Error estimates for interpolation of rough data using the scattered shifts of a radial basis function
aa r X i v : . [ m a t h . NA ] M a y Error estimates for interpolation of rough data using thescattered shifts of a radial basis function
R. A. Brownlee Department of Mathematics, University of Leicester, Leicester LE1 7RH, England
Abstract
The error between appropriately smooth functions and their radial basis functioninterpolants, as the interpolation points fill out a bounded domain in IR d , is a wellstudied artifact. In all of these cases, the analysis takes place in a natural functionspace dictated by the choice of radial basis function—the native space. The nativespace contains functions possessing a certain amount of smoothness. This paper es-tablishes error estimates when the function being interpolated is conspicuously rough. MSC2000:
Keywords:
Scattered data interpolation, radial basis functions, error estimates, roughfunctions.
Short title:
Error estimates for interpolation of rough data. This author was supported by a studentship from the Engineering and Physical Sciences ResearchCouncil. Introduction
In this paper we are interested in interpolation of a finite scattered data set
A ⊂ IR d bytranslates of a single basis function. Of the differing set ups to this problem, the onepreferred in this paper is the following variational formulation. Firstly, we require a spaceof continuous functions Z which carries a seminorm. The minimal norm interpolant to f : A →
IR on A from Z is the function Sf ∈ Z which agrees with f on A and hassmallest seminorm amongst all other interpolants to f on A from Z . The particular spacewe shall be concerned with is Z m (IR d ) := (cid:26) f ∈ S ′ : d D α f ∈ L , loc (IR d ) , Z IR d w ( x ) | ( d D α f )( x ) | d x < ∞ , | α | = m (cid:27) , which carries the seminorm | f | m := (cid:18) X | α | = m c α Z IR d w ( x ) | ( d D α f )( x ) | d x (cid:19) / , f ∈ Z m (IR d ) . The constants c α are chosen so that P | α | = m c α x α = | x | m , for all x ∈ IR d . The notation S ′ is used to denote the usual Schwartz space of distributions. The space Z m (IR d ) ischristened the native space . The weight function function w : IR d → IR is initially chosento satisfy(W0) w ∈ C (IR d \ w ( x ) > x = 0;(W2) 1 /w ∈ L , loc (IR d );(W3) there is a positive µ ∈ IR such that ( w ( x )) − = O ( | x | − µ ) as | x | → ∞ .A consequence of (W0)–(W3) is that Z m (IR d ) is complete with respect to | · | m , and if m + µ − d/ > Z m (IR d ) is embedded in the continuous functions (see [6]). As thetitle of this work suggests, we expect this set up to admit minimal norm interpolants ofthe form ( Sf )( x ) := X a ∈A b a ψ ( x − a ) , for x ∈ IR d , (1.1)for an appropriate basis function ψ . We are not disappointed, but for brevity we omit thedetails which are well presented in [6]. The coefficients b a in (1.1) are determined by theinterpolation equations ( Sf )( a ) = f ( a ), a ∈ A . In some situations it may be necessaryto append a polynomial p onto (1.1) and take up the ensuing extra degrees of freedom bysatisfying the side conditions: X a ∈A b a q ( a ) = 0 , whenever q is a polynomial of the same degree (or less) as p . The archetypal scenariothe author has in mind is w ( x ) = | x | µ for x ∈ IR d , where µ < d/
2. This leads tominimal norm interpolants of the form (1.1) modulo a polynomial of degree m . Here, the2 adial basis function is ψ : x
7→ | x | m +2 µ − d log | x | if 2 m + 2 µ − d is an even integer or ψ : x
