# On weighted average interpolation with cardinal splines

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## Splines

J. López-Salazar • G. Pérez-Villalón

Abstract Given a sequence of data {y„}„ez with polynomial growth and an odd number d, Schoenberg proved that there exists a unique cardinal spline / of degree d with polynomial growth such that f(n) = yn for all n e Z. In this work, we show that this result also holds if we consider weighted average data f * h(n) = yn, whenever the average function h satisfles some light conditions. In particular, the interpolation result is valid if we consider cell-average data f_ " f(x)dx = y„ with 0 < a < 1/2. The case of even degree d is also studied.

Keywords spline • cardinal interpolation • average sampling

### 1 Introduction

During the last forty years, the spaces of splines have become one of the most useful func-tion spaces in applied mathematics. Within that área, this paper is devoted to the topic of interpolation with spline functions. Let fid be the central B-spline of degree ¿ G N given by

fid = AÍ-i/2,1/2] * • • • * #[-i/2,i/2] (d + 1 terms),

where #[-1/2,1/2] denotes the characteristic function of the interval [—1/2, 1/2] and the sym-bol * denotes the integral convolution. In this work we consider the space Sd generated by the integer shifts of the B-spline fid. That is, a function / belongs to Sd if and only if there is a unique sequence {a^ltez in C such that

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As is well-known, if d is odd, then Sd coincides with the space of all functions / e Cd~l (R) such that f\[k,k+i] is a polynomial of degree not exceeding d for each k e Z. If d is even, then Sd coincides with the space of all functions / e Cd_1(]R) such that f\[k-i/2,k+i/2\ is a

polynomial of degree not exceeding d for every k e Z.

Given a sequence of real or complex numbers {y„}„ez, there is a unique linear spline

f £ Si such that / ( « ) = y„ for every n e Z , which is the function obtained by linear in-terpolation between every pair of consecutive data. On the contrary, for d > 2, there are inflnitely many splines / e Sd such that / ( « ) = y„ for n e Z. However, Schoenberg proved in the seventies that if the function / is required to have some growth conditions, then the interpolation problem has a unique solution, as the following theorem shows.

Theorem 1 (Shoenberg [9]) Let a > 0. If{y„}„ez is a sequence in C such that y„ = O (\n \a)

as n -> ±oo, then there is a unique function f e Sd such that f(n) = yn for all n e Z and

f(x) = O(\x\a) as x^±oo.

Because of physical reasons, the available data often are not the valúes of a function / at n, but weighted averages near n. That is,

### /

1/2 j-n+1/2

f(n — x)h(x)dx = l f(x)h(n — x)dx,

-1/2 Jn-1/2

where the average function h, with support in [—1/2, 1/2], reflects the characteristic of the acquisition device. Note that [—1/2, 1/2] is the máximum possible support of h without overlap between the samples. The average interpolation problem / * h(n) = y„ has been studied for band-limited functions and for shift invariant spaces in [1-8, 11, 12] and [13].

The aim of this paper is to prove the following theorem.

Theorem 2 Let h : R -> R be a measurable function that satisfles the following properties: (a) h(x)>OforallxeR.

(b) The support of h is contained in [—1/2, 1/2]. (c) O < f_1/2h(x)dx < oo andO < fQ h(x)dx < oo.

Let ¿ G N and a > 0. If {y„}„ez is a sequence in C such that y„ = 0(\n\a) as n —»• ± o o ,

then there is a unique function f e Sd such that f * h(n) = y„ for all n e Z and f(x) = 0 ( | x |a) asx ^ ± o o .

Theorem 2 was proven in [5] for degree d = 1, 2, 3, 4. In [6] and [7], it was proved with-out limitation on the degree, but with more restrictive conditions on the average function h.

Our approach to Theorem 2 is based on the following result.

Theorem 3 (Pérez and Portal [5]) Let d e N and let h be a function with the properties given in Theorem 2. Let us assume that all the zeros of the function

## G(.t) = J2[Pd*Hk)]r

k keZ

are simple and none ofthem is on the unit árele [z e C : |z| = 1}. If a > O and {y„}„ez is a

sequence in C such that y„ = 0(\n\a) as n —»• ± o o , then there is a unique function f e Sd

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In order to apply Theorem 3, all the results presen ted below will be used to pro ve that the zeros of G satisfy the required conditions.

After this paper was flnished, the authors knew that the same result had been obtained independently and at the same time by Ponnaian and Shanmugam in [8]. Although both papers use Theorem 3, the way those authors follow to study the zeros of the G function is different from ours. Ponnaian and Shanmugam base their arguments on the study that they carry out about the roots of the exponential Euler splines. In particular, they need to obtain a recursion relation for those exponential splines. Moreover, they consider four different cases on the degree d. On the contrary, our arguments may be more direct, since we apply already known properties of the Euler-Frobenius polynomials and only have to consider even and odd degree cases separately.

