About the determination of extremes of Hölder functions

Costica Mustata
(original), Approximation Theory, paper

Abstract

Authors

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

Keywords

Paper coordinates

C. Mustăţa, About the determination of extremes of Hölder functions, Seminar of Functional Analysis and Numerical Methods, Preprint Nr. 1 (1983), 107-116.

PDF

https://ictp.acad.ro/wp-content/uploads/2025/07/1983-Mustata-UBB-Seminar-About-the-determination-of-extremes-of-Holder-functions.pdf

About this paper

Journal
Publisher Name
DOI
Print ISSN
Online ISSN

google scholar link

[1] G. Aronsson, Extension of functions satisfying Lipschitz conditions, Arkiv for Mathematik  6, 28 (1967), 551-561.
[2] J. Czipser and L. Geher, Extension of functions satisfying a Lipschitz condition, Acta Math. Acad. Sci. Hungar 6 (1955), 213-220.
[3] R. Levi and M.D., Rice, The Approximation of Uniformly Continuous Mappings by Lipschitz and Holder Mappings (prerint), 1980.
[4] E.J. McShane, Extension  of range of functions, Bull. Amer. Math.Soc. 40 (1934), 837-842.
[5] C. Mustata, Best Approximation and Unique Extension of Lipschitz Functions, Journal Approx. Theory 19 (1977), 222-230.
[6] B. Shubert, A sequential method seeking the global maximum of a  function, SIAM J. Numer. Anal. 9 (1972), 379-388.
[7] The Otto Dunkel Memorial Problem Book, New York (1957).

Paper (preprint) in HTML form

1983-Mustata-UBB-Seminar-About-the-determination-of-extremes-of-Holder-functions
−
About the determination of extremes of Hölder functions

byCostica Mustata

Iot ( X , d ) ( X , d ) (X,d)(\mathrm{X}, \mathrm{d})(X,d) be a metric space and α ∈ ( 0 , 1 ] α ∈ ( 0 , 1 ] alpha in(0,1]\alpha \in(0,1]α∈(0,1]. A sunction δ : X → B δ : X → B delta:XrarrB\delta: \mathrm{X} \rightarrow \mathrm{B}δ:X→B is called Hölder of eless α α alpha\alphaα on X X XXX if there exists K ⩾ 0 K ⩾ 0 K >= 0K \geqslant 0K⩾0 eveh that
(x) | x ( x ) − x ( y ) | ≤ x 0 d α ( x , y ) , (x) | x ( x ) − x ( y ) | ≤ x 0 d α ( x , y ) , {:(x)|x(x)-x(y)| <= x_(0)d^(alpha)(x","y)",":}\begin{equation*} |x(x)-x(y)| \leq x_{0} d^{\alpha}(x, y), \tag{x} \end{equation*}(x)|x(x)−x(y)|≤x0dα(x,y),
Lor all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X.
Put
(2) ‖ I ‖ α , x = sup { | f ( x ) − f ( y ) | / a α ( x , y ) : x , y ∈ X , x ∣ y } (2) ‖ I ‖ α , x = sup | f ( x ) − f ( y ) | / a α ( x , y ) : x , y ∈ X , x ∣ y {:(2)||I||_(alpha,x)=s u p{|f(x)-f(y)|//a^(alpha)(x,y):x,y in X,quad x∣y}:}\begin{equation*} \|I\|_{\alpha, x}=\sup \left\{|f(x)-f(y)| / a^{\alpha}(x, y): x, y \in X, \quad x \mid y\right\} \tag{2} \end{equation*}(2)‖I‖α,x=sup{|f(x)−f(y)|/aα(x,y):x,y∈X,x∣y}
Then ‖ f ‖ α , X ‖ f ‖ α , X ||f||_(alpha,X)\|f\|_{\alpha, X}‖f‖α,X is the smallest constant Z Z Z\mathbb{Z}Z. for which the inequality (1) holds anditis called the Holder noxy of 1 .
Denote by Λ α ( x , d ) Λ α ( x , d ) Lambda_(alpha)(x,d)\Lambda_{\alpha}(x, d)Λα(x,d) the set of Hölder functions of class α α alpha\alphaα on X X XXX [3]. Then Λ α ( X , d ) Λ α ( X , d ) Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d)Λα(X,d) is a vector lattice, that is, it is closed under the operations of addition, multiplication by ecalars and formation of supremum and infimum of two of its elemonts.
For a nonvoid subset X X XXX of X X XXX, the Hölder norm. ‖ f ‖ α , X ‖ f ‖ α , X ||f||_(alpha,X)\|f\|_{\alpha, X}‖f‖α,X and the space Λ α ( Y , d ) Λ α ( Y , d ) Lambda_(alpha)(Y,d)\Lambda_{\alpha}(Y, d)Λα(Y,d) are defined similarly .
THEOREM 1. Ist ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a motric space, I ⊂ X I ⊂ X I sub XI \subset XI⊂X and α ∈ ( 0 , 1 ] α ∈ ( 0 , 1 ] alpha in(0,1]\alpha \in(0,1]α∈(0,1]. If I ∈ Λ α ( Y , d ) I ∈ Λ α ( Y , d ) I inLambda_(alpha)(Y,d)I \in \Lambda_{\alpha}(Y, d)I∈Λα(Y,d) then the functions
F 1 ( x ) = inf { f ( y ) + ‖ f ‖ α , y ⋅ d α ( x , y ) : y ∈ Y } , x ∈ Y F 1 ( x ) = inf f ( y ) + ‖ f ‖ α , y ⋅ d α ( x , y ) : y ∈ Y , x ∈ Y F_(1)(x)=i n f{f(y)+||f||_(alpha,y)*d^(alpha)(x,y):quad y in Y},quad x in YF_{1}(x)=\inf \left\{f(y)+\|f\|_{\alpha, y} \cdot d^{\alpha}(x, y): \quad y \in Y\right\}, \quad x \in YF1(x)=inf{f(y)+‖f‖α,y⋅dα(x,y):y∈Y},x∈Y
and
(3) I 2 ( x ) = sup { f ( y ) − ‖ f ‖ α , Y ⋅ d α ˙ ( x , y ) : z , y ∈ Y } , x ∈ X (3) I 2 ( x ) = sup f ( y ) − ‖ f ‖ α , Y ⋅ d α ˙ ( x , y ) : z , y ∈ Y , x ∈ X {:(3)I_(2)(x)=s u p{f(y)-||f||_(alpha,Y)*d^(alpha^(˙))(x,y):z,y in Y}","quad x in X:}\begin{equation*} I_{2}(x)=\sup \left\{f(y)-\|f\|_{\alpha, Y} \cdot d^{\dot{\alpha}}(x, y): z, y \in Y\right\}, \quad x \in X \tag{3} \end{equation*}(3)I2(x)=sup{f(y)−‖f‖α,Y⋅dα˙(x,y):z,y∈Y},x∈X
ax extensions of I I III, 1.2.
a) F 1 | Y = F 2 | Y = 1 F 1 Y = F 2 Y = 1 F_(1)|Y=F_(2)|Y=1F_{1}\left|Y=F_{2}\right| Y=1F1|Y=F2|Y=1,
and
b) ‖ F 1 ‖ α , X = ‖ F 2 ‖ α , X = ‖ x ‖ α , I F 1 α , X = F 2 α , X = ‖ x ‖ α , I ||F_(1)||_(alpha,X)=||F_(2)||_(alpha,X)=||x||_(alpha,I)\left\|F_{1}\right\|_{\alpha, X}=\left\|F_{2}\right\|_{\alpha, X}=\|x\|_{\alpha, I}‖F1‖α,X=‖F2‖α,X=‖x‖α,I.
Theorem 1 follows from Corollary 1.2 in [3] (see also [1],[2]). Let ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a compact metric space and let I I III be in Λ α ( X , d ) Λ α ( X , d ) Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d)Λα(X,d). If Y Y YYY is a subset of X X XXX and q > ‖ P | Y ‖ α , Y q > P Y α , Y q > ||P|_(Y)||_(alpha,Y)q>\left\|\left.P\right|_{Y}\right\|_{\alpha, Y}q>‖P|Y‖α,Y (kere I | Y I Y I|_(Y)\left.I\right|_{Y}I|Y denotes the restriction of I I I\mathcal{I}I to Y Y YYY ), then the functions
U ( x ) = i n f { f ( y ) + q ⋅ d α ( x , y ) : y ∈ I } , x ∈ X U ( x ) = i n f f ( y ) + q ⋅ d α ( x , y ) : y ∈ I , x ∈ X U(x)=inf{f(y)+q*d^(alpha)(x,y):y in I},x in XU(x)=i n f\left\{f(y)+q \cdot d^{\alpha}(x, y): y \in I\right\}, x \in XU(x)=inf{f(y)+q⋅dα(x,y):y∈I},x∈X
and
(4) u ( x ) = sup { f ( y ) − q ⋅ d α ( x , y ) : y ∈ Y } , x ∈ X (4) u ( x ) = sup f ( y ) − q ⋅ d α ( x , y ) : y ∈ Y , x ∈ X {:(4)u(x)=s u p{f(y)-q*d^(alpha)(x,y):y in Y}","x in X:}\begin{equation*} u(x)=\sup \left\{f(y)-q \cdot d^{\alpha}(x, y): y \in Y\right\}, x \in X \tag{4} \end{equation*}(4)u(x)=sup{f(y)−q⋅dα(x,y):y∈Y},x∈X
(8) M f = sup { f ( x ) : x ∈ X } . B f = { x ∈ X : P ( x ) = M f } (8) M f = sup { f ( x ) : x ∈ X } . B f = x ∈ X : P ( x ) = M f {:(8)M_(f)=s u p{f(x):x in X}.quadB_(f)={x in X:P(x)=M_(f)}:}\begin{equation*} M_{f}=\sup \{f(x): x \in X\} . \quad B_{f}=\left\{x \in X: P(x)=M_{f}\right\} \tag{8} \end{equation*}(8)Mf=sup{f(x):x∈X}.Bf={x∈X:P(x)=Mf}
and for q > ‖ f | Y ‖ α , X q > f Y α , X q > ||f|_(Y)||_(alpha,X)q>\left\|\left.f\right|_{Y}\right\|_{\alpha, X}q>‖f|Y‖α,X we have
A meximum (respectively a minimum) point of a function fi X → R X → R X rarr RX \rightarrow RX→R is a point X ∗ ∈ X X ∗ ∈ X X^(**)in XX^{*} \in XX∗∈X such that
f ( x ∗ ) ⩾ f ( x ) ( respectively f ( x ∗ ) ⩽ f ( x ) ) f x ∗ ⩾ f ( x )  respectively  f x ∗ ⩽ f ( x ) f(x^(**)) >= f(x)quad(" respectively "f(x^(**)) <= f(x)quad)f\left(x^{*}\right) \geqslant f(x) \quad\left(\text { respectively } f\left(x^{*}\right) \leqslant f(x) \quad\right)f(x∗)⩾f(x)( respectively f(x∗)⩽f(x))
for all x ∈ X x ∈ X x inXx \in \mathbb{X}x∈X.
For a bounded real function f f fff on X X XXX put
are extensions of f | Y f Y f|_(Y)\left.f\right|_{Y}f|Y which belong to Λ α ( X , d ) Λ α ( X , d ) Lambda_(alpha)(X,d)\Lambda_{\alpha}(X, d)Λα(X,d) and have the norms q q qqq (see [2]). If F 1 F 1 F_(1)F_{1}F1 and F 2 F 2 F_(2)F_{2}F2 denotes the functions defined by (3), then
(5) F 2 ( x ) ⩽ f ( x ) ⩽ F 1 ( x ) , x ∈ X (5) F 2 ( x ) ⩽ f ( x ) ⩽ F 1 ( x ) , x ∈ X {:(5)F_(2)(x) <= f(x) <= F_(1)(x)","quad x in X:}\begin{equation*} F_{2}(x) \leqslant f(x) \leqslant F_{1}(x), \quad x \in X \tag{5} \end{equation*}(5)F2(x)⩽f(x)⩽F1(x),x∈X
and for q > ‖ f | Y ‖ α , X q > f Y α , X q > ||f|_(Y)||_(alpha,X)q>\left\|\left.f\right|_{Y}\right\|_{\alpha, X}q>‖f|Y‖α,X we have
(6) u ( x ) ⩽ P 2 ( x ) ⩽ I ( x ) ⩽ P 1 ( x ) ⩽ U ( x ) , x ∈ X (6) u ( x ) ⩽ P 2 ( x ) ⩽ I ( x ) ⩽ P 1 ( x ) ⩽ U ( x ) , x ∈ X {:(6)u(x) <= P_(2)(x) <= I(x) <= P_(1)(x) <= U(x)","quad x in X:}\begin{equation*} u(x) \leqslant P_{2}(x) \leqslant I(x) \leqslant P_{1}(x) \leqslant U(x), \quad x \in X \tag{6} \end{equation*}(6)u(x)⩽P2(x)⩽I(x)⩽P1(x)⩽U(x),x∈X

