Return to Article Details Extension of bounded linear functionals and best approximation in spaces with asymmetric norm

EXTENSION OF BOUNDED LINEAR FUNCTIONALS AND BEST APPROXIMATION IN SPACES WITH ASYMMETRIC NORM

S. COBZAŞ ∗ ∗ ^(**){ }^{*}∗ and C. MUSTĂŢA † † ^(†){ }^{\dagger}†

Abstract

The present paper is concerned with the characterization of the elements of best approximation in a subspace Y Y YYY of a space with asymmetric norm, in terms of some linear functionals vanishing on Y Y YYY. The approach is based on some extension results, proved in Section 3, for bounded linear functionals on such spaces. Also, the well known formula for the distance to a hyperplane in a normed space is extended to the nonsymmetric case.

MSC 2000. 41A65.
Keywords. Spaces with asymmetric norm, best approximation, Hahn-Banach theorem, characterization of best approximation.

1. INTRODUCTION

Let X X XXX be a real vector space. An asymmetric seminorm on X X XXX is a positive sublinear functional p : X → [ 0 , ∞ ) p : X → [ 0 , ∞ ) p:X rarr[0,oo)p: X \rightarrow[0, \infty)p:X→[0,∞), i.e. p p ppp satisfies the conditions:
(AN1) p ( x ) ≥ 0 p ( x ) ≥ 0 p(x) >= 0p(x) \geq 0p(x)≥0,
(AN2) p ( t x ) = t p ( x ) , t ≥ 0 p ( t x ) = t p ( x ) , t ≥ 0 quad p(tx)=tp(x),t >= 0\quad p(t x)=t p(x), t \geq 0p(tx)=tp(x),t≥0,
(AN3) p ( x + y ) ≤ p ( x ) + p ( y ) p ( x + y ) ≤ p ( x ) + p ( y ) p(x+y) <= p(x)+p(y)p(x+y) \leq p(x)+p(y)p(x+y)≤p(x)+p(y),
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X. The function p ¯ : X → [ 0 , ∞ ) p ¯ : X → [ 0 , ∞ ) bar(p):X rarr[0,oo)\bar{p}: X \rightarrow[0, \infty)p¯:X→[0,∞) defined by p ¯ ( x ) = p ( − x ) , x ∈ X p ¯ ( x ) = p ( − x ) , x ∈ X bar(p)(x)=p(-x),x in X\bar{p}(x)=p(-x), x \in Xp¯(x)=p(−x),x∈X, is another positive sublinear functional on X X XXX, called the conjugate of p p ppp, and
p s ( x ) = max { p ( x ) , p ( − x ) } , x ∈ X , p s ( x ) = max { p ( x ) , p ( − x ) } , x ∈ X , p^(s)(x)=max{p(x),p(-x)},x in X,p^{s}(x)=\max \{p(x), p(-x)\}, x \in X,ps(x)=max{p(x),p(−x)},x∈X,
is a seminorm on X X XXX. The inequalities
| p ( x ) − p ( y ) | ≤ p s ( x − y ) and | p ¯ ( x ) − p ¯ ( y ) | ≤ p s ( x − y ) | p ( x ) − p ( y ) | ≤ p s ( x − y )  and  | p ¯ ( x ) − p ¯ ( y ) | ≤ p s ( x − y ) |p(x)-p(y)| <= p^(s)(x-y)quad" and "quad| bar(p)(x)- bar(p)(y)| <= p^(s)(x-y)|p(x)-p(y)| \leq p^{s}(x-y) \quad \text { and } \quad|\bar{p}(x)-\bar{p}(y)| \leq p^{s}(x-y)|p(x)−p(y)|≤ps(x−y) and |p¯(x)−p¯(y)|≤ps(x−y)
hold for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X. If the seminorm p s p s p^(s)p^{s}ps is a norm on X X XXX then we say that p p ppp is an asymmetric norm on X X XXX. This means that, beside (AN1)-(AN3), it satisfies also the condition
(AN4) p ( x ) = 0 p ( x ) = 0 p(x)=0p(x)=0p(x)=0 and p ( − x ) = 0 p ( − x ) = 0 p(-x)=0p(-x)=0p(−x)=0 imply x = 0 x = 0 x=0x=0x=0.
The pair ( X , p X , p X,pX, pX,p ), where X X XXX is a linear space and p p ppp is an asymmetric seminorm on X X XXX is called a space with asymmetric seminorm, respectively a space with asymmetric norm, if p p ppp is an asymmetric norm.
An asymmetric seminorm p p ppp generates a topology τ p τ p tau_(p)\tau_{p}τp on X X XXX, having as a basis of neighborhoods of a point x ∈ X x ∈ X x in Xx \in Xx∈X the open p p ppp-balls
B p ′ ( x , r ) = { x ′ ∈ X : p ( x ′ − x ) < r } , r > 0 B p ′ ( x , r ) = x ′ ∈ X : p x ′ − x < r , r > 0 B_(p)^(')(x,r)={x^(')in X:p(x^(')-x) < r},r > 0B_{p}^{\prime}(x, r)=\left\{x^{\prime} \in X: p\left(x^{\prime}-x\right)<r\right\}, r>0Bp′(x,r)={x′∈X:p(x′−x)<r},r>0
The family of closed p p ppp-balls
B p ( x , r ) = { x ′ ∈ X : p ( x ′ − x ) ≤ r } , r > 0 B p ( x , r ) = x ′ ∈ X : p x ′ − x ≤ r , r > 0 B_(p)(x,r)={x^(')in X:p(x^(')-x) <= r},r > 0B_{p}(x, r)=\left\{x^{\prime} \in X: p\left(x^{\prime}-x\right) \leq r\right\}, r>0Bp(x,r)={x′∈X:p(x′−x)≤r},r>0
generates the same topology.
Denote by B p = B p ( 0 , 1 ) B p = B p ( 0 , 1 ) B_(p)=B_(p)(0,1)B_{p}=B_{p}(0,1)Bp=Bp(0,1) the closed unit ball of ( X , p X , p X,pX, pX,p ) and by B p ′ = B p ′ ( 0 , 1 ) B p ′ = B p ′ ( 0 , 1 ) B_(p)^(')=B_(p)^(')(0,1)B_{p}^{\prime}=B_{p}^{\prime}(0,1)Bp′=Bp′(0,1) its open unit ball.
The topology τ p τ p tau_(p)\tau_{p}τp is translation invariant, i.e. the addition + : X × X → X + : X × X → X +:X xx X rarr X+: X \times X \rightarrow X+:X×X→X is continuous, but the multiplication by scalars ⋅ R × X → X ⋅ R × X → X *Rxx X rarr X\cdot \mathbb{R} \times X \rightarrow X⋅R×X→X need not be continuous. For instance, in the space
C 0 [ 0 , 1 ] = { x ∈ C [ 0 , 1 ] : ∫ 0 1 x ( t ) d t = 0 } C 0 [ 0 , 1 ] = x ∈ C [ 0 , 1 ] : ∫ 0 1   x ( t ) d t = 0 C_(0)[0,1]={x in C[0,1]:int_(0)^(1)x(t)dt=0}C_{0}[0,1]=\left\{x \in C[0,1]: \int_{0}^{1} x(t) \mathrm{d} t=0\right\}C0[0,1]={x∈C[0,1]:∫01x(t)dt=0}
with the asymmetric seminorm p ( x ) = max x ( [ 0 , 1 ] ) p ( x ) = max x ( [ 0 , 1 ] ) p(x)=max x([0,1])p(x)=\max x([0,1])p(x)=maxx([0,1]), the multiplication by scalars is not continuous at t 0 = − 1 t 0 = − 1 t_(0)=-1t_{0}=-1t0=−1 and x 0 = 0 x 0 = 0 x_(0)=0x_{0}=0x0=0. Indeed, the ball B p ( 0 , 1 ) B p ( 0 , 1 ) B_(p)(0,1)B_{p}(0,1)Bp(0,1) is a neighborhood of 0 = ( − 1 ) 0 0 = ( − 1 ) 0 0=(-1)00=(-1) 00=(−1)0, but − B ( 0 , r ) ⊈ B ( 0 , 1 ) − B ( 0 , r ) ⊈ B ( 0 , 1 ) -B(0,r)⊈B(0,1)-B(0, r) \nsubseteq B(0,1)−B(0,r)⊈B(0,1) for any r > 0 r > 0 r > 0r>0r>0, because the functions x n x n x_(n)x_{n}xn defined by
x n ( t ) = { ( n − 1 ) ( n t − 1 ) , for 0 ≤ t ≤ 1 n n n − 1 ( t − 1 n ) , for 1 n ≤ t ≤ 1 x n ( t ) = ( n − 1 ) ( n t − 1 ) ,       for  0 ≤ t ≤ 1 n n n − 1 t − 1 n ,       for  1 n ≤ t ≤ 1 x_(n)(t)={[(n-1)(nt-1)","," for "0 <= t <= (1)/(n)],[(n)/(n-1)(t-(1)/(n))","," for "(1)/(n) <= t <= 1]:}x_{n}(t)= \begin{cases}(n-1)(n t-1), & \text { for } 0 \leq t \leq \frac{1}{n} \\ \frac{n}{n-1}\left(t-\frac{1}{n}\right), & \text { for } \frac{1}{n} \leq t \leq 1\end{cases}xn(t)={(n−1)(nt−1), for 0≤t≤1nnn−1(t−1n), for 1n≤t≤1
is in B p ( 0 , 1 ) B p ( 0 , 1 ) B_(p)(0,1)B_{p}(0,1)Bp(0,1) for all n n nnn, while p ( − x n ) = n − 1 > r p − x n = n − 1 > r p(-x_(n))=n-1 > rp\left(-x_{n}\right)=n-1>rp(−xn)=n−1>r for large n n nnn (see [2]).
The topology τ p τ p tau_(p)\tau_{p}τp could not be Hausdorff even if p p ppp is an asymmetric norm on X X XXX. Necessary and sufficient conditions in order that τ p τ p tau_(p)\tau_{p}τp be Hausdorff were given in [8].
In this paper we shall study some best approximation problems in spaces with asymmetric seminorm. The significance of asymmetric norms for best approximation problems was first emphasized by Krein and Nudel'man (see [10, Ch. 9, § 5]). In the spaces C ( T ) C ( T ) C(T)C(T)C(T) and L r , 1 ≤ r < ∞ L r , 1 ≤ r < ∞ L_(r),1 <= r < ooL_{r}, 1 \leq r<\inftyLr,1≤r<∞, one considers asymmetric norms defined through a pair w = ( w + , w − ) w = w + , w − w=(w_(+),w_(-))w=\left(w_{+}, w_{-}\right)w=(w+,w−)of nonnegative upper semicontinuous functions, called weight functions, via the formula ‖ f | w = max { w + ( t ) f + ( t ) − w − ( t ) f − ( t ) : t ∈ T } ‖ f w = max w + ( t ) f + ( t ) − w − ( t ) f − ( t ) : t ∈ T ||f|_(w)=max{w_(+)(t)f_(+)(t)-w_(-)(t)f_(-)(t):t in T}\|\left. f\right|_{w}=\max \left\{w_{+}(t) f_{+}(t)-w_{-}(t) f_{-}(t): t \in T\right\}‖f|w=max{w+(t)f+(t)−w−(t)f−(t):t∈T}, where f + , f − f + , f − f_(+),f_(-)f_{+}, f_{-}f+,f−are the positive, respectively negative part of f f fff. In the case of the spaces L r L r L_(r)L_{r}Lr the above formula is adapted to the corresponding integral norm. The approximation in such spaces is called sign-sensitive approximation and it is studied in a lot of papers, following the ideas from the symmetric case (see [1, 4, 5, 9, 18, 19, 20] and the references given in these papers). There are also papers concerning existence results, mainly generic, for best approximation in abstract spaces with asymmetric norms, see [3, 11, 12, 17].
In [14, 16], there were studied the relations between the existence of best approximation and uniqueness of the extension of bounded linear functionals on spaces with asymmetric norm. In 13, 15] similar problems were considered
within the framework of spaces of semi-Lipschitz functions on an asymmetric metric space (called quasi-metric space).
The present paper is concerned with the characterization of the elements of best approximation in a subspace Y Y YYY of a space with asymmetric norm in terms of some linear functionals vanishing on Y Y YYY. The approach is based on some extension results, proved in Section 3, for bounded linear functionals on such spaces. Also, the well known formula for the distance to a hyperplane in a normed space is extended to the nonsymmetric case. For the case of normed spaces see 21 .

