Applied Mathematics
Best approximation and hierarchical bases
Vladimir L.Vaskevich
Sobolev Institute of Mathematics, SB RAS, Novosibirsk, Russia e-mail:[email protected]
Received: October 31, 2001
Summary.
In the present paper we extend the denition of a hi-erarchical basis to the case of Banach spaces with higher smooth-ness order elements. Hierarchical bases are similar to the well-known Faber{Schauder system. We prove that there exists a hierarchical basis of the Banach space under consideration. We study the inter-play between hierarchical bases and solutions to specic problems of best approximation. The construction of some hierarchical basis is nally described.Key words:
hierarchical bases, reexive Banach spaces, best ap-proximation, reproducing mappings, extremal functions of cubature formulas, splines of ane varietiesMathematics Subject Classication (1991): 41A55, 46N05, 65D32
1. Introduction
Solutions to most types of problems in applied mathematics are mem-bers of a given Banach space and we nd it convenient to look for the solution to a problem of such kind in the form of a convergent series with respect to a given basis of the initial Banach space. In par-ticular, in the spaceC01] of continuous functions with domain 01] the well-known Faber|Schauder sequence constitutes a basis (see, e.g., 9, p. 227]) and for each function fromC01] the corresponding series is convergent in the norm ofC01]. It was shown in 4] and 5]
that representation of functions in the form of the Faber|Schauder series for problems like interpolation and numerical quadrature has a long tradition. For partial dierential equations a similar approach was studied in 1], 4], 5], and 17].
It seems to be several properties of the Faber|Schauder basis which are of crucial importance for problems in the theory of approx-imation and numerical analysis. In particular, for a given continuous function every coecient of the corresponding series with respect to the Faber|Schauder system is completely determined by the values of the initial function in the nite subset of its domain. But we can not use the Faber|Schauder system as a basis in case when the functions of a given Banach space has derivatives of order greater than 1. In this event it is natural to ask
Does there exist a basis like the Faber|Schauder system for a given Banach space with functional members of high order of smoothness? Throughout the sequel we call bases like the Faber|Schauder system the hierarchical bases.
In general the hierarchical basis can inherit the properties of the Faber|Schauder system only in part. An example is as follows.
Let Rn be a bounded domain and let B be a Banach space
of harmonic functions with domain. Ifuis a member of some basis of B then the support of u can not be in the interior of . For, were it otherwise, we would nd that u is identically equal to 0 a contradiction.
Nevertheless, we can dene a hierarchical basis in a separable Ba-nach space of rather general type and the hierarchical basis dened in such a way inherits some important properties of the Faber| Schauder system. The goal of the present paper is to describe the construction of some hierarchical basis in a separable reexive Banach space. By this way we also establish the existence of a hierarchical basis in the space under consideration.
Let Rnbe a bounded domain with suciently smooth
bound-ary and let the origin be in the interior of.
The setting is a separable reexive Banach spaceX=X() and a reexive Banach spaceY =X , dual toX. The membersofXare real valued continuous functions with domain. Let X be embedded in the Banach spaceC() of functions which are continuous in and let the embedding be linear and bounded. Hence the conventional Dirac delta function(x) is a member ofY.
We also assume that for a given nite subset F of there exists a member u(x) of X such that the values of u(x) at points of F
are prescribed real numbers. It will be true if, for example, every polynomial belongs toX.
Furthermore, letX be astrictly normedlinear space, i.e.,
ku+vjXk=kujXk+kvjXk
implies that u =tv for somet0 or else v = 0. This constraint on
the norm is easily seen to be equivalent to the geometric condition that the unit ball ofX be rotund. SinceXis strictly normed it follows thatY =X is asmoothly normed reexive Banach space 8, p.173]. Let N () =k jYk. By a denition, Y is smoothly normed exactly
when
(N 0
G (l)m) = limt!0
1
t(N (l+tm);N (l))
exists for allml2Y,kljYk= 1, and denes a functionalN
0
G(l) in
Y =X. The functionalN 0
G (l) is a Gateaux dierential of the norm
N () atl.
Let Y be also a strictly normed linear space. Then there exists a Gateaux dierential of the normN() =kjXkat all unit vectors
u2X.
Examples of strictly and smoothly normed spaces are Hilbert spaces and the Sobolev spacesW(m)
p () 1< p <1. By a denition,
'(x) with domainbelongs toW(m)
p () i'have all derivatives up
to orderm locally integrable and
'jW (m) p () p = Z fj'j 2+ X jj=m m! !jD'j 2 gp= 2dx < 1:
The integral here spreads over, and summation is taken over some multi-indices= (12:::n) with integer coecients,
! =1!2!:::n! jj= n X j=1 j D'= @x1 @m' 1 @x 2 2 :::@x n n : For mp > n W(m)
p () is embedded in C() and the embedding is
linear and bounded. Let = f kg
1
k=0 be a sequence of nite subsets of 0 = fx~ (0) j j j = 12::: (0)g, k = k ;1 fx~ (k) j j j = 12::: (k)g, ~ x(k) j 2= k
;1, k = 12:::. We assume that the union of all k is
dense in. The sequence =f kg 1
k=0 is said to be a multigrid in
k is a k{levelof and vectors ~x(k)
Given a multigrid , we introduce the sequence Nod(k) j of subsets of by putting Nod(0) 1 = and fork= 012::: Nod(k) j = Nod(k) j;1 fx~ (k) j;1 g j= 23::: (k) Nod(k) (k)+1= Nod (k+1) 1 = k: If k < k1 or (k=k1 and j j 1), then k;1 Nod (k) j Nod(k 1 ) j1 k 1: LetH =fh (k) j 2Xjk0 j= 1::: (k)gbe a countable subset of X.
Denition 1.
