Posts by Costica Mustata

Costica Mustata

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

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C. Mustăţa, On the best approximation in metric spaces, Anal. Numér. Théor. Approx., 4 (1975) 1, 45-50.

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https://ictp.acad.ro/jnaat/journal/article/view/1975-vol4-no1-art5/Mustata

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Mathematica-Revue d’Analyse Numer. Theor.Approx.

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Romanian Academy

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https://ictp.acad.ro/jnaat/journal/article/view/1975-vol4-no1-art5

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MR # 82c: 46052

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[1] Dunford, N. and Schwartz, J. T., Linear Operators. I. New York (1958)
[2] Johnson, J.A., Banach Spaces of Lipschitz functions and Vector-Valued Lipschitz Functions, Trans. Amer. Math. soc., 148, 147-169 (1970).
[3] Michael, E., A short prof of the Arens-Eells embedding theorem, Proc. Amer. Math. Soc., 15, 415-416 (1964).
[4] Pantelidis, G., Approximationstheorie fur metrische lineare Raume, Math. Ann., 184, 30-48 (1969).
[5] Singer, I., Cea mai bună aproximare în spații vectoriale normate prin elemente din subspații vectoriale. Ed. Acad. București, 1967.
[6] Vlasov, L.P., Approximationye svojstva mnojestv v linejnyh normirovannyh prostranstvah, Uspechi Mat. Nauk., 18, 6, 4-66 (1973).

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1975-Mustata-On the best approximation in metric spaces-Jnaat

ON THE BEST APPROXIMATION IN METRIC SPACES

byCOSTICĂ MUSTĂTA(Cluj-Napoca)