2. BOUNDED LINEAR MAPPINGS AND THE DUAL OF A SPACE WITH ASYMMETRIC SEMINORM

Let ( X , p ) ( X , p ) (X,p)(X, p)(X,p) and ( Y , q ) ( Y , q ) (Y,q)(Y, q)(Y,q) be spaces with asymmetric seminorms and A : X → Y A : X → Y A:X rarr YA: X \rightarrow YA:X→Y a linear mapping. The mapping A A AAA is called bounded (or semi-Lipschitz) if there exists L ≥ 0 L ≥ 0 L >= 0L \geq 0L≥0 such that
(2.1) q ( A x ) ≤ L p ( x ) , for all x ∈ X (2.1) q ( A x ) ≤ L p ( x ) ,  for all  x ∈ X {:(2.1)q(Ax) <= Lp(x)","quad" for all "x in X:}\begin{equation*} q(A x) \leq L p(x), \quad \text { for all } x \in X \tag{2.1} \end{equation*}(2.1)q(Ax)≤Lp(x), for all x∈X
It was shown in [6] (see also [7]) that the boundedness of the linear mapping A A AAA is equivalent to its continuity with respect to the topologies τ p τ p tau_(p)\tau_{p}τp and τ q τ q tau_(q)\tau_{q}τq. Denoting by L b ( X , Y ) L b ( X , Y ) L_(b)(X,Y)L_{b}(X, Y)Lb(X,Y) the set of all bounded linear mapping from ( X , p X , p X,pX, pX,p ) to ( Y , q ) ( Y , q ) (Y,q)(Y, q)(Y,q), it turns out that L b ( X , Y ) L b ( X , Y ) L_(b)(X,Y)L_{b}(X, Y)Lb(X,Y) is not necessarily a linear space but rather a convex cone in the vector space L a ( X , Y ) L a ( X , Y ) L_(a)(X,Y)L_{a}(X, Y)La(X,Y) of all linear mappings from X X XXX to Y Y YYY, i.e.
λ ≥ 0 and A , B ∈ L b ( X , Y ) ⇒ A + B ∈ L b ( X , Y ) and λ A ∈ L b ( X , Y ) . λ ≥ 0  and  A , B ∈ L b ( X , Y ) ⇒ A + B ∈ L b ( X , Y )  and  λ A ∈ L b ( X , Y ) . lambda >= 0" and "A,B inL_(b)(X,Y)=>A+B inL_(b)(X,Y)" and "lambda A inL_(b)(X,Y).\lambda \geq 0 \text { and } A, B \in L_{b}(X, Y) \Rightarrow A+B \in L_{b}(X, Y) \text { and } \lambda A \in L_{b}(X, Y) .λ≥0 and A,B∈Lb(X,Y)⇒A+B∈Lb(X,Y) and λA∈Lb(X,Y).
For instance, in the space X = C 0 [ 0 , 1 ] X = C 0 [ 0 , 1 ] X=C_(0)[0,1]X=C_{0}[0,1]X=C0[0,1] considered in the previous section, the linear functional φ ( x ) = x ( 1 ) , x ∈ C 0 [ 0 , 1 ] φ ( x ) = x ( 1 ) , x ∈ C 0 [ 0 , 1 ] varphi(x)=x(1),x inC_(0)[0,1]\varphi(x)=x(1), x \in C_{0}[0,1]φ(x)=x(1),x∈C0[0,1], is bounded because φ ( x ) ≤ p ( x ) , x ∈ X φ ( x ) ≤ p ( x ) , x ∈ X varphi(x) <= p(x),x in X\varphi(x) \leq p(x), x \in Xφ(x)≤p(x),x∈X, but the functional − φ − φ -varphi-\varphi−φ is not bounded. Taking x n ( t ) = 1 − n t n − 1 x n ( t ) = 1 − n t n − 1 x_(n)(t)=1-nt^(n-1)x_{n}(t)=1-n t^{n-1}xn(t)=1−ntn−1 we have p ( x n ) = 1 p x n = 1 p(x_(n))=1p\left(x_{n}\right)=1p(xn)=1 for all n n nnn, but − φ ( x n ) = n − 1 → ∞ − φ x n = n − 1 → ∞ -varphi(x_(n))=n-1rarr oo-\varphi\left(x_{n}\right)=n-1 \rightarrow \infty−φ(xn)=n−1→∞ for n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞ (see [2]).
As in the case of bounded linear mapping between normed linear spaces, one can define an asymmetric seminorm on L b ( X , Y ) L b ( X , Y ) L_(b)(X,Y)L_{b}(X, Y)Lb(X,Y) by the formula
(2.2) ‖ A ∣= sup { q ( A x ) : x ∈ X , p ( x ) ≤ 1 } . (2.2) ‖ A ∣= sup { q ( A x ) : x ∈ X , p ( x ) ≤ 1 } . {:(2.2)||A∣=s u p{q(Ax):x in X","p(x) <= 1}.:}\begin{equation*} \| A \mid=\sup \{q(A x): x \in X, p(x) \leq 1\} . \tag{2.2} \end{equation*}(2.2)‖A∣=sup{q(Ax):x∈X,p(x)≤1}.
It is not difficult to see that ‖ ⋅ ∣ ‖ ⋅ ∣ ||*∣\| \cdot \mid‖⋅∣ is an asymmetric seminorm on the cone L b ( X , Y ) L b ( X , Y ) L_(b)(X,Y)L_{b}(X, Y)Lb(X,Y) which has properties similar to those of the usual norm:
Proposition 2.1. Let ( X , p X , p X,pX, pX,p ) and ( Y , q Y , q Y,qY, qY,q ) be spaces with asymmetric seminorms and A ∈ L b ( X , Y ) A ∈ L b ( X , Y ) A inL_(b)(X,Y)A \in L_{b}(X, Y)A∈Lb(X,Y). Then
  1. ∀ x ∈ X q ( A x ) ≤ ‖ A ∣ ⋅ p ( x ) ∀ x ∈ X q ( A x ) ≤ ‖ A ∣ ⋅ p ( x ) AA x in X quad q(Ax) <= ||A∣*p(x)\forall x \in X \quad q(A x) \leq \| A \mid \cdot p(x)∀x∈Xq(Ax)≤‖A∣⋅p(x), and ‖ A ∣ ‖ A ∣ ||A∣\| A \mid‖A∣ is the smallest number L ≥ 0 L ≥ 0 L >= 0L \geq 0L≥0 for which the inequality (2.1) holds.
  2. ‖ A | = sup { q ( A x ) p ( x ) : x ∈ X , p ( x ) > 0 } ‖ A = sup q ( A x ) p ( x ) : x ∈ X , p ( x ) > 0 ||A|=s u p{(q(Ax))/(p(x)):x in X,p(x) > 0}:}\| A \left\lvert\,=\sup \left\{\frac{q(A x)}{p(x)}: x \in X, p(x)>0\right\}\right.‖A|=sup{q(Ax)p(x):x∈X,p(x)>0}.
Proof. 1) If p ( x ) = 0 p ( x ) = 0 p(x)=0p(x)=0p(x)=0 then, by the boundedness of A A AAA, q ( A x ) = 0 = ‖ A ∣ p ( x ) q ( A x ) = 0 = ‖ A ∣ p ( x ) q(Ax)=0=||A∣p(x)q(A x)=0=\| A \mid p(x)q(Ax)=0=‖A∣p(x). If p ( x ) > 0 p ( x ) > 0 p(x) > 0p(x)>0p(x)>0 then p ( ( 1 / p ( x ) ) x ) = 1 p ( ( 1 / p ( x ) ) x ) = 1 p((1//p(x))x)=1p((1 / p(x)) x)=1p((1/p(x))x)=1 and
q ( A ( 1 p ( x ) x ) ) ≤ ‖ A ‖ ⟺ q ( A x ) ≤ ‖ A | ⋅ p ( x ) . q A 1 p ( x ) x ≤ ‖ A ‖ ⟺ q ( A x ) ≤ ‖ A ⋅ p ( x ) . {:q(A((1)/(p(x))x)) <= ||A||Longleftrightarrow q(Ax) <= ||A|*p(x).\left.q\left(A\left(\frac{1}{p(x)} x\right)\right) \leq\|A\| \Longleftrightarrow q(A x) \leq \| A \right\rvert\, \cdot p(x) .q(A(1p(x)x))≤‖A‖⟺q(Ax)≤‖A|⋅p(x).
If q ( A x ) ≤ L p ( x ) , ∀ x ∈ X q ( A x ) ≤ L p ( x ) , ∀ x ∈ X q(Ax) <= Lp(x),AA x in Xq(A x) \leq L p(x), \forall x \in Xq(Ax)≤Lp(x),∀x∈X, for some L ≥ 0 L ≥ 0 L >= 0L \geq 0L≥0, then q ( A x ) ≤ L q ( A x ) ≤ L q(Ax) <= Lq(A x) \leq Lq(Ax)≤L for all x ∈ X x ∈ X x in Xx \in Xx∈X with p ( x ) ≤ 1 p ( x ) ≤ 1 p(x) <= 1p(x) \leq 1p(x)≤1, implying ‖ A ∣≤ L ‖ A ∣≤ L ||A∣≤L\| A \mid \leq L‖A∣≤L.
2) Follows from the facts that q ( A x ) = 0 q ( A x ) = 0 q(Ax)=0q(A x)=0q(Ax)=0 if p ( x ) = 0 p ( x ) = 0 p(x)=0p(x)=0p(x)=0 and
1 p ( x ) q ( A x ) = q ( A ( 1 p ( x ) ) ) 1 p ( x ) q ( A x ) = q A 1 p ( x ) (1)/(p(x))q(Ax)=q(A((1)/(p(x))))\frac{1}{p(x)} q(A x)=q\left(A\left(\frac{1}{p(x)}\right)\right)1p(x)q(Ax)=q(A(1p(x)))
if p ( x ) > 0 p ( x ) > 0 p(x) > 0p(x)>0p(x)>0.

