Rev. Anal. Num´er. Th´eor. Approx., vol. 33 (2004) no. 1, pp. 39–50 ictp.acad.ro/jnaat
EXTENSION OF BOUNDED LINEAR FUNCTIONALS AND BEST APPROXIMATION IN SPACES WITH ASYMMETRIC NORM
S. COBZAS¸∗and C. MUST ˘AT¸ A†
Abstract. The present paper is concerned with the characterization of the ele- ments of best approximation in a subspaceY of a space with asymmetric norm, in terms of some linear functionals vanishing onY. 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
LetX be a real vector space. Anasymmetric seminorm on X is a positive sublinear functional p:X→[0,∞), i.e. p satisfies the conditions:
(AN1) p(x)≥0,
(AN2) p(tx) =tp(x), t≥0, (AN3) p(x+y)≤p(x) +p(y),
for allx, y∈X.The function ¯p:X →[0,∞) defined by ¯p(x) =p(−x), x∈X, is another positive sublinear functional on X, called theconjugate ofp, and
ps(x) = max{p(x), p(−x)}, x∈X, is a seminorm on X. The inequalities
|p(x)−p(y)| ≤ps(x−y) and |¯p(x)−p(y)| ≤¯ ps(x−y)
hold for allx, y∈X. If the seminormpsis a norm on X then we say thatp is anasymmetric normonX. This means that, beside (AN1)–(AN3), it satisfies also the condition
(AN4) p(x) = 0 and p(−x) = 0 imply x= 0.
The pair (X, p), whereXis a linear space andpis an asymmetric seminorm on X is called a space with asymmetric seminorm, respectively a space with asymmetric norm, ifp is an asymmetric norm.
∗“Babe¸s-Bolyai” University, Faculty of Mathematics and Computer Science, 400084 Cluj- Napoca, Romania, e-mail: [email protected].
†“T. Popoviciu” Institute of Numerical Analysis, O.P 1, C.P. 68, Cluj-Napoca, Romania, e-mail: [email protected].
An asymmetric seminormpgenerates a topologyτp onX, having as a basis of neighborhoods of a point x∈X the openp-balls
B0p(x, r) ={x0 ∈X :p(x0−x)< r}, r >0.
The family of closed p-balls
Bp(x, r) ={x0 ∈X :p(x0−x)≤r}, r >0, generates the same topology.
Denote byBp =Bp(0,1) the closed unit ball of (X, p) and byBp0 =Bp0(0,1) its open unit ball.
The topologyτp is translation invariant, i.e. the addition + :X×X →X is continuous, but the multiplication by scalars · :R×X → X need not be continuous. For instance, in the space
C0[0,1] =nx∈C[0,1] : Z 1
0
x(t)dt= 0o
with the asymmetric seminorm p(x) = maxx([0,1]), the multiplication by scalars is not continuous at t0 =−1 andx0 = 0. Indeed, the ball Bp(0,1) is a neighborhood of 0 = (−1)0,but−B(0, r)*B(0,1) for any r >0, because the functions xn defined by
xn(t) =
((n−1)(nt−1), for 0≤t≤ n1,
n
n−1(t−n1), for n1 ≤t≤1,
is in Bp(0,1) for all n, whilep(−xn) =n−1> r for largen(see [2]).
The topology τp could not be Hausdorff even if p is an asymmetric norm on X. Necessary and sufficient conditions in order thatτp 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 Nudel0man (see [10, Ch. 9, § 5]). In the spaces C(T) and Lr, 1 ≤ r < ∞, one consid- ers asymmetric norms defined through a pair w = (w+, w−) of nonnega- tive upper semicontinuous functions, called weight functions, via the formula kf|w= max{w+(t)f+(t)−w−(t)f−(t) :t∈T}, where f+, f− are the positive, respectively negative part of f. In the case of the spaces Lr the above for- mula 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 subspaceY of a space with asymmetric norm in terms of some linear functionals vanishing onY. The approach is based on some ex- tension 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) and (Y, q) be spaces with asymmetric seminorms andA:X →Y a linear mapping. The mapping A is called bounded (or semi-Lipschitz) if there existsL≥0 such that
(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 is equivalent to its continuity with respect to the topologies τp and τq. Denoting by Lb(X, Y) the set of all bounded linear mapping from (X, p) to (Y, q), it turns out that Lb(X, Y) is not necessarily a linear space but rather a convex cone in the vector space La(X, Y) of all linear mappings fromX to Y, i.e.