A maximum (respectively a minimum) point of a function
if X → R X → R X rarr RX \rightarrow RX→R is a point x ∗ ∈ X x ∗ ∈ X x^(**)in Xx^{*} \in Xx∗∈X such that
x ∈ X x ∈ X x in Xx \in Xx∈X.

∴ ∴ :.\therefore∴
(9) η f = Lnf { f ( x ) : x ∈ I } , e f = { x ∈ I , f ( x ) = a f } (9) η f = Lnf { f ( x ) : x ∈ I } , e f = x ∈ I , f ( x ) = a f {:(9)eta_(f)=Lnf{f(x):x in I}","quade_(f)={x in I,f(x)=a_(f)}:}\begin{equation*} \eta_{f}=\operatorname{Lnf}\{f(x): x \in I\}, \quad e_{f}=\left\{x \in I, f(x)=a_{f}\right\} \tag{9} \end{equation*}(9)ηf=Lnf{f(x):x∈I},ef={x∈I,f(x)=af}
Iot now ( X , d X , d X,dX, dX,d ) be a compact metric space and lot f f fff be a function in Λ α ( X 0 , d ) Λ α X 0 , d Lambda_(alpha)(X_(0),d)\Lambda_{\alpha}\left(X_{0}, d\right)Λα(X0,d).
We define now inductively two sequences ( z n ) n ≥ 0 z n n ≥ 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n≥0 and ( M n ) n ≥ 0 M n n ≥ 0 (M_(n))_(n >= 0)\left(M_{n}\right)_{n \geq 0}(Mn)n≥0 of points in X X XXX and of real numbers, respectively, as follows :
Ift Q > ‖ & ‖ α , I Q > ‖ & ‖ α , I Q > ||&||_(alpha,I)Q>\|\&\|_{\alpha, I}Q>‖&‖α,I be fired and lot x 0 x 0 x_(0)x_{0}x0 be a fixed point in x x xxx. Let
(20) v 0 ( x ) = P ( x 0 ) + q ⋅ d α ( x 0 x 0 ) , z ∈ X (20) v 0 ( x ) = P x 0 + q ⋅ d α x 0 x 0 , z ∈ X {:(20)v^(0)(x)=P(x_(0))+q*d^(alpha)(x_(0)x_(0))quad","quad z in X:}\begin{equation*} v^{0}(x)=P\left(x_{0}\right)+q \cdot d^{\alpha}\left(x_{0} x_{0}\right) \quad, \quad z \in X \tag{20} \end{equation*}(20)v0(x)=P(x0)+q⋅dα(x0x0),z∈X
the greatest extension of f f fff obtained from (4) for I I = { x 0 } x 0 {x_(0)}\left\{x_{0}\right\}{x0} and 10t
M 0 = sup { π 0 ( x ) , x ∈ X } . M 0 = sup π 0 ( x ) , x ∈ X . M_(0)=s u p{pi^(0)(x),x in X}.M_{0}=\sup \left\{\pi^{0}(x), x \in X\right\} .M0=sup{π0(x),x∈X}.
Suppose now that for a natural number n ⩾ 1 n ⩾ 1 n >= 1n \geqslant 1n⩾1, the points z 0 , x 1 , … , z n − 1 z 0 , x 1 , … , z n − 1 z_(0),x_(1),dots,z_(n-1)z_{0}, x_{1}, \ldots, z_{n-1}z0,x1,…,zn−1 and the number M 0 , M 1 , … , M n − 1 M 0 , M 1 , … , M n − 1 M_(0),M_(1),dots,M_(n-1)M_{0}, M_{1}, \ldots, M_{n-1}M0,M1,…,Mn−1 were defined. Let σ n − 1 σ n − 1 sigma^(n-1)\sigma^{n-1}σn−1 be the greatest extension of I | Y I Y I|_(Y)\left.\mathcal{I}\right|_{Y}I|Y obtained from (4) for Y = { x 0 , x 1 , … , x n − 1 } Y = x 0 , x 1 , … , x n − 1 Y={x_(0),x_(1),dots,x_(n-1)}Y=\left\{x_{0}, x_{1}, \ldots, x_{n-1}\right\}Y={x0,x1,…,xn−1}. Put y n = sup { 0 n − 1 ( x ) y n = sup 0 n − 1 ( x ) y_(n)=s u p{0^(n-1)(x):}y_{n}=\sup \left\{0^{n-1}(x)\right.yn=sup{0n−1(x); 0 x ∈ X } 0 x ∈ X } 0x in X}0 x \in X\}0x∈X} and lot x n x n x_(n)x_{n}xn be a point in X X XXX such that U n − 1 ( z a ) = ξ a U n − 1 z a = ξ a U^(n-1)(z_(a))=xi_(a)U^{n-1}\left(z_{a}\right)=\xi_{a}Un−1(za)=ξa.
The properties of the so defined sequences ( x a ) a ≥ 0 x a a ≥ 0 (x_(a))_(a >= 0)\left(x_{a}\right)_{a \geq 0}(xa)a≥0 and. ( y a ) n ⩾ 0 y a n ⩾ 0 (y_(a))_(n >= 0)\left(y_{a}\right)_{n \geqslant 0}(ya)n⩾0 are described in the following theorem:
TH80REM 2. Let ( I , d I , d I,dI, dI,d ) be a compact metric space and let f ∈ Λ α ( x , d ) f ∈ Λ α ( x , d ) f inLambda_(alpha)(x,d)f \in \Lambda_{\alpha}(x, d)f∈Λα(x,d). For a fired q > ‖ f ‖ α , x q > ‖ f ‖ α , x q > ||f||_(alpha,x)q>\|f\|_{\alpha, x}q>‖f‖α,x let the sequences ( x Δ ) n = 0 x Δ n = 0 (x_(Delta))_(n=0)\left(x_{\Delta}\right)_{n=0}(xΔ)n=0 and ( M a ) n > 0 M a n > 0 (M_(a))_(n > 0)\left(M_{a}\right)_{n>0}(Ma)n>0 be defined as above. Then
a) lim a → ∞ M a = M b lim a → ∞   M a = M b lim_(a rarr oo)M_(a)=M_(b)\lim _{a \rightarrow \infty} M_{a}=M_{b}lima→∞Ma=Mb :
b) lim n → ∞ lim n → ∞   lim_(n rarr oo)\lim _{n \rightarrow \infty}limn→∞ inf { a ( x , x n ) , x ∈ E f } = 0 a x , x n , x ∈ E f = 0 {a(x,x_(n)),x inE_(f)}=0\left\{a\left(x, x_{n}\right), x \in \mathbb{E}_{f}\right\}=0{a(x,xn),x∈Ef}=0;
0）The sequence ( I ( z a ) ) a ⩾ 0 I z a a ⩾ 0 (I(z_(a)))_(a >= 0)\left(I\left(z_{a}\right)\right)_{a \geqslant 0}(I(za))a⩾0 has the number If ase as
Zinit point－