### 2 The Roots of G when d Is Even

The foliowing lemma gives the degree of the Laurent polynomial G.

Lemma 1 Let h be a function with the properties given in Theorem 2. For d eN and t e Z , it holds

### (i) p

d*h(k) = 0if\k\>d-f.

### (ii) p

d*h(k)>0if\k\<d-^.

Proof Let us recall that pd is a continuous function such that pd{x) > 0 if

•1+1 d±

2 ' 2

d+1 ^ „ ^ d+1

and pd(x) = 0 if x i ( - ^ á^k). The support of h is contained in [ - 1 / 2 , 1/2], so

,-1/2

### pd*h(k)= I pd(k — x)h(x)dx.

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If \k\ > ^ and x e [ - 1 / 2 , 1/2], then pd(k - x) = 0, so pd *h(k) = 0.

Since fQ h(x)dx > 0, there is e e (0, 1/2) such that fs h(x)dx > 0. Let

í 1 d+1 1 M = mini pd(x) : — < x < e i > 0.

I f 0 < i t < ^±i a n d x e [ e , 1/2], then/fc-x e [-±, ^ - e]. Henee

,.1/2 ^.1/2

Pd*h(k)> pd(k - x)h{x)dx > M • / h(x)dx>0.

Similarly, using that f°1/2 h(x)dx > 0, we obtain that pd * h(k) > 0 if - ?±± < k < 0. D

Let us now define the splines Tí d that will be extensively used throughout the paper. For

each t e C\{0} and each d eN, the symbol Tí d denotes the function

### rtJ{x) = YJt-kPÁx-k).

keZ

It is easy to check that if x e R and n e Z, then

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Moreover, if d > 2, then

M

M

1

### (x + i/2). (2)

Property (2) can be deduced from the following known fact:

P'd(x) = fa-iix + 1/2) - ^ _ i ( x - 1/2).

Lemma 2 For any f < 0 and any d e N, thefunction Tíjd has a unique root in [—1/2, 1/2).

Proof By (1), Tí j d(l/2) = f_ 1Tí d(—1/2). Since f < 0, there are two possibilities:

M

M

### (i/2) = o

or

Tíjd (—1/2) and Tíjd (1/2) have different sign.

The case d = 1 can be easily deduced in any of the above possibilities, having in mind that T(j i (x) is a piecewise linear spline with knots in Z and Tt, i (0) = 1.

We now assume that the result holds for some d G N; that is, there is a unique aíjd e

[ - 1 / 2 , 1/2) such that Tí>d(aí>d) = 0. The spline Tí>d+1 is a Cd-function on R. By (2),

rM + i M = (i - f)Tí>d(x + 1/2) = (1 - O f - ' r ^ í x - 1/2).

If Tí j d + 1 has a local extremum at a point x0 € (—1/2, 1/2), then T /d + 1 (x0) = 0, so Tíjd(x0 +

1/2) = Tí d( x0 — 1/2) = 0. Consequently, Tí d + 1 has at most one local extremum x0 on the

i n t e r v a l ( - l / 2 , 1/2):

xo = at¡d- 1/2 if at4 e (0, 1/2)

or

_xo = at,d+ l / 2 if «í,d G [ - 1 / 2 , 0).

That implies the following consequences:

1. If Tí > d + 1( - l / 2 ) = Tí > d + 1(l/2) = 0, then Tí>d+1(x) > 0 for every x G ( - 1 / 2 , 1/2) or

Tí d + 1( x ) < 0 for every x G (—1/2, 1/2). In this case, —1/2 is the unique root of Tí d + 1

on [ - 1 / 2 , 1/2).

2. If Tíjd+1(—1/2) and Tí j d + 1(l/2) have different sign, then there is aí j d +i G (—1/2, 1/2)

such that Tí d + 1( aí d + 1) = 0. As Tí d + 1 has at most one local extremum on (—1/2, 1/2),

it follows that Tí j d + 1(x) ^ 0 for every x G [—1/2, 1/2), x ^ «í>d+i.

This concludes the proof for d + 1, so the result holds for every d e N. D

In the proof of the following lemmas we use the Euler-Frobenius polynomials, which are deflned in terms of the forward B-splines

/ d+l\ Qd+i(x) = pd\x — I.

Here Qd+i (x) ^ 0 if and only if x G (0, d + 1). The Euler-Frobenius polynomial of degree d — 1 is the function

d - l

### nd(t) = d\J2Qd+i(k + i)tk.

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Then 77d is a monic polynomial of degree d — 1 whose roots are all simple and negative. If

A.1 < • • • < Xd_! are the roots of 77d and /¿i < • • • < n¿i-2 a r e the roots of /7d_!, then

A.1 < ¡A-i < X2 < H2 < A.3 < • • • < Xd_2 < Md-2 < A.d_i.