Bounded linear functionals on a space with asymmetric norm

As in the case of normed spaces, the cone of bounded linear functional on a space with asymmetric seminorm will play a key role in various problems concerning these spaces.
On the space R R R\mathbb{R}R of real numbers, consider the asymmetric seminorm u ( α ) = max { α , 0 } u ( α ) = max { α , 0 } u(alpha)=max{alpha,0}u(\alpha)= \max \{\alpha, 0\}u(α)=max{α,0} and denote by R u R u R_(u)\mathbb{R}_{u}Ru the space R R R\mathbb{R}R equipped with the topology τ u τ u tau_(u)\tau_{u}τu generated by u u uuu. It is the topology generated by the intervals of the form ( − ∞ , a ) , a ∈ R ( − ∞ , a ) , a ∈ R (-oo,a),a inR(-\infty, a), a \in \mathbb{R}(−∞,a),a∈R. A neighborhood basis of a point a ∈ R u a ∈ R u a inR_(u)a \in \mathbb{R}_{u}a∈Ru is formed by the intervals ( − ∞ , a + ϵ ) , ϵ > 0 ( − ∞ , a + ϵ ) , ϵ > 0 (-oo,a+epsilon),epsilon > 0(-\infty, a+\epsilon), \epsilon>0(−∞,a+ϵ),ϵ>0. The seminorm conjugate to u u uuu is u ¯ ( α ) = u ( − α ) = max { − α , 0 } u ¯ ( α ) = u ( − α ) = max { − α , 0 } bar(u)(alpha)=u(-alpha)=max{-alpha,0}\bar{u}(\alpha)=u(-\alpha)= \max \{-\alpha, 0\}u¯(α)=u(−α)=max{−α,0}, and u s ( α ) = max { u ( α ) , u ( − α ) } = | α | u s ( α ) = max { u ( α ) , u ( − α ) } = | α | u^(s)(alpha)=max{u(alpha),u(-alpha)}=|alpha|u^{s}(\alpha)=\max \{u(\alpha), u(-\alpha)\}=|\alpha|us(α)=max{u(α),u(−α)}=|α|. The continuity of a linear functional φ : ( X , p ) → ( R , u ) φ : ( X , p ) → ( R , u ) varphi:(X,p)rarr(R,u)\varphi:(X, p) \rightarrow(\mathbb{R}, u)φ:(X,p)→(R,u) with respect to the topologies τ p τ p tau_(p)\tau_{p}τp and τ u τ u tau_(u)\tau_{u}τu will be called ( p , u ) ( p , u ) (p,u)(p, u)(p,u)-continuity. It is easily seen that the ( p , u ) ( p , u ) (p,u)(p, u)(p,u)-continuity of a linear functional φ : ( X , τ p ) → ( R , u ) φ : X , τ p → ( R , u ) varphi:(X,tau_(p))rarr(R,u)\varphi:\left(X, \tau_{p}\right) \rightarrow(\mathbb{R}, u)φ:(X,τp)→(R,u) is equivalent to its upper semi-continuity as a functional from ( X , τ p ) X , τ p (X,tau_(p))\left(X, \tau_{p}\right)(X,τp) to ( R , | | ) ( R , | | ) (R,||)(\mathbb{R},| |)(R,||). This is equivalent to the fact that for every α ∈ R α ∈ R alpha inR\alpha \in \mathbb{R}α∈R the set { x ∈ X : φ ( x ) ≥ α } { x ∈ X : φ ( x ) ≥ α } {x in X:varphi(x) >= alpha}\{x \in X: \varphi(x) \geq \alpha\}{x∈X:φ(x)≥α} is closed in ( X , τ p ) X , τ p (X,tau_(p))\left(X, \tau_{p}\right)(X,τp) and has consequence the fact that, for every τ p τ p tau_(p)\tau_{p}τp-compact subset Y Y YYY of X X XXX, the functional φ φ varphi\varphiφ is upper bounded on Y Y YYY and there exists y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y such that φ ( y 0 ) = sup φ ( Y ) φ y 0 = sup φ ( Y ) varphi(y_(0))=s u p varphi(Y)\varphi\left(y_{0}\right)=\sup \varphi(Y)φ(y0)=supφ(Y). Also, the linear functional φ φ varphi\varphiφ is ( p , u ) ( p , u ) (p,u)(p, u)(p,u)-continuous if and only if it is p p ppp-bounded, i.e. there exists L ≥ 0 L ≥ 0 L >= 0L \geq 0L≥0 such that
(2.3) ∀ x ∈ X φ ( x ) ≤ L p ( x ) (2.3) ∀ x ∈ X φ ( x ) ≤ L p ( x ) {:(2.3)AA x in X quad varphi(x) <= Lp(x):}\begin{equation*} \forall x \in X \quad \varphi(x) \leq L p(x) \tag{2.3} \end{equation*}(2.3)∀x∈Xφ(x)≤Lp(x)
Denote by X p b X p b X_(p)^(b)X_{p}^{b}Xpb ( X b X b X^(b)X^{b}Xb when it is no danger of confusion) the cone of all bounded linear functionals on the space with asymmetric seminorm ( X , p X , p X,pX, pX,p ) and call it the asymmetric dual of ( X , p ) ( X , p ) (X,p)(X, p)(X,p). It follows that the functional
‖ φ | = ‖ φ | p = sup { φ ( x ) : x ∈ X , p ( x ) ≤ 1 } φ | = ‖ φ | p = sup { φ ( x ) : x ∈ X , p ( x ) ≤ 1 } ||varphi|=||varphi|_(p)=s u p{varphi(x):x in X,p(x) <= 1}:}\left\|\varphi|=\| \varphi|_{p}=\sup \{\varphi(x): x \in X, p(x) \leq 1\}\right.‖φ|=‖φ|p=sup{φ(x):x∈X,p(x)≤1}
is an asymmetric seminorm on X b X b X^(b)X^{b}Xb.
We shall need the following simple properties of this seminorm.
Proposition 2.2. If φ φ varphi\varphiφ is a bounded linear functional on a space with asymmetric seminorm ( X , p X , p X,pX, pX,p ), with p ≠ 0 p ≠ 0 p!=0p \neq 0p≠0, then:
  1. ‖ φ ∣ ‖ φ ∣ ||varphi∣\| \varphi \mid‖φ∣ is the smallest of the numbers L ≥ 0 L ≥ 0 L >= 0L \geq 0L≥0 for which the inequality (2.3) holds;
  2. We have:
‖ φ ∣ = sup { φ ( x ) / p ( x ) : x ∈ X , p ( x ) > 0 } = sup { φ ( x ) : x ∈ X , p ( x ) < 1 } = sup { φ ( x ) : x ∈ X , p ( x ) = 1 } ‖ φ ∣ = sup { φ ( x ) / p ( x ) : x ∈ X , p ( x ) > 0 } = sup { φ ( x ) : x ∈ X , p ( x ) < 1 } = sup { φ ( x ) : x ∈ X , p ( x ) = 1 } {:[||varphi∣=s u p{varphi(x)//p(x):x in X","p(x) > 0}],[=s u p{varphi(x):x in X","p(x) < 1}],[=s u p{varphi(x):x in X","p(x)=1}]:}\begin{aligned} \| \varphi \mid & =\sup \{\varphi(x) / p(x): x \in X, p(x)>0\} \\ & =\sup \{\varphi(x): x \in X, p(x)<1\} \\ & =\sup \{\varphi(x): x \in X, p(x)=1\} \end{aligned}‖φ∣=sup{φ(x)/p(x):x∈X,p(x)>0}=sup{φ(x):x∈X,p(x)<1}=sup{φ(x):x∈X,p(x)=1}
  1. If φ ≠ 0 φ ≠ 0 varphi!=0\varphi \neq 0φ≠0 then | | φ | > 0 | | φ | > 0 ||varphi| > 0||\varphi|>0||φ|>0.
Also, if φ ≠ 0 φ ≠ 0 varphi!=0\varphi \neq 0φ≠0 and φ ( x 0 ) = ‖ φ ∣ φ x 0 = ‖ φ ∣ varphi(x_(0))=||varphi∣\varphi\left(x_{0}\right)=\| \varphi \midφ(x0)=‖φ∣ for some x 0 ∈ B p x 0 ∈ B p x_(0)inB_(p)x_{0} \in B_{p}x0∈Bp, then p ( x 0 ) = 1 p x 0 = 1 p(x_(0))=1p\left(x_{0}\right)=1p(x0)=1.
Proof. We shall prove the assertions 2) and 3), the first one being a particular case of the corresponding result for linear mappings.
Supposing c := sup { φ ( x ) : p ( x ) < 1 } < ‖ φ ∣ c := sup { φ ( x ) : p ( x ) < 1 } < ‖ φ ∣ c:=s u p{varphi(x):p(x) < 1} < ||varphi∣c:=\sup \{\varphi(x): p(x)<1\}<\| \varphi \midc:=sup{φ(x):p(x)<1}<‖φ∣, let x 0 ∈ X , p ( x 0 ) = 1 x 0 ∈ X , p x 0 = 1 x_(0)in X,p(x_(0))=1x_{0} \in X, p\left(x_{0}\right)=1x0∈X,p(x0)=1, be such that c < φ ( x 0 ) ≤ ‖ φ ∣ c < φ x 0 ≤ ‖ φ ∣ c < varphi(x_(0)) <= ||varphi∣c<\varphi\left(x_{0}\right) \leq \| \varphi \midc<φ(x0)≤‖φ∣. Then there is a number α , 0 < α < 1 α , 0 < α < 1 alpha,0 < alpha < 1\alpha, 0<\alpha<1α,0<α<1, such that φ ( α x 0 ) = α φ ( x 0 ) > c φ α x 0 = α φ x 0 > c varphi(alphax_(0))=alpha varphi(x_(0)) > c\varphi\left(\alpha x_{0}\right)=\alpha \varphi\left(x_{0}\right)>cφ(αx0)=αφ(x0)>c, in contradiction to the definition of c c ccc.
Let's show now that ‖ φ ∣= sup { φ ( x ) : p ( x ) = 1 } ‖ φ ∣= sup { φ ( x ) : p ( x ) = 1 } ||varphi∣=s u p{varphi(x):p(x)=1}\| \varphi \mid=\sup \{\varphi(x): p(x)=1\}‖φ∣=sup{φ(x):p(x)=1}. Suppose again that β := sup { φ ( x ) : p ( x ) = 1 } < ‖ φ ∣ β := sup { φ ( x ) : p ( x ) = 1 } < ‖ φ ∣ beta:=s u p{varphi(x):p(x)=1} < ||varphi∣\beta:=\sup \{\varphi(x): p(x)=1\}<\| \varphi \midβ:=sup{φ(x):p(x)=1}<‖φ∣, and choose x 0 ∈ X x 0 ∈ X x_(0)in Xx_{0} \in Xx0∈X such that p ( x 0 ) < 1 p x 0 < 1 p(x_(0)) < 1p\left(x_{0}\right)<1p(x0)<1 and φ ( x 0 ) > β φ x 0 > β varphi(x_(0)) > beta\varphi\left(x_{0}\right)>\betaφ(x0)>β. Putting x 1 = ( 1 / p ( x 0 ) ) x 0 x 1 = 1 / p x 0 x 0 x_(1)=(1//p(x_(0)))x_(0)x_{1}=\left(1 / p\left(x_{0}\right)\right) x_{0}x1=(1/p(x0))x0, it follows p ( x 1 ) = 1 p x 1 = 1 p(x_(1))=1p\left(x_{1}\right)=1p(x1)=1 and