7→ | x | m +2 µ − d otherwise.It is of central importance to understand the behaviour of the error between a function f : Ω → IR and its interpolant as the set A becomes dense in a bounded domain Ω. Themeasure of density we employ is the fill-distance h := sup x ∈ Ω min a ∈A | x − a | . One findsthat there is a positive constant γ ( m ), independent of h , such that for all f ∈ Z m (IR d ), k f − Sf k L (Ω) = O ( h γ ( m ) ) , as h → . It is natural to ask: what happens if the function being approximated does not lie in Z m (IR d )? It may well be that f lies in Z k (IR d ), where k < m and k + µ − d/ >
0. Thecondition k + µ − d/ > f ( a ) exists for each a ∈ A , so Sf certainly exists.It is tempting to conjecture that the new error estimate should be k f − Sf k L (Ω) = O ( h γ ( k ) ) , as h → . We are conjecturing the same approximation order as if we had instead approximated f with the minimal norm interpolant to f on A from Z k (IR d ). This is precisely whathappens in the case w = 1, which was considered by Brownlee & Light in [2]. In thiswork, with the aid of a recent result from [1] (Lemma 2.5), we employ the technique usedby Brownlee & Light to extend their work to more general weight functions. Theorem 3.5is the definitive result we obtain. The interested reader may enjoy consulting the relatedpapers [7, 8, 9, 10].To close this section we introduce some notation that will be employed throughout thepaper. A domain is understood to be a connected open set. The support of a function φ : IR d → IR, denoted by supp ( φ ), is defined to be the closure of the set { x ∈ IR d : φ ( x ) =0 } . We make much use of the linear space Π m (IR d ) which consists of all polynomials ofdegree at most m in d variables. We fix ℓ as the dimension of this space. Finally, when wewrite b f we mean the Fourier transform of f . The context will clarify whether the Fouriertransform is the natural one on L (IR d ), b f ( x ) := (2 π ) − d/ R IR d f ( t )e − i xt d t , or one of itsseveral extensions to L (IR d ) or S ′ . In this section we gather a number of useful results, chiefly about the sorts of extensionswhich can be carried out on our native spaces. This will first require us to establishthe notion of local native spaces. To do this, we rewrite the seminorm | f | m in its directform —that is, without the Fourier transform of f appearing explicitly. Let us demandthat w satisfies the following additional axioms:(W4) w ( y ) = w ( − y ) for all y ∈ IR d ;(W5) w (0) = 0 and b w ( x ) ≤ x ∈ IR d ;(W6) b w is a measurable function and for any neighbourhood N of the origin, b w ∈ L (IR d \ N ); 3W7) | b w ( y ) | = O ( | y | λ ) as y → λ + d + 2 > | f | m = − X | α | = m c α Z IR d Z IR d b w ( x − y ) | ( D α f )( x ) − ( D α f )( y ) | d x d y, f ∈ Z m (IR d ) . (2.1)The notation C m (IR d ) is used for the space of compactly supported m -times continuouslydifferentiable functions on IR d . Now, let us define the following space for a domain Ω ⊂ IR d , X m (Ω) := n f | Ω : f ∈ C m (IR d ) , | f | m, Ω < ∞ o , where | f | m, Ω := (cid:18) − X | α | = m c α Z Ω Z Ω b w ( x − y ) | ( D α f )( x ) − ( D α f )( y ) | d x d y (cid:19) / , f ∈ X m (Ω) . A norm is placed on X m (Ω) via k f k m, Ω := (cid:16) k f k W m (Ω) + | f | m, Ω (cid:17) / , f ∈ X m (Ω) . The notation X m (Ω) denotes the completion of X m (Ω) with respect to k · k m, Ω , while Y m (Ω) denotes the completion of X m (Ω) with respect to | · | m, Ω . It is these spaces thatwe call the local native spaces .We are nearly ready to state our first extension theorem, but first it is necessary to takeon board four additional axioms and introduce an important type of bounded domain: Definition 2.1.