Let ( X , d ) ( X , d ) (X,d)(X, d)(X,d) be a metric space and x 0 x 0 x_(0)x_{0}x0 a fixed point in X X XXX. The set
(1) X 0 ‡ ‡ = { f : X → R , sup x ≠ y x , y ∈ X | f ( x ) − f ( y ) | d ( x , y ) < ∞ , f ( x 0 ) = 0 } , (1) X 0 ‡ ‡ = f : X → R , sup x ≠ y x , y ∈ X   | f ( x ) − f ( y ) | d ( x , y ) < ∞ , f x 0 = 0 , {:(1)X_(0)^(‡‡)={f:X rarr R,s u p_({:[x!=y],[x","y in X]:})(|f(x)-f(y)|)/(d(x,y)) < oo,f(x_(0))=0}",":}\begin{equation*} X_{0}^{\ddagger \ddagger}=\left\{f: X \rightarrow R, \sup _{\substack{x \neq y \\ x, y \in X}} \frac{|f(x)-f(y)|}{d(x, y)}<\infty, f\left(x_{0}\right)=0\right\}, \tag{1} \end{equation*}(1)X0‡‡={f:X→R,supx≠yx,y∈X|f(x)−f(y)|d(x,y)<∞,f(x0)=0},
with the usual operation of addition and multiplication by real scalars, normed by
(2) ‖ f ‖ X = sup x ≠ y x , y ∈ X | f ( x ) − f ( y ) | d ( x , y ) , f ∈ X 0 # (2) ‖ f ‖ X = sup x ≠ y x , y ∈ X   | f ( x ) − f ( y ) | d ( x , y ) , f ∈ X 0 # {:(2)||f||_(X)=s u p_({:[x!=y],[x","y in X]:})(|f(x)-f(y)|)/(d(x,y))","f inX_(0)^(#):}\begin{equation*} \|f\|_{X}=\sup _{\substack{x \neq y \\ x, y \in X}} \frac{|f(x)-f(y)|}{d(x, y)}, f \in X_{0}^{\#} \tag{2} \end{equation*}(2)‖f‖X=supx≠yx,y∈X|f(x)−f(y)|d(x,y),f∈X0#
is a Banach space (even a conjugate Banach space [2]).
The space X 0 # X 0 # X_(0)^(#)X_{0}^{\#}X0# plays, with respect to X X XXX, in many ways, the same role as the conjugate E ∗ E ∗ E^(**)E^{*}E∗ of a normed linear space E E EEE, with respect to E E EEE. In this paper we give further details in this direction.
For ∅ ≠ Y ⊆ X ∅ ≠ Y ⊆ X O/!=Y sube X\emptyset \neq Y \subseteq X∅≠Y⊆X and x ∈ X x ∈ X x in Xx \in Xx∈X we denote by d ( x , Y ) d ( x , Y ) d(x,Y)d(x, Y)d(x,Y) the distance from x x xxx to Y Y YYY, i.e.
(3) d ( x , Y ) = inf y ∈ Y d ( x , y ) (3) d ( x , Y ) = inf y ∈ Y   d ( x , y ) {:(3)d(x","Y)=i n f_(y in Y)d(x","y):}\begin{equation*} d(x, Y)=\inf _{y \in Y} d(x, y) \tag{3} \end{equation*}(3)d(x,Y)=infy∈Yd(x,y)
Proposition 1. Let Y ⊂ X , x 0 ∈ Y Y ⊂ X , x 0 ∈ Y Y sub X,x_(0)in YY \subset X, x_{0} \in YY⊂X,x0∈Y and x 1 ∈ X − Y x 1 ∈ X − Y x_(1)in X-Yx_{1} \in X-Yx1∈X−Y such that
(4) d ( x 1 , Y ) = q > 0 (4) d x 1 , Y = q > 0 {:(4)d(x_(1),Y)=q > 0:}\begin{equation*} d\left(x_{1}, Y\right)=q>0 \tag{4} \end{equation*}(4)d(x1,Y)=q>0
Proof. For every y ∈ f ( − 1 ) ( 0 ) y ∈ f ( − 1 ) ( 0 ) y inf^((-1))(0)y \in f^{(-1)}(0)y∈f(−1)(0) and x ∈ X , | f ( x ) | = | f ( x ) − f ( y ) | ≦ x ∈ X , | f ( x ) | = | f ( x ) − f ( y ) | ≦ x in X,|f(x)|=|f(x)-f(y)| <=x \in X,|f(x)|=|f(x)-f(y)| \leqqx∈X,|f(x)|=|f(x)−f(y)|≦
Then there is f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# such that
(5) f | Y = 0 , f ( x 1 ) = 1 , ‖ f ‖ X = 1 q (5) f Y = 0 , f x 1 = 1 , ‖ f ‖ X = 1 q {:(5)f|_(Y)=0","f(x_(1))=1","||f||_(X)=(1)/(q):}\begin{equation*} \left.f\right|_{Y}=0, f\left(x_{1}\right)=1,\|f\|_{X}=\frac{1}{q} \tag{5} \end{equation*}(5)f|Y=0,f(x1)=1,‖f‖X=1q
Proof. We will show that a function with the required properties is given by
(6) f ( x ) = 1 q d ( x , Y ) (6) f ( x ) = 1 q d ( x , Y ) {:(6)f(x)=(1)/(q)d(x","Y):}\begin{equation*} f(x)=\frac{1}{q} d(x, Y) \tag{6} \end{equation*}(6)f(x)=1qd(x,Y)
Indeed, f ( x 0 ) = 0 f x 0 = 0 f(x_(0))=0f\left(x_{0}\right)=0f(x0)=0, because x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y. For x , z ∈ X x , z ∈ X x,z in Xx, z \in Xx,z∈X we have
