}+(1-\lambda) F_{2} \in E_{s}(f), then lambdaF_(1)|_(Y)+(1-lambda)F_(2)|_(Y)=f\left.\lambda F_{1}\right|_{Y}+\left.(1-\lambda) F_{2}\right|_{Y}=f. Since ff is extremal it follows that F_(1)|_(Y)=F_(2)|_(Y)=£\left.F_{1}\right|_{Y}=\left.F_{2}\right|_{Y}=£ and ||F_(1)||_(s)=||F_(2)||_(s)=1\left\|F_{1}\right\|_{s}=\left\|F_{2}\right\|_{s}=1, i.e. F_(1),F_(2)inE_(s)(£)F_{1}, F_{2} \in E_{s}(£). We have shown that E_(s)(f)E_{s}(f) is a face of the unit ball of Lambda_(alpha)(x,d)\Lambda_{\alpha}(x, d).
Conversely, suppose that ||f||_(s)=1\|f\|_{s}=1 and ff is not an extremal element of the unit ball of Lambda_(alpha)(Y,d)\Lambda_{\alpha}(\mathrm{Y}, \mathrm{d}). Then there exists f_(1),f_(2)f_{1}, f_{2} in Lambda_(alpha)(Y,d),f_(1)!=f_(2),||f_(i)||_(s)=1,i=1,2\Lambda_{\alpha}(Y, d), f_{1} \neq f_{2},\left\|f_{i}\right\|_{s}=1, i=1,2, and lambda in(0,1)\lambda \in(0,1) such that lambdaf_(1)+(1-lambda)f_(2)=f\lambda f_{1}+(1-\lambda) f_{2}=f. If F_(i)inE_(s)(f_(i))F_{i} \in E_{s}\left(f_{i}\right), i=1,2i=1,2, then lambdaF_(1)^(')||_(Y)+(1-lambda)F_(2)^(')||_(Y)=f\lambda F_{1}^{\prime}\left\|_{Y}+(1-\lambda) F_{2}^{\prime}\right\|_{Y}=f and 1=||lambdaF_(1)^(')||_(Y)++(1-lambda)F_(2)^(')||_(Y)||_(s) <= ||lambdaF_(1)+((1^(˙))-lambda)F_(2)^(')||_(s) <= 11=\left\|\lambda F_{1}^{\prime}\right\|_{Y}+ +(1-\lambda) F_{2}^{\prime}\left\|_{Y}\right\|_{s} \leq\left\|\lambda F_{1}+(\dot{1}-\lambda) F_{2}^{\prime}\right\|_{s} \leq 1, which show that lambdaF_(1)^(')+(1-lambda)F_(2)^(')inE_(8)(f)\lambda F_{1}^{\prime}+(1-\lambda) F_{2}^{\prime} \in E_{8}(f). Consequently E_(g)(f)E_{g}(f) is not a face of the unit ball of Lambda_(alpha)(x,d)\Lambda_{\alpha}(x, d).
THEOREM 7. Let ( X,d\mathrm{X}, \mathrm{d} ) be a metric space of finite diameter, Y a subset of X and suppose Lambda_(alpha)(Y,d)\Lambda_{\alpha}(\mathrm{Y}, \mathrm{d}) and Lambda_(alpha)(X,d)\Lambda_{\alpha}(\mathrm{X}, \mathrm{d}) endoved with the "max-norm". Then -
a) For every F inE_(s)(f),||F||_(m)=||f||_(m)F \in \mathrm{E}_{\mathrm{s}}(\mathrm{f}),\|\mathrm{F}\|_{\mathrm{m}}=\|\mathrm{f}\|_{\mathrm{m}};
E_(m)(f,={F inLambda_(alpha)(X,d):F||_(Y)=f,||F||_(m)=||f||_(m)},:}E_{m}\left(f,=\left\{F \in \Lambda_{\alpha}(X, d): F\left\|_{Y}=f,\right\| F\left\|_{m}=\right\| f \|_{m}\right\},\right.