λ≥0 andA, B∈Lb(X, Y) ⇒ A+B ∈Lb(X, Y) andλA∈Lb(X, Y).
For instance, in the space X =C0[0,1] considered in the previous section, the linear functional ϕ(x) = x(1), x ∈ C0[0,1], is bounded because ϕ(x) ≤ p(x), x∈X,but the functional−ϕis not bounded. Takingxn(t) = 1−ntn−1 we have p(xn) = 1 for all n, but−ϕ(xn) =n−1→ ∞ forn→ ∞ (see [2]).
As in the case of bounded linear mapping between normed linear spaces, one can define an asymmetric seminorm onLb(X, Y) by the formula
(2.2) kA|= sup{q(Ax) :x∈X, p(x)≤1}.
It is not difficult to see that k · | is an asymmetric seminorm on the cone Lb(X, Y) which has properties similar to those of the usual norm:
Proposition 2.1. Let (X, p) and (Y, q) be spaces with asymmetric semi- norms and A∈Lb(X, Y).Then
1) ∀x∈X q(Ax)≤ kA| ·p(x),
and kA| is the smallest number L ≥ 0 for which the inequality (2.1) holds.
2) kA|= supq(Ax)p(x) :x∈X, p(x)>0 .
Proof. 1) Ifp(x) = 0 then, by the boundedness ofA, q(Ax) = 0 =kA|p(x).
Ifp(x)>0 thenp((1/p(x))x) = 1 and
q(A(p(x)1 x))≤ kA| ⇐⇒ q(Ax)≤ kA| ·p(x).
If q(Ax)≤Lp(x), ∀x ∈X, for some L≥0, then q(Ax) ≤L for allx ∈X withp(x)≤1,implyingkA| ≤L.
2) Follows from the facts that q(Ax) = 0 ifp(x) = 0 and
1
p(x)q(Ax) =q(A(p(x)1 ))
ifp(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 spaceRof real numbers, consider the asymmetric seminormu(α) = max{α,0} and denote by Ru the space R equipped with the topology τu generated by u. It is the topology generated by the intervals of the form (−∞, a), a ∈ R. A neighborhood basis of a point a ∈ Ru is formed by the intervals (−∞, a+), >0.The seminorm conjugate to uis ¯u(α) =u(−α) = max{−α,0},and us(α) = max{u(α), u(−α)}=|α|.The continuity of a linear functional ϕ: (X, p)→(R, u) with respect to the topologiesτp and τu will be called (p, u)-continuity. It is easily seen that the (p, u)-continuity of a linear functional ϕ: (X, τp) →(R, u) is equivalent to its upper semi-continuity as a functional from (X, τp) to (R,| |). This is equivalent to the fact that for every α ∈ R the set {x ∈ X :ϕ(x) ≥ α} is closed in (X, τp) and has consequence the fact that, for every τp-compact subset Y of X, the functional ϕis upper bounded on Y and there existsy0∈Y such that ϕ(y0) = supϕ(Y).Also, the linear functionalϕis (p, u)-continuous if and only if it isp-bounded, i.e. there existsL≥0 such that
(2.3) ∀x∈X ϕ(x)≤Lp(x).
Denote byXp[ (X[when it is no danger of confusion) the cone of all bounded linear functionals on the space with asymmetric seminorm (X, p) and call it theasymmetric dualof (X, p). It follows that the functional
kϕ|=kϕ|p = sup{ϕ(x) :x∈X, p(x)≤1}
is an asymmetric seminorm on X[.
We shall need the following simple properties of this seminorm.