| d ( x , Y ) − d ( z , Y ) | ⩽ d ( x , z ) | d ( x , Y ) − d ( z , Y ) | ⩽ d ( x , z ) |d(x,Y)-d(z,Y)| <= d(x,z)|d(x, Y)-d(z, Y)| \leqslant d(x, z)|d(x,Y)−d(z,Y)|⩽d(x,z)
and by the definition of f f fff, it follows that
(7) ‖ f ‖ X ≦ 1 q < ∞ (7) ‖ f ‖ X ≦ 1 q < ∞ {:(7)||f||_(X) <= (1)/(q) < oo:}\begin{equation*} \|f\|_{X} \leqq \frac{1}{q}<\infty \tag{7} \end{equation*}(7)‖f‖X≦1q<∞
This means that f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0#.
Evidently, f | Y = 0 f Y = 0 f|_(Y)=0\left.f\right|_{Y}=0f|Y=0 and f ( x 1 ) = 1 f x 1 = 1 f(x_(1))=1f\left(x_{1}\right)=1f(x1)=1. Since d ( x 1 , Y ) = q > 0 d x 1 , Y = q > 0 d(x_(1),Y)=q > 0d\left(x_{1}, Y\right)=q>0d(x1,Y)=q>0, then there is a sequence ( y n ) n ∈ N ⊂ Y y n n ∈ N ⊂ Y (y_(n))_(n in N)sub Y\left(y_{n}\right)_{n \in N} \subset Y(yn)n∈N⊂Y such that d ( x 1 , y n ) → d ( x 1 , Y ) d x 1 , y n → d x 1 , Y d(x_(1),y_(n))rarr d(x_(1),Y)d\left(x_{1}, y_{n}\right) \rightarrow d\left(x_{1}, Y\right)d(x1,yn)→d(x1,Y) when y → ∞ y → ∞ y rarr ooy \rightarrow \inftyy→∞, It follows that we can find an increasing sequence of natural numbers { n k } n k {n_(k)}\left\{n_{k}\right\}{nk}. such that d ( x 1 , y n h ) ≦ d ( x 1 , Y ) + 1 k d x 1 , y n h ≦ d x 1 , Y + 1 k d(x_(1),y_(n_(h))) <= d(x_(1),Y)+(1)/(k)d\left(x_{1}, y_{n_{h}}\right) \leqq d\left(x_{1}, Y\right)+\frac{1}{k}d(x1,ynh)≦d(x1,Y)+1k. Then
| d ( y n k ′ , Y ) − d ( x 1 , Y ) | d ( y n k ′ ! x 1 ) = d ( x 1 , Y ) d ( y 1 k ′ , x 1 ) ≧ d ( x 1 , y n k ) − 1 k d ( x 1 , y n k ) → 1 d y n k ′ , Y − d x 1 , Y d y n k ′ ! x 1 = d x 1 , Y d y 1 k ′ , x 1 ≧ d x 1 , y n k − 1 k d x 1 , y n k → 1 (|d(y_(n_(k^('))),Y)-d(x_(1),Y)|)/(d(y_(n_(k^(')))!x_(1)))=(d(x_(1),Y))/(d(y_(1_(k^('))),x_(1))) >= (d(x_(1),y_(n_(k)))-(1)/(k))/(d(x_(1),y_(n_(k))))rarr1\frac{\left|d\left(y_{n_{k^{\prime}}}, Y\right)-d\left(x_{1}, Y\right)\right|}{d\left(y_{n_{k^{\prime}}}!x_{1}\right)}=\frac{d\left(x_{1}, Y\right)}{d\left(y_{1_{k^{\prime}}}, x_{1}\right)} \geqq \frac{d\left(x_{1}, y_{n_{k}}\right)-\frac{1}{k}}{d\left(x_{1}, y_{n_{k}}\right)} \rightarrow 1|d(ynk′,Y)−d(x1,Y)|d(ynk′!x1)=d(x1,Y)d(y1k′,x1)≧d(x1,ynk)−1kd(x1,ynk)→1
From the above inequality we obtain
(8) ‖ f ‖ X ≧ 1 q (8) ‖ f ‖ X ≧ 1 q {:(8)||f||_(X) >= (1)/(q):}\begin{equation*} \|f\|_{X} \geqq \frac{1}{q} \tag{8} \end{equation*}(8)‖f‖X≧1q
By (7) and (8) it follows that ‖ f ‖ X = 1 q ‖ f ‖ X = 1 q ||f||_(X)=(1)/(q)\|f\|_{X}=\frac{1}{q}‖f‖X=1q, which completes the proof of the proposition.
For f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# we denote
(9)
f ( − 1 ) ( 0 ) = { x ∈ X , f ( x ) = 0 } f ( − 1 ) ( 0 ) = { x ∈ X , f ( x ) = 0 } f^((-1))(0)={x in X,f(x)=0}f^{(-1)}(0)=\{x \in X, f(x)=0\}f(−1)(0)={x∈X,f(x)=0}
Proposition 2. Let f ∈ X 0 # − { θ } f ∈ X 0 # − { θ } f inX_(0)^(#)-{theta}f \in X_{0}^{\#}-\{\theta\}f∈X0#−{θ}. Then
(10)
d ( x , f ( − 1 ) ( 0 ) ) ≧ | f ( x ) | ‖ f ‖ X d x , f ( − 1 ) ( 0 ) ≧ | f ( x ) | ‖ f ‖ X d(x,f^((-1))(0)) >= (|f(x)|)/(||f||_(X))d\left(x, f^{(-1)}(0)\right) \geqq \frac{|f(x)|}{\|f\|_{X}}d(x,f(−1)(0))≧|f(x)|‖f‖X
for every x ∈ X x ∈ X x in Xx \in Xx∈X.
≦ ‖ f ‖ X d ( x , y ) ≦ ‖ f ‖ X d ( x , y ) <= ||f||_(X)d(x,y)\leqq\|f\|_{X} d(x, y)≦‖f‖Xd(x,y). Therefore