Let Ω and Ω be domains in IR d , and Φ a bijection from Ω to Ω .We say that Φ is m -smooth if, writing Φ( x ) = ( φ ( x , . . . , x d ) , . . . , φ d ( x , . . . , x d )) and Φ − ( x ) = Ψ( x ) = ( ψ ( x , . . . , x d ) , . . . , ψ d ( x , . . . , x d )) , then the functions φ , . . . , φ d belongto C m (Ω ) and ψ , . . . , ψ d belong to C m (Ω ) . Let Φ be a bijection from IR d to IR d . Wesay Φ is locally m -smooth if Φ is m -smooth on every bounded domain in IR d . (W8) for every locally ( m + 1)-smooth map φ on IR d , and every bounded subset Ω of IR d ,there is a C > b w ( φ ( x ) − φ ( y )) ≤ C b w ( x − y ), for all x, y ∈ Ω;(W9) there exists a constant C > x = ( x ′ , x d ) ∈ IR d and y = ( x ′ , y d ) ∈ IR d with | x d | ≥ | y d | , then b w ( x ) ≤ C b w ( y ).(W10) R A b w < A has positive measure;(W11) b w ( y ) = b w ( − y ) for all y ∈ IR d . Definition 2.2.
Let B = { ( y , y , . . . , y d ) ∈ IR d : | y j | < , ≤ j ≤ d } , and set B + = { y ∈ B : y = ( y ′ , y d ) and y d > } and B = { y ∈ B : y = ( y ′ , y n ) and y n = 0 } . A boundedconvex domain Ω in IR d with boundary ∂ Ω will be called a V-domain if the following allhold: A1) there exist open sets G , . . . , G N ⊂ IR d such that ∂ Ω ⊂ S Nj =1 G j ; (A2) there exist locally (m+1)-smooth maps φ j : IR d → IR d such that φ j ( B ) = G j , φ j ( B + ) = G j ∩ Ω and φ j ( B ) = G j ∩ ∂ Ω , j = 1 , . . . , N ;(A3) let Ω δ be the set of all points in Ω whose distance from ∂ Ω is less than δ . Thenfor some δ > , Ω δ ⊂ N [ j =1 φ j (cid:18)(cid:26) ( y , y , . . . , y d ) ∈ IR d : | y j | < m + 1 , ≤ j ≤ d (cid:27)(cid:19) . The definition of a V-domain is taken from a paper by Light & Vail [5] in whichextension theorems for our local native spaces are considered.
Theorem 2.3 (Light & Vail [5]) . Let Ω ⊂ IR d be a V-domain. Let b w : IR d → IR satisfy(W6)–(W11). Then there exists a continuous linear operator L : X m (Ω) → X m (IR d ) suchthat for all f ∈ X m (Ω) ,1. Lf = f on Ω ;2. supp ( Lf ) is compact and independent of f ;3. k Lf k m, IR d ≤ K k f k m, Ω , for some positive constant K = K (Ω) independent of f . A feature of the construction of the extension operator in Theorem 2.3 is that Lf canbe chosen to be supported on any compact subset of IR d containing Ω. For details of theconstruction, the reader should consult [5]. Also at our disposal is a seminorm version ofTheorem 2.3: Theorem 2.4 (Light & Vail [5]) . Let Ω ⊂ IR d be a V-domain. Let b w : IR d → IR satisfy(W6)–(W11). Given f ∈ Y m (Ω) , there exists a function f Ω ∈ Y m (IR d ) such that:1. f Ω = f on Ω ;2. | f Ω | m, IR d ≤ C | f | m, Ω , for some positive constant C = C (Ω) independent of f . It is convenient for us to be able to work with a norm on X m (Ω) that is equivalent to k · k m, Ω . Lemma 2.5 (Brownlee & Levesley [1]) . Let Ω ⊂ IR d be a V-domain. Let w : IR d → IR satisfy (W0)–(W12) and let m + µ − d/ > . Let b , . . . , b ℓ ∈ Ω be unisolvent with respectto Π m (IR d ) . Define a norm on X m (Ω) via k f k Ω := (cid:18) | f | m, Ω + ℓ X i =1 | f ( b i ) | (cid:19) / , f ∈ X m (Ω) . There are positive constants K and K such that for all f ∈ X m (Ω) , K k f k m, Ω ≤ k f k Ω ≤ K k f k m, Ω . K (Ω) in the statement of Theorem 2.3 can be understoodfor simple choices of Ω. To realise this, we require that the weight function satisfies onefurther and final axiom:(W12) there exists C , C > C h λ b w ( x ) ≤ b w ( hx ) ≤ C h λ b w ( x ), for all h > x ∈ IR d .Now, an elementary change of variables gives us: Lemma 2.6.