Moreover, X1Xd_1 = X2Xd_2 = • • • = 1, so if <¿ is even, then

^-d/2 — —

1-For the proof of these facts, see Schoenberg [10, pp. 391-392].

Lemma 3 Let d e N be even and let k\ < • • • < kd_\ be the roots of nd. Then sign(r,j > d(x)) = (-l)>+f

for every x e (—1/2, 1/2) and every j e { 1 , . . . , d — 1}.

Proof Let X be any root of 77d. Then

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¿fceZ ¿fceZ

+ 1

d + 1

X1- !

¿fceZ di

d

### (X) = o.

By Lemma 2, T^jd only has one root on [—1/2, 1/2), so the sign of T^jd is constant on

(—1/2, 1/2). Since 7ij d(—1/2) = 0, we obtain the following alternatives:

TX4{x) > 0 f o r a l l x e ( - 1 / 2 , 1/2) i f T ¿d( - l / 2 ) > 0

or

YKd{x) < 0 f o r a l l x e ( - 1 / 2 , 1/2) if T £d( - l / 2 ) < 0.

We now study the sign of T[d{-1/2). By (2),

d

á

k

### P*-Ák)

¿fceZ ^ ' ¿fceZ

( I - A . ) ; .1- !

### ^A.*e¿(* + i):

V ~ ? -77^!(X). ( á - 1 ) !

Since X < 0 and x e (—1/2, 1/2), we have

sign(T,,d(x)) = s i g n ( T £d( - l / 2 ) ) = ( - I )1" ! • sign(/7d_1(X))

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Let us flrst assume that d = 2. Since Yl\ (f) = 1 for every t e R, we have

si gn(T,,2(x)) = ( - 1 )1- ! . sign(77i(A.)) = 1

for all x e (—1/2, 1/2). That pro ves the statement of the Lemma when d = 2.

If d is an even integer bigger than 2, A.i < • • • < kd_i are the roots of 77d and /¿i < • • • <

/¿d-2 are the roots of 77d_!, then

A.1 < ni < X2 < n2 < k3 < • • • < Xd_2 < Md-2 < A.d_i.

As 77d_! is monic and its roots are simple, it can be written as

d

d

## -2).

Then 77¿_i(A.i) > 0, nd^(k2) < 0, ... 77¿_i(A.¿_i) > 0. That is,

sign(77d_1(^)) = ( - l )j + 1

for every j = l,... ,d — 1. Therefore, if x e (—1/2, 1/2), then

signfa.,„(*)) = ( - I )1" ! • sign(77d_1(^))

= ( - i )1- i . ( - i y ' +1 = ( - i y ' + l

This completes the proof. D

Theorem 4 Let h be afunction with the properties given in Theorem 2. If d e N is even, then the function

## G(.t) = J2[Pd*Hk)]r

k

¿fceZ

has d roots which are simple, negative and differentfrom — 1.

Proof We will study the roots of the function

### F(t) = tál2G(t) = YX^*h(k)]ti-k.

¿fceZ

By Lemma 1, F is a polynomial of degree d:

### F(t) = pd *h(~^\td + • • • + pd *h(^ - \X + pd *h{^\.

Since the coefflcients of F are all positive, we have F(0) > 0 and limí^_c o F(t) = +oo.

Thus, F and G have exactly the same roots. The function fid is even, so if t ^ 0, then

,-1/2 ,-1/2 T r * pd(k - x)h(x)dx = td/2Tt¡d(x)h(x)dx.

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Let A.1 < • • • < Ad_i be the roots of 77d, which are all negative. By Lemma 3,

úgn(kf2rXj,d(x)) = ( - l )d / 2 • (-l)j+di = (-l)j (5)

for every x e (—1/2, 1/2) and every j e ¡ l , . . . , ¿ - l ¡ . By (4) and (5) and having in mind that h is non-negative, we have

F ( A . i ) < 0 , F ( A2) > 0 , ... F ( Ad_ i ) < 0 .

Since limí^_c o F(t) = +oo and F(0) > 0, there are

such that F(si) = 0 , . . . , F(sd) = 0. As F is a polynomial of degree d, it cannot have other roots. By (3), it is known that Ad/2 = — 1, so F(— 1) ^ 0. Therefore, si ^ —l,...,s¿ ^

- 1 . D

Remark 1 Theorem 2 with d even is now a direct consequence of Theorems 3 and 4.