d ( x , f ( − 1 ) ( 0 ) ) ≧ | f ( x ) | ‖ f ‖ X d x , f ( − 1 ) ( 0 ) ≧ | f ( x ) | ‖ f ‖ X d(x,f^((-1))(0)) >= (|f(x)|)/(||f||_(X))d\left(x, f^{(-1)}(0)\right) \geqq \frac{|f(x)|}{\|f\|_{X}}d(x,f(−1)(0))≧|f(x)|‖f‖X
and the proposition is proved.
Definition 1. A subset Y Y YYY of the metric space X X XXX is called proximinal if for cyery x ∈ X x ∈ X x in Xx \in Xx∈X there is an element y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y such that
d ( x , y 0 ) = d ( x , Y ) d x , y 0 = d ( x , Y ) d(x,y_(0))=d(x,Y)d\left(x, y_{0}\right)=d(x, Y)d(x,y0)=d(x,Y)
If, for all x ∈ X x ∈ X x in Xx \in Xx∈X the element y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y verifying (11) is unique, then the set Y Y YYY is called chebyshevian. An element y 0 ∈ X y 0 ∈ X y_(0)in Xy_{0} \in Xy0∈X, verifying (11) is called element of best approximation of x x xxx by elements of Y Y YYY.
Proposition 3. Let f ∈ X 0 # − { θ } f ∈ X 0 # − { θ } f inX_(0)^(#)-{theta}f \in X_{0}^{\#}-\{\theta\}f∈X0#−{θ}. If for every x ∈ X − f ( − 1 ) ( 0 ) x ∈ X − f ( − 1 ) ( 0 ) x in X-f^((-1))(0)x \in X-f^{(-1)}(0)x∈X−f(−1)(0) there is an element y x ∈ f ( − 1 ) ( 0 ) y x ∈ f ( − 1 ) ( 0 ) y_(x)inf^((-1))(0)y_{x} \in f^{(-1)}(0)yx∈f(−1)(0) such that
(12) | f ( x ) − f ( y x ) | = ‖ f ‖ X d ( x , y x ) (12) f ( x ) − f y x = ‖ f ‖ X d x , y x {:(12)|f(x)-f(y_(x))|=||f||_(X)d(x,y_(x)):}\begin{equation*} \left|f(x)-f\left(y_{x}\right)\right|=\|f\|_{X} d\left(x, y_{x}\right) \tag{12} \end{equation*}(12)|f(x)−f(yx)|=‖f‖Xd(x,yx)
then f ( − 1 ) ( 0 ) f ( − 1 ) ( 0 ) f^((-1))(0)f^{(-1)}(0)f(−1)(0) is proximinal.
Proof. Let x ∈ X − f ( − 1 ) ( 0 ) x ∈ X − f ( − 1 ) ( 0 ) x in X-f^((-1))(0)x \in X-f^{(-1)}(0)x∈X−f(−1)(0). Since f ( − 1 ) ( 0 ) f ( − 1 ) ( 0 ) f^((-1))(0)f^{(-1)}(0)f(−1)(0) is closed it follows that 0 << d ( x , f ( − 1 ) ( 0 ) ) ≦ d ( x , y ) 0 << d x , f ( − 1 ) ( 0 ) ≦ d ( x , y ) 0<<d(x,f^((-1))(0)) <= d(x,y)0< <d\left(x, f^{(-1)}(0)\right) \leqq d(x, y)0<<d(x,f(−1)(0))≦d(x,y), for all y ∈ f ( − 1 ) ( 0 ) y ∈ f ( − 1 ) ( 0 ) y inf^((-1))(0)y \in f^{(-1)}(0)y∈f(−1)(0). Now, let y x y x y_(x)y_{x}yx be an element of f ( − 1 ) ( 0 ) f ( − 1 ) ( 0 ) f^((-1))(0)f^{(-1)}(0)f(−1)(0) for which (12) holds. Then for every y ∈ f ( − 1 ) ( 0 ) y ∈ f ( − 1 ) ( 0 ) y inf^((-1))(0)y \in f^{(-1)}(0)y∈f(−1)(0),
‖ f ‖ X = | f ( x ) − f ( y x ) | d ( x , y x ) ≧ | f ( x ) − f ( y ) | d ( x , y ) , ‖ f ‖ X = f ( x ) − f y x d x , y x ≧ | f ( x ) − f ( y ) | d ( x , y ) , ||f||_(X)=(|f(x)-f(y_(x))|)/(d(x,y_(x))) >= (|f(x)-f(y)|)/(d(x,y)),\|f\|_{X}=\frac{\left|f(x)-f\left(y_{x}\right)\right|}{d\left(x, y_{x}\right)} \geqq \frac{|f(x)-f(y)|}{d(x, y)},‖f‖X=|f(x)−f(yx)|d(x,yx)≧|f(x)−f(y)|d(x,y),
that is
| f ( x ) | d ( x , y x ) ≥ | f ( x ) | d ( x , y ) | f ( x ) | d x , y x ≥ | f ( x ) | d ( x , y ) (|f(x)|)/(d(x,y_(x))) >= (|f(x)|)/(d(x,y))\frac{|f(x)|}{d\left(x, y_{x}\right)} \geq \frac{|f(x)|}{d(x, y)}|f(x)|d(x,yx)≥|f(x)|d(x,y)
Therefore, d ( x , y x ) ≦ d ( x , y ) d x , y x ≦ d ( x , y ) d(x,y_(x)) <= d(x,y)d\left(x, y_{x}\right) \leqq d(x, y)d(x,yx)≦d(x,y) and, taking the infimum relatively to y y yyy we get d ( x , y x ) = d ( x , f ( − 1 ) ( 0 ) ) d x , y x = d x , f ( − 1 ) ( 0 ) d(x,y_(x))=d(x,f^((-1))(0))d\left(x, y_{x}\right)=d\left(x, f^{(-1)}(0)\right)d(x,yx)=d(x,f(−1)(0)).
In the following proposition we give a characterization of the elements of best approximation.