Proof．Since π n ⩽ π n − 1 π n ⩽ π n − 1 pi^(n) <= pi^(n-1)\pi^{n} \leqslant \pi^{n-1}πn⩽πn−1 for B = 1 , 2 , … B = 1 , 2 , … B=1,2,dotsB=1,2, \ldotsB=1,2,… ，it rollows that the sequence ( 第 ) a ⩾ 0 (  第  ) a ⩾ 0 (" 第 ")_(a >= 0)(\text { 第 })_{a \geqslant 0}( 第 )a⩾0 is nondecreasing ．By（6） y a = v n − 1 ( x a ) ⩾⩾ f ( x a ) ⩾ min { f ( x ) y a = v n − 1 x a ⩾⩾ f x a ⩾ min { f ( x ) y_(a)=v^(n-1)(x_(a))⩾⩾f(x_(a)) >= min{f(x)y_{a}=v^{n-1}\left(x_{a}\right) \geqslant \geqslant f\left(x_{a}\right) \geqslant \min \{f(x)ya=vn−1(xa)⩾⩾f(xa)⩾min{f(x) s x ∈ I } x ∈ I } x in I}x \in I\}x∈I} so that the sequence ( h a ) m ⩾ 0 18 s h a m ⩾ 0 18 s (h_(a))_(m >= 0)(18 )/(s)\left(h_{a}\right)_{m \geqslant 0} \frac{18}{s}(ha)m⩾018s also bounded ．Therefore there exiets M = 1 ± 3 Ig M = 1 ± 3 Ig M=1+-3IgM=1 \pm 3 \mathrm{Ig}M=1±3Ig ．By（6） f ( x ) ≤ f n ( x ) ≤ I , f ( x ) ≤ f n ( x ) ≤ I , f(x) <= f^(n)(x) <= I,quadf(x) \leq f^{n}(x) \leq I, \quadf(x)≤fn(x)≤I, for all x ∈ X x ∈ X x in Xx \in Xx∈X and a ∈ I a ∈ I a in Ia \in Ia∈I so that
(21) f ( x ) ⩽ 贴, for all x ∈ x (21) f ( x ) ⩽  贴, for all  x ∈ x {:(21)f(x) <= " 贴, for all "x in x:}\begin{equation*} f(x) \leqslant \text { 贴, for all } x \in x \tag{21} \end{equation*}(21)f(x)⩽ 贴, for all x∈x
The metric space z z zzz being compact，the sequence ( z n ) n ≥ 0 z n n ≥ 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n≥0 contains a subsequence ( z n k ) k ⩾ 0 z n k k ⩾ 0 (z_(n_(k)))_(k >= 0)\left(z_{n_{k}}\right)_{k \geqslant 0}(znk)k⩾0 convergent to a point z ∈ X z ∈ X z in Xz \in Xz∈X ． Since the function if is continuous it follows that
(12) f ( x n k ) → f ( z 1 ) , k → ∞ (12) f x n k → f z 1 , k → ∞ {:(12)f(x_(n_(k)))rarr f(z_(1))quad","quad k rarr oo:}\begin{equation*} f\left(x_{n_{k}}\right) \rightarrow f\left(z_{1}\right) \quad, \quad k \rightarrow \infty \tag{12} \end{equation*}(12)f(xnk)→f(z1),k→∞
But，for k ⩾ 1 k ⩾ 1 k >= 1k \geqslant 1k⩾1 ，
| v n k − 1 ( z ) − n n k − 1 | = | v n k − 1 ( z ) − v n k − 1 ( z n k ) | ≤ q ⋅ d α ( z v z n k ) → 0 v n k − 1 ( z ) − n n k − 1 = v n k − 1 ( z ) − v n k − 1 z n k ≤ q ⋅ d α z v z n k → 0 |v^(n_(k)-1)(z)-n_(n_(k)-1)|=|v^(n_(k)-1)(z)-v^(n_(k)-1)(z_(n_(k)))| <= q*d^(alpha)(z_(v)z_(n_(k)))rarr0\left|v^{n_{k}-1}(z)-n_{n_{k}-1}\right|=\left|v^{n_{k}-1}(z)-v^{n_{k}-1}\left(z_{n_{k}}\right)\right| \leq q \cdot d^{\alpha}\left(z_{v} z_{n_{k}}\right) \rightarrow 0|vnk−1(z)−nnk−1|=|vnk−1(z)−vnk−1(znk)|≤q⋅dα(zvznk)→0
for k → ∞ k → ∞ k rarr ook \rightarrow \inftyk→∞ ，and M k − 1 → M M k − 1 → M M_(k^(-1))rarr MM_{k^{-1}} \rightarrow MMk−1→M for k → ∞ k → ∞ k rarr ook \rightarrow \inftyk→∞ ，so that
(13) v n k − 1 ( z ) → I , for k → ∞ (13) v n k − 1 ( z ) → I ,  for  k → ∞ {:(13)v^(n_(k)-1)(z)rarrIquad","" for "krarr oo:}\begin{equation*} \mathrm{v}^{\mathrm{n}_{k}-1}(\mathrm{z}) \rightarrow \mathrm{I} \quad, \text { for } \mathrm{k} \rightarrow \infty \tag{13} \end{equation*}(13)vnk−1(z)→I, for k→∞
By the relation
| ∇ n k ( z ) − f ( z n k ) | = | ∇ n k ( z ) − v n k ( z n k ) | ≤ q ⋅ d α ( z , x n k ) → 0 , k → ∞ ∇ n k ( z ) − f z n k = ∇ n k ( z ) − v n k z n k ≤ q ⋅ d α z , x n k → 0 , k → ∞ |grad^(n_(k))(z)-f(z_(n_(k)))|=|grad^(n_(k))(z)-v^(n_(k))(z_(n_(k)))| <= q*d^(alpha)(z,x_(n_(k)))rarr0,k rarr oo\left|\nabla^{n_{k}}(z)-f\left(z_{n_{k}}\right)\right|=\left|\nabla^{n_{k}}(z)-v^{n_{k}}\left(z_{n_{k}}\right)\right| \leq q \cdot d^{\alpha}\left(z, x_{n_{k}}\right) \rightarrow 0, k \rightarrow \infty|∇nk(z)−f(znk)|=|∇nk(z)−vnk(znk)|≤q⋅dα(z,xnk)→0,k→∞, and by（12）it follows that
(14) U n k ( z ) → f ( z ) , k → ∞ (14) U n k ( z ) → f ( z ) , k → ∞ {:(14)U^(n)k(z)rarr f(z)quad","k rarr oo:}\begin{equation*} U^{n} k(z) \rightarrow f(z) \quad, k \rightarrow \infty \tag{14} \end{equation*}(14)Unk(z)→f(z),k→∞
Therefore，is in the inequalities
f ( g ) ≤ π n e − 1 ( z ) ≤ π n z − 1 ( z ) , x ⩾ 1 f ( g ) ≤ π n e − 1 ( z ) ≤ π n z − 1 ( z ) , x ⩾ 1 f(g) <= pi^(n)e^(-1)(z) <= pi^(n)z-1(z)quad,x >= 1f(g) \leq \pi^{n} e^{-1}(z) \leq \pi^{n} z-1(z) \quad, x \geqslant 1f(g)≤πne−1(z)≤πnz−1(z),x⩾1
we lot k → ∞ k → ∞ k rarr ook \rightarrow \inftyk→∞ one obtains l ( g ) ≤ l ≤ l ( g ) l ( g ) ≤ l ≤ l ( g ) l(g) <= l <= l(g)l(g) \leq l \leq l(g)l(g)≤l≤l(g) ，so that f ( z ) = M f ( z ) = M f(z)=Mf(z)=Mf(z)=M ． Taking into account（11）it follows that
盖 ≡ I ( x ) ≡ max { f ( x ) ; x ∈ X } .  盖  ≡ I ( x ) ≡ max { f ( x ) ; x ∈ X } . " 盖 "-=I(x)-=max{f(x)quad;quad x inX}.\text { 盖 } \equiv \mathcal{I}(x) \equiv \max \{f(x) \quad ; \quad x \in \mathbb{X}\} . 盖 ≡I(x)≡max{f(x);x∈X}.