Proposition2.2. Ifϕis a bounded linear functional on a space with asym- metric seminorm (X, p), withp6= 0, then:
1) kϕ|is the smallest of the numbersL≥0 for which the inequality (2.3) holds;
2) We have:
kϕ|= 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};
3) If ϕ6= 0 then kϕ|>0.
Also, if ϕ6= 0 and ϕ(x0) =kϕ|for some x0 ∈Bp,then p(x0) = 1.
Proof. We shall prove the assertions 2) and 3), the first one being a partic- ular case of the corresponding result for linear mappings.
Supposing c := sup{ϕ(x) : p(x) < 1} < kϕ|, let x0 ∈ X, p(x0) = 1, be such that c < ϕ(x0) ≤ kϕ|.Then there is a number α,0 < α < 1,such that ϕ(αx0) =αϕ(x0)> c, in contradiction to the definition ofc.
Let’s show now that kϕ| = sup{ϕ(x) : p(x) = 1}. Suppose again that β := sup{ϕ(x) :p(x) = 1}<kϕ|,and choose x0 ∈X such thatp(x0)<1 and ϕ(x0)> β.Puttingx1 = (1/p(x0))x0, it followsp(x1) = 1 and
ϕ(x1) = p(x1
0)ϕ(x0)> ϕ(x0)> β, a contradiction.
3) Because ϕ(x) ≤ kϕ|p(x), the equality kϕ| = 0 implies ϕ(x) ≤ 0 and
−ϕ(x) =ϕ(−x)≤0,i.e. ϕ(x) = 0 for allx∈X.
Suppose now that that for ϕ6= 0 there exists x0 ∈X, with 0< p(x0)<1, such thatϕ(x0) =kϕ|. Thenα:= 1/p(x0)>1, x1 =αx0∈Bp and
kϕ| ≥ϕ(x1) =αϕ(x0) =αkϕ|,
a contradiction, becausekϕ|>0.
An immediate consequence of the preceding result is the following one. We agree to call a linear functional (p,p)-bounded if it is both¯ p- and ¯p-bounded.
Proposition 2.3. Let ϕ6= 0 be a linear functional on a space with asym- metric seminorm (X, p).
1) If ϕis (p,p)-bounded then¯
ϕ(rBp0) = (−rkϕ|p¯, rkϕ|p) and ϕ(rBp0¯) = (−rkϕ|p, rkϕ|p¯)
where Bp0 ={x∈X:p(x)<1}, Bp0¯={x∈X : ¯p(x)<1} and r >0.
2) If ϕis p-bounded but not p-bounded then¯ ϕ(rBp0) = (−∞, rkϕ|p).
Proof. Obviously that it is sufficient to give the proof only forr= 1.
Suppose that ϕis (p,p)-bounded. Then, by Proposition 2.2,¯ supϕ(Bp0) =kϕ|p
and
infϕ(Bp0) =−sup{ϕ(−x) :p(x)<1}=−sup{ϕ(x0) :p(−x0)<1}=−kϕ|p¯.
Also, by the assertion 3) of Proposition 2.2, ϕ(x) < kϕ|p and ϕ(x) > −kϕ|p¯ for anyx∈Bp0.
BecauseB0pis convex,ϕ(B0p) will be a convex subset ofR, that is an interval, and the above considerations show that
ϕ(Bp0) = (infϕ(Bp0),supϕ(Bp0)) = (−kϕ|p¯,kϕ|p).
Ifϕ isp-bounded and
sup{ϕ(x); ¯p(x)<1}=∞.
then
inf{ϕ(x0) :p(x0)<1}=−sup{ϕ(x) :p(−x)<1}=−∞.
Reasoning like above, one obtains
ϕ(Bp0) = (−∞,kϕ|p).
3. EXTENSION RESULTS FOR BOUNDED LINEAR FUNCTIONALS
In this section we shall prove the analogs of some well known extension re- sults 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) will be a space with asymmetric seminorm.