Proposition 4. Let Y Y YYY be a subset of X X XXX such that x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y, and let x ∈ X − Y x ∈ X − Y x in X-Yx \in X-Yx∈X−Y. Then y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is an clement of best approximation for x x xxx by elcments of Y Y YYY, if and only if there is an f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# such that
  1. ‖ f ‖ X = 1 ‖ f ‖ X = 1 ||f||_(X)=1\|f\|_{X}=1‖f‖X=1
  2. f | Y = 0 f Y = 0 f|_(Y)=0\left.f\right|_{Y}=0f|Y=0
  3. | f ( x ) − f ( y 0 ) | = d ( x , y 0 ) f ( x ) − f y 0 = d x , y 0 |f(x)-f(y_(0))|=d(x,y_(0))\left|f(x)-f\left(y_{0}\right)\right|=d\left(x, y_{0}\right)|f(x)−f(y0)|=d(x,y0).
Proof. If x ∈ X − Y x ∈ X − Y x in X-Yx \in X-Yx∈X−Y and y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is an element of best approximation for x x xxx by elements of Y Y YYY, then from the proof of proposition 1 it follows that the function
(13)
f ( x ) = d ( x , Y ) f ( x ) = d ( x , Y ) f(x)=d(x,Y)f(x)=d(x, Y)f(x)=d(x,Y)
has all the required properties.
Conversely, if f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# is such that the conditions 1), 2), 3) hold, then for every y ∈ Y y ∈ Y y in Yy \in Yy∈Y,
d ( x , y 0 ) = | f ( x ) − f ( y 0 ) | = | f ( x ) − f ( y ) | ≦ ‖ f ‖ X d ( x , y ) = d ( x , y ) , d x , y 0 = f ( x ) − f y 0 = | f ( x ) − f ( y ) | ≦ ‖ f ‖ X d ( x , y ) = d ( x , y ) , d(x,y_(0))=|f(x)-f(y_(0))|=|f(x)-f(y)| <= ||f||_(X)d(x,y)=d(x,y),d\left(x, y_{0}\right)=\left|f(x)-f\left(y_{0}\right)\right|=|f(x)-f(y)| \leqq\|f\|_{X} d(x, y)=d(x, y),d(x,y0)=|f(x)−f(y0)|=|f(x)−f(y)|≦‖f‖Xd(x,y)=d(x,y),
which completes the proof of the proposition.
Proposition 5. Let Y Y YYY be a proximinal subset of X , x 0 ∈ Y X , x 0 ∈ Y X,x_(0)in YX, x_{0} \in YX,x0∈Y, and x ∈ X − Y x ∈ X − Y x in X-Yx \in X-Yx∈X−Y. Let y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y be an element of best approximation of x x xxx by elements of Y Y YYY. The following conditions are equivalent:
i) y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is the only element of best approximation of x x xxx.
ii) There is no y ∈ Y , y ≠ y 0 y ∈ Y , y ≠ y 0 y in Y,y!=y_(0)y \in Y, y \neq y_{0}y∈Y,y≠y0 and f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# such that
a) ‖ f ‖ X = 1 ‖ f ‖ X = 1 ||f||_(X)=1\|f\|_{X}=1‖f‖X=1
b) f ( y 0 ) = f ( y ) f y 0 = f ( y ) f(y_(0))=f(y)f\left(y_{0}\right)=f(y)f(y0)=f(y)
c) | f ( x ) − f ( y ) | = d ( x , y ) | f ( x ) − f ( y ) | = d ( x , y ) |f(x)-f(y)|=d(x,y)|f(x)-f(y)|=d(x, y)|f(x)−f(y)|=d(x,y).
Proof. Let us suppose that i) holds and that there is y ∈ Y , y ≠ y 0 y ∈ Y , y ≠ y 0 y in Y,y!=y_(0)y \in Y, y \neq y_{0}y∈Y,y≠y0 and f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# such that a), b), c) hold. Then
d ( x , y ) = | f ( x ) − f ( y ) | ≦ | f ( x ) − f ( y 0 ) | + | f ( y 0 ) − f ( y ) | = | f ( x ) − f ( y 0 ) | = d ( x , y 0 ) . d ( x , y ) = | f ( x ) − f ( y ) | ≦ f ( x ) − f y 0 + f y 0 − f ( y ) = f ( x ) − f y 0 = d x , y 0 . d(x,y)=|f(x)-f(y)| <= |f(x)-f(y_(0))|+|f(y_(0))-f(y)|=|f(x)-f(y_(0))|=d(x,y_(0)).d(x, y)=|f(x)-f(y)| \leqq\left|f(x)-f\left(y_{0}\right)\right|+\left|f\left(y_{0}\right)-f(y)\right|=\left|f(x)-f\left(y_{0}\right)\right|=d\left(x, y_{0}\right) .d(x,y)=|f(x)−f(y)|≦|f(x)−f(y0)|+|f(y0)−f(y)|=|f(x)−f(y0)|=d(x,y0).