To prove b）observe that is contrary，then there exist ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0 and an infinits subset J J JJJ of N N NNN such that
(15) inp { a ( x , x j ) ; x ∈ I l } > ε (15) inp a x , x j ; x ∈ I l > ε {:(15)inp{a(x,x_(j));x inI_(l)} > epsi:}\begin{equation*} \operatorname{inp}\left\{a\left(x, x_{j}\right) ; x \in I_{l}\right\}>\varepsilon \tag{15} \end{equation*}(15)inp{a(x,xj);x∈Il}>ε
for all j ∈ J j ∈ J j in Jj \in Jj∈J ．The space X X XXX being compact there exists a subse－ quence ( x j k ) k ⩾ 0 x j k k ⩾ 0 (x_(j_(k)))_(k >= 0)\left(x_{j_{k}}\right)_{k \geqslant 0}(xjk)k⩾0 of ( x j ) j ∈ J x j j ∈ J (x_(j))_(j in J)\left(x_{j}\right)_{j \in J}(xj)j∈J converging to a point y ∈ X y ∈ X y in Xy \in Xy∈X ．But then，ropeating the above arguments will follow that y ∈ K f y ∈ K f y inK_(f)y \in K_{f}y∈Kf ， which contradicts（15）．
The affination c）follows from（12）．
Remarks．1）In the case X ≈ [ a , b ] X ≈ [ a , b ] X~~[a,b]X \approx[a, b]X≈[a,b] and α = 1 α = 1 alpha=1\alpha=1α=1 a similar result in proved in［6］．
2）If the extensions v n v n v^(n)v^{n}vn are replaced by the extensions u n u n u^(n)u^{n}un and g Ω = inf { a i A ( x ) i x ∈ X } , u n A ( x a + j ) = m a g Ω = inf a i A ( x ) i x ∈ X , u n A x a + j = m a g_(Omega)=i n f{a^(i_(A))(x)quad i quad x in X},u^(n^(A))(x_(a+j))=m_(a)g_{\Omega}=\inf \left\{a^{i_{A}}(x) \quad i \quad x \in X\right\}, u^{n^{A}}\left(x_{a+j}\right)=m_{a}gΩ=inf{aiA(x)ix∈X},unA(xa+j)=ma ，then one obtains a procedure to find the minimum． g f g f g_(f)g_{f}gf of a function f ∈ Λ α ( X , d f ∈ Λ α ( X , d f inLambda_(alpha)(X,df \in \Lambda_{\alpha}(X, df∈Λα(X,d
Exaaple．Let X = [ 0 , 1 ] , d ( x , y ) = | x − y | X = [ 0 , 1 ] , d ( x , y ) = | x − y | X=[0,1],d(x,y)=|x-y|X=[0,1], d(x, y)=|x-y|X=[0,1],d(x,y)=|x−y| and
f ( x ) = x ⋅ sin ⁡ i x , if x ∈ ( 0 , 1 ] = 0 , if x = 0 f ( x ) = x ⋅ sin ⁡ i x ,  if  x ∈ ( 0 , 1 ] = 0 ,  if  x = 0 {:[f(x)=x*sin((i)/(x))","," if "x in(0","1]],[=0,","" if "x=0]:}\begin{array}{cl} f(x)=x \cdot \sin \frac{i}{x}, & \text { if } x \in(0,1] \\ =0 & , \text { if } x=0 \end{array}f(x)=x⋅sin⁡ix, if x∈(0,1]=0, if x=0
It is known that f ∈ Λ α ( α , d ) f ∈ Λ α ( α , d ) f inLambda_(alpha)(alpha,d)f \in \Lambda_{\alpha}(\alpha, d)f∈Λα(α,d) if and only if α ∈ ( 0 , 1 2 ] α ∈ 0 , 1 2 alpha in(0,(1)/(2)]\alpha \in\left(0, \frac{1}{2}\right]α∈(0,12] (see [7], Probiem 153) and in this caee
‖ x ‖ α D x ≤ [ 1 + 2.1 π ( 1 + 2 π ) + 2 π ] 1 / 2 < 4 . ‖ x ‖ α D x ≤ [ 1 + 2.1 π ( 1 + 2 π ) + 2 π ] 1 / 2 < 4 . ||x||_(alpha D)x <= [1+2.1 pi(1+2pi)+2pi]^(1//2) < 4.\|x\|_{\alpha D} x \leq[1+2.1 \pi(1+2 \pi)+2 \pi]^{1 / 2}<4 .‖x‖αDx≤[1+2.1π(1+2π)+2π]1/2<4.
We apply now Theorem 2 to find the global maximum of the function f f fff on [ 0 , 1 / π ] [ 0 , 1 / π ] [0,1//pi][0,1 / \pi][0,1/π] for α = 1 / 2 α = 1 / 2 alpha=1//2\alpha=1 / 2α=1/2.
Step 0. Take x 0 = 1 / 2 π x 0 = 1 / 2 π x_(0)=1//2pi\mathrm{x}_{0}=1 / 2 \pix0=1/2π.
The greatest extension of f | { x 0 } f x 0 f|_({x_(0)})\left.f\right|_{\left\{x_{0}\right\}}f|{x0} to [ 0 , 1 / π ] [ 0 , 1 / π ] [0,1//pi][0,1 / \pi][0,1/π] is U 0 ( x ) = f ( x 0 ) + 4 | x − x 0 | 1 / 2 U 0 ( x ) = f x 0 + 4 x − x 0 1 / 2 U^(0)(x)=f(x_(0))+4|x-x_(0)|^(1//2)U^{0}(x)=f\left(x_{0}\right)+4\left|x-x_{0}\right|^{1 / 2}U0(x)=f(x0)+4|x−x0|1/2
The set of points of local maximum of the function U 0 U 0 U^(0)U^{0}U0 is
D 0 = { ( 0 , 4 / 2 π ) , ( 2 / π , 4 / 2 π ) } D 0 = { ( 0 , 4 / 2 π ) , ( 2 / π , 4 / 2 π ) } D_(0)={(0,4//sqrt(2pi)),(2//pi,4//sqrt(2pi))}D_{0}=\{(0,4 / \sqrt{2 \pi}),(2 / \pi, 4 / \sqrt{2 \pi})\}D0={(0,4/2π),(2/π,4/2π)}.
Step 1. Take x 1 = 1 / π x 1 = 1 / π x_(1)=1//pix_{1}=1 / \pix1=1/π, and let U 1 U 1 U^(1)U^{1}U1 be the greatest extension of f ∣ { x 0 , x 1 } f ∣ x 0 , x 1 f∣{x_(0),x_(1)}f \mid\left\{x_{0}, x_{1}\right\}f∣{x0,x1} io [ 0 , 1 / π ] [ 0 , 1 / π ] [0,1//pi][0,1 / \pi][0,1/π], that is U 1 ( x ) = min { f ( x k ) + 4 | x − x k | 1 / 2 k , k ∈ { 0 , 1 } } U 1 ( x ) = min f x k + 4 x − x k 1 / 2 k , k ∈ { 0 , 1 } U^(1)(x)=min{f(x_(k))+4|x-x_(k)|^(1//2)quad k,k in{0,1}}U^{1}(x)=\min \left\{f\left(x_{k}\right)+4\left|x-x_{k}\right|^{1 / 2} \quad k, k \in\{0,1\}\right\}U1(x)=min{f(xk)+4|x−xk|1/2k,k∈{0,1}}