φ ( x 1 ) = 1 p ( x 0 ) φ ( x 0 ) > φ ( x 0 ) > β , φ x 1 = 1 p x 0 φ x 0 > φ x 0 > β , varphi(x_(1))=(1)/(p(x_(0)))varphi(x_(0)) > varphi(x_(0)) > beta,\varphi\left(x_{1}\right)=\frac{1}{p\left(x_{0}\right)} \varphi\left(x_{0}\right)>\varphi\left(x_{0}\right)>\beta,φ(x1)=1p(x0)φ(x0)>φ(x0)>β,
a contradiction.
3) Because φ ( x ) ≤ ‖ φ ∣ p ( x ) φ ( x ) ≤ ‖ φ ∣ p ( x ) varphi(x) <= ||varphi∣p(x)\varphi(x) \leq \| \varphi \mid p(x)φ(x)≤‖φ∣p(x), the equality ‖ φ ∣= 0 ‖ φ ∣= 0 ||varphi∣=0\| \varphi \mid=0‖φ∣=0 implies φ ( x ) ≤ 0 φ ( x ) ≤ 0 varphi(x) <= 0\varphi(x) \leq 0φ(x)≤0 and − φ ( x ) = φ ( − x ) ≤ 0 − φ ( x ) = φ ( − x ) ≤ 0 -varphi(x)=varphi(-x) <= 0-\varphi(x)=\varphi(-x) \leq 0−φ(x)=φ(−x)≤0, i.e. φ ( x ) = 0 φ ( x ) = 0 varphi(x)=0\varphi(x)=0φ(x)=0 for all x ∈ X x ∈ X x in Xx \in Xx∈X.
Suppose now that that for φ ≠ 0 φ ≠ 0 varphi!=0\varphi \neq 0φ≠0 there exists x 0 ∈ X x 0 ∈ X x_(0)in Xx_{0} \in Xx0∈X, with 0 < p ( x 0 ) < 1 0 < p x 0 < 1 0 < p(x_(0)) < 10<p\left(x_{0}\right)<10<p(x0)<1, such that φ ( x 0 ) = ‖ φ ∣ φ x 0 = ‖ φ ∣ varphi(x_(0))=||varphi∣\varphi\left(x_{0}\right)=\| \varphi \midφ(x0)=‖φ∣. Then α := 1 / p ( x 0 ) > 1 , x 1 = α x 0 ∈ B p α := 1 / p x 0 > 1 , x 1 = α x 0 ∈ B p alpha:=1//p(x_(0)) > 1,x_(1)=alphax_(0)inB_(p)\alpha:=1 / p\left(x_{0}\right)>1, x_{1}=\alpha x_{0} \in B_{p}α:=1/p(x0)>1,x1=αx0∈Bp and
‖ φ | ≥ φ ( x 1 ) = α φ ( x 0 ) = α ‖ φ | φ ≥ φ x 1 = α φ x 0 = α ‖ φ ||varphi| >= varphi(x_(1))=alpha varphi(x_(0))=alpha||varphi|:}\left\|\varphi\left|\geq \varphi\left(x_{1}\right)=\alpha \varphi\left(x_{0}\right)=\alpha \| \varphi\right|\right.‖φ|≥φ(x1)=αφ(x0)=α‖φ|
a contradiction, because ‖ φ ∣> 0 ‖ φ ∣> 0 ||varphi∣>0\| \varphi \mid>0‖φ∣>0.
An immediate consequence of the preceding result is the following one. We agree to call a linear functional ( p , p ¯ p , p ¯ p, bar(p)p, \bar{p}p,p¯ )-bounded if it is both p p ppp - and p ¯ p ¯ bar(p)\bar{p}p¯-bounded.
Proposition 2.3. Let φ ≠ 0 φ ≠ 0 varphi!=0\varphi \neq 0φ≠0 be a linear functional on a space with asymmetric seminorm ( X , p ) ( X , p ) (X,p)(X, p)(X,p).
  1. If φ φ varphi\varphiφ is ( p , p ¯ ) ( p , p ¯ ) (p, bar(p))(p, \bar{p})(p,p¯)-bounded then
φ ( r B p ′ ) = ( − r ‖ φ | p ¯ , r ‖ φ | p ) and φ ( r B p ¯ ′ ) = ( − r ‖ φ | p , r ‖ φ | p ¯ ) φ r B p ′ = − r φ p ¯ , r φ p  and  φ r B p ¯ ′ = − r φ p , r φ p ¯ varphi(rB_(p)^('))=(-r|| varphi|_( bar(p)),r||varphi|_(p))quad" and "quad varphi(rB_( bar(p))^('))=(-r|| varphi|_(p),r||varphi|_( bar(p)))\varphi\left(r B_{p}^{\prime}\right)=\left(-\left.r\left\|\left.\varphi\right|_{\bar{p}}, r\right\| \varphi\right|_{p}\right) \quad \text { and } \quad \varphi\left(r B_{\bar{p}}^{\prime}\right)=\left(-\left.r\left\|\left.\varphi\right|_{p}, r\right\| \varphi\right|_{\bar{p}}\right)φ(rBp′)=(−r‖φ|p¯,r‖φ|p) and φ(rBp¯′)=(−r‖φ|p,r‖φ|p¯)
where B p ′ = { x ∈ X : p ( x ) < 1 } , B p ¯ ′ = { x ∈ X : p ¯ ( x ) < 1 } B p ′ = { x ∈ X : p ( x ) < 1 } , B p ¯ ′ = { x ∈ X : p ¯ ( x ) < 1 } B_(p)^(')={x in X:p(x) < 1},B_( bar(p))^(')={x in X: bar(p)(x) < 1}B_{p}^{\prime}=\{x \in X: p(x)<1\}, B_{\bar{p}}^{\prime}=\{x \in X: \bar{p}(x)<1\}Bp′={x∈X:p(x)<1},Bp¯′={x∈X:p¯(x)<1} and r > 0 r > 0 r > 0r>0r>0.
2) If φ φ varphi\varphiφ is p p ppp-bounded but not p ¯ p ¯ bar(p)\bar{p}p¯-bounded then
φ ( r B p ′ ) = ( − ∞ , r ‖ φ | p ) φ r B p ′ = − ∞ , r ‖ φ p varphi(rB_(p)^('))=(-oo,r|| varphi|_(p))\varphi\left(r B_{p}^{\prime}\right)=\left(-\infty, r \|\left.\varphi\right|_{p}\right)φ(rBp′)=(−∞,r‖φ|p)
Proof. Obviously that it is sufficient to give the proof only for r = 1 r = 1 r=1r=1r=1.
Suppose that φ φ varphi\varphiφ is ( p , p ¯ ) ( p , p ¯ ) (p, bar(p))(p, \bar{p})(p,p¯)-bounded. Then, by Proposition 2.2,
sup φ ( B p ′ ) = ‖ φ | p sup φ B p ′ = ‖ φ p s u p varphi(B_(p)^('))=|| varphi|_(p)\sup \varphi\left(B_{p}^{\prime}\right)=\|\left.\varphi\right|_{p}supφ(Bp′)=‖φ|p
and
inf φ ( B p ′ ) = − sup { φ ( − x ) : p ( x ) < 1 } = − sup { φ ( x ′ ) : p ( − x ′ ) < 1 } = − ‖ φ | p ¯ inf φ B p ′ = − sup { φ ( − x ) : p ( x ) < 1 } = − sup φ x ′ : p − x ′ < 1 = − ‖ φ p ¯ i n f varphi(B_(p)^('))=-s u p{varphi(-x):p(x) < 1}=-s u p{varphi(x^(')):p(-x^(')) < 1}=-|| varphi|_( bar(p))\inf \varphi\left(B_{p}^{\prime}\right)=-\sup \{\varphi(-x): p(x)<1\}=-\sup \left\{\varphi\left(x^{\prime}\right): p\left(-x^{\prime}\right)<1\right\}=-\|\left.\varphi\right|_{\bar{p}}infφ(Bp′)=−sup{φ(−x):p(x)<1}=−sup{φ(x′):p(−x′)<1}=−‖φ|p¯
Also, by the assertion 3) of Proposition 2.2, φ ( x ) < ‖ φ | p φ ( x ) < ‖ φ p varphi(x) < || varphi|_(p)\varphi(x)<\|\left.\varphi\right|_{p}φ(x)<‖φ|p and φ ( x ) > − ‖ φ | p ¯ φ ( x ) > − ‖ φ p ¯ varphi(x) > -|| varphi|_( bar(p))\varphi(x)>-\|\left.\varphi\right|_{\bar{p}}φ(x)>−‖φ|p¯ for any x ∈ B p ′ x ∈ B p ′ x inB_(p)^(')x \in B_{p}^{\prime}x∈Bp′.
Because B p ′ B p ′ B_(p)^(')B_{p}^{\prime}Bp′ is convex, φ ( B p ′ ) φ B p ′ varphi(B_(p)^('))\varphi\left(B_{p}^{\prime}\right)φ(Bp′) will be a convex subset of R R R\mathbb{R}R, that is an interval, and the above considerations show that
φ ( B p ′ ) = ( inf φ ( B p ′ ) , sup φ ( B p ′ ) ) = ( − ‖ φ | p ¯ , ‖ φ | p ) . φ B p ′ = inf φ B p ′ , sup φ B p ′ = − φ p ¯ , φ p . varphi(B_(p)^('))=(i n f varphi(B_(p)^(')),s u p varphi(B_(p)^(')))=(-|| varphi|_( bar(p)),||varphi|_(p)).\varphi\left(B_{p}^{\prime}\right)=\left(\inf \varphi\left(B_{p}^{\prime}\right), \sup \varphi\left(B_{p}^{\prime}\right)\right)=\left(-\left.\left\|\left.\varphi\right|_{\bar{p}},\right\| \varphi\right|_{p}\right) .φ(Bp′)=(infφ(Bp′),supφ(Bp′))=(−‖φ|p¯,‖φ|p).
If φ φ varphi\varphiφ is p p ppp-bounded and
sup { φ ( x ) ; p ¯ ( x ) < 1 } = ∞ sup { φ ( x ) ; p ¯ ( x ) < 1 } = ∞ s u p{varphi(x); bar(p)(x) < 1}=oo\sup \{\varphi(x) ; \bar{p}(x)<1\}=\inftysup{φ(x);p¯(x)<1}=∞
then
inf { φ ( x ′ ) : p ( x ′ ) < 1 } = − sup { φ ( x ) : p ( − x ) < 1 } = − ∞ inf φ x ′ : p x ′ < 1 = − sup { φ ( x ) : p ( − x ) < 1 } = − ∞ i n f{varphi(x^(')):p(x^(')) < 1}=-s u p{varphi(x):p(-x) < 1}=-oo\inf \left\{\varphi\left(x^{\prime}\right): p\left(x^{\prime}\right)<1\right\}=-\sup \{\varphi(x): p(-x)<1\}=-\inftyinf{φ(x′):p(x′)<1}=−sup{φ(x):p(−x)<1}=−∞
Reasoning like above, one obtains
φ ( B p ′ ) = ( − ∞ , ‖ φ | p ) . φ B p ′ = − ∞ , ‖ φ p . varphi(B_(p)^('))=(-oo,|| varphi|_(p)).\varphi\left(B_{p}^{\prime}\right)=\left(-\infty, \|\left.\varphi\right|_{p}\right) .φ(Bp′)=(−∞,‖φ|p).