Therefore, y y yyy is also an element of best approximation of x x xxx, which condradicts i).
Now, let suppose that the condition i) is not accomplished. Then there is y ∈ Y , y ≠ y 0 y ∈ Y , y ≠ y 0 y in Y,y!=y_(0)y \in Y, y \neq y_{0}y∈Y,y≠y0 such that
d ( x , y ) = d ( x , y 0 ) = d ( x , Y ) d ( x , y ) = d x , y 0 = d ( x , Y ) d(x,y)=d(x,y_(0))=d(x,Y)d(x, y)=d\left(x, y_{0}\right)=d(x, Y)d(x,y)=d(x,y0)=d(x,Y)
By proposition 4, there is f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# such that ‖ f ‖ X = 1 , f | X = 0 ‖ f ‖ X = 1 , f X = 0 ||f||_(X)=1,f|_(X)=0\|f\|_{X}=1,\left.f\right|_{X}=0‖f‖X=1,f|X=0 and ∣ f ( x ) ∣ f ( x ) ∣f(x)\mid f(x)∣f(x) - − f ( y ) ∣= d ( x , y ) − f ( y ) ∣= d ( x , y ) -f(y)∣=d(x,y)-f(y) \mid=d(x, y)−f(y)∣=d(x,y). From f | Y = 0 f Y = 0 f|_(Y)=0\left.f\right|_{Y}=0f|Y=0 it follows that f ( y 0 ) = 0 = f ( y ) f y 0 = 0 = f ( y ) f(y_(0))=0=f(y)f\left(y_{0}\right)=0=f(y)f(y0)=0=f(y). Therefore the condition a), b), c) hold.
Let Y ⊂ X Y ⊂ X Y sub XY \subset XY⊂X and x 0 ∈ Y x 0 ∈ Y x_(0)in Yx_{0} \in Yx0∈Y. Let us denote
(14) Y ⊥ = { f | f ∈ X 0 # , f | Y = 0 } . (14) Y ⊥ = f f ∈ X 0 # , f Y = 0 . {:(14)Y^(_|_)={f|f inX_(0)^(#),f|_(Y)=0}.:}\begin{equation*} Y^{\perp}=\left\{f\left|f \in X_{0}^{\#}, f\right|_{Y}=0\right\} . \tag{14} \end{equation*}(14)Y⊥={f|f∈X0#,f|Y=0}.
For x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X we denote
(15) d Y ⊥ ( x , y ) = sup f ∈ Y ⊥ − { 0 } | f ( x ) − f ( y ) | ‖ f ‖ X . (15) d Y ⊥ ( x , y ) = sup f ∈ Y ⊥ − { 0 }   | f ( x ) − f ( y ) | ‖ f ‖ X . {:(15)d_(Y^(_|_))(x","y)=s u p_(f inY^(_|_)-{0})(|f(x)-f(y)|)/(||f||_(X)).:}\begin{equation*} d_{Y^{\perp}}(x, y)=\sup _{f \in Y^{\perp}-\{0\}} \frac{|f(x)-f(y)|}{\|f\|_{X}} . \tag{15} \end{equation*}(15)dY⊥(x,y)=supf∈Y⊥−{0}|f(x)−f(y)|‖f‖X.
We have the following inequality:
(16) d Y ⊥ ( x , y ) ≦ d ( x , y ) . (16) d Y ⊥ ( x , y ) ≦ d ( x , y ) . {:(16)d_(Y^(_|_))(x","y) <= d(x","y).:}\begin{equation*} d_{Y^{\perp}}(x, y) \leqq d(x, y) . \tag{16} \end{equation*}(16)dY⊥(x,y)≦d(x,y).
Indeed, for all f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# and for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X,
| f ( x ) − f ( y ) | ≦ ‖ f ‖ X d ( x , y ) , | f ( x ) − f ( y ) | ≦ ‖ f ‖ X d ( x , y ) , |f(x)-f(y)| <= ||f||_(X)d(x,y),|f(x)-f(y)| \leqq\|f\|_{X} d(x, y),|f(x)−f(y)|≦‖f‖Xd(x,y),
so that, for f ≠ θ f ≠ θ f!=thetaf \neq \thetaf≠θ,
| f ( x ) − f ( y ) | ‖ f ‖ X ≦ d ( x , y ) , | f ( x ) − f ( y ) | ‖ f ‖ X ≦ d ( x , y ) , (|f(x)-f(y)|)/(||f||_(X)) <= d(x,y),\frac{|f(x)-f(y)|}{\|f\|_{X}} \leqq d(x, y),|f(x)−f(y)|‖f‖X≦d(x,y),
and
sup f ∈ X 0 # − { θ } | f ( x ) − f ( y ) | ‖ f ‖ X ≦ d ( x , y ) . sup f ∈ X 0 # − { θ }   | f ( x ) − f ( y ) | ‖ f ‖ X ≦ d ( x , y ) . s u p_(f inX_(0)^(#)-{theta})(|f(x)-f(y)|)/(||f||_(X)) <= d(x,y).\sup _{f \in X_{0}^{\#}-\{\theta\}} \frac{|f(x)-f(y)|}{\|f\|_{X}} \leqq d(x, y) .supf∈X0#−{θ}|f(x)−f(y)|‖f‖X≦d(x,y).
Therefore