Let Ω be a measurable subset of IR d . Let w : IR d → IR be a measurablefunction that is nonpositive almost everywhere and satisfies (W11). Define the mapping σ : IR d → IR d by σ ( x ) = a + h ( x − t ) , where h > , and a , t , x ∈ IR d . Then there exists aconstant K , K > , independent of Ω , such that for all f ∈ Y m ( σ (Ω)) , K ≤ | f ◦ σ | m, Ω h m − λ/ − d | f | m,σ (Ω) ≤ K . We are now ready to state the key result of this section, but before doing this let usmake a simple observation. Look at the unisolvent points b , . . . , b ℓ in the statement ofLemma 2.5. Since X m (Ω) can be embedded in C (Ω), it makes sense to talk about theinterpolation operator P : X m (Ω) → Π m (IR d ) based on these points. Lemma 2.7.
Let w : IR d → IR satisfy (W0)–(W12). Let B be any ball of radius h andcentre a ∈ IR d , and let f ∈ X m ( B ) . Whenever b , . . . , b ℓ ∈ IR d are unisolvent with respectto Π m (IR d ) let P b : C (IR d ) → Π m (IR d ) be the Lagrange interpolation operator on b , . . . , b ℓ .Then there exists c = ( c , . . . , c ℓ ) ∈ B ℓ and g ∈ X m (IR d ) such that1. g ( x ) = ( f − P c f )( x ) for all x ∈ B ;2. g ( x ) = 0 for all | x − a | > h ;3. there exists a C > , independent of f and B , such that | g | m, IR d ≤ C | f | m,B .Furthermore, c , . . . , c ℓ can be arranged so that c = a .Proof. Let B be the unit ball in IR d and let B = 2 B . Let b , . . . , b ℓ ∈ B be unisolventwith respect to Π m (IR d ). Define σ ( x ) = h − ( x − a ) for all x ∈ IR d . Set c i = σ − ( b i )for i = 1 , . . . , ℓ so that c , . . . , c ℓ ∈ B are unisolvent with respect to Π m (IR d ). Take f ∈ X m ( B ). Then ( f − P c f ) ◦ σ − ∈ X m ( B ). Set F = ( f − P c f ) ◦ σ − . Let F B beconstructed as an extension to F on B . By Theorem 2.3 and the remark following it, wecan assume F B is supported on B . Define g = F B ◦ σ ∈ X m (IR d ). Let x ∈ B . Since σ ( B ) = B there is a y ∈ B such that x = σ − ( y ). Then, g ( x ) = ( F B ◦ σ )( x ) = F B ( y ) = (( f − P c f ) ◦ σ − )( y ) = ( f − P c f )( x ) . Also, for x ∈ IR d with | x − a | > h , we have | σ ( x ) | >
2. Since F B is supported on B , g ( x ) = 0 for | x − a | > h . Hence, g satisfies properties and . By Theorem 2.3 there isa K , independent of f and B , such that k F B k m,B ≤ k F B k m, IR d ≤ K k F k m,B .