3. EXTENSION RESULTS FOR BOUNDED LINEAR FUNCTIONALS

In this section we shall prove the analogs of some well known extension results for linear functional in normed spaces. The main tool is the Hahn-Banach extension theorem for linear functionals dominated by sublinear functionals.
Throughout this section ( X , p X , p X,pX, pX,p ) will be a space with asymmetric seminorm.
Proposition 3.1. Let Y Y YYY be a subspace of X X XXX and φ 0 : Y → R φ 0 : Y → R varphi_(0):Y rarrR\varphi_{0}: Y \rightarrow \mathbb{R}φ0:Y→R a bounded linear functional. Then there exists a bounded linear functional φ : X → R φ : X → R varphi:X rarrR\varphi: X \rightarrow \mathbb{R}φ:X→R such that
φ | Y = φ 0 and ‖ φ | = ‖ φ 0 | . φ Y = φ 0  and  φ = ‖ φ 0 . varphi|_(Y)=varphi_(0)quad" and "quad||varphi|=||varphi_(0)|.:}\left.\varphi\right|_{Y}=\varphi_{0} \quad \text { and } \quad\left\|\varphi\left|=\| \varphi_{0}\right| .\right.φ|Y=φ0 and ‖φ|=‖φ0|.
Proof. The functional q ( x ) = ‖ φ 0 ∣ p ( x ) , x ∈ X q ( x ) = ‖ φ 0 ∣ p ( x ) , x ∈ X q(x)=||varphi_(0)∣p(x),x in Xq(x)=\| \varphi_{0} \mid p(x), x \in Xq(x)=‖φ0∣p(x),x∈X, is sublinear and φ 0 ( y ) ≤ q ( y ) , y ∈ Y φ 0 ( y ) ≤ q ( y ) , y ∈ Y varphi_(0)(y) <= q(y),y in Y\varphi_{0}(y) \leq q(y), y \in Yφ0(y)≤q(y),y∈Y. By the Hahn-Banach extension theorem there exists a linear functional φ : X → R φ : X → R varphi:X rarrR\varphi: X \rightarrow \mathbb{R}φ:X→R such that
φ | Y = φ 0 and ∀ x ∈ X φ ( x ) ≤ ‖ φ 0 ∣ p ( x ) . φ Y = φ 0  and  ∀ x ∈ X φ ( x ) ≤ ‖ φ 0 ∣ p ( x ) . varphi|_(Y)=varphi_(0)quad" and "quad AA x in X varphi(x) <= ||varphi_(0)∣p(x).\left.\varphi\right|_{Y}=\varphi_{0} \quad \text { and } \quad \forall x \in X \varphi(x) \leq \| \varphi_{0} \mid p(x) .φ|Y=φ0 and ∀x∈Xφ(x)≤‖φ0∣p(x).
The second of the above relations implies that φ φ varphi\varphiφ is bounded and ‖ φ | ≤ ‖ φ 0 | φ ≤ ‖ φ 0 ||varphi| <= ||varphi_(0)|:}\left\|\varphi\left|\leq \| \varphi_{0}\right|\right.‖φ|≤‖φ0|. Since
‖ φ | = sup { φ ( x ) : x ∈ X , p ( x ) ≤ 1 } ≥ sup { φ ( y ) : y ∈ Y , p ( y ) ≤ 1 } = ‖ φ 0 | φ = sup { φ ( x ) : x ∈ X , p ( x ) ≤ 1 } ≥ sup { φ ( y ) : y ∈ Y , p ( y ) ≤ 1 } = ‖ φ 0 ||varphi|=s u p{varphi(x):x in X,p(x) <= 1} >= s u p{varphi(y):y in Y,p(y) <= 1}=||varphi_(0)|:}\left\|\varphi\left|=\sup \{\varphi(x): x \in X, p(x) \leq 1\} \geq \sup \{\varphi(y): y \in Y, p(y) \leq 1\}=\| \varphi_{0}\right|\right.‖φ|=sup{φ(x):x∈X,p(x)≤1}≥sup{φ(y):y∈Y,p(y)≤1}=‖φ0|
it follows ‖ φ | = ‖ φ 0 | φ = ‖ φ 0 ||varphi|=||varphi_(0)|:}\left\|\varphi\left|=\| \varphi_{0}\right|\right.‖φ|=‖φ0|.
We agree to call a functional φ φ varphi\varphiφ satisfying the conclusions of Proposition 3.1 a norm preserving extension of φ 0 φ 0 varphi_(0)\varphi_{0}φ0.
Proposition 3.2. If x 0 x 0 x_(0)x_{0}x0 is a point in X X XXX such that p ( x 0 ) > 0 p x 0 > 0 p(x_(0)) > 0p\left(x_{0}\right)>0p(x0)>0 then there exists a bounded linear functional φ : X → R φ : X → R varphi:X rarrR\varphi: X \rightarrow \mathbb{R}φ:X→R such that
‖ φ ∣= 1 and φ ( x 0 ) = p ( x 0 ) . ‖ φ ∣= 1  and  φ x 0 = p x 0 . ||varphi∣=1quad" and "quad varphi(x_(0))=p(x_(0)).\| \varphi \mid=1 \quad \text { and } \quad \varphi\left(x_{0}\right)=p\left(x_{0}\right) .‖φ∣=1 and φ(x0)=p(x0).
Proof. Let Y := R x 0 Y := R x 0 Y:=Rx_(0)Y:=\mathbb{R} x_{0}Y:=Rx0 and let φ 0 : Y → R φ 0 : Y → R varphi_(0):Y rarrR\varphi_{0}: Y \rightarrow \mathbb{R}φ0:Y→R be defined by φ 0 ( t x 0 ) = t p ( x 0 ) φ 0 t x 0 = t p x 0 varphi_(0)(tx_(0))=tp(x_(0))\varphi_{0}\left(t x_{0}\right)=t p\left(x_{0}\right)φ0(tx0)=tp(x0), t ∈ R t ∈ R t inRt \in \mathbb{R}t∈R. It follows that φ 0 φ 0 varphi_(0)\varphi_{0}φ0 is linear and
φ 0 ( t x 0 ) = t p ( x 0 ) = p ( t x 0 ) φ 0 t x 0 = t p x 0 = p t x 0 varphi_(0)(tx_(0))=tp(x_(0))=p(tx_(0))\varphi_{0}\left(t x_{0}\right)=t p\left(x_{0}\right)=p\left(t x_{0}\right)φ0(tx0)=tp(x0)=p(tx0)
for t > 0 t > 0 t > 0t>0t>0 and
φ 0 ( t x 0 ) = t p ( x 0 ) ≤ 0 ≤ p ( t x 0 ) φ 0 t x 0 = t p x 0 ≤ 0 ≤ p t x 0 varphi_(0)(tx_(0))=tp(x_(0)) <= 0 <= p(tx_(0))\varphi_{0}\left(t x_{0}\right)=t p\left(x_{0}\right) \leq 0 \leq p\left(t x_{0}\right)φ0(tx0)=tp(x0)≤0≤p(tx0)
for t ≤ 0 t ≤ 0 t <= 0t \leq 0t≤0. Again, the Hahn-Banach extension theorem yields a linear functional φ : X → R φ : X → R varphi:X rarrR\varphi: X \rightarrow \mathbb{R}φ:X→R, such that
φ | Y = φ 0 and ∀ x ∈ X φ ( x ) ≤ p ( x ) . φ Y = φ 0  and  ∀ x ∈ X φ ( x ) ≤ p ( x ) . varphi|_(Y)=varphi_(0)quad" and "quad AA x in X varphi(x) <= p(x).\left.\varphi\right|_{Y}=\varphi_{0} \quad \text { and } \quad \forall x \in X \varphi(x) \leq p(x) .φ|Y=φ0 and ∀x∈Xφ(x)≤p(x).
It follows ‖ φ ∣≤ 1 , φ ( x 0 ) = p ( x 0 ) ‖ φ ∣≤ 1 , φ x 0 = p x 0 ||varphi∣≤1,varphi(x_(0))=p(x_(0))\| \varphi \mid \leq 1, \varphi\left(x_{0}\right)=p\left(x_{0}\right)‖φ∣≤1,φ(x0)=p(x0), and, since p ( ( 1 / p ( x 0 ) ) x 0 ) = 1 p 1 / p x 0 x 0 = 1 p((1//p(x_(0)))x_(0))=1p\left(\left(1 / p\left(x_{0}\right)\right) x_{0}\right)=1p((1/p(x0))x0)=1,
‖ φ | ≥ φ ( 1 p ( x 0 ) x 0 ) = 1 ‖ φ ≥ φ 1 p x 0 x 0 = 1 ||varphi| >= varphi((1)/(p(x_(0)))x_(0))=1:}\| \varphi \left\lvert\, \geq \varphi\left(\frac{1}{p\left(x_{0}\right)} x_{0}\right)=1\right.‖φ|≥φ(1p(x0)x0)=1
i.e. | | φ | = 1 | | φ | = 1 ||varphi|=1||\varphi|=1||φ|=1.
This last proposition has as consequence the following useful result.
Corollary 3.3. If p ( x 0 ) > 0 p x 0 > 0 p(x_(0)) > 0p\left(x_{0}\right)>0p(x0)>0, then
p ( x 0 ) = sup { φ ( x 0 ) : φ ∈ X b , ‖ φ ∣≤ 1 } . p x 0 = sup φ x 0 : φ ∈ X b , ‖ φ ∣≤ 1 . p(x_(0))=s u p{varphi(x_(0)):varphi inX^(b),||varphi∣≤1}.p\left(x_{0}\right)=\sup \left\{\varphi\left(x_{0}\right): \varphi \in X^{b}, \| \varphi \mid \leq 1\right\} .p(x0)=sup{φ(x0):φ∈Xb,‖φ∣≤1}.
Proof. Denote by s s sss the supremum in the right hand side of the above formula. Since φ ( x 0 ) ≤ ‖ φ ∣ p ( x 0 ) ≤ p ( x 0 ) φ x 0 ≤ ‖ φ ∣ p x 0 ≤ p x 0 varphi(x_(0)) <= ||varphi∣p(x_(0)) <= p(x_(0))\varphi\left(x_{0}\right) \leq \| \varphi \mid p\left(x_{0}\right) \leq p\left(x_{0}\right)φ(x0)≤‖φ∣p(x0)≤p(x0) for every φ ∈ X b , ‖ φ ∣≤ 1 φ ∈ X b , ‖ φ ∣≤ 1 varphi inX^(b),||varphi∣≤1\varphi \in X^{b}, \| \varphi \mid \leq 1φ∈Xb,‖φ∣≤1, it follows s ≤ p ( x 0 ) s ≤ p x 0 s <= p(x_(0))s \leq p\left(x_{0}\right)s≤p(x0). Choosing φ ∈ X p b φ ∈ X p b varphi inX_(p)^(b)\varphi \in X_{p}^{\mathrm{b}}φ∈Xpb as in Proposition 3.2, it follows p ( x 0 ) = φ ( x 0 ) ≤ s p x 0 = φ x 0 ≤ s p(x_(0))=varphi(x_(0)) <= sp\left(x_{0}\right)=\varphi\left(x_{0}\right) \leq sp(x0)=φ(x0)≤s.
The next extension result involves the distance from a point to a set in an asymmetric seminormed space. Let Y Y YYY be a nonempty subset of an asymmetric seminormed space ( X , p ) ( X , p ) (X,p)(X, p)(X,p). Due to the asymmetry of the seminorm p p ppp we have to consider two distances from a point x ∈ X x ∈ X x in Xx \in Xx∈X to Y Y YYY, namely
(3.1) d p ( x , Y ) = inf { p ( y − x ) : y ∈ Y } (3.1) d p ( x , Y ) = inf { p ( y − x ) : y ∈ Y } {:(3.1)d_(p)(x","Y)=i n f{p(y-x):y in Y}:}\begin{equation*} d_{p}(x, Y)=\inf \{p(y-x): y \in Y\} \tag{3.1} \end{equation*}(3.1)dp(x,Y)=inf{p(y−x):y∈Y}
and
(3.2) d p ( Y , x ) = inf { p ( x − y ) : y ∈ Y } . (3.2) d p ( Y , x ) = inf { p ( x − y ) : y ∈ Y } . {:(3.2)d_(p)(Y","x)=i n f{p(x-y):y in Y}.:}\begin{equation*} d_{p}(Y, x)=\inf \{p(x-y): y \in Y\} . \tag{3.2} \end{equation*}(3.2)dp(Y,x)=inf{p(x−y):y∈Y}.