We call H a -hierarchical system inX i for all k,j, and x~(m) i 2 k, x~ (m) i 6= ~x (k) j , the function h(k) j equals 0 at x~(m) i and the value of h(k)
j at x~(k)
j equals 1.
Hence, His a {hierarchical system in Xi the following equal-ities hold
(1:1) h(k)
j (~x(m)
l ) = ljmk m= 01:::k l= 12::: (k):
Here lj is the conventional Kronecker delta.
Given the nite subset k ofand an integerj,j= 12::: (k),
we can apply the Lagrange interpolation formula over the set k of
nodes and nd h(k)
j 2 X such that (1.1) holds. Hence the set of
{hierarchical systems of X is not empty. Obviously, every nite subsequence of a {hierarchical systemH is linearly independent.
In 12] and 2] it was shown that hierarchical systems in spaces like Sobolev spaces may be constructed as sequences of interpolatingDm{
splines. The properties of {hierarchical systems in Hilbert spaces were studied in 3]. Cubature formulas based on hierarchical systems were constructed in 15] and 16].
Denition 2.
If a -hierarchical system H in X is a basis of X, thenH is called a hierarchical basis.An example of a hierarchical basis for the space C01] and the Sobolev space W1
201] simultaneously is the well-known Faber|
Schauder system.
Problem 1.
Given a separable Banach space X, nd a -hierarchi-cal basis of X.Alongside Problem 1, it stands to reason to consider another prob-lem that is posed in the theory of approximation. To be more pre-cise, we bear in mind the problem of best approximation in a Banach space. Before stating the problem, we introduce a few designations.
Since Nod(k)
j it follows that for all ~x(m)
i 2 Nod
(k)
j the
func-tional(x;x~ (m)
i ) is a member ofY. Evidently the linear span
L(k) j = spanf(x;x~ (m) i )jx~ (m) i 2Nod (k) j g
is a nite dimensional closed subspace ofY.
Problem 2.
Given an arbitrary nonzero functional l(0) 12 Y, nd
an element of best approximation to l(0)
1 from L (k)
j .
As is known, every closed convex subset of a reexive strictly normed Banach space is a Chebyshev set (see, e.g.,10, p.104]). It means that every element of the Banach space has exactly one ele-ment of best approximation from the set under consideration. Con-sequently, there exists a unique solution to Problem 2.
The theme of our presentation up to this point may be described as a study of the interplay between the solutions of Problems 1 and 2.
To begin with, we take l(0) 1 2 Y, l (0) 1 6 = 0. For example, l(0) 1 may be the indicator (x) of . To l(0)
1 and every vector fc (m) i jx~ (m) i 2 Nod(k)
j g of real numbers, we assign the associated sequence of error
functionals by putting l(k) j =l(0) 1 ; X ~ x(m) i 2No d (k) j c(m) i (x;x~ (m) i ) k= 01::: j= 12::: (k):
We call the sequencefl (k)
j gtheerror multifunctional. The
correspond-ing sequence of cubature formulas is said to be a multicubature for-mula. LetY(k) j be a at parallel toL(k) j Y(k) j =l(0) 1 +L (k) j . ThenY(k) j is
an ane variety and dimY(k) j =N =N(kj) = (0) + (1) ++ (k;1) +j;1: If k < k1 or (k=k1 and j j 1), thenY (k) j Y(k 1 ) j1 .
For given integerskandj, we use the symbol(k)
jopt(x) to designate
the element of best approximationtol(0)
1 fromL (k)
j , and coecients of
the expansion of(k)
jopt(x) with respect to delta functions(x ;x~ (m) i ) we denote by c(m) iopt=c (m) iopt(jk), i.e., (k) jopt(x) = k;1 X m=0 (m) X i=1 c(m) iopt(x ;x~ (m) i ) + j;1 X i=1 c(k) iopt(x ;x~ (k) i ): Letl(k) jopt=l (0) 1 ; (k)
jopt. The corresponding cubature formula is said to
be X-optimal on the set Nod(k)
j of nodes 14]. The normkl (k) jopt jYk equals E(l(0) 1 L (k)
j ), where E(wN) is the distance from w 2 Y to
a linear subspaceN of Y.
Let l2Y andu2X. If the following equalities hold
(1:2) kljYk
2 = (lu) =
kujXk 2
thenuis said to bean extremal function forl14] and lis said to be
a generated extremal function for u.
By the reexivity ofXand James Theorem (see, e.g., 11, p.236]), there exists an extremal function for an arbitraryl 2Y. Since X is
a strictly normed space it follows that for a given functional l 2 Y
an extremal function u 2 X is unique. By the same reasons, for
a given function u 2 X there exists a unique generated extremal
function. Throughout the sequel we denote the extremal function for
l(k)
joptby u (k)
jopt.
In Section5 we discuss how to transform the set
fu (k) jopt(x)u (k) j+1opt(x):::u (k) (k)opt(x) g
into the functionh(k)
j (x) of some -hierarchical basis ofX and it is
the main result of the paper.
2. Extremal Functions and Reproducing Mappings
Let l(0) 1 2Y, l (0) 1 6 = 0, and letu(0)1 be the extremal function for l (0) 1 . Then (2:1) kl (0) 1 jYk 2 = (l (0) 1 u (0) 1 ) = ku (0) 1 jXk 2:
As our next step, we consider the properties ofu(0) 1 .
Theorem 1.