then there exists widetilde(F)_(1)\widetilde{F}_{1} and widetilde(F)_(2)\widetilde{F}_{2} in E_(m)(f)\mathrm{E}_{m}(f) such that
(4.9)quad tilde(F)_(1)(x) >= F(x) >= tilde(F)_(2)(x),x in X,(4.9) \quad \tilde{F}_{1}(x) \geq F(x) \geq \tilde{F}_{2}(x), x \in X,
for every F inE_(m)(f)F \in \mathrm{E}_{m}(f);
c) The set E_(m)(f)E_{m}(f) is a convex subset of the ball (in the norm ||||_(m):}\left\|\|_{m}\right. ) of center 0 and radius ||f||_(m)\| f \|_{m} in Lambda_(alpha)(X,d)\Lambda_{\alpha}(\mathrm{X}, \mathrm{d});
d) The functions widetilde(F)_(1)\widetilde{\mathrm{F}}_{1} and widetilde(F)_(2)\widetilde{\mathrm{F}}_{2} satisfying (4.9) are extremal elements of the set E_(m)(f)\mathrm{E}_{\mathrm{m}}(\mathrm{f});
e) The function ff is an extremal element of the unit ball of Lambda_(alpha)(Y,d)\Lambda_{\alpha}(\mathrm{Y}, \mathrm{d}) if and only if E_(m)(f)\mathrm{E}_{\mathrm{m}}(\mathrm{f}) is a face of the unit ball of A_(a)(x,d)A_{a}(x, d).
Proof. a) If F inE_(s)(f)F \in E_{s}(f) then, by Theorem 6. a),
||F||_(alpha,x)=||f||_(alpha,y)quad" and "||F||_(u,x)=||f||_(u,y),\|F\|_{\alpha, x}=\|f\|_{\alpha, y} \quad \text { and }\|F\|_{u, x}=\|f\|_{u, y},
so that quad||F||_(m)=||f||_(m)\quad\|F\|_{m}=\|f\|_{m}.
b) Let
H^(˙)_(1)(x)=i n f{f(y)+||f||_(m)d^(a)(x,y):y in y},x in X.\dot{H}_{1}(x)=\inf \left\{f(y)+\|f\|_{m} d^{a}(x, y): y \in y\right\}, x \in X .
Then (see [2]) the function H_(1)H_{1} has the properties
that is widetilde(F)_(1)inE_(m)(f)\widetilde{F}_{1} \in E_{m}(f).
Similarly, by truncating the function
H_(2)(x)=s u p{f(y)-gamma f||_(m)d^(alpha)(x,y):y in y},x in XH_{2}(x)=\sup \left\{f(y)-\gamma f \|_{m} d^{\alpha}(x, y): y \in y\right\}, x \in X
one obtains the function
{:[(4.11) tilde(F)_(2)(x)=H_(2)(x)" if "H_(2)(x) >= -||f||_(m)","],[=-||f||_(m)" if "H_(2)(x) < -||f||_(m)","]:}\begin{align*}
\tilde{F}_{2}(x) & =H_{2}(x) & \text { if } & H_{2}(x) \geq-\|f\|_{m}, \tag{4.11}\\
& =-\|f\|_{m} & \text { if } & H_{2}(x)<-\|f\|_{m},
\end{align*}
which is an extension of ff with respect to the "max-norm", i.e. quad widetilde(F)_(2)inE_(m)(f)\quad \widetilde{F}_{2} \in E_{m}(f).
The inequalities (4.9) can be obtained reasoning like in the proof of assertion b) in Theorem 2.
The proofs of c) and d) are similar to the proofs of assertion a) and c) of Theorem 2 and the proof of e) is similar to the proof of e) in Theorem 6.
From Theorem 6 and 7 obtains the following corolary:
COROLLARY 1. If (X,d)(\mathrm{X}, \mathrm{d}) is a metric space of finite diameter, YY is a subset of XX and f inLambda_(alpha)(Y,d)f \in \Lambda_{\alpha}(Y, d), then
a) quadE_(g)(f)subE_(m)(f)\quad E_{g}(f) \subset E_{m}(f);
b) quad tilde(F)_(1)(x) >= bar(F)_(1)(x) >= bar(F)_(2)(x) >= tilde(F)_(2)(x),x in X\quad \tilde{F}_{1}(x) \geq \bar{F}_{1}(x) \geq \bar{F}_{2}(x) \geq \tilde{F}_{2}(x), x \in X,
where the functions tilde(F)_(1), tilde(F)_(2), bar(F)_(1), bar(F)_(2)\tilde{F}_{1}, \tilde{F}_{2}, \bar{F}_{1}, \bar{F}_{2} are defined by (4.3), (4.4), (4.10) and (4.11) , respectively.