The points of local maximum of the function 0 2 0 2 0^(2)0^{2}02 is D 1 = { ( x o 1 , U 1 ( x o 1 ) ) , ( 0 , U 1 ( 0 ) ) } D 1 = x o 1 , U 1 x o 1 , 0 , U 1 ( 0 ) D_(1)={(x_(o1),U^(1)(x_(o1))),(0,U^(1)(0))}D_{1}=\left\{\left(x_{o 1}, U^{1}\left(x_{o 1}\right)\right),\left(0, U^{1}(0)\right)\right\}D1={(xo1,U1(xo1)),(0,U1(0))}
Where x o 1 x o 1 x_(o1)x_{o 1}xo1 is the solution of the equation
U 0 ( x ) = U 2 ( x ) U 0 ( x ) = U 2 ( x ) U^(0)(x)=U^(2)(x)U^{0}(x)=U^{2}(x)U0(x)=U2(x)
belonging to [ x 0 ; x 1 ] x 0 ; x 1 [x_(0);x_(1)]\left[x_{0} ; x_{1}\right][x0;x1] and let x 2 = 0 x 2 = 0 x_(2)=0x_{2}=0x2=0.
Step 2. Let π 2 π 2 pi^(2)\pi^{2}π2 be the greatest extension of l ′ ∣ { x 0 , x 1 , x 2 } l ′ ∣ x 0 , x 1 , x 2 l^(')∣{x_(0),x_(1),x_(2)}l^{\prime} \mid\left\{x_{0}, x_{1}, x_{2}\right\}l′∣{x0,x1,x2} to [ 0 , 2 / π ] [ 0 , 2 / π ] [0,2//pi][0,2 / \pi][0,2/π]. Let
D 2 = { ( x 12 , π 2 ( x 12 ) ) , ( x 01 , π 2 ( x 01 ) ) } D 2 = x 12 , π 2 x 12 , x 01 , π 2 x 01 D_(2)={(x_(12),pi^(2)(x_(12))),(x_(01),pi^(2)(x_(01)))}D_{2}=\left\{\left(x_{12}, \pi^{2}\left(x_{12}\right)\right),\left(x_{01}, \pi^{2}\left(x_{01}\right)\right)\right\}D2={(x12,π2(x12)),(x01,π2(x01))}
be the set of points of local maximum of the function σ 2 σ 2 sigma^(2)\sigma^{2}σ2 and let x 3 = x 01 x 3 = x 01 x_(3)=x_(01)x_{3}=x_{01}x3=x01.
Step a . Let U n U n U^(n)\mathrm{U}^{\mathrm{n}}Un be the greatest extension of the function f ∣ { x 0 , x 1 , … , x n } f ∣ x 0 , x 1 , … , x n f∣{x_(0),x_(1),dots,x_(n)}f \mid\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}f∣{x0,x1,…,xn} to [ 0 , 1 / π ] [ 0 , 1 / π ] [0,1//pi][0,1 / \pi][0,1/π] and let x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1 be the greatest of the abscisses of the poits of the global 昷aximum of U n U n U^(n)U^{n}Un.
To stop the algorithm one can proceeds in two ways :
a) the algorithm stops after a fixed number of iterations &
b) the algorithm stops when the difference between the global maximum U n ( x M ) U n x M U^(n)(x_(M))U^{n}\left(x_{M}\right)Un(xM) of the function U n U n U^(n)U^{n}Un and f ( x M ) f x M f(x_(M))f\left(x_{M}\right)f(xM) is smaller than a fixed number ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0.
We have programmed this algorithm in the language BASIC with n = 300 n = 300 n=300n=300n=300 iterations.
The program is the following . 8
1 OPEN'LP: 'FOROUTPUT ASFILEITO WRITW
10 INPUT X % = X % + 18 Y % = 2 ∗ Z % X % = X % + 18 Y % = 2 ∗ Z % X%=X%+18Y%=2^(**)Z%\mathrm{X} \%=\mathrm{X} \%+18 \mathrm{Y} \%=2{ }^{*} \mathrm{Z} \%X%=X%+18Y%=2∗Z%
20 DIM X ( Y % ) , F 1 ( X % ) , FZ ( X % ) 20 DIM X ( Y % ) , F 1 ( X % ) , FZ ( X % ) 20DIMX(Y%),F1(X%),FZ(X%)20 \mathrm{DIM} \mathrm{X}(\mathrm{Y} \%), \mathrm{F} 1(\mathrm{X} \%), \mathrm{FZ}(\mathrm{X} \%)20DIMX(Y%),F1(X%),FZ(X%)
25 M = 1 ; P % = 0 25 M = 1 ; P % = 0 25M=1;P%=025 \mathrm{M}=1 ; \mathrm{P} \%=025M=1;P%=0
30 DEF FN F ( X ) F ( X ) F(X)F(X)F(X)
31 IF X = 0 X = 0 X=0X=0X=0 THEN FNF = 0 = 0 =0=0=0 ELSE 33
32 GO TO 35
33 FNE = X ∗ SIN ( 1 . / X ) 33 FNE = X ∗ SIN ( 1 . / X ) 33FNE=X^(**)SIN(1.//X)33 \mathrm{FNE}=\mathrm{X}^{*} \mathrm{SIN}(1 . / \mathrm{X})33FNE=X∗SIN(1./X)
35 FNEND
40 DEF FN P ( X , U ) = F N P ( U ) + 3.355 ∗ SQR ( A B S ( X ! U ) ) P ( X , U ) = F N P ( U ) + 3.355 ∗ SQR ( A B S ( X ! U ) ) P(X,U)=FNP(U)+3.355^(**)SQR(ABS(X!U))P(X, U)=F N P(U)+3.355^{*} \operatorname{SQR}(A B S(X!U))P(X,U)=FNP(U)+3.355∗SQR(ABS(X!U))
50 DEF WH B（X，Y）
60 A = I + X A = I + X A=I+X\mathrm{A}=\mathrm{I}+\mathrm{X}A=I+X
70 D = ( HEP ( Y ) − HSE ( X ) ) / 3.3558 D = D ∗ D D = ( HEP ( Y ) − HSE ( X ) ) / 3.3558 D = D ∗ D D=(HEP(Y)-HSE(X))//3.3558 D=D^(**)DD=(\operatorname{HEP}(Y)-\operatorname{HSE}(X)) / 3.3558 D=D^{*} DD=(HEP(Y)−HSE(X))/3.3558D=D∗D \＆富＝X－X
75 D3 = D ∗ ( E + E − D ) = D ∗ ( E + E − D ) =D^(**)(E+E-D)=D^{*}(E+E-D)=D∗(E+E−D)
80 BI＝（A＋SQR（D3））／2．
90 II 82 >= X 2 82 >= X 2 82>=X282>=X 282>=X2 Am 81 <= Y 81 <= Y 81<=Y81<=Y81<=Y GO 20110
100 82 = 1 − B 2 82 = 1 − B 2 82=1-B282=1-B 282=1−B2
110 FITB = 81 = 81 =81=81=81
120 FINETD