d Y ⊥ ( x , y ) = sup f ∈ Y ⊥ − { θ } | f ( x ) − f ( x ) | ‖ f ‖ X ≦ sup f ∈ X n # − { θ } | f ( x ) − f ( y ) | ‖ f ‖ X ≦ d ( x , y ) . d Y ⊥ ( x , y ) = sup f ∈ Y ⊥ − { θ }   | f ( x ) − f ( x ) | ‖ f ‖ X ≦ sup f ∈ X n # − { θ }   | f ( x ) − f ( y ) | ‖ f ‖ X ≦ d ( x , y ) . d_(Y^(_|_))(x,y)=s u p_(f inY^(_|_)-{theta})(|f(x)-f(x)|)/(||f||_(X)) <= s u p_(f inX_(n)^(#)-{theta})(|f(x)-f(y)|)/(||f||_(X)) <= d(x,y).d_{Y^{\perp}}(x, y)=\sup _{f \in Y^{\perp}-\{\theta\}} \frac{|f(x)-f(x)|}{\|f\|_{X}} \leqq \sup _{f \in X_{n}^{\#}-\{\theta\}} \frac{|f(x)-f(y)|}{\|f\|_{X}} \leqq d(x, y) .dY⊥(x,y)=supf∈Y⊥−{θ}|f(x)−f(x)|‖f‖X≦supf∈Xn#−{θ}|f(x)−f(y)|‖f‖X≦d(x,y).
Proposition 6. Let Y ⊂ X Y ⊂ X Y sub XY \subset XY⊂X and y 0 ∈ Y , x ∈ X − Y y 0 ∈ Y , x ∈ X − Y y_(0)in Y,x in X-Yy_{0} \in Y, x \in X-Yy0∈Y,x∈X−Y. Then, y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is an element of best approximation for x x xxx by elements of Y Y YYY if and only if
(17) d Y ⊥ ( x , y 0 ) = d ( x , y 0 ) . (17) d Y ⊥ x , y 0 = d x , y 0 . {:(17)d_(Y^(_|_))(x,y_(0))=d(x,y_(0)).:}\begin{equation*} d_{Y^{\perp}}\left(x, y_{0}\right)=d\left(x, y_{0}\right) . \tag{17} \end{equation*}(17)dY⊥(x,y0)=d(x,y0).
Proof. Let y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y be an element of best approximation for x x xxx. Then, by Proposition 4 it follows that there exist an element f ∈ Y ⊥ f ∈ Y ⊥ f in Y _|_f \in Y \perpf∈Y⊥ such that ‖ f ‖ X = 1 ‖ f ‖ X = 1 ||f||_(X)=1\|f\|_{X}=1‖f‖X=1 and | f ( x ) − f ( y ) | = d ( x , y 0 ) | f ( x ) − f ( y ) | = d x , y 0 |f(x)-f(y)|=d(x,y_(0))|f(x)-f(y)|=d\left(x, y_{0}\right)|f(x)−f(y)|=d(x,y0). We have
d Y ⊥ ( x , y 0 ) = sup g ∈ Y ⊥ − { θ } | g ( x ) − g ( y 0 ) | ‖ g ‖ X ≧ | f ( x ) − f ( y 0 ) | ‖ f ‖ X = d ( x , y 0 ) , d Y ⊥ x , y 0 = sup g ∈ Y ⊥ − { θ }   g ( x ) − g y 0 ‖ g ‖ X ≧ f ( x ) − f y 0 ‖ f ‖ X = d x , y 0 , d_(Y^(_|_))(x,y_(0))=s u p_(g inY^(_|_)-{theta})(|g(x)-g(y_(0))|)/(||g||_(X)) >= (|f(x)-f(y_(0))|)/(||f||_(X))=d(x,y_(0)),d_{Y^{\perp}}\left(x, y_{0}\right)=\sup _{g \in Y^{\perp}-\{\theta\}} \frac{\left|g(x)-g\left(y_{0}\right)\right|}{\|g\|_{X}} \geqq \frac{\left|f(x)-f\left(y_{0}\right)\right|}{\|f\|_{X}}=d\left(x, y_{0}\right),dY⊥(x,y0)=supg∈Y⊥−{θ}|g(x)−g(y0)|‖g‖X≧|f(x)−f(y0)|‖f‖X=d(x,y0),
and, because of d Y ⊥ ( x , y 0 ) ≦ d ( x , y 0 ) d Y ⊥ x , y 0 ≦ d x , y 0 d_(Y^(_|_))(x,y_(0)) <= d(x,y_(0))d_{Y^{\perp}}\left(x, y_{0}\right) \leqq d\left(x, y_{0}\right)dY⊥(x,y0)≦d(x,y0) we have (17).
Conversely, if (17) holds, then for all y ∈ Y y ∈ Y y in Yy \in Yy∈Y we have:
d ( x , y 0 ) = d Y ⊥ ( x , y 0 ) = sup f ∈ Y ⊥ − { θ } | f ( x ) − f ( y 0 ) | ‖ f ‖ X = sup f ∈ Y ⊥ − { θ } | f ( x ) − f ( y ) | ‖ f ‖ X = = d Y ⊥ ( x , y ) ≦ d ( x , y ) . d x , y 0 = d Y ⊥ x , y 0 = sup f ∈ Y ⊥ − { θ }   f ( x ) − f y 0 ‖ f ‖ X = sup f ∈ Y ⊥ − { θ }   | f ( x ) − f ( y ) | ‖ f ‖ X = = d Y ⊥ ( x , y ) ≦ d ( x , y ) . {:[d(x,y_(0))=d_(Y^(_|_))(x,y_(0))=s u p_(f inY^(_|_)-{theta})(|f(x)-f(y_(0))|)/(||f||_(X))=s u p_(f inY^(_|_)-{theta})(|f(x)-f(y)|)/(||f||_(X))=],[=d_(Y^(_|_))(x","y) <= d(x","y).]:}\begin{aligned} d\left(x, y_{0}\right)=d_{Y^{\perp}}\left(x, y_{0}\right)= & \sup _{f \in Y^{\perp}-\{\theta\}} \frac{\left|f(x)-f\left(y_{0}\right)\right|}{\|f\|_{X}}=\sup _{f \in Y^{\perp}-\{\theta\}} \frac{|f(x)-f(y)|}{\|f\|_{X}}= \\ & =d_{Y^{\perp}}(x, y) \leqq d(x, y) . \end{aligned}d(x,y0)=dY⊥(x,y0)=supf∈Y⊥−{θ}|f(x)−f(y0)|‖f‖X=supf∈Y⊥−{θ}|f(x)−f(y)|‖f‖X==dY⊥(x,y)≦d(x,y).
Hence y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is an element of best approximation for x x xxx by elements of Y Y YYY.