5. In the following we shall give a procedure to find the global extrema of a function f inLambda_(alpha)(X,d)f \in \Lambda_{\alpha}(X, d) by using the extensions of the restriction of ff to some finite subset of XX.
Let ( X,dX, d ) be a compact metric space and let ff be in Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d). If YY is a subset of XX and q >= ||||_(Y)||_(alpha,Y)q \geq\| \|_{Y} \|_{\alpha, Y} (here ||_(Y)\|_{Y} denotes the restriction of ff to YY ), then the functions
{:[(5.1)U_(q)(x)=i n f{f(y)+qd^(alpha)(x,y):y in Y}","x in X","],[u_(q)(x)=s u p{f(y)-qd^(alpha)(x,y):y in Y}","x in X","]:}\begin{align*}
& U_{q}(x)=\inf \left\{f(y)+q d^{\alpha}(x, y): y \in Y\right\}, x \in X, \tag{5.1}\\
& u_{q}(x)=\sup \left\{f(y)-q d^{\alpha}(x, y): y \in Y\right\}, x \in X,
\end{align*}
are extensions of f|_(Y)\left.f\right|_{Y} which belongs to Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d) and have the Hölder norms at most qq (see [2]). If uu and uu denotes the
functions defined by (5.1) for q=||f||_(alpha,x)q=\|f\|_{\alpha, x} then
(5.2) quad u(x) <= f(x) <= U(x),x in X\quad u(x) \leq f(x) \leq U(x), x \in X,
and for q > ||f||_(alpha,x)q>\|f\|_{\alpha, x} we have
(5.3) quadu_(q)(x) <= u(x) <= f(x) <= U(x) <= U_(q)(x),x in X\quad u_{q}(x) \leq u(x) \leq f(x) \leq U(x) \leq U_{q}(x), x \in X.
A maximum (respectively a minimum) point of a function f:X rarr Rf: X \rightarrow R is a point x^(**)in Xx^{*} \in X such that
(5.4) quad f(x^(**)) >= f(x)quad\quad f\left(x^{*}\right) \geq f(x) \quad (respectively f(x^(**)) <= f(x)f\left(x^{*}\right) \leq f(x) )
for all x in Xx \in X.
For a bounded real function ff on XX put
(5.5)
{:[M_(f)=s u p{f(x):x in X}","m_(f)={f(x):x in X}],[E_(f)={x in X:f(x)=M_(f)}","e_(f)={x in X:f(x)=m_(f)}]:}\begin{aligned}
& M_{f}=\sup \{f(x): x \in X\}, m_{f}=\{f(x): x \in X\} \\
& E_{f}=\left\{x \in X: f(x)=M_{f}\right\}, e_{f}=\left\{x \in X: f(x)=m_{f}\right\}
\end{aligned}
Let now (X,d)(X, d) be a compact metric and let ff be a function in Lambda_(alpha)(x,d)\Lambda_{\alpha}(x, d).
We define now inductively two sequences (x_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0} and (M_(n))_(n >= 0)\left(M_{n}\right)_{n \geq 0} of points in XX and of real numbers, respectively, as follows:
Let q >= ||f||_(alpha,x)q \geq\|f\|_{\alpha, x} be fixed and let x_(0)x_{0} be a fixed point in XX. Let u^(0)(x)=f(x_(0))+qd^(alpha)(x,x_(0)),x in Xu^{0}(x)=f\left(x_{0}\right)+q d^{\alpha}\left(x, x_{0}\right), x \in X, the greatest extension of ff obtained from (5.1) for Y={x_(0)}Y=\left\{x_{0}\right\} and let
M_(0)=s u p{U^(0)(x):x in X}:M_{0}=\sup \left\{U^{0}(x): x \in X\right\}:
Let x_(1)in Xx_{1} \in X be a point with u^(0)(x_(1))=M_(0)u^{0}\left(x_{1}\right)=M_{0}.