130 X ( 1 ) = 0 & X ( 2 ) = 1 . / P I & X ( 3 ) = 1 . / ( 2 . ∗ P I ) X ( 1 ) = 0 & X ( 2 ) = 1 . / P I & X ( 3 ) = 1 . / 2 . ∗ P I X(1)=0&X(2)=1.//PI&X(3)=1.//(2.^(**)PI)X(1)=0 \& X(2)=1 . / P I \& X(3)=1 . /\left(2 .{ }^{*} P I\right)X(1)=0&X(2)=1./PI&X(3)=1./(2.∗PI)
140 I\％ ≈ 3 : J % = 1 : P 2 ( J % ) = MP ( X ( 1 ) , X ( 3 ) ) , M 1 ( J % ) ≈ X ( 1 ) ≈ 3 : J % = 1 : P 2 ( J % ) = MP ( X ( 1 ) , X ( 3 ) ) , M 1 ( J % ) ≈ X ( 1 ) ~~3:J%=1:P2(J%)=MP(X(1),X(3)),M1(J%)~~X(1)\approx 3: J \%=1: P 2(J \%)=\operatorname{MP}(X(1), X(3)), \operatorname{M1(J\% )} \approx X(1)≈3:J%=1:P2(J%)=MP(X(1),X(3)),M1(J%)≈X(1)
150 J % = J % + 1 : E 2 ( J % ) = MPP ( X ( 2 ) , X ( 3 ) ) , FI ( J % ) = X ( 2 ) J % = J % + 1 : E 2 ( J % ) = MPP ( X ( 2 ) , X ( 3 ) ) , FI ( J % ) = X ( 2 ) J%=J%+1:E2(J%)=MPP(X(2),X(3)),FI(J%)=X(2)J \%=J \%+1: \operatorname{E2}(J \%)=\operatorname{MPP}(X(2), X(3)), \operatorname{FI}(J \%)=X(2)J%=J%+1:E2(J%)=MPP(X(2),X(3)),FI(J%)=X(2)
160 Σ σ 日 = 188 ⇒ F 2 ( 1 ) Σ σ 日 = 188 ⇒ F 2 ( 1 ) Sigmasigma_(日)=188=>F2(1)\Sigma \sigma_{日}=188 \Rightarrow \mathrm{~F} 2(1)Σσ日=188⇒ F2(1)
3.70 BOR E\％＝2 20 J % 20 J % 20J%20 \mathrm{~J} \mathrm{\%}20 J%
180
190 MEXYS K\％
200 D 1 = X ( 1 ) − X ( 2 ) D 1 = X ( 1 ) − X ( 2 ) D1=X(1)-X(2)D 1=X(1)-X(2)D1=X(1)−X(2) \＆ V 1 % = 0 V 1 % = 0 V1%=0V 1 \%=0V1%=0 ； V 2 % = 0 V 2 % = 0 V2%=0V 2 \%=0V2%=0 ：D2＝－D1，CaI1（If）
210 FOR K % = 1 K % = 1 K%=1K \%=1K%=1 TO \％
220 Tax（K\％）－0
230 IF T＜0 THEST 240 ETBE 250
240 If T < D 1 T < D 1 T < D1\mathrm{T}<\mathrm{D} 1T<D1 THEN ； 250 ET8S D 1 = T 1 D 1 = T 1 D1=T_(1)\mathrm{D} 1=\mathrm{T}_{1}D1=T1 ： 71 % = 71 % = 71%=71 \%=71%= K
250 II m 2 > 0 m 2 > 0 (m)/(2) > 0\frac{m}{2}>0m2>0 सादता 2010 STSE 1020
1010 IF T D2 THEN 1020 ELSE D2＝T \＆V2\％－K\％
1020 NEXT K\％
1030 IF V1\％＝0 THEN 1070
1040 姐 = I % : I % = I % + 1 = I % : I % = I % + 1 =I%:I%=I%+1=I \%: I \%=I \%+1=I%:I%=I%+1
1050 X ( I % ) = MNS ( X ( V 1 % ) , 0 ) X ( I % ) = MNS ( X ( V 1 % ) , 0 ) X(I%)=MNS(X(V1%),0)\mathrm{X}(\mathrm{I} \%)=\operatorname{MNS}(\mathrm{X}(\mathrm{V} 1 \%), 0)X(I%)=MNS(X(V1%),0)
50 DEF WH B（X，Y） 60 A=I+X 70 D=(HEP(Y)-HSE(X))//3.3558 D=D^(**)D \＆富＝X－X 75 D3 =D^(**)(E+E-D) 80 BI＝（A＋SQR（D3））／2． 90 II 82>=X2 Am 81<=Y GO 20110 100 82=1-B2 110 FITB =81 120 FINETD 130 X(1)=0&X(2)=1.//PI&X(3)=1.//(2.^(**)PI) 140 I\％~~3:J%=1:P2(J%)=MP(X(1),X(3)),M1(J%)~~X(1) 150 J%=J%+1:E2(J%)=MPP(X(2),X(3)),FI(J%)=X(2) 160 Sigmasigma_(日)=188=>F2(1) 3.70 BOR E\％＝2 20J% 180 https://cdn.mathpix.com/cropped/2025_09_08_ff27670f11cd2d1aba5ag-5.jpg?height=64&width=815&top_left_y=996&top_left_x=361 190 MEXYS K\％ 200 D1=X(1)-X(2) \＆V1%=0 ；V2%=0 ：D2＝－D1，CaI1（If） 210 FOR K%=1 TO \％ 220 Tax（K\％）－0 230 IF T＜0 THEST 240 ETBE 250 240 If T < D1 THEN ； 250 ET8S D1=T_(1) ： 71%= K 250 II (m)/(2) > 0 सादता 2010 STSE 1020 1010 IF T D2 THEN 1020 ELSE D2＝T \＆V2\％－K\％ 1020 NEXT K\％ 1030 IF V1\％＝0 THEN 1070 1040 姐 =I%:I%=I%+1 1050 X(I%)=MNS(X(V1%),0)| 50 DEF WH B（X，Y） | | | :--- | :--- | | 60 | $\mathrm{A}=\mathrm{I}+\mathrm{X}$ | | 70 | $D=(\operatorname{HEP}(Y)-\operatorname{HSE}(X)) / 3.3558 D=D^{*} D$ \＆富＝X－X | | 75 | D3 $=D^{*}(E+E-D)$ | | 80 | BI＝（A＋SQR（D3））／2． | | 90 | II $82>=X 2$ Am $81<=Y$ GO 20110 | | 100 | $82=1-B 2$ | | 110 | FITB $=81$ | | 120 | FINETD | | 130 | $X(1)=0 \& X(2)=1 . / P I \& X(3)=1 . /\left(2 .{ }^{*} P I\right)$ | | 140 | I\％$\approx 3: J \%=1: P 2(J \%)=\operatorname{MP}(X(1), X(3)), \operatorname{M1(J\% )} \approx X(1)$ | | 150 | $J \%=J \%+1: \operatorname{E2}(J \%)=\operatorname{MPP}(X(2), X(3)), \operatorname{FI}(J \%)=X(2)$ | | 160 | $\Sigma \sigma_{日}=188 \Rightarrow \mathrm{~F} 2(1)$ | | 3.70 | BOR E\％＝2 $20 \mathrm{~J} \mathrm{\%}$ | | 180 | ![](https://cdn.mathpix.com/cropped/2025_09_08_ff27670f11cd2d1aba5ag-5.jpg?height=64&width=815&top_left_y=996&top_left_x=361) | | 190 | MEXYS K\％ | | 200 | $D 1=X(1)-X(2)$ \＆$V 1 \%=0$ ；$V 2 \%=0$ ：D2＝－D1，CaI1（If） | | 210 | FOR $K \%=1$ TO \％ | | 220 | Tax（K\％）－0 | | 230 | IF T＜0 THEST 240 ETBE 250 | | 240 | If $\mathrm{T}<\mathrm{D} 1$ THEN ； 250 ET8S $\mathrm{D} 1=\mathrm{T}_{1}$ ： $71 \%=$ K | | 250 | II $\frac{m}{2}>0$ सादता 2010 STSE 1020 | | 1010 | IF T D2 THEN 1020 ELSE D2＝T \＆V2\％－K\％ | | 1020 | NEXT K\％ | | 1030 | IF V1\％＝0 THEN 1070 | | 1040 | 姐 $=I \%: I \%=I \%+1$ | | 1050 | $\mathrm{X}(\mathrm{I} \%)=\operatorname{MNS}(\mathrm{X}(\mathrm{V} 1 \%), 0)$ |