Observe that d p ( Y , x ) = d p ¯ ( x , Y ) d p ( Y , x ) = d p ¯ ( x , Y ) d_(p)(Y,x)=d_( bar(p))(x,Y)d_{p}(Y, x)=d_{\bar{p}}(x, Y)dp(Y,x)=dp¯(x,Y), where p ¯ p ¯ bar(p)\bar{p}p¯ is the seminorm conjugate to p p ppp.
Proposition 3.4. Let Y Y YYY be a subspace of a space with asymmetric seminorm ( X , p X , p X,pX, pX,p ) and x 0 ∈ X x 0 ∈ X x_(0)in Xx_{0} \in Xx0∈X. Denote by d ¯ d ¯ bar(d)\bar{d}d¯ the distance d p ¯ ( x 0 , Y ) d p ¯ x 0 , Y d_( bar(p))(x_(0),Y)d_{\bar{p}}\left(x_{0}, Y\right)dp¯(x0,Y) and suppose d ¯ > 0 d ¯ > 0 bar(d) > 0\bar{d}>0d¯>0.
Then there exists a p p ppp-bounded linear functional φ : X → R φ : X → R varphi:X rarrR\varphi: X \rightarrow \mathbb{R}φ:X→R such that
( i ) φ | Y = 0 , ( i i ) ‖ φ ∣= 1 , and ( i i i ) φ ( x 0 ) = d ¯ ( i ) φ Y = 0 , ( i i ) ‖ φ ∣= 1 ,  and  ( i i i ) φ x 0 = d ¯ (i)varphi|_(Y)=0,quad(ii)||varphi∣=1,quad" and "quad(iii)varphi(x_(0))= bar(d)\left.(i) \varphi\right|_{Y}=0, \quad(i i) \| \varphi \mid=1, \quad \text { and } \quad(i i i) \varphi\left(x_{0}\right)=\bar{d}(i)φ|Y=0,(ii)‖φ∣=1, and (iii)φ(x0)=d¯
If d = d p ( x 0 , Y ) > 0 d = d p x 0 , Y > 0 d=d_(p)(x_(0),Y) > 0d=d_{p}\left(x_{0}, Y\right)>0d=dp(x0,Y)>0 then there exists ψ ∈ X p b ψ ∈ X p b psi inX_(p)^(b)\psi \in X_{p}^{b}ψ∈Xpb such that
( j ) ψ | Y = 0 , ( j j ) ‖ ψ ∣= 1 , ( j j j ) ψ ( − x 0 ) = d ( j ) ψ Y = 0 , ( j j ) ‖ ψ ∣= 1 , ( j j j ) ψ − x 0 = d (j)psi|_(Y)=0,quad(jj)||psi∣=1,quad(jjj)psi(-x_(0))=d\left.(j) \psi\right|_{Y}=0, \quad(j j) \| \psi \mid=1, \quad(j j j) \psi\left(-x_{0}\right)=d(j)ψ|Y=0,(jj)‖ψ∣=1,(jjj)ψ(−x0)=d
Proof. Suppose that d ¯ = d p ¯ ( x 0 , Y ) > 0 d ¯ = d p ¯ x 0 , Y > 0 bar(d)=d_( bar(p))(x_(0),Y) > 0\bar{d}=d_{\bar{p}}\left(x_{0}, Y\right)>0d¯=dp¯(x0,Y)>0, so that x 0 ∉ Y x 0 ∉ Y x_(0)!in Yx_{0} \notin Yx0∉Y. Let Z := Y + ˙ R x 0 Z := Y + ˙ R x 0 Z:=Y+^(˙)Rx_(0)Z:=Y \dot{+} \mathbb{R} x_{0}Z:=Y+˙Rx0 ( + ˙ + ˙ +^(˙)\dot{+}+˙ stands for the direct sum) and let ψ 0 : Z → R ψ 0 : Z → R psi_(0):Z rarrR\psi_{0}: Z \rightarrow \mathbb{R}ψ0:Z→R be defined by
ψ 0 ( y + t x 0 ) = t , y ∈ Y , t ∈ R ψ 0 y + t x 0 = t , y ∈ Y , t ∈ R psi_(0)(y+tx_(0))=t,y in Y,t inR\psi_{0}\left(y+t x_{0}\right)=t, y \in Y, t \in \mathbb{R}ψ0(y+tx0)=t,y∈Y,t∈R
Then ψ 0 ψ 0 psi_(0)\psi_{0}ψ0 is linear, ψ 0 ( y ) = 0 , ∀ y ∈ Y ψ 0 ( y ) = 0 , ∀ y ∈ Y psi_(0)(y)=0,AA y in Y\psi_{0}(y)=0, \forall y \in Yψ0(y)=0,∀y∈Y, and ψ 0 ( x 0 ) = 1 ψ 0 x 0 = 1 psi_(0)(x_(0))=1\psi_{0}\left(x_{0}\right)=1ψ0(x0)=1. For t > 0 t > 0 t > 0t>0t>0 we have
p ( y + t x 0 ) = t p ( x 0 + t − 1 y ) ≥ t d ¯ = d ¯ ⋅ ψ 0 ( y + t x 0 ) p y + t x 0 = t p x 0 + t − 1 y ≥ t d ¯ = d ¯ ⋅ ψ 0 y + t x 0 p(y+tx_(0))=tp(x_(0)+t^(-1)y) >= t bar(d)= bar(d)*psi_(0)(y+tx_(0))p\left(y+t x_{0}\right)=t p\left(x_{0}+t^{-1} y\right) \geq t \bar{d}=\bar{d} \cdot \psi_{0}\left(y+t x_{0}\right)p(y+tx0)=tp(x0+t−1y)≥td¯=d¯⋅ψ0(y+tx0)
so that
ψ 0 ( y + t x 0 ) = t ≤ 1 d p ( y + t x 0 ) . ψ 0 y + t x 0 = t ≤ 1 d p y + t x 0 . psi_(0)(y+tx_(0))=t <= (1)/(d)p(y+tx_(0)).\psi_{0}\left(y+t x_{0}\right)=t \leq \frac{1}{d} p\left(y+t x_{0}\right) .ψ0(y+tx0)=t≤1dp(y+tx0).
Since this inequality obviously holds for t ≤ 0 t ≤ 0 t <= 0t \leq 0t≤0, it follows ‖ ψ 0 ∣≤ 1 / d ¯ ‖ ψ 0 ∣≤ 1 / d ¯ ||psi_(0)∣≤1// bar(d)\| \psi_{0} \mid \leq 1 / \bar{d}‖ψ0∣≤1/d¯. Let ( y n ) y n (y_(n))\left(y_{n}\right)(yn) be a sequence in Y Y YYY such that p ( x 0 − y n ) → d ¯ p x 0 − y n → d ¯ p(x_(0)-y_(n))rarr bar(d)p\left(x_{0}-y_{n}\right) \rightarrow \bar{d}p(x0−yn)→d¯ for n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞ and p ( x 0 − y n ) > 0 p x 0 − y n > 0 p(x_(0)-y_(n)) > 0p\left(x_{0}-y_{n}\right)>0p(x0−yn)>0 for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N. Then
‖ ψ 0 | ≥ ψ 0 ( x 0 − y n p ( x 0 − y n ) ) = 1 p ( x 0 − y n ) → 1 d ‖ ψ 0 ≥ ψ 0 x 0 − y n p x 0 − y n = 1 p x 0 − y n → 1 d ||psi_(0)| >= psi_(0)((x_(0)-y_(n))/(p(x_(0)-y_(n))))=(1)/(p(x_(0)-y_(n)))rarr(1)/(d):}\| \psi_{0} \left\lvert\, \geq \psi_{0}\left(\frac{x_{0}-y_{n}}{p\left(x_{0}-y_{n}\right)}\right)=\frac{1}{p\left(x_{0}-y_{n}\right)} \rightarrow \frac{1}{d}\right.‖ψ0|≥ψ0(x0−ynp(x0−yn))=1p(x0−yn)→1d
implying ‖ ψ 0 ∣≥ 1 / d ¯ ‖ ψ 0 ∣≥ 1 / d ¯ ||psi_(0)∣≥1// bar(d)\| \psi_{0} \mid \geq 1 / \bar{d}‖ψ0∣≥1/d¯. Therefore ‖ ψ 0 ∣= 1 / d ¯ ‖ ψ 0 ∣= 1 / d ¯ ||psi_(0)∣=1// bar(d)\| \psi_{0} \mid=1 / \bar{d}‖ψ0∣=1/d¯.
If ψ ¯ : X → R ψ ¯ : X → R bar(psi):X rarrR\bar{\psi}: X \rightarrow \mathbb{R}ψ¯:X→R is a linear functional such that
ψ ¯ | Z = ψ 0 and ‖ ψ ¯ | = ‖ ψ 0 | ψ ¯ Z = ψ 0  and  ψ ¯ = ‖ ψ 0 ( bar(psi))|_(Z)=psi_(0)quad" and "quad||( bar(psi))|=||psi_(0)|:}\left.\bar{\psi}\right|_{Z}=\psi_{0} \quad \text { and } \quad\left\|\bar{\psi}\left|=\| \psi_{0}\right|\right.ψ¯|Z=ψ0 and ‖ψ¯|=‖ψ0|
then the linear functional φ = d ¯ ⋅ ψ ¯ φ = d ¯ ⋅ ψ ¯ varphi= bar(d)* bar(psi)\varphi=\bar{d} \cdot \bar{\psi}φ=d¯⋅ψ¯ fulfills all the requirements of the proposition.
Suppose now d = d p ( x 0 , Y ) > 0 d = d p x 0 , Y > 0 d=d_(p)(x_(0),Y) > 0d=d_{p}\left(x_{0}, Y\right)>0d=dp(x0,Y)>0, and let Z := Y + R x 0 Z := Y + R x 0 Z:=Y+Rx_(0)Z:=Y+\mathbb{R} x_{0}Z:=Y+Rx0. Define ψ 0 : Z → R ψ 0 : Z → R psi_(0):Z rarrR\psi_{0}: Z \rightarrow \mathbb{R}ψ0:Z→R by
ψ 0 ( y + t x 0 ) = − t ⟺ ψ 0 ( y − t x 0 ) = t for y ∈ Y and t ∈ R . ψ 0 y + t x 0 = − t ⟺ ψ 0 y − t x 0 = t  for  y ∈ Y  and  t ∈ R . psi_(0)(y+tx_(0))=-t Longleftrightarrowpsi_(0)(y-tx_(0))=t quad" for "y in Y" and "t inR.\psi_{0}\left(y+t x_{0}\right)=-t \Longleftrightarrow \psi_{0}\left(y-t x_{0}\right)=t \quad \text { for } y \in Y \text { and } t \in \mathbb{R} .ψ0(y+tx0)=−t⟺ψ0(y−tx0)=t for y∈Y and t∈R.
Then ψ 0 ψ 0 psi_(0)\psi_{0}ψ0 is linear and, for t > 0 t > 0 t > 0t>0t>0, we have
p ( y − t x 0 ) = t p ( 1 t y − x 0 ) ≥ t d = d ⋅ ψ 0 ( y − t x 0 ) , p y − t x 0 = t p 1 t y − x 0 ≥ t d = d ⋅ ψ 0 y − t x 0 , p(y-tx_(0))=tp((1)/(t)y-x_(0)) >= td=d*psi_(0)(y-tx_(0)),p\left(y-t x_{0}\right)=t p\left(\frac{1}{t} y-x_{0}\right) \geq t d=d \cdot \psi_{0}\left(y-t x_{0}\right),p(y−tx0)=tp(1ty−x0)≥td=d⋅ψ0(y−tx0),
so that
ψ 0 ( y − t x 0 ) ≤ 1 d p ( y − t x 0 ) , ψ 0 y − t x 0 ≤ 1 d p y − t x 0 , psi_(0)(y-tx_(0)) <= (1)/(d)p(y-tx_(0)),\psi_{0}\left(y-t x_{0}\right) \leq \frac{1}{d} p\left(y-t x_{0}\right),ψ0(y−tx0)≤1dp(y−tx0),
for t > 0 t > 0 t > 0t>0t>0. Since this inequality is obviously true if ψ 0 ( y − t x 0 ) = t ≤ 0 ψ 0 y − t x 0 = t ≤ 0 psi_(0)(y-tx_(0))=t <= 0\psi_{0}\left(y-t x_{0}\right)=t \leq 0ψ0(y−tx0)=t≤0, it follows that ψ 0 ψ 0 psi_(0)\psi_{0}ψ0 is bounded and ‖ ψ 0 ∣≤ 1 / d ‖ ψ 0 ∣≤ 1 / d ||psi_(0)∣≤1//d\| \psi_{0} \mid \leq 1 / d‖ψ0∣≤1/d. Choosing a sequence ( y n ′ y n ′ y_(n)^(')y_{n}^{\prime}yn′ ) in Y Y YYY such that p ( y n ′ − x 0 ) → d p y n ′ − x 0 → d p(y_(n)^(')-x_(0))rarr dp\left(y_{n}^{\prime}-x_{0}\right) \rightarrow dp(yn′−x0)→d and p ( y n ′ − x 0 ) > 0 p y n ′ − x 0 > 0 p(y_(n)^(')-x_(0)) > 0p\left(y_{n}^{\prime}-x_{0}\right)>0p(yn′−x0)>0 for all n n nnn, and reasoning like above one obtains the inequality ‖ ψ 0 ∣≥ 1 / d ‖ ψ 0 ∣≥ 1 / d ||psi_(0)∣≥1//d\| \psi_{0} \mid \geq 1 / d‖ψ0∣≥1/d, so that ‖ ψ 0 ∣= 1 / d ‖ ψ 0 ∣= 1 / d ||psi_(0)∣=1//d\| \psi_{0} \mid=1 / d‖ψ0∣=1/d. Extending ψ 0 ψ 0 psi_(0)\psi_{0}ψ0 to a functional ψ 1 ∈ X p b ψ 1 ∈ X p b psi_(1)inX_(p)^(b)\psi_{1} \in X_{p}^{b}ψ1∈Xpb of the same norm, and letting ψ = d ⋅ ψ 1 ψ = d ⋅ ψ 1 psi=d*psi_(1)\psi=d \cdot \psi_{1}ψ=d⋅ψ1, one obtains the wanted functional ψ ψ psi\psiψ.