There is a unique elementu=u(x) of best approxima-tion to zero element of X fromV =fv2X j(l (0) 1 v) = kl (0) 1 k 2 g:
The function u is a unique extremal function for l(0)
1 in X. If M is
the kernel ofl(0)
1 and E(wN)is the distance from w
2X to a linear subspace N of X, then E(uM) =E(0V) = kl (0) 1 jYk=kujXk: Proof. Let v 2 X, (l (0) 1 v ) = d 6 = 0, and = kl (0) 1 k 2=d. Then v0 = v 2 V. Consequently, V = v 0+M 6 = , and V is a closed
convex subset ofX. Whence and from the reexivity of X, we infer thatV is a Chebyshev subset ofX, and there exists a unique element
u=u(x) of best approximation to zero element ofX fromV. By the denition,
kuk= minfkvkjv2Vg= minfku;vkjv2Mg
and zero element ofX is an element of best approximation toufrom
M. Whence and from the well-known theorem of characterization of elements of best approximation 13, p.2] it follows that there exists
f0 2Y such that (2:2) kf 0 k= 1 kuk= (f 0u) (f0v) = 0 8v2M: Ifv 2Xand (l (0) 1 v ) =d 6
= 0, then for8w2Xwe havew=v +v,
where = (l(0)
1 w)=d and v
2 M. By the third equality of (2.2),
(f0w) = (f0u ) = (l (0) 1 w), where = (f 0u )=d. Therefore, f0 =l (0) 1 and (f 0u) = (l (0) 1 u) = kl (0) 1 k
2. By the second equality
of (2.2),=kuk=kl (0) 1
k
2. Considering this, we derive >0, and, by
the rst equality of (2.2), = 1=kl (0) 1 k. Hence kuk =kl (0) 1 k, and we
arrive at the sought relations (2.1) for the functionu(0) 1 =u.
From thestrict convexity ofXit is immediatethat there is aunique extremal function forl(0)
1 inX. For, were it otherwise, we would nd
at least two extremal functions u1 and u2 with the same norm and
their half-sumu12= (u1+u2)=2 would then have the norm less than
each of them. In this event,
kl (0) 1 k (l(0) 1 u 12) ku 12 k >kl (0) 1 k a contradiction. ut
Applying Theorem 1 to the space Y, we obtain
Theorem 2.
There is a unique element l of best approximation to zero element of Y fromV =fm2Y j(mu (0) 1 ) = ku (0) 1 k 2 g:
The functional l is a unique generated extremal function for u(0) 1 in
Y. If M = fm 2 Y j (mu (0) 1 ) = 0
g and E(lN ) is the distance
from l2Y to a linear subspace N of Y, then
E(lM ) =E(0V ) =ku (0) 1 jXk=kljYk: Let l(0) 1 2 Y, u (0) 1 2 X, u (0) 1 6
= 0, and (2.1) holds. Then we can dene the mapping : Y ! X by (l
(0) 1 ) = u
(0)
1 . We also assume
that (0) = 0. By the denition,(l(0)
1 ) is the extremal function for
l(0) 1
2Y. By Theorem 1, it follows that is a single-valued mapping
with domainY and for l2Y and 2Rwe have k(l)jXk
2= (l(l)) =
kljYk
2 (l) =(l):
Together with , we consider a mapping : X ! Y, dual to .
Let u(0) 1 2 X, u (0) 1 6 = 0, l(0) 1
2 Y, and (2.1) holds. We assume that
(u(0) 1 ) = l
(0)
1 and (0) = 0. By Theorem 2, it follows that is
a single-valued mapping with domainX and
k (u)jYk
2= ( (u)u) =
kujXk
2 (u) = (u)
8u2X:
Letl2Y andu2X. By Theorems 1 and 2l= (u) iu=(l).
In particular forl2Y and u2X
l= ((l)) and u=( (u)):
Hence : Y ! X and : X ! Y are reciprocal and surjective
mappings. The image of the sphere of radius R under the mapping
is the sphere of the same radius R. Conversely, the image of the sphere of radius R under the mapping is the sphere of the same radiusR.
IfXis a Hilbert space, then= andis said to bereproducing mapping of X 2, p.23]. We also nd it convenient to use the term \reproducing mapping" in the case of a Banach space. To be more precise, we call (resp. ) the reproducing mapping of the Banach space Y (resp. X). In 8, p.174] is calledthe norm-duality map.
Because of developments in the abstract theory of convex pro-gramming, it is possible to readily characterize the extremal function
for an arbitrary functional l(0) 1 2 Y, l (0) 1 6 = 0. By hypothesis X is a smoothly normed space. In this case let
(2:3) (N0
G(v)w) = limt!0
1
t(N(v+tw);N(v))
this is dened (by assumption)wheneverv6=0 andN 0
G(v) isaGateaux
dierential of the normN() =kjXkatvN 0
G(v)2Y. We have
Theorem 3.
Let l(0)1 be a nonzero element of Y M is the kernel of
l(0) 1 u 2 X and (l (0) 1 u) = kl (0) 1 k
2. The function u is extremal for
l(0) 1 i (N
0
G(u)w) = 0for 8w2M. Moreover, the extremal function
u2X for l (0)
1 is the solution to the following problem
(2:4) 8 > > > < > > > : N0 G(u) = 1 kl (0) 1 k l(0) 1 (l(0) 1 u) = kl (0) 1 k 2:
Conversely, every solution u 2 X to (2.4) is the extremal function
for l(0)
1 . There is a unique solution to (2.4).
Proof. LetV =fv2Xj(l (0) 1 v) = kl (0) 1 k 2 g=v 0+M andu 2V. By
Theorem 1,u is the extremal function for l(0) 1 i kuk= minfkvkjv2Vg:
Consequently, u is anR-spline interpolant ofV, withRthe identity map on X 6, p.576]. Whence and from Corollary 3.1 6, p.584] it follows thatu2V is the extremal function forl
(0) 1 i (N
0
G(u)w) = 0
for8w2M.
Let u be the extremal function for l(0) 1 , u
2 X. If ' 2 X then
'=u+w, wherew2M and 2R. Considering this, we derive
(N0
G(u)') =(N0
G(u)u)+ (N0
G(u)w):
Since (2.3) holds, we have (N0
G(u)u) = limt!0 1
t(ku+tuk;kuk) =kuk.