1070 II V2\％कासारा 1140
1080 正自3\％ ← 1 ← 1 larr1\leftarrow 1←1
1090 X ( I % ) = M B ( C , X ( V 2 % ) ) 1090 X ( I % ) = M B ( C , X ( V 2 % ) ) 1090 X(I%)=MB(C,X(V2%))1090 X(I \%)=M B(C, X(V 2 \%))1090X(I%)=MB(C,X(V2%))
3100 IT V1\％＝0 THEN 2110 ELEES 1120

1120 J % = 3 % + 1 % = 3 % + 1 %=3%+1\%=3 \%+1%=3%+1
1125 제 ग\％－8\％－1 O2 I\％I F⿻丷木－1 स्पष्म 2150
D Y．MYTENUEISE1135：P\％ = 7 % = 7 % =7%=7 \%=7%
1135 ए J % = X % J % = X % J%=X%J \%=X \%J%=X% OR I % = X % I % = X % I%=X%I \%=X \%I%=X% शन्सरण 1150
1140 II ABS ( ABS ( ABS(\operatorname{ABS}(ABS( F2（J\％）－FNE ( X ( I % ) ) ) < 1 . E − 3 ( X ( I % ) ) ) < 1 . E − 3 (X(I%))) < 1.E-3(\mathrm{X}(\mathrm{I} \%)))<1 . \mathrm{E}-3(X(I%)))<1.E−3 THEN 1150 ETSZ 160