Proposition 3.1. Let Y be a subspace of X and ϕ0 : Y → R a bounded linear functional. Then there exists a bounded linear functional ϕ : X → R such that
ϕ|Y =ϕ0 and kϕ|=kϕ0|.
Proof. The functional q(x) = kϕ0|p(x), x ∈ X, is sublinear and ϕ0(y) ≤ q(y), y ∈ Y. By the Hahn-Banach extension theorem there exists a linear functional ϕ:X →Rsuch that
ϕ|Y =ϕ0 and ∀x∈X ϕ(x)≤ kϕ0|p(x).
The second of the above relations implies thatϕis bounded andkϕ| ≤ kϕ0|.
Since
kϕ|= sup{ϕ(x) :x∈X, p(x)≤1} ≥sup{ϕ(y) :y∈Y, p(y)≤1}=kϕ0|,
it followskϕ|=kϕ0|.
We agree to call a functionalϕsatisfying the conclusions of Proposition 3.1 a norm preserving extension of ϕ0.
Proposition 3.2. If x0 is a point in X such that p(x0) > 0 then there exists a bounded linear functional ϕ:X→R such that
kϕ|= 1 and ϕ(x0) =p(x0).
Proof. Let Y := Rx0 and let ϕ0 : Y → R be defined by ϕ0(tx0) = tp(x0), t∈R.It follows that ϕ0 is linear and
ϕ0(tx0) =tp(x0) =p(tx0) fort >0 and
ϕ0(tx0) =tp(x0)≤0≤p(tx0)
fort≤0. Again, the Hahn-Banach extension theorem yields a linear functional ϕ:X →R,such that
ϕ|Y =ϕ0 and ∀x∈X ϕ(x)≤p(x).
It follows kϕ| ≤1, ϕ(x0) =p(x0),and, sincep((1/p(x0))x0) = 1, kϕ| ≥ϕ(p(x1
0)x0) = 1,
i.e. kϕ|= 1.
This last proposition has as consequence the following useful result.
Corollary 3.3. If p(x0)>0, then
p(x0) = sup{ϕ(x0) :ϕ∈X[, kϕ| ≤1}.
Proof. Denote by s the supremum in the right hand side of the above formula. Since ϕ(x0) ≤ kϕ|p(x0) ≤ p(x0) for every ϕ ∈ X[, kϕ| ≤ 1, it follows s ≤ p(x0). Choosing ϕ ∈ Xp[ as in Proposition 3.2, it follows
p(x0) =ϕ(x0)≤s.
The next extension result involves the distance from a point to a set in an asymmetric seminormed space. LetY be a nonempty subset of an asymmetric seminormed space (X, p). Due to the asymmetry of the seminorm p we have to consider two distances from a pointx∈X toY, namely
(3.1) dp(x, Y) = inf{p(y−x) :y∈Y} and
(3.2) dp(Y, x) = inf{p(x−y) :y∈Y}.
Observe thatdp(Y, x) =dp¯(x, Y),where ¯p is the seminorm conjugate top.
Proposition 3.4. Let Y be a subspace of a space with asymmetric semi- norm (X, p) and x0 ∈ X. Denote by d¯the distance dp¯(x0, Y) and suppose d >¯ 0.
Then there exists ap-bounded linear functional ϕ:X →Rsuch that (i)ϕ|Y = 0, (ii)kϕ|= 1, and (iii)ϕ(x0) = ¯d.
If d=dp(x0, Y)>0 then there exists ψ∈Xp[ such that (j) ψ|Y = 0, (jj) kψ|= 1, (jjj) ψ(−x0) =d.
Proof. Suppose that ¯d=dp¯(x0, Y)>0, so thatx0 ∈/Y. LetZ :=Y uRx0
(u stands for the direct sum) and let ψ0 :Z →Rbe defined by ψ0(y+tx0) =t, y∈Y, t∈R.
Then ψ0 is linear, ψ0(y) = 0, ∀y∈Y, and ψ0(x0) = 1.Fort >0 we have p(y+tx0) =tp(x0+t−1y)≥td¯= ¯d·ψ0(y+tx0),
so that
ψ0(y+tx0) =t≤ 1¯
dp(y+tx0).