6e have seen in Lemma 2.5 that if we endow X m ( B ) and X m ( B ) with the norms k v k B i = (cid:18) | v | m,B i + ℓ X i =1 | v ( b i ) | (cid:19) / , i = 1 , , then k · k B i and k · k m,B i are equivalent for i = 1 ,
2. Thus, there are constants K and K ,independent of f and B , such that k F B k B ≤ K k F B k m,B ≤ K K k F k m,B ≤ K K K k F k B . Set C = K K K . Since F B ( b i ) = F ( b i ) = ( f − P c f )( σ − ( b i )) = ( f − P c f )( c i ) = 0 for i = 1 , . . . , ℓ , it follows that | F B | m,B ≤ C | F | m,B . Thus, | g ◦ σ − | m, IR d ≤ C | ( f − P c f ) ◦ σ − | m,B . Now, Lemma 2.6 can be employed twice to provide us with constants C and C >
0, independent of f and B , such that | g | m, IR d ≤ C h d + λ/ − m | g ◦ σ − | m, IR d ≤ C C h d + λ/ − m | ( f − P c f ) ◦ σ − | m,B ≤ C C C | f − P c f | m,B . Finally, we observe that | f − P c f | m,B = | f | m,B to complete the first part of the proof. Theremaining part follows by selecting b = 0 and choosing b , . . . , b ℓ accordingly in the aboveconstruction. In this section we establish the error estimate conjectured in the introduction. We beginwith a function f in Z k (IR d ). We want to estimate k f − S m f k L (Ω) , (3.1)where S m is the minimal norm interpolation operator from Z m (IR d ) on A and m > k .The essence of the proof is as follows. Firstly, by adjusting f , we obtain a function ˜ f ,still in Z k (IR d ), with seminorm in Z k (IR d ) not too far from that of f . We then smooth˜ f by convolving it with a function φ ∈ C ∞ (IR d ). The key feature of the adjustment of f to F := φ ∗ ˜ f is that F ( a ) = f ( a ) for every point a ∈ A (Theorem 3.4). This enablesus to replace S m f with S m F in (3.1). Furthermore, it follows that F ∈ Z m (IR d ) so wecan employ an existing L -error estimate to F − S m F . The remaining part of the error, f − F , is easily dealt with as it vanishes on A . Finally, Lemma 3.1 takes us back to anerror estimate in Z k (IR d ). Lemma 3.1.
Let w : IR d → IR satisfy (W0) and (W1). Let k ≤ m and φ ∈ C ∞ (IR d ) .For each h > let φ h ( x ) = h − d φ ( x/h ) for x ∈ IR d . Then there exists a constant C > ,independent of h , such that for all f ∈ Z k (IR d ) , | φ h ∗ f | m, IR d ≤ Ch k − m | f | k, IR d .Proof. The case w = 1 is established in [2]. The proof for this more general set up doesnot differ substantially so is omitted. 7 emma 3.2 (Brownlee & Light [2]) . Suppose φ ∈ C ∞ (IR d ) is supported on the unit balland satisfies Z IR d φ ( x ) d x = 1 and Z IR d φ ( x ) x α d x = 0 , for all < | α | ≤ k. For each ε > and x ∈ IR d , let φ ε ( x ) = ε − d φ ( x/ε ) . Let B be any ball of radius h andcentre a ∈ IR d . For a fixed p ∈ Π k (IR d ) let f be a mapping from IR d to IR such that f ( x ) = p ( x ) for all x ∈ B . Then ( φ ε ∗ f )( a ) = p ( a ) for all ε ≤ h . Definition 3.3.
Let Ω be an open, bounded subset of IR d . Let A be a set of points in Ω .The quantity h := sup x ∈ Ω inf a ∈A | x − a | is called the fill-distance of A in Ω . The separationof A is given by the quantity q := min a,b ∈A a = b | a − b | . The quantity h/q will be called themesh-ratio of A . Theorem 3.4.