But (N0
G(u)w) = 0 for8w2M, and we obtain= (N 0
G(u)')=kuk.
Inserting this equality in (l(0)
1 ') =(l (0)
1 u) and considering thatu
is the extremal function for l(0)
1 , we nd that 1 kl (0) 1 k (l(0) 1 ') = ( l(0) 1 u) kuk 2 (N 0 G(u)') = (N0 G(u)'):
Hence,u is actually a solution to (2.4).
Assume now that u is a solution to (2.4). Then (N0
G(u)w) = 1 kl (0) 1 k (l(0) 1 w) = 0 for
8w2M. In this event, as we know, the function
u is extremal forl(0) 1 .
By Theorem 1 there is a unique extremal function for l(0) 1 inX.
Hence, there is a unique solution of (2.4). ut
By hypotheses of Theorem 3, the image of l(0) 1
2 Y under the
mapping is the unique solution to (2.4).
Lemma 1.
Let X be a Hilbert space with the inner product ()Xand letl(0)
1 be a nonzero member ofY. The extremal functionu (0) 1 for
l(0)
1 satises the following equalities
(2:5) (l(0) 1 ') = (u (0) 1 ')X 8'2X: Thus, u(0)
1 is the member of X associated to the given functional by
virtue of the Riesz Theorem on the general form of a bounded linear functional.
Proof. LetXbe a Hilbert space. In this event the Gateaux dierential
N0 G(v) of the normN() at u (0) 1 is dened by (N0 G(u(0) 1 )w) = 1 ku (0) 1 k (u(0) 1 w)X 8w2X:
Whence and from Theorem 3 it follows that (2.5) holds. ut
The following theorem is dual to Theorem 3.
Theorem 4.
Letu(0) 1 be a nonzero element of X M =fm2Y j(mu (0) 1 ) = 0 g be the annihilator of fu (0) 1 g X l2Y, and(lu (0) 1 ) = ku (0) 1 k 2. Thenl is a generated extremal function for u(0)
1 i (mN
0
G(l)) = 0 for
8m2M . Moreover, the generated extremal function l2Y for u (0) 1
is the solution to the following problem
(2:6) 8 > > > < > > > : N 0 G(l) = 1 ku (0) 1 k u(0) 1 (lu(0) 1 ) = ku (0) 1 k 2:
Conversely, every solution l 2 Y to (2.6) is the generated extremal
function for u(0)
By hypotheses of Theorem 4, the image of u(0) 1
2 Y under the
mapping is the unique solution to (2.6).
Theorem 5.
The reproducing mappings and are demicontinu-ous and the following inequalities hold( (u); (v)u;v)0 8uv2X
((l);(m)l;m)0 8lm2Y
i.e., and are monotone. If for each nonzerov2X (resp.l2Y)
the functional N0
G(v) (resp. N 0
G(l)) is the Frechet dierential of the norm atv (resp.l), then (resp.) is continuous.
Proof. To begin with, we consider the reproducing mapping . Let
u2X,u6= 0 andl= (u). By (2.4), the following equalities hold
l=klkN 0 G(u) =kukN 0 G(u) =N0 G(kuku):
In terms of 8, p.174] is the norm-duality map from X into Y. There are proofs of the monotonicity inequality for and the demi-continuity of in 8, p.174].
Let for each nonzero v2 X the functional N 0
G(v) be the Frechet
dierential ofkjXkat v. Under this hypothesis, there is a proof of
the continuity of in 7, p.149].
By the same way, we establish the properties of . ut
If there is a Frechet dierential of kj Xk at v 2 X,v 6= 0, and
a Frechet dierential of kjYk atl 2Y, l6= 0, then it follows from
Theorem 5 that and are homeomorphisms ofX and Y. Since the ane variety Y(k)
j of error functionals is an
unbound-ed subset of Y it follows that the image X(k)
j of Y(k)
j under is
an unbounded subset of X. If k < k1 or (k = k1 and j j
1), then
X(k)
j X(k1)
j1 . Let be continuous. SinceY (k)
j is a closed subset of
Y it follows thatX(k)
j is a closed subset ofX.
There is a one-to-one correspondence(k)
j of topologicalspaceX(k)
j
onto RN(jk), where N(jk) = dimY (k)
j . The denition of (k)
j is as
follows.
Let H be a hierarchical system in X, h(m)
i 2 H, ~x (m) i 2 Nod (k) j , and u2X (k)
j . Then we assume that
c(m) i (u) = (l(0) 1 ; (u)h (m) i ):
Let(k) j (u) =fc (m) i jx~ (m) i 2Nod (k) j g2RN (jk). Examine that (k) j is
actually a one-to-one correspondence ofX(k)
j ontoRN(jk). Let u1 2 X (k) j , u2 2 X (k) j , and (k) j (u1) = (k) j (u2). Then l1 = (u1) 2Y (k) j , l2 = (u2) 2 Y (k) j , and for 8x~ (m) i 2Nod (k) j we have (l1 ;l 2h (m) i ) = 0. Considering thatl1 ;l 2 2L (k) j , we arrive atl1 =l2. Hence u1 =(l1) =(l2) =u2. If is continuous then(k)
j is also continuous. In this eventX(k)
j
is a topological variety of dimensionN(jk) = dimY(k)
j and X(k)
j is
homeomorphic to the ane varietyY(k)
j .
In case of a Hilbert space the extremal function for l(k)
j 2 Y
(k)
j
is given by the formula u(k)
j (x) = u(0) 1 (x) ; P ~ x(m) i 2No d (k) j c (m) i U(m) i (x), whereU(m)
i (x) is the extremal function for(x;x~ (m)
i ). IfU(x) is the
extremal function for the Dirac delta function(x) andU(x;x~ (m) i )2 X, then U(m) i (x) =U(x;~x (m) i ).