1151 PRI \＃1，\＆PRI \＆1，＂B－AU YACUT＂B J\％＂JIERATII＂
1155 P R I ≠ 1 , 8 P R I ⇒ 1 , ∘ X ′ & T A B ( 11 ) ; ′ F ( X ) ′ & T A B ( 21 ) ; ∘ P ( X ) ′ & T A B ( 33 ) g ′ X ′ & & TAB ( 43 ) ; ′ P ( x ) ′ 1155 P R I ≠ 1 , 8 P R I ⇒ 1 , ∘ X ′ & T A B ( 11 ) ; ′ F ( X ) ′ & T A B ( 21 ) ; ∘ P ( X ) ′ & T A B ( 33 ) g ′ X ′ & & TAB ( 43 ) ; ′ P ( x ) ′ 1155 PRI!=1,8PRI=>1,^(@)X^(')&TAB(11);^(')F(X)^(')&TAB(21);^(@)P(X)^(')&TAB(33)g^(')X^(')&quad&TAB(43);^(')P(x)^(')1155 P R I \neq 1,8 P R I \Rightarrow 1,{ }^{\circ} X^{\prime} \& T A B(11) ;{ }^{\prime} F(X)^{\prime} \& T A B(21) ;{ }^{\circ} P(X)^{\prime} \& T A B(33) g^{\prime} X^{\prime} \& \quad \& \operatorname{TAB}(43) ;{ }^{\prime} P(x)^{\prime}1155PRI≠1,8PRI⇒1,∘X′&TAB(11);′F(X)′&TAB(21);∘P(X)′&TAB(33)g′X′&&TAB(43);′P(x)′ \＆ 2 AB ( 53 ) ; ′ P ( x ) ′ ( 53 ) ; ′ P ( x ) ′ (53);^(')P(x)^(')(53) ;{ }^{\prime} P(x)^{\prime}(53);′P(x)′ ；
TAB（207）； ′ F ( X ) ′ ′ F ( X ) ′ ^(')F(X)^('){ }^{\prime} F(X)^{\prime}′F(X)′ ；TAB（117）； ′ P ( X ) ′ ′ P ( X ) ′ ^(')P(X)^('){ }^{\prime} P(X)^{\prime}′P(X)′ \＆PRI \＃ I 。 I 。  I_("。 ")I_{\text {。 }}I。 
FIX（II（K\％））；E2（K\％）；FOR K\％＝1 TO J\％－1
1180 PRI \＃1，8 PRI \＃1，8 PRI \＃1， ∘ ∘ ^(@){ }^{\circ}∘ SOL APROXINATIVA ′ ′ ′ X = ′ g F 1 ( P % ) ′ ′ ′ X = ′ g F 1 ( P % ) ^(')^(')^(')X=^(')gF1(P%){ }^{\prime}{ }^{\prime}{ }^{\prime} \mathrm{X}={ }^{\prime} \mathrm{g} F 1(\mathrm{P} \%)′′′X=′gF1(P%) ；\＆ ′ F ( X ) = ′ ; F N P ( F I ( P % ) ; ′ P ( X ) = ′ ξ F 2 ( P % ) ′ F ( X ) = ′ ; F N P F I ( P % ) ; ′ P ( X ) = ′ ξ F 2 ( P % ) ^(')F(X)=^(');FNP(FI(P%);^(')P(X)=^(')xi F2(P%):}{ }^{\prime} F(X)={ }^{\prime} ; F N P\left(F I(P \%) ;^{\prime} P(X)={ }^{\prime} \xi F 2(P \%)\right.′F(X)=′;FNP(FI(P%);′P(X)=′ξF2(P%) ；＇DELMA＝＇\＆\＆ ABS（B2（P\％）－FNF（FI（P\％）））
11840 = 1 . / 71 11840 = 1 . / 71 11840=1.//7111840=1 . / 7111840=1./71（P\％）
1185PRT \＃1， 2 PRI \＃1，＂VATOARRA DERIVATTEL ： I ′ ( X ¯ ) = # % I ′ ( X ¯ ) = # % I^(')( bar(X))=^(#)%I^{\prime}(\bar{X})={ }^{\#} \%I′(X¯)=#% SIW ( U ¯ ) − U ∗ cOS ( U ) ( U ¯ ) − U ∗ cOS ( U ) ( bar(U))-U^(**)cOS(U)(\bar{U})-U^{*} \operatorname{cOS}(U)(U¯)−U∗cOS(U) 1190 푼（）
One can show that in this case the sequence ( x a ) a j e x a a j e (x_(a))_(aje)\left(x_{a}\right)_{a j e}(xa)aje conver－ ges to the maximum point z ∗ = 0.12944 z ∗ = 0.12944 z^(**)=0.12944z^{*}=0.12944z∗=0.12944（the solution of the equate tion tg ( 1 / x ) = 1 / x , x ∈ ( 1 / 3 π , 1 / 2 π ) ) tg ( 1 / x ) = 1 / x , x ∈ ( 1 / 3 π , 1 / 2 π ) ) tg(1//x)=1//x,x in(1//3pi,1//2pi))\operatorname{tg}(1 / x)=1 / x, x \in(1 / 3 \pi, 1 / 2 \pi))tg(1/x)=1/x,x∈(1/3π,1/2π)) ．
The maximum of f f fff is ξ f = 2 ( x ∗ ) = 0.128374 ξ f = 2 x ∗ = 0.128374 xi_(f)=2(x^(**))=0.128374\xi_{f}=2\left(x^{*}\right)=0.128374ξf=2(x∗)=0.128374 ．Arter 300 ite－ ration we have obtained the following results \＆
x 300 = 0.129982 x 300 = 0.129982 x_(300)=0.129982x_{300}=0.129982x300=0.129982
f ( x 300 ) = 0.128309 f x 300 = 0.128309 f(x_(300))=0.128309f\left(x_{300}\right)=0.128309f(x300)=0.128309
M 229 ≈ v 299 ( x 300 ) = 0.144181 M 229 ≈ v 299 x 300 = 0.144181 M_(229)~~v^(299)(x_(300))=0.144181M_{229} \approx v^{299}\left(x_{300}\right)=0.144181M229≈v299(x300)=0.144181
The errors are in this case
M S − & ( x 300 ) = 0.000065 x 300 − x ∗ = 0.000537 M S − & x 300 = 0.000065 x 300 − x ∗ = 0.000537 {:[M_(S)-&(x_(300))=0.000065],[x_(300)-x^(**)=0.000537]:}\begin{aligned} & M_{S}-\&\left(x_{300}\right)=0.000065 \\ & x_{300}-x^{*}=0.000537 \end{aligned}MS−&(x300)=0.000065x300−x∗=0.000537

Refferences

1．G．ARONSSON，Extension of functions aatisfying Lipschits con－ ditions ，Arkiv ior Matheratik 6， 28 （1967），551－561．
2．J．CZIPSER and I．GEHER，Ritersion of functions satisfying a Iipschitz condition，Acta．Math．Acad．Sci，Hungar 6 （1955）， 213－220．
3．B．LEVY and M．D．RICE ，The Approximation of Uniformly Continuous Mappings by Lipschitz and Hölder Mappings，（preprint），1980．
4．E．J．McSHANS，Extension of range of functions，Bull，Amer．Math． Soc． 40 （1934），837－842．
5．O．MUSTATA，Best Approximation and Unique Extension of Lipachits Functions，Journal Approx．Theory 19 （1977），222－230．
6．B．SHUBERT，A sequential aethod seeking the global maximum of a function，SIAM J．Humer．Anal．9（1972）， 379 − 388 379 − 388 379-388379-388379−388.
7．The Otto Dunkel Memorial Problem Book，New Yort（1957）, Normal cone valued metrics and nonconvex vector
1983

Related Posts