Let w : IR d → IR satisfy (W0)–(W12). Let k + µ − d/ > and m ≥ k .Let A be a finite subset of IR d of separation q > . Then for all f ∈ X k (IR d ) there existsan F ∈ X m (IR d ) such that1. F ( a ) = f ( a ) for all a ∈ A ;2. there exists a C > , independent of f and q , with | F | k, IR d ≤ C | f | k, IR d and | F | m, IR d ≤ Cq k − m | f | k, IR d .Proof. Take f ∈ X k (IR d ). For each a ∈ A let B a ⊂ IR d denote the ball of radius δ = q/ a . For each B a let g a be constructed in accordance with Lemma 2.7. That is,for each a ∈ A take c ′ = ( c , . . . , c ℓ ) ∈ B ℓ − a and g a ∈ X k (IR d ) such that1. a, c , . . . , c ℓ are unisolvent with respect to Π k (IR d )2. g a ( x ) = ( f − P ( a,c ′ ) f )( x ) for all x ∈ B a ;3. P ( a,c ′ ) f ∈ Π k (IR d ) and ( P ( a,c ′ ) f )( a ) = f ( a );4. g a ( x ) = 0 for all | x − a | > δ ;5. there exists a C >
0, independent of f and B a , such that | g a | k, IR d ≤ C | f | k,B a .Note that if a = b , then supp ( g a ) does not intersect supp ( g b ), because if x ∈ supp ( g a )then | x − b | > | b − a | − | x − a | ≥ q − δ = 6 δ. U = S b ∈A supp ( g b ), then writing IR d = (IR d \ U ) ∪ U we obtain (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A g a (cid:12)(cid:12)(cid:12)(cid:12) k, IR d = − X | α | = k c α Z IR d Z IR d b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y = − X | α | = k c α (cid:18)Z IR d \ U Z IR d \ U b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y + 2 Z IR d \ U Z U b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y + Z U Z U b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y (cid:19) . (3.2)We shall now consider each of the double integrals in (3.2) separately. Firstly, the integralover (IR d \ U ) × (IR d \ U ) is zero because P a ∈A g a is supported on U . Next, using theobservation above regarding the support of g a , a ∈ A , it follows that − X | α | = k c α Z IR d \ U Z U b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y = X b ∈A − X | α | = k c α Z IR d \ U Z supp ( g b ) b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A ( D α g a )( x ) (cid:12)(cid:12)(cid:12)(cid:12) d x d y = X b ∈A − X | α | = k c α Z IR d \ U Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) | d x d y = X b ∈A − X | α | = k c α Z IR d \ U Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g b )( y ) | d x d y ≤ X b ∈A | g b | k, IR d . (3.3)Before calculating the final integral let us examine the following expression for b ∈ A and α ∈ ZZ d + with | α | = m , X c ∈A c = b Z supp ( g c ) Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g c )( y ) | d x d y = X c ∈A c = b Z supp ( g c ) Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g b )( y ) + ( D α g c )( x ) − ( D α g c )( y ) | d x d y ≤ X c ∈A c = b Z supp ( g c ) Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g b )( y ) | d x d y + 2 X c ∈A c = b Z supp ( g c ) Z supp ( g b ) b w ( x − y ) | ( D α g c )( x ) − ( D α g c )( y ) | d x d y Z IR d Z IR d b w ( x − y ) | ( D α g b )( x ) − ( D α g b )( y ) | d x d y + 2 X c ∈A c = b Z IR d Z supp ( g b ) b w ( x − y ) | ( D α g c )( x ) − ( D α g c )( y ) | d x d y. (3.4)Finally, using the observation regarding the support of g a , a ∈ A , once again and (3.4) itfollows that − X | α | = k c α Z U Z U b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y = X b ∈A X c ∈A − X | α | = k c α Z supp ( g c ) Z supp ( g b ) b w ( x − y ) (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A (( D α g a )( x ) − ( D α g a )( y )) (cid:12)(cid:12)(cid:12)(cid:12) d x d y = X b ∈A X c ∈A − X | α | = k c α Z supp ( g c ) Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g c )( y ) | d x d y = X b ∈A − X | α | = k c α Z supp ( g b ) Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g b )( y ) | d x d y + X b ∈A X c ∈A c = b − X | α | = k c α Z supp ( g c ) Z