Since this inequality obviously holds for t ≤0 , it follows kψ0| ≤ 1/d.¯ Let (yn) be a sequence inY such thatp(x0−yn)→d¯forn→ ∞and p(x0−yn)>0 for all n∈N. Then
kψ0| ≥ψ0 x0−yn
p(x0−yn)
= p(x 1
0−yn) → 1¯
d, implying kψ0| ≥1/d.¯ Thereforekψ0|= 1/d.¯
If ¯ψ:X →Ris a linear functional such that ψ|¯Z =ψ0 and kψ|¯ =kψ0|
then the linear functional ϕ= ¯d·ψ¯fulfills all the requirements of the propo- sition.
Suppose now d=dp(x0, Y)>0,and let Z :=Y uRx0.Define ψ0 :Z →R by
ψ0(y+tx0) =−t ⇐⇒ ψ0(y−tx0) =t fory∈Y andt∈R. Then ψ0 is linear and, fort >0,we have
p(y−tx0) =tp(1ty−x0)≥td=d·ψ0(y−tx0), so that
ψ0(y−tx0)≤ 1dp(y−tx0),
for t > 0. Since this inequality is obviously true if ψ0(y−tx0) = t ≤ 0, it follows that ψ0 is bounded and kψ0| ≤ 1/d. Choosing a sequence (yn0) in Y such thatp(yn0 −x0)→dandp(y0n−x0)>0 for alln, and reasoning like above one obtains the inequality kψ0| ≥1/d, so that kψ0|= 1/d.Extending ψ0 to a functional ψ1∈Xp[ of the same norm, and letting ψ=d·ψ1, one obtains the
wanted functionalψ.
4. APPLICATIONS TO BEST APPROXIMATION
Let (X, p) be a space with asymmetric seminorm andY a nonempty subset ofX. By the asymmetry of the seminormpwe have to distinct two “distances”
from a pointx∈X to the subsetY, as given by (3.1) and (3.2).
Sincedp(Y, x) =dp¯(x, Y), we shall use the notationdp¯(x, Y) for the distance (3.2).
An element y0 ∈ Y such thatp(x−y0) = ¯p(y0−x) =dp¯(x, Y) is called a p-nearest point¯ toxinY, and an elementy1∈Y such thatp(y1−x) =dp(x, Y) will be called ap-nearest point tox inY.
By Proposition 3.4, we obtain the following characterization of ¯p-nearest points.
Proposition 4.1. Let (X, p) be a space with asymmetric seminorm, Y a subspace of X andx0 a point inX such that d¯=dp¯(x0, Y)>0.
An elementy0 ∈Y is a p-nearest point to¯ x0 inY if and only if there exists a bounded linear functional ϕ:X→R such that
(i)ϕ|Y = 0, (ii)kϕ|= 1, (iii)ϕ(x0) =p(x0−y0).
Proof. Suppose that y0 ∈ Y is such that p(x0−y0) = d =dp¯(x0, Y) > 0.
By Proposition 3.4, there exists ϕ ∈ Xp[, kϕ| = 1, such that ϕ|Y = 0 and ϕ(x0) =d=p(x0−y0).
Conversely, if for y0 ∈ Y there exists ϕ ∈ Xp[ satisfying the conditions (i)–(iii), then for everyy ∈Y,
p(x0−y)≥ϕ(x0−y) =ϕ(x0−y0) =p(x0−y0),
implying 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 be a subspace of a space with asymmetric semi- norm (X, p).If dp(Y, x0)>0 then the following duality formula holds:
dp(Y, x0) = sup{ψ(x0) :ψ∈Y⊥, kψ| ≤1}, where Y⊥ ={ϕ∈Xp[:ϕ|Y = 0}.