Lemma 2.
The norm of the extremal functionu(k)jopt for the optimal
error functional l(k)
jopt is less than the norm of an arbitrary element
of X(k)
j
u(k)
jopt= arg min
fkvjXkjv 2X (k)
j g:
There is a unique element ofX(k)
j with this property. Proof. Letu2X (k) j u6=u (k) jopt. Thenl= (u) 2Y (k) j andl6=l (k) jopt. Hence kujXk=kljYk>kl (k) jopt jYk=ku (k) jopt jXk. ut
Let us establish the additional properties ofu(k)
jopt. To this end, we
apply the following
Theorem 6.
10, p.116] Let Y1 be a closed linear subspace of Y,l(0) 1
2Y nY
1, andopt 2Y
1. The functionaloptis the element of best
approximation tol(0)
1 fromY
1 i there exists a memberf0 of Y such
that kf 0 jY k= 1 kl (0) 1 ; opt jYk=f 0(l (0) 1 ) f 0(c) = 0 8c2Y 1:
Theorem 7.
A functionu2X (k)j is extremal for optimal error func-tionall(k) jopt i (2:7) u(~x(p) s ) = 0 8x~ (p) s 2Nod (k) j :
Proof. By the reexivity of X, it follows that for 8f 2 Y there
is a function u 2 X such that f(l) = l(u) for 8l 2 Y. Applying
Theorem 6 to the space Y and the subspace
Y1 =L (k) j = spanf(x;x~ (m) i )jx~ (m) i 2Nod (k) j g
and considering that the element of best approximation to l(0) 1 from
L(k)
j is denoted by(k)
jopt, we conclude that there is a functionu 0 2X such that (2:8) ku 0 k= 1 kl (0) 1 ; (k) jopt jYk= (l (0) 1 u 0) (cu0) = 0 8c2L (k) j :
The third condition of (2.8) holds i (2:9) u0(~x (p) s ) = 0 8x~ (p) s 2Nod (k) j :
Considering this, we write the second condition of (2.8) in the equiv-alent form kl (k) jopt jYk=kl (0) 1 ; (k) jopt jYk= (l (0) 1 ; (k) joptu 0): Ifv(x) =kl (k) jopt jYku
0(x), then it follows from the last equality that kl (k) jopt jYk 2= (l (k) joptv) = kvjXk
2. Thus,v is the extremal function
for l(k)
jopt, and v =u (k)
jopt. Whence and from (2.9) we infer that (2.7)
holds.
Assume now that u 2 X (k) j , (2.7) holds, and c(m) i = c(m) i (u) are local coordinates of u 2 X (k) j , i.e., c(m) i = c(m)
i (u) are entries of the
vector (k) j (u) 2 RN (jk). Let = P ~ x(m) i 2No d (k) j c (m) i (x;x~ (m) i ). Then 2 L (k) j , and (k) j ((l(0) 1 ; )) = (k)
j (u). Hence the function u =
(l(0) 1 ; ) is extremal forl (0) 1 ; , and (2:10) (l(0) 1 ; u) =kl (0) 1 ; jYk 2= kujXk 2: The function u0(x) = 1 kujXk
u(x) belongs to the unit sphere of X
and, by (2.7), satises (cu0) = 0 for
8c 2L (k)
j . Whence and from
(2.10) it follows that (l(0) 1 u 0) = (l (0) 1 ; u 0) = 1 kujXk (l(0) 1 ; u)
= 1 kujXk kl (0) 1 ; jYk 2 = kl (0) 1 ; jYk:
Thus for given 2L (k)
j there is a memberu0 of X =Y such that
(2.8) holds. By Theorem 6, is the element of best approximation tol(0) 1 fromL (k) j , and l(0) 1 ; =l (k) jopt. Consequently, u=(l (k) jopt) = u(k) jopt. u t
As is well known, for a given set of nodes there is a unique optimal cubature formula in a Hilbert space and the corresponding optimal extremal function equals 0 at all nodes of this formula. Moreover, in case of Hilbert spaces like Sobolev spaces, the system (2.7) is a start-ing point of the algorithm of constructstart-ing the (unknown) weights of an optimal cubature formula (see, e.g., 14, Chapter9]).
Let M(k) j =fv 2Xj(l (0) 1 v) = 0 ((x ;x~ (m) i )v) = 08x~ (m) i 2Nod (k) j g
be a closed linear subspace ofX and letV(k)
j =u(k) jopt+M (k) j be the at parallel toM(k) j .
Theorem 8.
Let the norm N() = k j Xk be a twice continuous(Frechet) dierentiable functional on Xnf0g and let d 2N(
) be the
second Frechet dierential of this norm. The norm of the extremal function u(k)
jopt for the optimal error functional l (k)
jopt is less than the
norm of an arbitrary element of V(k)
j (2:11) u(k) jopt(x) = argmin fkvjXkjv2V (k) j g:
There is a unique element ofV(k)
j with this property. Proof. Let v 2V (k) j ,v 6=u (k) jopt, and 0 t 1. The function '(t) = k(1;t)u (k) jopt+tv
j Xk is twice dierentiable and, by the Taylor
formula, we have (2:12) '(t) ='(0) +t'0 (0) + 12'00() 20t] where'0(0) = (N0(u (k) jopt)v ;u (k) jopt) and '00() =d2N(u (k) jopt+(v ;u (k) jopt))(v ;u (k) joptv ;u (k) jopt):
Using (2.12) together with (2.4), we obtain '0(0) = (N0(u (k) jopt)v ;u (k) jopt) = 1 kl (k) jopt jYk (l(k) joptv ;u (k) jopt): Sincev;u (k) joptbelongs toM (k) j andM(k)
j is embedded into the kernel
of l(k)
jopt, it follows that '
0(0) = 0.