supp ( g b ) b w ( x − y ) | ( D α g b )( x ) − ( D α g c )( y ) | d x d y ≤ X b ∈A | g b | k, IR d + 2 X b ∈A | g b | k, IR d + 2 X c ∈A | g c | k, IR d ≤ X b ∈A | g b | k, IR d . (3.5)Substituting (3.3) and (3.5) into (3.2) we find (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A g a (cid:12)(cid:12)(cid:12)(cid:12) k, IR d ≤ X a ∈A | g a | k, IR d . Hence, applying Condition 5 to the above inequality we have (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A g a (cid:12)(cid:12)(cid:12)(cid:12) k, IR d ≤ C X a ∈A | f | k,B a ≤ C | f | k, IR d . Now set H = f − P a ∈A g a . It then follows from Condition 1 that H ( x ) = ( P ( a,c ′ ) f )( x ) forall x ∈ B a , and from Condition 3 that H ( a ) = f ( a ) for all a ∈ A . Let φ ∈ C ∞ (IR d ) besupported on the unit ball and enjoy the properties Z IR d φ ( x ) d x = 1 and Z IR d φ ( x ) x α d x = 0 , for all 0 < | α | ≤ k. Now set F = φ δ ∗ H . Using Lemma 3.1, there is a constant C >
0, independent of q and10 , such that | F | m, IR d ≤ C δ k − m ) (cid:12)(cid:12)(cid:12)(cid:12) f − X a ∈A g a (cid:12)(cid:12)(cid:12)(cid:12) k, IR d ≤ C δ k − m ) (cid:18) | f | k, IR d + (cid:12)(cid:12)(cid:12)(cid:12)X a ∈A g a (cid:12)(cid:12)(cid:12)(cid:12) k, IR d (cid:19) ≤ C (1 + 7 C ) δ k − m ) | f | k, IR d . Similarly, there is a constant C >
0, independent of q and f , such that | F | k, IR d ≤ C (cid:12)(cid:12)(cid:12)(cid:12) f − X a ∈A g a (cid:12)(cid:12)(cid:12)(cid:12) k, IR d ≤ C (1 + 7 C ) | f | k, IR d . Thus | F | m, IR d ≤ Cq k − m | f | k, IR d and | F | k, IR d ≤ C | f | k, IR d for some appropriate constant C >
0. Since F = φ δ ∗ H and H | B a ∈ Π k (IR d ) for each a ∈ A , it follows from Lemma 3.2that F ( a ) = H ( a ) = f ( a ) for all a ∈ A . Theorem 3.5.
Let Ω ⊂ IR d be a V-domain and let w : IR d → IR be a measurable functionsatisfying (W0)–(W12). Let k + µ − d/ > and m ≥ k . For each h > , let A h bea finite, Π m (IR d ) -unisolvent subset of Ω with fill-distance h . Assume also that there is aquantity ρ > such that the mesh-ratio of each A h is bounded by ρ for all h > . For eachmapping f : A h → IR , let S hm f be the minimal norm interpolant to f on A h from Z m (IR d ) .Then there exists a constant C > , independent of h , such that for all f ∈ Y k (Ω) , k f − S hm f k L (Ω) ≤ Ch k − λ/ − d/ | f | k, Ω , as h → . Proof.
Take f ∈ X k (IR d ). Construct F in accordance with Theorem 3.4 and set G = f − F .Then F ( a ) = f ( a ) and G ( a ) = 0 for all a ∈ A h . Furthermore, there is a constant C > f and h , such that | F | m, IR d ≤ C (cid:16) hρ (cid:17) k − m | f | k, IR d , | G | k, IR d ≤ | f | k, IR d + | F | k, IR d ≤ (1 + C ) | f | k, IR d . (3.6)Thus S hm f = S hm F and S hk G = 0, where we have adopted the obvious notation for S hk .Hence, k f − S hm f k L (Ω) = k ( F + G ) − S hm F k L (Ω) ≤ k F − S hm F k L (Ω) + k G − S hk G k L (Ω) . Now, employing the error estimate in [1], there are positive constants C > C > h and f , such that k f − S hm f k L (Ω) ≤ C h m − λ/ − d/ | F | m, Ω + C h k − λ/ − d/ | G | k, Ω , as h → k f − S hm f k L (Ω) ≤ C h k − λ/ − d/ | f | k, IR d , as h →
0, (3.7)for some appropriate C >
0. In particular, (3.7) holds for all f ∈ X k (IR d ). As Y k (IR d )is a dense linear subspace of X k (IR d ) then (3.7) extends to hold for all f ∈ Y k (IR d ) usinga standard normed space argument [3, Page 180]. To complete the proof we now let f ∈ Y k (Ω) and define f Ω in accordance with Theorem 2.4. It follows that there is a C > k f − S hm f k L (Ω) ≤ C h k − λ/ − d/ | f Ω | k, IR d ≤ C C h k − λ/ − d/ | f | k, Ω , as h → cknowledgements It is a pleasure to acknowledge that this paper is a generalisation of joint work with WillLight. The author would like to dedicate this paper to Will’s memory.
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