4. APPLICATIONS TO BEST APPROXIMATION

Let ( X , p ) ( X , p ) (X,p)(X, p)(X,p) be a space with asymmetric seminorm and Y Y YYY a nonempty subset of X X XXX. By the asymmetry of the seminorm p p ppp we have to distinct two "distances" from a point x ∈ X x ∈ X x in Xx \in Xx∈X to the subset Y Y YYY, as given by (3.1) and (3.2).
Since d p ( Y , x ) = d p ¯ ( x , Y ) d p ( Y , x ) = d p ¯ ( x , Y ) d_(p)(Y,x)=d_( bar(p))(x,Y)d_{p}(Y, x)=d_{\bar{p}}(x, Y)dp(Y,x)=dp¯(x,Y), we shall use the notation d p ¯ ( x , Y ) d p ¯ ( x , Y ) d_( bar(p))(x,Y)d_{\bar{p}}(x, Y)dp¯(x,Y) for the distance (3.2).
An element y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y such that p ( x − y 0 ) = p ¯ ( y 0 − x ) = d p ¯ ( x , Y ) p x − y 0 = p ¯ y 0 − x = d p ¯ ( x , Y ) p(x-y_(0))= bar(p)(y_(0)-x)=d_( bar(p))(x,Y)p\left(x-y_{0}\right)=\bar{p}\left(y_{0}-x\right)=d_{\bar{p}}(x, Y)p(x−y0)=p¯(y0−x)=dp¯(x,Y) is called a p ¯ p ¯ bar(p)\bar{p}p¯-nearest point to x x xxx in Y Y YYY, and an element y 1 ∈ Y y 1 ∈ Y y_(1)in Yy_{1} \in Yy1∈Y such that p ( y 1 − x ) = d p ( x , Y ) p y 1 − x = d p ( x , Y ) p(y_(1)-x)=d_(p)(x,Y)p\left(y_{1}-x\right)=d_{p}(x, Y)p(y1−x)=dp(x,Y) will be called a p p ppp-nearest point to x x xxx in Y Y YYY.
By Proposition 3.4, we obtain the following characterization of p ¯ p ¯ bar(p)\bar{p}p¯-nearest points.
Proposition 4.1. Let ( X , p ) ( X , p ) (X,p)(X, p)(X,p) be a space with asymmetric seminorm, Y Y YYY a subspace of X X XXX and x 0 x 0 x_(0)x_{0}x0 a point in X X XXX such that d ¯ = d p ¯ ( x 0 , Y ) > 0 d ¯ = d p ¯ x 0 , Y > 0 bar(d)=d_( bar(p))(x_(0),Y) > 0\bar{d}=d_{\bar{p}}\left(x_{0}, Y\right)>0d¯=dp¯(x0,Y)>0.
An element y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is a p ¯ p ¯ bar(p)\bar{p}p¯-nearest point to x 0 x 0 x_(0)x_{0}x0 in Y Y YYY if and only if there exists a bounded linear functional φ : X → R φ : X → R varphi:X rarrR\varphi: X \rightarrow \mathbb{R}φ:X→R such that
(i) φ | Y = 0 φ Y = 0 varphi|_(Y)=0\left.\varphi\right|_{Y}=0φ|Y=0,
(ii) ‖ φ ∣= 1 ‖ φ ∣= 1 ||varphi∣=1\| \varphi \mid=1‖φ∣=1,
(iii) φ ( x 0 ) = p ( x 0 − y 0 ) φ x 0 = p x 0 − y 0 varphi(x_(0))=p(x_(0)-y_(0))\varphi\left(x_{0}\right)=p\left(x_{0}-y_{0}\right)φ(x0)=p(x0−y0).
Proof. Suppose that y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is such that p ( x 0 − y 0 ) = d = d p ¯ ( x 0 , Y ) > 0 p x 0 − y 0 = d = d p ¯ x 0 , Y > 0 p(x_(0)-y_(0))=d=d_( bar(p))(x_(0),Y) > 0p\left(x_{0}-y_{0}\right)=d=d_{\bar{p}}\left(x_{0}, Y\right)>0p(x0−y0)=d=dp¯(x0,Y)>0. By Proposition 3.4, there exists φ ∈ X p b , ‖ φ ∣= 1 φ ∈ X p b , ‖ φ ∣= 1 varphi inX_(p)^(b),||varphi∣=1\varphi \in X_{p}^{b}, \| \varphi \mid=1φ∈Xpb,‖φ∣=1, such that φ | Y = 0 φ Y = 0 varphi|_(Y)=0\left.\varphi\right|_{Y}=0φ|Y=0 and φ ( x 0 ) = d = p ( x 0 − y 0 ) φ x 0 = d = p x 0 − y 0 varphi(x_(0))=d=p(x_(0)-y_(0))\varphi\left(x_{0}\right)=d=p\left(x_{0}-y_{0}\right)φ(x0)=d=p(x0−y0).
Conversely, if for y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y there exists φ ∈ X p b φ ∈ X p b varphi inX_(p)^(b)\varphi \in X_{p}^{b}φ∈Xpb satisfying the conditions (i)-(iii), then for every y ∈ Y y ∈ Y y in Yy \in Yy∈Y,
p ( x 0 − y ) ≥ φ ( x 0 − y ) = φ ( x 0 − y 0 ) = p ( x 0 − y 0 ) , p x 0 − y ≥ φ x 0 − y = φ x 0 − y 0 = p x 0 − y 0 , p(x_(0)-y) >= varphi(x_(0)-y)=varphi(x_(0)-y_(0))=p(x_(0)-y_(0)),p\left(x_{0}-y\right) \geq \varphi\left(x_{0}-y\right)=\varphi\left(x_{0}-y_{0}\right)=p\left(x_{0}-y_{0}\right),p(x0−y)≥φ(x0−y)=φ(x0−y0)=p(x0−y0),
implying p ( x 0 − y 0 ) = d p ¯ ( x 0 , Y ) p x 0 − y 0 = d p ¯ x 0 , Y p(x_(0)-y_(0))=d_( bar(p))(x_(0),Y)p\left(x_{0}-y_{0}\right)=d_{\bar{p}}\left(x_{0}, Y\right)p(x0−y0)=dp¯(x0,Y).
Another consequence of Proposition 3.4 is the following duality formula for best approximation:
Proposition 4.2. Let Y Y YYY be a subspace of a space with asymmetric seminorm ( X , p ) ( X , p ) (X,p)(X, p)(X,p). If d p ( Y , x 0 ) > 0 d p Y , x 0 > 0 d_(p)(Y,x_(0)) > 0d_{p}\left(Y, x_{0}\right)>0dp(Y,x0)>0 then the following duality formula holds:
d p ( Y , x 0 ) = sup { ψ ( x 0 ) : ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 } d p Y , x 0 = sup ψ x 0 : ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 d_(p)(Y,x_(0))=s u p{psi(x_(0)):psi inY^(_|_),||psi∣≤1}d_{p}\left(Y, x_{0}\right)=\sup \left\{\psi\left(x_{0}\right): \psi \in Y^{\perp}, \| \psi \mid \leq 1\right\}dp(Y,x0)=sup{ψ(x0):ψ∈Y⊥,‖ψ∣≤1}
where Y ⊥ = { φ ∈ X p b : φ | Y = 0 } Y ⊥ = φ ∈ X p b : φ Y = 0 Y^(_|_)={varphi inX_(p)^(b): varphi|_(Y)=0}Y^{\perp}=\left\{\varphi \in X_{p}^{b}:\left.\varphi\right|_{Y}=0\right\}Y⊥={φ∈Xpb:φ|Y=0}.
Proof. For any ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 psi inY^(_|_),||psi∣≤1\psi \in Y^{\perp}, \| \psi \mid \leq 1ψ∈Y⊥,‖ψ∣≤1 and any y ∈ Y y ∈ Y y in Yy \in Yy∈Y, we have:
ψ ( x 0 ) = ψ ( x 0 − y ) ≤ p ( x 0 − y ) , ψ x 0 = ψ x 0 − y ≤ p x 0 − y , psi(x_(0))=psi(x_(0)-y) <= p(x_(0)-y),\psi\left(x_{0}\right)=\psi\left(x_{0}-y\right) \leq p\left(x_{0}-y\right),ψ(x0)=ψ(x0−y)≤p(x0−y),
implying sup { ψ ( x 0 ) : ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 } ≤ d p ( Y , x 0 ) sup ψ x 0 : ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 ≤ d p Y , x 0 s u p{psi(x_(0)):psi inY^(_|_),||psi∣≤1} <= d_(p)(Y,x_(0))\sup \left\{\psi\left(x_{0}\right): \psi \in Y^{\perp}, \| \psi \mid \leq 1\right\} \leq d_{p}\left(Y, x_{0}\right)sup{ψ(x0):ψ∈Y⊥,‖ψ∣≤1}≤dp(Y,x0). If we choose ψ ψ psi\psiψ to be the functional φ φ varphi\varphiφ given by Proposition 3.4, then we obtain the reverse inequality: d p ( Y , x 0 ) = φ ( x 0 ) ≤ sup { ψ ( x 0 ) : ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 } d p Y , x 0 = φ x 0 ≤ sup ψ x 0 : ψ ∈ Y ⊥ , ‖ ψ ∣≤ 1 d_(p)(Y,x_(0))=varphi(x_(0)) <= s u p{psi(x_(0)):psi inY^(_|_),||psi∣≤1}d_{p}\left(Y, x_{0}\right)=\varphi\left(x_{0}\right) \leq \sup \left\{\psi\left(x_{0}\right): \psi \in Y^{\perp}, \| \psi \mid \leq 1\right\}dp(Y,x0)=φ(x0)≤sup{ψ(x0):ψ∈Y⊥,‖ψ∣≤1}.