By a denition, the second Frechet dierential d2N of N( ) is a
continuous symmetric bilinear function onX X. Considering that
N() is convex on X, we show that d
2N is positive semi-denite for
everyv2X.
Let u2X,v2X, and
f(t) = (1;t)kuk+tkvk;k(1;t)u+tvk:
It is evident that f(t) 0 and f(0) = f(1) = 0. Hence f 0(0)
0.
Applying the Taylor formula to f(t), we derive that
f(1) =f(0) +f0 (0) + 12f00() () ; 1 2f00() =f0(0) 0 where 201]. Sincef 00() = ;d 2N(u+(v ;u))(v;uv;u), we have d2N(u+(v ;u))(v;uv;u)0:
Putting in (2.12)t= 1, we observe that
kvjXk=ku (k) jopt jXk+ 12d 2N(u (k) jopt+(v ;u (k) jopt))(v ;u (k) joptv ;u (k) jopt): Hence kvjXkku (k) jopt jXk, and (2.11) holds. Since V(k)
j is a closed convex subset of X, it follows that V(k)
j is
a Chebyshev set, and there is a unique function with (2.11). ut
3. Extremal Functions and Splines of A ne Varieties
Let M be a closed linear subspace of X. Assume that M has nite codimension inX. Given an elementu2X, we consider V =u+M
a at parallel toM.
Denition 3.
6, p.576]Letu02X, and
u0 = argmin
fkvjXkjv2Vg:
Following 6], we can now consider the spline operator S : X ! X
of M. By a denition, for everyu 2X S(u) is the spline of u+M.
Whence and from Denition 3 it follows thatS =S(M) is the map-pingI;PM, where I is the identity map onXand PM is the metric
projection of X onto M. Some authors use for the metric projec-tionPM the termnormal projection, orbest approximation operator,
or nearest point map, or Chebyshev map. By a denition, for every
u 2 X PM(u) is the element of best approximation to u from M.
The spline operatorS of M is homogeneous, i.e.S(u) =S(u) for
82R. The spline operator S is linear (resp. continuous) iPM is
linear (resp. continuous). IfX is a Hilbert space, thenPM and S are
linear and continuous. The linearity of metric projections is an in-frequent phenomenon in non-Hilbert spaces. There are examples of linear metric projections in non-Hilbert spaces (see, e.g., 6, p.580]). Throughout the sequel we are interested in the spline operators of the following sequence of the closed linear subspaces ofX
M(k) j =fv2Xj(l (0) 1 v) = 0 ((x ;~x (m) i )v) = 08x~ (m) i 2Nod (k) j g:
We are also interested in the splines of ats ~V(k)
j =u(k) jopt+M (k) j where u(k) jopt = 1 ku (k) jopt jXk 2u (k) jopt. Evidently M (k)
j has the nite codimension.
The spline operator ofM(k)
j is denoted byS(k)
j S(k)
j =I;P
M(k)
j . Let
us show the validity of the following
Lemma 3.
Let k < k1 or (k = k1 and jj
1). Then the equality
holds (3:1) S(k 1 ) j1 S (k) j =S(k) j :
If the norm N() =kjXk is a twice continuous (Frechet)
dieren-tiable functional onXnf0g then S (k) j (u(k) jopt) =u (k) jopt. Proof. Since S(k 1 ) j1 S (k) j =S(k) j ;P M(k 1 ) j1 S(k)
j , it follows that (3.1) holds
i (3:2) 8u2X S (k) j (u)2kerP M(k 1 ) j1 : By Denition 3, ker PM(k 1 ) j1 =fw2Xjw=S (k 1 ) j1 (w) g, andwbelongs to the kernel ofPM(k 1 ) j1 i (3:3) 8v2M (k 1 ) j1 (N 0(w)v) = 0:
SinceS(k)
j (u) is the spline ofu+M(k)
j , it follows from Corollary 3.1 6,
p.584] that for allv2M (k)
j (N0(S (k)
j (u))v) = 0. Using this equality,
together with the fact that fork < k1 or (k=k1 and j j 1)M (k 1 ) j1 is a subset of M(k)
j , we arrive at the sought relation (3.3) where
w=S(k)
j (u). Consequently (3.2) is also valid.
If the norm N() = k j Xk is a twice continuous (Frechet)
dif-ferentiable functional onXnf0gthen (2.11) holds. Considering that
u(k)
jopt 2V~
(k)
j and using the homogeneity ofS(k)
j , we obtain
u(k)
jopt= arg min
fkvjXkjv2V~ (k) j g=S (k) j (u(k) jopt):
Finally, by the homogeneity ofS(k)
j , we come toS(k) j (u(k) jopt) =u (k) jopt. u t
4. Multigrid and Error Multifunctional
with Agreement Conditions
Throughout the sequel we assume that
(S)
the optimal error multifunctional consists of the pairwise distinct functionals.This assumption is not superuous. An example is as follows. Let be the unit ball of Rn and let X be a space of harmonic
functions with domain. If members ofX are continuous functions in the closure of , then Mean Value Theorem implies that
((x)u(x)) = ( 1
jj
(x)u(x)) for 8u2X:
Now take an arbitrary multigrid and assume that ~x(0)
1 = 0. In this
event each optimal error functional is identically 0. In general the initial functional l(0)
1 does not agree with a linear
combination of Dirac delta functions, and the hypothesis (
S
) seems to be very natural. We now dwell in more detail on the explanation of this claim. Let Nod(k) j Nod(k 1 ) j1 . Thenl (k)j is an error functional with nodes
in Nod(k 1
)
j1 and the weights of l (k)
j at the points of Nod(k 1 ) j1 nNod (k) j equal 0. Hence,kl (k 1 ) j1 opt jX k kl (k) jopt jX k. Ifl (k 1 ) j1 opt=l (k) jopt, then kl (k) jopt jYk=kl (k) j+1opt jYk==kl (k 1 ) j1;1opt jYk=kl (k 1 ) j1 opt jYk:
Since the optimal cubature formula with the given set Nod(k 1
)
j1 of
nodes is unique it follows that
l(k) jopt=l (k) j+1opt= =l (k 1 ) j1;1opt=l (k 1 ) j1 opt:
Thus we extend the set Nod(k)
j of nodes to Nod(k 1
)
j1 and with it all
the norm of the optimal error functional does not decrease. Hence some levels of the initial multigrid contain of a few \irrelevant" nodes and the hypothesis (
S
) means that we do not consider a multi-grid with the \irrelevant" nodes. In this event we say that and the optimal error multifunctional satisfy the agreement conditions. In particular, under assumption (S
) for (kj)6= (k1j1) we have (4:1) Nod(k) j Nod(k 1 ) j1 = ) kl (k 1 ) j1 opt jYk<kl (k) jopt jYk:
Lemma 4.
