Return to Article Details On the interpolation in linear normed spaces using multiple nodes

On the Interpolation in Linear Normed Spaces
using multiple nodes

Adrian Diaconu

April 10, 2015.

Babeş-Bolyai University, Faculty of Mathematics and Computer Science, M. Kogălniceanu St, No. 1, 3400 Cluj-Napoca, Romania, e-mail: adiaconu@math.ubbcluj.ro.

Dedicated to prof. I. Păvăloiu on the occasion of his 75th anniversary

In the papers [ 2 ] , [ 3 ] , [ 4 ] , [ 6 ] , [ 7 ] we indicated a method of extending the notion of interpolation polynomial to the case of a non-linear mapping f:XY where X and Y are linear spaces with special structures. This extension offers the possibility to establish, in this general and abstract case as well, the main properties known in the case of the interpolation of real functions.

To switch to the case using multiple nodes, case that compulsorily uses the notion of Fréchet differential of the first order as well as of higher orders, we will point out the definition and certain properties of these differentials. On this basis we can present the manner of building an abstract interpolation polynomial with multiple nodes.

MSC. 41A05, 46B70.

Keywords. interpolation in linear normed spaces, multiple nodes.

1 Introduction

The topic of the interpolation of the functions defined between linear spaces or between linear normed spaces has been approached by Păvăloiu, I. in [ 9 ] , [ 10 ] , [ 11 ] Prenter, M. in [ 12 ] , Argyros, I. K. [ 1 ] , Makarov, V. L., Hlobistov, V. V. [ 8 ] and by myself in [ 2 ] , [ 3 ] , [ 4 ] , [ 6 ] , [ 7 ] .

We will recall the elements of the construction of the abstract interpolation polynomial with simple nodes.

Let us consider X and Y two linear spaces and f:XY a non-linear mapping.

We note by L(X,Y) the set of the linear mappings from X to Y and by (X,Y) the subset of L(X,Y) that contains linear and continuous mappings from X to Y.

For n2 we introduce the set Ln(X,Y)=L(X,Ln1(X,Y)) with
L1(X,Y)=L(X,Y) and similarly (Xn,Y)=(X,(Xn1,Y)) with
(X1,Y)=(X,Y).

For n=2 the set L2(X,Y) represents the set of the bi-linear mappings from X×X to Y.

Let θX and θY be the null elements of the space X and Y respectively. We will note by Θn the null element of the space Ln(X,Y). For n=1 we will use the notation Θ.

Let us consider now the bilinear mapping BL2(Y,Y) that verifies
B(u,v)=B(v,u) for any u,vY and B(B(u,v),w)=B(u,B(v,w)) for any u,v,wY.

We will now suppose the following properties:

  • there exists u0Y the identity element of the semi-group (Y,B) and as well Y0Y with u0Y0 so that (Y0,B) form a group.

  • there exists X0X and the linear and bijective mapping U0:X0Y0.

  • there exists the linear mapping U:XY so that U|X0=U0.

Using the mappings UL(X,Y) and BL2(Y,Y) we will introduce the sequence (An)nN with An:XnY where AnLn(X,Y) so that:

A1(y)=U(y),for\ yX;An(y1,,yn)=B(An1(y1,,yn1),U(yn)),for\ y1,,ynX.
1

We consider now the points x0,x1,,xnX0 and the mapping:

w0,n:XY,w0,n(x)=An+1(xx0,,xxn)
2

and for any i{0,1,,n} the mappings:

w0,n(xi)L(X,Y),w0,n(xi)h=An+1(xix0,,xixi1,xixi+1,,xixn,h),
3

noting that w0,n(xi) represents the Fréchet differential of the mapping defined by (???), evidently in the case where X and Y are linear normed spaces.

A first result from [ 2 ] , [ 7 ] shows that the restrictions to X0 of the mappings (???) have values in Y0 and are bijective, so we can speak of

w0,n(xi)1:Y0X0

and this mapping can be prolonged through linearity to sp(Y0) and in the case where Y has a topological structure (for example it is a linear normed space), the prolongation can be extended to cl(sp(Y0)).

If we denote by Y1 the maximal subspace from Y to which the introduced mappings can be extended and we suppose that for any i{0,1,...,n} we have f(xi)Y1 and we can define the mappings:

L(x0,x1,...,xn;f):XY;L(x0,x1,...,xn;f)(x)==i=0nAn+1(xx0,...,xxi1,xxi+1,...,xxn;Zn,i),

where:

Zn,i=w0,n(xi)1f(xi)

and these mappings verify, for any i=0,n, the equalities:

L(x0,x1,...,xn;f)(xi)=f(xi).
5

Because An+1Ln+1(X,Y), there exists D0Y and for any k{1,2,, n} the mappings Dk:XkY so that:

Dkxk=Dk(x,,xk times),

we will have:

L(x0,x1,,xn;f)(x)=\medskipDnxn+Dn1xn1++D1x+D0.
6

The relation (???) is the expression of the character of an abstract polynomial of the mapping (???), so this relation together with the relation (???) prove the fact that this mapping can be denominated an abstract interpolation polynomial.