Remarks.

1 0 1 0 1^(0)1^{0}10. Let ( X , d X , d X,dX, dX,d ) be a linear metric space, the metric d d ddd being translation invariant, and x 0 = θ ∈ X x 0 = θ ∈ X x_(0)=theta in Xx_{0}=\theta \in Xx0=θ∈X. If Y Y YYY is a subspace of X X XXX, then one can choose the function f f fff in : roposition 1 such that f ∈ C X f ∈ C X f inC_(X)f \in C_{X}f∈CX, where C X C X C_(X)C_{X}CX denotes the ccne of subadditive function in X 0 # X 0 # X_(0)^(#)X_{0}^{\#}X0# [4]. The subadditivity of function f f fff follows from the proof of Proposition 2.1 [4]. If X X XXX is a normed linear space, then X 0 # ⊃ C X ⊃ X ∗ X 0 # ⊃ C X ⊃ X ∗ X_(0)^(#)supC_(X)supX^(**)X_{0}^{\#} \supset C_{X} \supset X^{*}X0#⊃CX⊃X∗. If Y Y YYY is a subspace of X X XXX, then Proposition 1 holds with f ∈ X ∗ ( [ 1 ] f ∈ X ∗ ( [ 1 ] f inX^(**)([1]f \in X^{*}([1]f∈X∗([1], Lemma 12, p. 64).
2 0 2 0 2^(0)2^{0}20. Simple example show that the inequality (10) in Proposition 2 can be strict. Let X = [ − 1 , 10 ] ⊂ R X = [ − 1 , 10 ] ⊂ R X=[-1,10]sub RX=[-1,10] \subset RX=[−1,10]⊂R with the usual metric d ( x , y ) = | x − y | d ( x , y ) = | x − y | d(x,y)=|x-y|d(x, y)=|x-y|d(x,y)=|x−y|, x 0 = 0 ∈ R x 0 = 0 ∈ R x_(0)=0in Rx_{0}=0 \in Rx0=0∈R and
f ( x ) = { 0 x ∈ [ − 1 , 0 ] x x ∈ ( 0 , 1 ] 1 x ∈ ( 1 , 10 ] f ( x ) = 0      x ∈ [ − 1 , 0 ] x      x ∈ ( 0 , 1 ] 1      x ∈ ( 1 , 10 ] f(x)={[0,x in[-1","0]],[x,x in(0","1]],[1,x in(1","10]]:}f(x)= \begin{cases}0 & x \in[-1,0] \\ x & x \in(0,1] \\ 1 & x \in(1,10]\end{cases}f(x)={0x∈[−1,0]xx∈(0,1]1x∈(1,10]
4 - L'analyse numérique et la théorie de l'approximation - Tome 4, No. 1/1975
Then f ( − 1 ) ( 0 ) = [ − 1 , 0 ] f ( − 1 ) ( 0 ) = [ − 1 , 0 ] f^((-1))(0)=[-1,0]f^{(-1)}(0)=[-1,0]f(−1)(0)=[−1,0] and for all x ∈ [ 2 , 10 ] x ∈ [ 2 , 10 ] x in[2,10]x \in[2,10]x∈[2,10],
d ( x , [ − 1 , 0 ] ) > 1 = | f ( x ) | ‖ f ‖ X d ( x , [ − 1 , 0 ] ) > 1 = | f ( x ) | ‖ f ‖ X d(x,[-1,0]) > 1=(|f(x)|)/(||f||_(X))d(x,[-1,0])>1=\frac{|f(x)|}{\|f\|_{X}}d(x,[−1,0])>1=|f(x)|‖f‖X
If X X XXX is a metric space, Y Y YYY a closed subset of X , x 0 ∈ Y X , x 0 ∈ Y X,x_(0)in YX, x_{0} \in YX,x0∈Y, then for every function f ∈ X 0 # f ∈ X 0 # f inX_(0)^(#)f \in X_{0}^{\#}f∈X0# of the form f ( x ) = λ d ( x , Y ) , λ ∈ R f ( x ) = λ d ( x , Y ) , λ ∈ R f(x)=lambda d(x,Y),lambda in Rf(x)=\lambda d(x, Y), \lambda \in Rf(x)=λd(x,Y),λ∈R, the relation (10) holds with the sign ,,=".
If X X XXX is a normed linear space and f ∈ X ∗ f ∈ X ∗ f inX^(**)f \in X^{*}f∈X∗, then (10) holds with the sign ,, =". (Ascoli's Theorem [6]).
3 0 3 0 3^(0)3^{0}30. If X X XXX is a normed linear space and f ∈ X ∗ f ∈ X ∗ f inX^(**)f \in X^{*}f∈X∗; then the condition (12) is equivalent to:
( J ) ( J ) (J)(\mathrm{J})(J). There is x 0 ∈ X x 0 ∈ X x_(0)in Xx_{0} \in Xx0∈X such that | f ( x 0 ) | = ‖ f ‖ ⋅ ‖ x 0 ‖ f x 0 = ‖ f ‖ ⋅ x 0 |f(x_(0))|=||f||*||x_(0)||\left|f\left(x_{0}\right)\right|=\|f\| \cdot\left\|x_{0}\right\||f(x0)|=‖f‖⋅‖x0‖.
Indeed, since X = f ( − 1 ) ( 0 ) ≠ R x 0 X = f ( − 1 ) ( 0 ) ≠ R x 0 X=f^((-1))(0)!=Rx_(0)X=f^{(-1)}(0) \neq R x_{0}X=f(−1)(0)≠Rx0, for every x ∈ X − f ( − 1 ) ( 0 ) x ∈ X − f ( − 1 ) ( 0 ) x in X-f^((-1))(0)x \in X-f^{(-1)}(0)x∈X−f(−1)(0) there is λ ∈ R λ ∈ R lambda in R\lambda \in Rλ∈R and y x ∈ f ( − 1 ) ( 0 ) y x ∈ f ( − 1 ) ( 0 ) y_(x)inf^((-1))(0)y_{x} \in f^{(-1)}(0)yx∈f(−1)(0) such that x = y x + λ x 0 x = y x + λ x 0 x=y_(x)+lambdax_(0)x=y_{x}+\lambda x_{0}x=yx+λx0. Then
| f ( x ) − f ( y x ) | = | f ( x ) − f ( x − λ x 0 ) | = | λ | ⋅ | f ( x 0 ) | = ‖ f ‖ ⋅ ‖ x 0 ‖ ⋅ | λ | = = ‖ f ‖ ⋅ ‖ 1 x − ( x − λ x 0 ) ‖ = ‖ f ‖ X ⋅ ‖ x − y λ ‖ ⋅ f ( x ) − f y x = f ( x ) − f x − λ x 0 = | λ | ⋅ f x 0 = ‖ f ‖ ⋅ x 0 ⋅ | λ | = = ‖ f ‖ ⋅ 1 x − x − λ x 0 = ‖ f ‖ X ⋅ x − y λ ⋅ {:[|f(x)-f(y_(x))|=|f(x)-f(x-lambdax_(0))|=|lambda|*|f(x_(0))|=||f||*||x_(0)||*|lambda|=],[=||f||*||_(1)x-(x-lambdax_(0))||=||f||_(X)*||x-y_(lambda)||*]:}\begin{aligned} \left|f(x)-f\left(y_{x}\right)\right| & =\left|f(x)-f\left(x-\lambda x_{0}\right)\right|=|\lambda| \cdot\left|f\left(x_{0}\right)\right|=\|f\| \cdot\left\|x_{0}\right\| \cdot|\lambda|= \\ & =\|f\| \cdot\left\|_{1} x-\left(x-\lambda x_{0}\right)\right\|=\|f\|_{X} \cdot\left\|x-y_{\lambda}\right\| \cdot \end{aligned}|f(x)−f(yx)|=|f(x)−f(x−λx0)|=|λ|⋅|f(x0)|=‖f‖⋅‖x0‖⋅|λ|==‖f‖⋅‖1x−(x−λx0)‖=‖f‖X⋅‖x−yλ‖⋅
Evidently, if f ∈ X ∗ ⊂ X 0 # f ∈ X ∗ ⊂ X 0 # f inX^(**)subX_(0)^(#)f \in X^{*} \subset X_{0}^{\#}f∈X∗⊂X0#, the norm (2) agrees with the usual norm of linear functionals).
The converse implication is obvious.
By a theorem of R. C. JAMES [6] it follows that (12) is a necessary and sufficient condition for f ( − 1 ) ( 0 ) f ( − 1 ) ( 0 ) f^((-1))(0)f^{(-1)}(0)f(−1)(0) to be proximinal.
4. Proposition 4 is analogous to Theorem 1.1, p. 16, [5] and to Proposition 2.1, [4], while Proposition 5 is analogous to the Theorem 3.1, p. 96, [5] and Proposition 4.1, [4].

REIERENCE

[1] Dunford, N. and Schwartz J. T., Linear Operators. 1. New York, (1958).
[2] Jolınson, J. A:, Banach Spaces of Lipschitz functions and Vector-Valued Lipschitz Functions. Trans. Amer. Matli. Soc., 148, 147-169 (1970).
[3] Michael, E., A short proof of the Arens-Eells embedding theorem. Proc. Amer. Math. Soc., 15, 415-416 (1964).
[4] Pantelidis, G., Approximationstheorie für metrische lineare Räume. Math. Ann., 184, 30-48 (1969).
[5] Singer, I., Cea mai bună aproximare in spatii vectoriale normate prin clemcnte din subspafii vectoriale. Ed. Acad., București, (1967).
[6] V1as ov, L. P., Approximationye svojstva mnojestv v linejnyh normirovannyh prostranstvah. Uspehi Mat. Nauk., 13, 6, 4-66 (1973).

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