The distance to a hyperplane

The well known formula for the distance to a closed hyperplane in a normed space has an analog in spaces with asymmetric seminorm. Remark that in this case we have to work with both of the distances d p d p d_(p)d_{p}dp and d p ¯ d p ¯ d_( bar(p))d_{\bar{p}}dp¯ given by 3.1 and (3.2).
Proposition 4.3. Let ( X , p X , p X,pX, pX,p ) be a space with asymmetric seminorm, φ ∈ X p b , φ ≠ 0 , c ∈ R φ ∈ X p b , φ ≠ 0 , c ∈ R varphi inX_(p)^(b),varphi!=0,c inR\varphi \in X_{p}^{b}, \varphi \neq 0, c \in \mathbb{R}φ∈Xpb,φ≠0,c∈R,
H = { x ∈ X : φ ( x ) = c } H = { x ∈ X : φ ( x ) = c } H={x in X:varphi(x)=c}H=\{x \in X: \varphi(x)=c\}H={x∈X:φ(x)=c}
the hyperplane corresponding to φ φ varphi\varphiφ and c c ccc, and
H < = { x ∈ X : φ ( x ) < c } and H > = { x ∈ X : φ ( x ) > c } H < = { x ∈ X : φ ( x ) < c }  and  H > = { x ∈ X : φ ( x ) > c } H^( < )={x in X:varphi(x) < c}quad" and "quadH^( > )={x in X:varphi(x) > c}H^{<}=\{x \in X: \varphi(x)<c\} \quad \text { and } \quad H^{>}=\{x \in X: \varphi(x)>c\}H<={x∈X:φ(x)<c} and H>={x∈X:φ(x)>c}
the open half-spaces determined by H H HHH.
  1. We have
(4.1) d p ¯ ( x 0 , H ) = φ ( x 0 ) − c ‖ φ ‖ (4.1) d p ¯ x 0 , H = φ x 0 − c ‖ φ ‖ {:(4.1)d_( bar(p))(x_(0),H)=(varphi(x_(0))-c)/(||varphi||):}\begin{equation*} d_{\bar{p}}\left(x_{0}, H\right)=\frac{\varphi\left(x_{0}\right)-c}{\|\varphi\|} \tag{4.1} \end{equation*}(4.1)dp¯(x0,H)=φ(x0)−c‖φ‖
for every x 0 ∈ H > x 0 ∈ H > x_(0)inH^( > )x_{0} \in H^{>}x0∈H>, and
(4.2) d p ( x 0 , H ) = c − φ ( x 0 ) ‖ φ ‖ (4.2) d p x 0 , H = c − φ x 0 ‖ φ ‖ {:(4.2)d_(p)(x_(0),H)=(c-varphi(x_(0)))/(||varphi||):}\begin{equation*} d_{p}\left(x_{0}, H\right)=\frac{c-\varphi\left(x_{0}\right)}{\|\varphi\|} \tag{4.2} \end{equation*}(4.2)dp(x0,H)=c−φ(x0)‖φ‖
for every x 0 ∈ H < x 0 ∈ H < x_(0)inH^( < )x_{0} \in H^{<}x0∈H<.
2) If there exists an element z 0 ∈ X z 0 ∈ X z_(0)in Xz_{0} \in Xz0∈X with p ( z 0 ) = 1 p z 0 = 1 p(z_(0))=1p\left(z_{0}\right)=1p(z0)=1 such that φ ( z 0 ) = ‖ φ ∣ φ z 0 = ‖ φ ∣ varphi(z_(0))=||varphi∣\varphi\left(z_{0}\right)= \| \varphi \midφ(z0)=‖φ∣, then every element in H > H > H^( > )H^{>}H>has a p ¯ p ¯ bar(p)\bar{p}p¯-nearest point in H H HHH and every element in H < H < H^( < )H^{<}H<has a p p ppp-nearest point in H H HHH.
If there is an element x 0 ∈ H > x 0 ∈ H > x_(0)inH^( > )x_{0} \in H^{>}x0∈H>having a p ¯ p ¯ bar(p)\bar{p}p¯-nearest point in H H HHH, or there is an element x 0 ′ ∈ H < x 0 ′ ∈ H < x_(0)^(')inH^( < )x_{0}^{\prime} \in H^{<}x0′∈H<having a p p ppp-nearest point in H H HHH, then there exists an element z 0 ∈ X , p ( z 0 ) = 1 z 0 ∈ X , p z 0 = 1 z_(0)in X,p(z_(0))=1z_{0} \in X, p\left(z_{0}\right)=1z0∈X,p(z0)=1, such that φ ( z 0 ) = ‖ φ ∣ φ z 0 = ‖ φ ∣ varphi(z_(0))=||varphi∣\varphi\left(z_{0}\right)=\| \varphi \midφ(z0)=‖φ∣. It follows that, in this case, every element in H > H > H^( > )H^{>}H>has a p ¯ p ¯ bar(p)\bar{p}p¯-nearest point in H H HHH, and every element in H < H < H^( < )H^{<}H<has a p p ppp-nearest point in H H HHH.
Proof. Let x 0 ∈ H > x 0 ∈ H > x_(0)inH^( > )x_{0} \in H^{>}x0∈H>. Then, for every h ∈ H , φ ( h ) = c h ∈ H , φ ( h ) = c h in H,varphi(h)=ch \in H, \varphi(h)=ch∈H,φ(h)=c, so that
φ ( x 0 ) − c = φ ( x 0 − h ) ≤ ‖ φ ∣ p ( x 0 − h ) , φ x 0 − c = φ x 0 − h ≤ ‖ φ ∣ p x 0 − h , varphi(x_(0))-c=varphi(x_(0)-h) <= ||varphi∣p(x_(0)-h),\varphi\left(x_{0}\right)-c=\varphi\left(x_{0}-h\right) \leq \| \varphi \mid p\left(x_{0}-h\right),φ(x0)−c=φ(x0−h)≤‖φ∣p(x0−h),
implying
d p ¯ ( x 0 , H ) ≥ φ ( x 0 ) − c ‖ φ ∣ d p ¯ x 0 , H ≥ φ x 0 − c ‖ φ ∣ d_( bar(p))(x_(0),H) >= (varphi(x_(0))-c)/(||varphi∣)d_{\bar{p}}\left(x_{0}, H\right) \geq \frac{\varphi\left(x_{0}\right)-c}{\| \varphi \mid}dp¯(x0,H)≥φ(x0)−c‖φ∣
By the assertion 2) of Proposition 2.2, there exists a sequence ( z n z n z_(n)z_{n}zn ) in X X XXX with p ( z n ) = 1 p z n = 1 p(z_(n))=1p\left(z_{n}\right)=1p(zn)=1, such that φ ( z n ) → ‖ φ ∣ φ z n → ‖ φ ∣ varphi(z_(n))rarr||varphi∣\varphi\left(z_{n}\right) \rightarrow \| \varphi \midφ(zn)→‖φ∣ and φ ( z n ) > 0 φ z n > 0 varphi(z_(n)) > 0\varphi\left(z_{n}\right)>0φ(zn)>0 for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N. Then
h n := x 0 − φ ( x 0 ) − c φ ( z n ) z n h n := x 0 − φ x 0 − c φ z n z n h_(n):=x_(0)-(varphi(x_(0))-c)/(varphi(z_(n)))z_(n)h_{n}:=x_{0}-\frac{\varphi\left(x_{0}\right)-c}{\varphi\left(z_{n}\right)} z_{n}hn:=x0−φ(x0)−cφ(zn)zn
belongs to H H HHH and
d p ¯ ( x 0 , H ) ≤ p ( x 0 − h n ) = φ ( x 0 ) − c φ ( z n ) → φ ( x 0 ) − c ‖ φ ‖ . d p ¯ x 0 , H ≤ p x 0 − h n = φ x 0 − c φ z n → φ x 0 − c ‖ φ ‖ . d_( bar(p))(x_(0),H) <= p(x_(0)-h_(n))=(varphi(x_(0))-c)/(varphi(z_(n)))rarr(varphi(x_(0))-c)/(||varphi||).d_{\bar{p}}\left(x_{0}, H\right) \leq p\left(x_{0}-h_{n}\right)=\frac{\varphi\left(x_{0}\right)-c}{\varphi\left(z_{n}\right)} \rightarrow \frac{\varphi\left(x_{0}\right)-c}{\|\varphi\|} .dp¯(x0,H)≤p(x0−hn)=φ(x0)−cφ(zn)→φ(x0)−c‖φ‖.
It follows d p ¯ ( x 0 , H ) ≥ ( φ ( x 0 ) − c ) / ‖ φ ∣ d p ¯ x 0 , H ≥ φ x 0 − c / ‖ φ ∣ d_( bar(p))(x_(0),H) >= (varphi(x_(0))-c)//||varphi∣d_{\bar{p}}\left(x_{0}, H\right) \geq\left(\varphi\left(x_{0}\right)-c\right) / \| \varphi \middp¯(x0,H)≥(φ(x0)−c)/‖φ∣, so that formula 4.1) holds.
To prove (4.2), observe that for h ∈ H h ∈ H h in Hh \in Hh∈H,
c − φ ( x 0 ′ ) = φ ( h − x 0 ′ ) ≤ ‖ φ ∣ p ( h − x 0 ′ ) , c − φ x 0 ′ = φ h − x 0 ′ ≤ ‖ φ ∣ p h − x 0 ′ , c-varphi(x_(0)^('))=varphi(h-x_(0)^(')) <= ||varphi∣p(h-x_(0)^(')),c-\varphi\left(x_{0}^{\prime}\right)=\varphi\left(h-x_{0}^{\prime}\right) \leq \| \varphi \mid p\left(h-x_{0}^{\prime}\right),c−φ(x0′)=φ(h−x0′)≤‖φ∣p(h−x0′),
implying
d p ( x 0 ′ , H ) ≥ c − φ ( x 0 ′ ) ‖ φ ∣ d p x 0 ′ , H ≥ c − φ x 0 ′ ‖ φ ∣ d_(p)(x_(0)^('),H) >= (c-varphi(x_(0)^(')))/(||varphi∣)d_{p}\left(x_{0}^{\prime}, H\right) \geq \frac{c-\varphi\left(x_{0}^{\prime}\right)}{\| \varphi \mid}dp(x0′,H)≥c−φ(x0′)‖φ∣
If the sequence ( z n z n z_(n)z_{n}zn ) is as above then
h n ′ := c − φ ( x 0 ′ ) φ ( z n ) z n + x 0 ′ h n ′ := c − φ x 0 ′ φ z n z n + x 0 ′ h_(n)^('):=(c-varphi(x_(0)^(')))/(varphi(z_(n)))z_(n)+x_(0)^(')h_{n}^{\prime}:=\frac{c-\varphi\left(x_{0}^{\prime}\right)}{\varphi\left(z_{n}\right)} z_{n}+x_{0}^{\prime}hn′:=c−φ(x0′)φ(zn)zn+x0′
belongs to H H HHH and
d p ( x 0 ′ , H ) ≤ p ( h n ′ − x 0 ′ ) = c − φ ( x 0 ′ ) φ ( z n ) → c − φ ( x 0 ′ ) ‖ φ ‖ , d p x 0 ′ , H ≤ p h n ′ − x 0 ′ = c − φ x 0 ′ φ z n → c − φ x 0 ′ ‖ φ ‖ , d_(p)(x_(0)^('),H) <= p(h_(n)^(')-x_(0)^('))=(c-varphi(x_(0)^(')))/(varphi(z_(n)))rarr(c-varphi(x_(0)^(')))/(||varphi||),d_{p}\left(x_{0}^{\prime}, H\right) \leq p\left(h_{n}^{\prime}-x_{0}^{\prime}\right)=\frac{c-\varphi\left(x_{0}^{\prime}\right)}{\varphi\left(z_{n}\right)} \rightarrow \frac{c-\varphi\left(x_{0}^{\prime}\right)}{\|\varphi\|},dp(x0′,H)≤p(hn′−x0′)=c−φ(x0′)φ(zn)→c−φ(x0′)‖φ‖,
so that d p ( x 0 ′ , H ) ≥ ( c − φ ( x 0 ′ ) ) / ‖ φ d p x 0 ′ , H ≥ c − φ x 0 ′ / ‖ φ d_(p)(x_(0)^('),H) >= (c-varphi(x_(0)^(')))//||varphid_{p}\left(x_{0}^{\prime}, H\right) \geq\left(c-\varphi\left(x_{0}^{\prime}\right)\right) / \| \varphidp(x0′,H)≥(c−φ(x0′))/‖φ, and formula (4.2) holds too.
2) Let z 0 ∈ X z 0 ∈ X z_(0)in Xz_{0} \in Xz0∈X be such that p ( z 0 ) = 1 p z 0 = 1 p(z_(0))=1p\left(z_{0}\right)=1p(z0)=1 and φ ( z 0 ) = ‖ φ ∣ φ z 0 = ‖ φ ∣ varphi(z_(0))=||varphi∣\varphi\left(z_{0}\right)=\| \varphi \midφ(z0)=‖φ∣. Then, for x 0 ∈ H > x 0 ∈ H > x_(0)inH^( > )x_{0} \in H^{>}x0∈H> and x 0 ′ ∈ H < x 0 ′ ∈ H < x_(0)^(')inH^( < )x_{0}^{\prime} \in H^{<}x0′∈H<, the elements
h 0 := x 0 − φ ( x 0 ) − c φ ( z 0 ) z 0 and h 0 ′ := c − φ ( x 0 ′ ) φ ( z 0 ) z 0 + x 0 ′ h 0 := x 0 − φ x 0 − c φ z 0 z 0  and  h 0 ′ := c − φ x 0 ′ φ z 0 z 0 + x 0 ′ h_(0):=x_(0)-(varphi(x_(0))-c)/(varphi(z_(0)))z_(0)quad" and "quadh_(0)^('):=(c-varphi(x_(0)^(')))/(varphi(z_(0)))z_(0)+x_(0)^(')h_{0}:=x_{0}-\frac{\varphi\left(x_{0}\right)-c}{\varphi\left(z_{0}\right)} z_{0} \quad \text { and } \quad h_{0}^{\prime}:=\frac{c-\varphi\left(x_{0}^{\prime}\right)}{\varphi\left(z_{0}\right)} z_{0}+x_{0}^{\prime}h0:=x0−φ(x0)−cφ(z0)z0 and h0′:=c−φ(x0′)φ(z0)z0+x0′
belong to H H HHH,
p ( x 0 − h 0 ) = φ ( x 0 ) − c ‖ φ ‖ = d p ¯ ( x 0 , H ) and p ( h n ′ − x 0 ′ ) = c − φ ( x 0 ′ ) ‖ φ ∣ = d p ( x 0 ′ , H ) . p x 0 − h 0 = φ x 0 − c ‖ φ ‖ = d p ¯ x 0 , H  and  p h n ′ − x 0 ′ = c − φ x 0 ′ ‖ φ ∣ = d p x 0 ′ , H . p(x_(0)-h_(0))=(varphi(x_(0))-c)/(||varphi||)=d_( bar(p))(x_(0),H)quad" and "quad p(h_(n)^(')-x_(0)^('))=(c-varphi(x_(0)^(')))/(||varphi∣)=d_(p)(x_(0)^('),H).p\left(x_{0}-h_{0}\right)=\frac{\varphi\left(x_{0}\right)-c}{\|\varphi\|}=d_{\bar{p}}\left(x_{0}, H\right) \quad \text { and } \quad p\left(h_{n}^{\prime}-x_{0}^{\prime}\right)=\frac{c-\varphi\left(x_{0}^{\prime}\right)}{\| \varphi \mid}=d_{p}\left(x_{0}^{\prime}, H\right) .p(x0−h0)=φ(x0)−c‖φ‖=dp¯(x0,H) and p(hn′−x0′)=c−φ(x0′)‖φ∣=dp(x0′,H).
If an element x 0 ∈ H > x 0 ∈ H > x_(0)inH^( > )x_{0} \in H^{>}x0∈H>has a p ¯ p ¯ bar(p)\bar{p}p¯-nearest point h 0 ∈ H h 0 ∈ H h_(0)in Hh_{0} \in Hh0∈H, then
p ( x 0 − h 0 ) = d p ¯ ( x 0 , H ) = φ ( x 0 ) − c ‖ φ ‖ = φ ( x 0 − h 0 ) ‖ φ ‖ . p x 0 − h 0 = d p ¯ x 0 , H = φ x 0 − c ‖ φ ‖ = φ x 0 − h 0 ‖ φ ‖ . p(x_(0)-h_(0))=d_( bar(p))(x_(0),H)=(varphi(x_(0))-c)/(||varphi||)=(varphi(x_(0)-h_(0)))/(||varphi||).p\left(x_{0}-h_{0}\right)=d_{\bar{p}}\left(x_{0}, H\right)=\frac{\varphi\left(x_{0}\right)-c}{\|\varphi\|}=\frac{\varphi\left(x_{0}-h_{0}\right)}{\|\varphi\|} .p(x0−h0)=dp¯(x0,H)=φ(x0)−c‖φ‖=φ(x0−h0)‖φ‖.
It follows that z 0 = ( x 0 − h 0 ) / p ( x 0 − h 0 ) z 0 = x 0 − h 0 / p x 0 − h 0 z_(0)=(x_(0)-h_(0))//p(x_(0)-h_(0))z_{0}=\left(x_{0}-h_{0}\right) / p\left(x_{0}-h_{0}\right)z0=(x0−h0)/p(x0−h0) satisfies the conditions p ( z 0 ) = 1 p z 0 = 1 p(z_(0))=1p\left(z_{0}\right)=1p(z0)=1 and φ ( z 0 ) = ‖ φ φ z 0 = ‖ φ varphi(z_(0))=||varphi\varphi\left(z_{0}\right)=\| \varphiφ(z0)=‖φ.
If an element x 0 ′ ∈ H < x 0 ′ ∈ H < x_(0)^(')inH^( < )x_{0}^{\prime} \in H^{<}x0′∈H<has a p p ppp-nearest point h 0 ′ h 0 ′ h_(0)^(')h_{0}^{\prime}h0′ in H H HHH, then z 0 ′ = ( h 0 ′ − x 0 ′ ) / p ( h 0 ′ − x 0 ′ ) z 0 ′ = h 0 ′ − x 0 ′ / p h 0 ′ − x 0 ′ z_(0)^(')=(h_(0)^(')-:}{:x_(0)^('))//p(h_(0)^(')-x_(0)^('))z_{0}^{\prime}=\left(h_{0}^{\prime}-\right. \left.x_{0}^{\prime}\right) / p\left(h_{0}^{\prime}-x_{0}^{\prime}\right)z0′=(h0′−x0′)/p(h0′−x0′) satisfies p ( z 0 ′ ) = 1 p z 0 ′ = 1 p(z_(0)^('))=1p\left(z_{0}^{\prime}\right)=1p(z0′)=1 and φ ( z 0 ′ ) = ‖ φ ∣ φ z 0 ′ = ‖ φ ∣ varphi(z_(0)^('))=||varphi∣\varphi\left(z_{0}^{\prime}\right)=\| \varphi \midφ(z0′)=‖φ∣.

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Received by the editors: March 28, 2004.
  1. *"Babeş-Bolyai" University, Faculty of Mathematics and Computer Science, 400084 ClujNapoca, Romania, e-mail: scobzas@math.ubbcluj.ro.
    † † ^(†){ }^{\dagger}† "T. Popoviciu" Institute of Numerical Analysis, O.P 1, C.P. 68, Cluj-Napoca, Romania, e-mail: cmustata@ictp.acad.ro.

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