Assuming(S)
the extremal functionu(k)jopt forl (k) joptis not equal to 0 at x~(k) j . Proof. Let u(k) jopt(~x (k) j ) = 0. Considering that u(k)
jopt equals 0 at the
nodes of Nod(k)
j , we come to the conclusion that forj (k);1 the
equalities hold (l(k) j+1optu (k) jopt) = (l (0) 1 u (k) jopt) = (l (k) joptu (k) jopt) = kl (k) jopt jYk 2:
Further, dividing both sides of this chains of the equalities by the normku (k) jopt jXk=kl (k) jopt jYk, we obtain kl (k) j+1opt jYk (l(k) j+1optu (k) jopt) ku (k) jopt jXk =kl (k) jopt jYk
which contradicts with (4.1). Ifj= (k), then we must considerl(k+1) 1opt
instead ofl(k)
j+1opt. u t
5. Constructing a -Hierarchical Basis
Let b(k)j =u(k)
jopt(~x (k)
j ). By Lemma 4,b(k)
j 6= 0, and we can deal with
the function 1
b(k)
j u
(k)
jopt(x). By Theorem 7, the function 1
b(k)
j u
(k)
jopt is a
member of the following ane variety
fv2Xj((x;x~ (k) j )v) = 1 ((x;x~ (m) i )v) = 08~x (m) i 2Nod (k) j g:
Let ~u(k)
jopt be the spline of this ane variety and let ~S (k)
j be the
corresponding spline operator, i.e., ~ u(k) jopt= ~S (k) j ( 1b(k) j u (k) jopt): By the denition of ~S(k) j , the function ~u(k)
joptagrees with the function 1
b(k)
j u
(k)
jopt(x) at the nodes of Nod (k) j fx~ (k) j g. Ifk < k 1 or (k=k1 and j j
1), then it follows in much the same way as in Lemma 3 that
(5:1) S~(k 1 ) j1 ~ S(k) j = ~S(k) j ~S(k) j (~u(k) jopt) = ~u (k) jopt:
Lemma 5.
For given integerskandj there is a unique functionh(k)j such that h(k) j 2 span fu~ (k) jopt(x)~u (k) j+1opt(x):::~u (k) (k)opt(x) g
and (1.1)holds. Assumingh~(k)
j = ~S(k) (k)(h (k) j ), the set H =f~h (k) j jk= 01:::j= 12::: (k)g
is a -hierarchical system of X. If for an arbitrary k the spline op-erator S~(k)
(k) is linear then h (k)
j = ~h(k)
j (x). Proof. Givenkandjwe seek a solutionh(k)
j to (1.1) in the form of the
linear combination of the splines ~u(k)
jopt, ~u (k)
j+1opt, :::, ~u (k)
(k)opt. To be
more precise, we assume that ~u(k)
(k)+1opt= ~u (k+1) 1opt fork= 012:::, and suppose (5:2) h(k) j = (k);j+1 X i=1 (jk) i u~(k) j+iopt ;u~ (k) jopt:
Examine that there are (jk)
i ,i= 12::: (k);j+ 1, such that h(k) j (~x(k) j+l) = 0 for l= 12::: (k) ;j (k);j+1 P i=1 (jk) i = 1:
By (2.7), we can write the last equalities in the equivalent form
l P i=1 (jk) i u~(k) j+iopt(~x (k) j+l) = ~u (k) jopt(~x (k) j+l) l= 12::: (k);j (k);j+1 P i=1 (jk) i = 1:
It is a system of linear equations with respect to unknown coe-cients ((jk)
1 ::: (jk)
(k);j+1). The matrix of this system is
subdiago-nal with 1 on the main diagosubdiago-nal. Hence, this matrix is non-singular and there exists a unique solution ((jk)
1 ::: (jk)
(k);j+1) to the system
under consideration. Thus, function h(k)
j is uniquely determined, and it is easy to show
that (1.1) holds andH is actually a -hierarchical system ofX. Let ~S(k)
(k), k = 012:::, be linear spline operators. By the
de-nition ofh(k)
j it follows from (5.1) that
h(k) j = (k);j+1 X i=1 (jk) i S~(k) j+i(~u (k) j+iopt) ;S~ (k) j (~u(k) jopt): Applying ~S(k)
(k)to both sides of the last equality and using (5.1) again,
we arrive at the sought equality ~h(k)
j =h(k)
j . ut
Theorem 9.
If there is a Frechet dierential of k j Yk at l 2 Y,l6= 0, and the spline operators S~ (k)
(k), k= 01:::, are linear then
H =f~h (k)
j (x)jk= 01:::j= 12::: (k)g
is a -hierarchical basis of X.