Suppose now that for a natural number n >= 1n \geq 1, the points x_(0),x_(1),dots,x_(n-1)x_{0}, x_{1}, \ldots, x_{n-1} and the numbers M_(0),M_(1),dots,M_(n-1)M_{0}, M_{1}, \ldots, M_{n-1} were defined. Let U^(n-1)U^{n-1} be the greatest extension of I_(Y)I_{Y} obtained from (5.1) for y={x_(0),x_(1),dots,x_(n-1)}y=\left\{x_{0}, x_{1}, \ldots, x_{n-1}\right\}. Put
M_(n)=s u p{U^(n-1)(x):x in X}M_{n}=\sup \left\{U^{n-1}(x): x \in X\right\}
and let x_(n)x_{n} be a point in xx such that u^(n-1)(x_(n))=M_(n)u^{n-1}\left(x_{n}\right)=M_{n}.
The properties of the so defined sequences (x_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0} and
Theorem a. Let (X,d)(\mathrm{X}, \mathrm{d}) be a compact metric space and let f inLambda_(alpha)(x,d)f \in \Lambda_{\alpha}(x, d). For a fixed, q >= ||f||_(alpha,x)q \geq\|f\|_{\alpha, x} let the sequences (x_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0} and (M_(n))_(n >= 0)\left(M_{n}\right)_{n \geq 0} be defined as above. Then
a) quadlim_(n rarr oo)M_(n)=M_(f)^(')\quad \lim _{n \rightarrow \infty} M_{n}=M_{f}^{\prime};
b) quadlim_(n rarr oo)[i n f{d(x,x_(n)):x inE_(f)}]=0\quad \lim _{n \rightarrow \infty}\left[\inf \left\{d\left(x, x_{n}\right): x \in E_{f}\right\}\right]=0;
c) The sequence (f(x_(n)))_(n >= 0)\left(f\left(x_{n}\right)\right)_{n \geq 0} has the number M_(f)M_{f} as a limit point.
Proor. Since U^(n) <= U^(n-1)U^{n} \leq U^{n-1} for n=1,2,dotsn=1,2, \ldots it follows that the sequence (M_(n))_(n >= 0)\left(M_{n}\right)_{n \geq 0} is nonincreasing. By (5.3), M_(n)=U^(n-1)(x_(n))≥≥f(x_(n)) >= min{f(x):x in X}M_{n}=U^{n-1}\left(x_{n}\right) \geq \geq f\left(x_{n}\right) \geq \min \{f(x): x \in X\} so that the sequence. (M_(n))_(n >= 0)\left(M_{n}\right)_{n \geq 0} is also bounded. Therefore there exists M=lim_(n rarr oo)M_(n)*By(5.3)M=\lim _{n \rightarrow \infty} M_{n} \cdot B y(5.3), f(x) <= U^(n)(x) <= M_(n)f(x) \leq U^{n}(x) \leq M_{n}, for all x in Nx \in N so that (5.6) quad f(x) <= M\quad f(x) \leq M, for all x in Xx \in X.
The metric space XX being compact, the sequence (x_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0} contains a subsequence (x_(n_(k)))_(k >= 0)\left(x_{n_{k}}\right)_{k \geq 0} converging to a point z in Xz \in X. Since the function ff is continuous it follows that
(5.7) quad f(x_(n_(k)))longrightarrow f(z),k longrightarrow oo\quad f\left(x_{n_{k}}\right) \longrightarrow f(z), k \longrightarrow \infty.