### 3 The Roots of G when d Is Odd

In order to study the roots of G when d is odd, we use the midpoint Euler-Frobenius poly-nomials introducedby Schoenberg [10, pp. 393-394]. Given d e N, the function

### Pd{t) = 2dd\YjQdJk+^)t

is a monic polynomial of degree d whose roots are all simple and negative. If Ai < • • • < Ad

are the roots of Pd and /¿i < • • • < /zd_i are the roots of P¿-i, then

Ai < /¿i < A2 < M2 < A3 < • • • < Ad_i < /zd_i < Ad.

Moreover, if d is odd, then

^(d+l)/2= — 1- (6)

Lemma 4 Let d e N be odd and letk\ < • • • <kd be the roots of Pd. Then

úgn(rXj,d(x)) = (-iy+ái1

for every x e (—1/2, 1/2) and every j e { 1 , . . . , d}.

Proof Let A be any root of Pd. Then

^ ' keZ ^ ' keZ ^ '

\-d

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By Lemma 2, T>L d only has one root on [—1/2, 1/2), so the sign of T>L d is constant on ( - 1 / 2 , 1 / 2 ) .

líd= l,then

A

ti

### 8i(-*) = l.

¿fceZ

Therefore, sign(Tx,i(x)) = 1 for every x e (—1/2, 1/2). That proves the statement of the Lemma when d = 1.

Let us now assume that d is any odd integer bigger than or equal to 3. Since 7i,d(—1/2) = 0 and the sign of T4 is constant on (—1/2, 1/2), we obtain the following alternatives:

TKd(x) > 0 f o r a l l x G ( - 1 / 2 , 1/2) if T £d( - l / 2 ) > 0

or

T.4{x) < 0 f o r a l l x e ( - 1 / 2 , 1/2) i f T ¿d( - l / 2 ) < 0.

We now study the sign of T[ d( - 1 / 2 ) . By (2),

Ti

á

kPd-i(k)

¿fceZ

k

d

k

d

d

## (k + i )

(l-k)k^ykkQd[k + - ) = -r-í —

v > ^ wy -r j 2d-1(d- 1)!

(1 - k) X-d

d

### -i{k).

Since X < 0 and x G (—1/2, 1/2), we have

sign(TM(*)) = s i g n ( T , 'd( - l / 2 ) ) = ( - l ) Y . sign(Pd_1(X)).

This study holds when X is any root of Pd.

If A.1 < • • • < kd are the roots of Pd and /¿i < • • • < H¿Í-I a r e the roots of P¿-i, then

A.1 < ni < k2 < n2 < k3 < • • • < kd-i < Hd-i < kd.

As Pd-i is monic and its roots are simple, it can be written as

Pd-x(t) = (t ~ Mi)(f - M2) • • • (f -

(¿d-l)-Then Pd-i(k{) > 0, Pd-i(k2) < 0,... Pd-i(kd) > 0. That is,

sign(Pd_1(^)) = ( - i y '+ 1

for every j = 1 , . . . , d. Therefore, if x G (—1/2, 1/2), then

sign(T,j>d(x)) = ( - l )1^ • sign(P¿_i(A.,-))

= ( _1) ¥ . ( - i y + i = ( - i y + ^ .

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Theorem 5 Let h be afunction with the properties given in Theorem 2. Ifd e N is odd, then the function

### G(t) = YJ[Pa*h(k)]t-k

¿fceZ

has d + 1 roots which are simple, negative and different from — 1.

Proof We will study the roots of the function

### H(t) = td-^G(t) =

YJ[^*h(k)Y^k-¿fceZ

By Lemma 1, H is a polynomial of degree d + 1:

Again the coefflcients of H ave all positive, which yields H(Q) > 0 and limí^_c o H(t) =

+oo. Henee, all zeros of H and G coincide. As in (4), if t ¿ O, then

td+Tt4{x)h{x)dx. (7)

•1/2

Let A.1 < • • • < k,¡ be the roots of the P¿, which are all negative. By Lemma 4,

g

1

(

(

### 8)

for every x e (—1/2, 1/2) and every j e { 1 , . . . , d}. By (7) and (8),

//(>.!)< O, H ( X2) > 0 , . . . H{Xd)<0.

Since //(O) > O and limí^_c o //(f) = +oo, there are

Si G ( - 0 0 , ki),S2 G (A.i,A.2), . . . , í d G (>-d-l,>-d),'S'd+l G (A.d,0)

such that H(si) = O , . . . , H(sd+{) = 0. Since / / is a polynomial of degree <¿ + 1, it does not have other roots. By (6), it is known that X(d+i)/2 = — 1, so H(—l) ^ O and therefore,

si=é-l,...,sd+i=é-l. D

Remark 2 Theorem 2 with d odd can be now deduced from Theorems 3 and 5.

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