Proof. Since the spline operators ~S(k)
(k), k = 01:::, are linear it
follows from Lemma5 that ~h(k)
j =h(k)
j . Hence it is sucient to expand
an arbitrary function ' 2 X in the series with respect to h (k)
j and
check the convergence of this series in the norm ofX.
For a given function ' 2 X and a positive integer m there are
coecients g(k) j of the sum m(x) = Pm k=0 (k) P j=1 g(k) j h(k) j (x) such that
m(x) ='(x) for8x2 m. Considering thatH is a -hierarchical
system by easy calculations we obtain the following recurrent rela-tions for g(k) j (see 15] and 16]) g(0) j = m(~x(0) j ) ='(~x(0) j ) j = 12::: (0) '0(x) = (0) P j=1 g(0) j h(0) j (x)
and further fork= 12:::m g(k) j = m(~x(k) j );'k ;1(~x (k) j ) ='(~x(k) j );'k ;1(~x (k) j ) j = 12::: (k) 'k(x) ='k;1(x) + (k) P j=1 g(k) j h(k) j (x): The coecientsg(k)
j are independent ofm. Hence the function m(x)
is a partial sum of the series 1 P k=0 (k) P j=1 g(k) j h(k)
j . This series is convergent
to '(x) in the norm of X i limm
!1
k'; m j Xk = 0. Let us show
that it is a valid equality.
By the denition of m(x) and fromthe property of ~S(m)
(m)it follows that ~S(m) (m)(') = ~S (m) (m)( m). The linearity of ~S (m) (m) together with
Lemma 5 and equalities (5.1) imply that ~ S(m) (m)( m) = m P k=0 (k) P j=1 g(k) j S~(m) (m)(h (k) j ) = Pm k=0 (k) P j=1 g(k) j S~(m) (m) ~ S(k) (k)(h (k) j ) = Pm k=0 (k) P j=1 g(k) j S~(k) (k)(h (k) j ) = Pm k=0 (k) P j=1 g(k) j h(k) j = m:
We now show that limm
!1 k'; mjXk= lim m!1 k';S~ (m) (m)(') jXk= 0. Let M(m)= fv2Xjv(~x (p) s ) = 08~x (p) s 2 mg and V(m)(') ='+M(m): Then V(m+1)(') V(m)('). Since ' 2 T 1 m=0V (m)(') it follows that
form= 01::: the inequalities hold (5:3) kS~ (m) (m)(') jXk kS~ (m+1) (m+1)(') jXk k'jXk:
Hence the bounded sequencefkS~ (m) (m)(') jXkg 1 m=0of norms increases to a nite limitA (5:4) mlim !1 kS~ (m) (m)(') jXk=A k'jXk<1:
Because X is reexive, there is a weak convergent subsequence of
fS~ (m)
(m)(') g
1
m=0. Let ' be the weak limit of the subsequence ' 2
fS~ (m)
(m)(') g
1
m=0 is weakly convergent to'. Then for 8x~
(k)
j 2 1
m=0 m
the equalities hold
'(~x(k) j ) = ((x;x~ (k) j )') = limm!1 ((x;x~ (k) j )S~(m) (m)(')) ='(~x (k) j ):
The dierence ';'is continuous in and equals 0 at the nodes of S
1
m=0 m. By the assumption, the set S
1
m=0 m is dense in. Hence
';'is identically equal to 0 in and the weak limit of the spline
sequence fS~ (m)
(m)(') g
1
m=0 coincides with '. Let us show that splines
~
S(m)
(m)(') converge to' in the norm ofX.
Since fS~ (m)
(m)(') g
1
m=0 is weakly convergent to ' it follows from
(5.4) that (') satises k'jXk 2 = ( (')') = lim m!1( (') ~ S(m) (m)(')) k (')jYk lim m!1 kS~ (m) (m)(') jXk Ak'jXk:
Consequently, k' j Xk A. This estimation together with (5.4)
implies (5:5) k'jXk=A= lim m!1 kS~ (m) (m)(') jXk: Put vm = 1 k ~ S(m) (m) (')jXk ~ S(m) (m)('). Then kvm j Xk = 1. Applying
(5.5), it is not hard to validate that fvmg 1
m=0 converges weakly to 1
k'jXk'. In particular, the equalities hold
lim m!1 ( 1 k (')jXk (')vm) = 1 k'jXkk (')jXk ( (')') = 1:
Given the memberl= ( 1
k (')jXk') of the unit sphere ofY, we have
pointed out the sequence fvmg 1
m=0 of members of the unit sphere
of X = Y such that limm
!1
(lvm) = 1. By the hypothesis, there is
a Frechet dierential = N 0
G(l) of the norm k j Yk at l 2 Y.
Considering that (2.6) holds we have N 0
G (l) = N 0
G ( ( 1
k'jXk')) = 1
k'jXk'. By the Shmulian criterion 7, p.147], fvmg
1
m=0must converge
to = 1
k'jXk' in the norm of X. The proof of the convergence in
If kvm ;k 9 0, then 9" > 0 and flmg 1
m=0 Y such that klmjYk= 1 and (lmvm;) 2"for8m1. Let
~ lm= 1"(kljYk;(lvm))lm=k~lmjYklm: Then k~lmjYk= 1 "(1;(lvm))!0 and kl+ ~lmjYk;kljYk;(~lm) k~lmjYk (l+ ~lmvm);1;(~lm) k~lmjYk = (lvm);1 +k~lmjYk(lmvm;) k~lm jYk =;"+ (lmvm;)"
which contradicts that is a Frechet dierential ofkjYk atl2Y.
Finally, the equality holds limm!1
kvm;k= 0. Whence and from
(5.5) the spline sequence fS~ (m)
(m)(') g
1
m=0 converges to'in the norm
of X. ut
Acknowledgements The author thanks Professor C.Zenger for helpful discus-sions.
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