But, for k >= 1k \geq 1, |U^(n_(k)-1)(z)-M_(n_(k)-1)|=|0^(n_(k)-1)(z)-0^(n_(k)-1)(x_(n_(k)))| <= qd^(alpha)(z,x_(n_(k)))rarr0\left|U^{n_{k}-1}(z)-M_{n_{k}-1}\right|=\left|0^{n_{k}-1}(z)-0^{n_{k}-1}\left(x_{n_{k}}\right)\right| \leq q d^{\alpha}\left(z, x_{n_{k}}\right) \rightarrow 0
for k longrightarrow ook \longrightarrow \infty, and M_(n_(k)-1)longrightarrow MM_{n_{k}-1} \longrightarrow M for k longrightarrow ook \longrightarrow \infty, so that (5.8) quadu^(n_(k)-1)(z)rarrM\quad \mathrm{u}^{\mathrm{n}_{\mathrm{k}}-1}(\mathrm{z}) \rightarrow \mathrm{M}, for krarr oo\mathrm{k} \rightarrow \infty.
By the relation |U^(n_(k))(z)-f(x_(n_(k)))|=|U^(n_(k))(z)-U^(n_(k))(x_(n_(k)))| <= qd^(alpha)(z,x_(n_(k)))longrightarrow0\left|U^{n_{k}}(z)-f\left(x_{n_{k}}\right)\right|=\left|U^{n_{k}}(z)-U^{n_{k}}\left(x_{n_{k}}\right)\right| \leq q d^{\alpha}\left(z, x_{n_{k}}\right) \longrightarrow 0, k rarr ook \rightarrow \infty, and by (5.7) it follows that (5.9) quadu^(n_(k))(z)longrightarrowf^(˙)(z),quad k longrightarrow oo\quad u^{n_{k}}(z) \longrightarrow \dot{f}(z), \quad k \longrightarrow \infty. Therefore, if in the inequalities
f(z) <= U^(n_(k)-1)(z) <= U^(n_(k-1))(z),quad k >= 1,f(z) \leq U^{n_{k}-1}(z) \leq U^{n_{k-1}}(z), \quad k \geq 1,
we let k longrightarrow ook \longrightarrow \infty one obtains f(z) <= M <= f(z)f(z) \leq M \leq f(z), so that f(z)=Mf(z)=M. Taking into account (5.6) it follows that
M=f(z)=max{f(x):x in X}.M=f(z)=\max \{f(x): x \in X\} .
To prove b) , observe that if contrary, then there exist > 0>0 and an infinite subset JJ of NN such that
{:(5.10)i n f{d(x,x_(j)):x inE_(f)} >= epsi",":}\begin{equation*}
\inf \left\{d\left(x, x_{j}\right): x \in E_{f}\right\} \geq \varepsilon, \tag{5.10}
\end{equation*}
for all j in Jj \in J. The space xx being compact there exists a subsequence (x_(j_(k)))_(k >= 0)\left(x_{j_{k}}\right)_{k \geq 0} of ( x_(j))_(j in J)\left.x_{j}\right)_{j \in J} converging to a point y in Xy \in X. But then, repeating the above arguments will follows that y inE_(f)y \in \mathrm{E}_{f}, which contradicts (5.10).
The affirmation c) follows from (5.7).
Remarks. 1) In the clase X=[a,b]\mathrm{X}=[\mathrm{a}, \mathrm{b}] and alpha=1\alpha=1 a similar result is proved in [7].
2) If the extensions u^(n)u^{n} are replaced by the extensions u^(n)u^{n} and m_(n)=i n f{u^(n)(x):x in x},u^(n)(x_(n+1))=m_(n)m_{n}=\inf \left\{u^{n}(x): x \in x\right\}, u^{n}\left(x_{n+1}\right)=m_{n}, then one obtains a procedure to find the minimum m_(f)m_{f} of a function f inLambda_(alpha)(x,d)f \in \Lambda_{\alpha}(x, d).
Example. Let x=[0,1],d(x,y)=|x-y|\mathrm{x}=[0,1], \mathrm{d}(\mathrm{x}, \mathrm{y})=|\mathrm{x}-\mathrm{y}| and
{:[f(x)=x*sin(1//x)","" if "x in(0","1]","],[=0","" if "x=0.]:}\begin{aligned}
f(x) & =x \cdot \sin (1 / x), \text { if } x \in(0,1], \\
& =0, \text { if } x=0 .
\end{aligned}
It is known that f inLambda_(alpha)(x,d)f \in \Lambda_{\alpha}(x, d) if and only if alpha in(0,1//2]\alpha \in(0,1 / 2] ( see [8], Problem 153 ) and in this case