Proof. For anyψ∈Y⊥, kψ| ≤1 and any y∈Y, we have:
ψ(x0) =ψ(x0−y)≤p(x0−y),
implying sup{ψ(x0) :ψ∈Y⊥,kψ| ≤1} ≤dp(Y, x0). If we choose ψ to be the functional ϕ given by Proposition 3.4, then we obtain the reverse inequality:
dp(Y, x0) =ϕ(x0)≤sup{ψ(x0) :ψ∈Y⊥,kψ| ≤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 distancesdp and dp¯given by (3.1) and (3.2).
Proposition 4.3. Let (X, p) be a space with asymmetric seminorm, ϕ ∈ Xp[, ϕ6= 0, c∈R,
H ={x∈X :ϕ(x) =c}
the hyperplane corresponding to ϕ andc, and
H<={x∈X :ϕ(x)< c} and H>={x∈X:ϕ(x)> c},
the open half-spaces determined byH.
1) We have
(4.1) dp¯(x0, H) = ϕ(xkϕ|0)−c for everyx0 ∈H>, and
(4.2) dp(x0, H) = c−ϕ(xkϕ|0) for everyx0 ∈H<.
2) If there exists an element z0 ∈ X with p(z0) = 1 such that ϕ(z0) = kϕ|, then every element in H> has a p-nearest point in¯ H and every element inH< has ap-nearest point in H.
If there is an elementx0∈H>having a p-nearest point in¯ H, or there is an element x00 ∈H< having a p-nearest point in H, then there exists an element z0 ∈X, p(z0) = 1, such that ϕ(z0) =kϕ|.It follows that, in this case, every element in H> has a p-nearest point in¯ H, and every element in H< has a p-nearest point in H.
Proof. Letx0 ∈H>. Then, for everyh∈H, ϕ(h) =c,so that ϕ(x0)−c=ϕ(x0−h)≤ kϕ|p(x0−h), implying
dp¯(x0, H)≥ ϕ(xkϕ|0)−c.
By the assertion 2) of Proposition 2.2, there exists a sequence (zn) in X with p(zn) = 1, such thatϕ(zn)→ kϕ| and ϕ(zn)>0 for alln∈N.Then
hn:=x0−ϕ(xϕ(z0)−c
n) zn belongs to H and
dp¯(x0, H)≤p(x0−hn) = ϕ(xϕ(z0)−c
n) → ϕ(xkϕ|0)−c. It follows dp¯(x0, H)≥(ϕ(x0)−c)/kϕ|,so that formula (4.1) holds.
To prove (4.2), observe that for h∈H,
c−ϕ(x00) =ϕ(h−x00)≤ kϕ|p(h−x00), implying
dp(x00, H)≥ c−ϕ(xkϕ|00). If the sequence (zn) is as above then
h0n:= c−ϕ(xϕ(z 00)
n) zn+x00 belongs to H and
dp(x00, H)≤p(h0n−x00) = c−ϕ(xϕ(z 00)
n) → c−ϕ(xkϕ|00), so that dp(x00, H)≥(c−ϕ(x00))/kϕ|, and formula (4.2) holds too.
2) Let z0 ∈X be such thatp(z0) = 1 and ϕ(z0) =kϕ|.Then, for x0 ∈H>
and x00 ∈H<, the elements h0:=x0−ϕ(xϕ(z0)−c
0) z0 and h00 := c−ϕ(xϕ(z 00)
0) z0+x00 belong toH,
p(x0−h0) = ϕ(xkϕ|0)−c =dp¯(x0, H) and p(h0n−x00) = c−ϕ(x
0 0)
kϕ| =dp(x00, H).
If an element x0 ∈H> has a ¯p-nearest point h0 ∈H, then p(x0−h0) =dp¯(x0, H) = ϕ(xkϕ|0)−c = ϕ(xkϕ|0−h0).
It follows thatz0 = (x0−h0)/p(x0−h0) satisfies the conditionsp(z0) = 1 and ϕ(z0) =kϕ|.
If an element x00 ∈ H< has a p-nearest point h00 in H, then z00 = (h00 − x00)/p(h00−x00) satisfiesp(z00) = 1 and ϕ(z00) =kϕ|.
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Received by the editors: March 28, 2004.