2 The Fréchet differential of a mapping. Special properties

Let us consider the linear normed spaces (X,X) and (Y,Y). We consider as well the nonlinear mapping f:XY with DX. For xD let be the set:

ED,x={hX:x+hD}.

First we will have the following:

Definition 1

The mapping f:DY admits a Fréchet differential in the point xD, if there exists T(X,Y) and ω:D×(ED,x{θX})Y so that:

f(x+h)f(x)=T(h)+hXω(x,h)

and:

limhθXω(x,h)Y=0.

For the mapping T(X,Y) we have:

Remark 2

We can easily prove that there exists at most a mapping T(X,Y) that corresponds to the requirements of the Definition 1.â–¡

In this way Definition 1 is completed with:

Definition 3

The mapping T(X,Y) from Definition 1 that is attached to the function f:DY and to the point x, is called the Fréchet differential of this mapping at the point x, and is denoted through f(x).

Now we have:

Remark 4

Usually there exists a subset AD so that for any uA, the function f:DY is Fréchet differentiable at every point u. In this case it is possible to define the function:

df:A(X,Y),df(u)=f(u).

The definition of the previous function allows for the introduction of differentials with higher orders.â–¡

Thus we have:

Definition 5

Besides the data from Definition 1 let us consider a number pN.

If:

  • there exists V a neighborhood of the point xD, so that for any uVD there exists the differential of the order p1 of the mapping f:DY at the point u, denoted by f(p1)(u)(Xp1,Y), so the function:

    dp1f:VD(Xp1,Y);(dp1f)(u)=f(p1)(u)
    7

    is defined;

  • the function defined by (???) is a differential (of the first order) at the point x; then we can say that the mapping f:DY admits a differential with the order p at the point x and in this case:

    f(p)(x):=(dp1f)(x)=(f(p1))(x)(X,(Xp1,Y))=(Xp,Y).

In the paper [ 5 ] , we have established certain properties of the Fréchet differentials of higher orders, which are relevant for the statements below.

We will recall some of these properties.

I) Let us consider the bilinear and symmetrical mapping BL2(Y,Y) together with the non-linear mappings f,g:DY with DX.

With the aid of this mappings we consider:

F:DY,F(x)=B(f(x),g(x)).
8

We have the following proposition:

Proposition 6

If the non-linear mappings f,g:DY admit Fréchet differential up to the order n, included, at the point xD, then the mapping introduced by (???) admits a Fréchet differential up to the same order n, at the same point x, and for any h1,,hnX we have:

\begin{equation} F^{\left( n\right) }\left( x\right) h_{1}...h_{n} ={\sum \limits _{k=0}^{n}} \; {\sum \limits _{1\leq i_{1}<... 9

where, for a fixed i1,i2,...,ikN with 1i1<...<ikn, we will choose {j1,...,jnk}{1,...,n}{i1,...,ik} with j1<...<jnk.

For the case h1=...=hn=h we have:

F(n)(x)hn=k=0n(nk)B(f(k)(x)hk,g(nk)(x)hnk)
10

We can notice that the equation (???) represents an extension of the well-known Leibnitz derivation formula.

II) It is necessary to generalize the property expressed by Proposition 6.

For this extension let us consider the sequence of mappings (Qm)mN,m2 with QmLm(X,Y) and:

Q2(u1,u2)=B(u1,u2);Qm(u1,...,um)=B(Qm1(u1,...,um1),um),mN,m2,
11

where by u1,...,um we have denoted arbitrary elements of Y.

We will now consider the natural numbers p,s with sp and the set of distinct elements:

H={x1,x2,...,xp}.

We introduce the set:

Cp,s(H)={(xi1,...,xis)Hs:1i1<i2<...<isp}

and obviously:

|Cp,s(H)|=p!s!(ps)!,

where |H| denotes the number of elements of the set H.

Let us now consider mN and α1,...,αmN so that α1+...+αm=n and to start with we denote:

{H1={1,2,...,n};p1=n;G1={(i1(1),...,iα1(1))H1α1:i1(1)<...<iα1(1)}.

For a fixed (i1(1),...,iα1(1))G1 we choose:

{H2={1,2,...,n}{i1(1),...,iα1(1)};p2=nα1;G2={(i1(2),...,iα2(2))H2α2:i1(2)<...<iα2(2)}.

For  kN,km and a fixed

(i1(1),...,iα1(1))G1,...,(i1(k1),...,iαk1(k1))Gk1

we choose:

{Hk={1,2,...,n}{i1(1),...,iα1(1),...,i1(k1),...,iαk1(k1)};pk=n(α1+...+αk1);\medskipGk={(i1(k),...,iαk(k))Hkαk:i1(k)<...<iαk(k)}.

Finally, for a fixed (i1(1),...,iα1(1))G1,...,(i1(m1),...,iαm1(m1))Gm1 we choose:

{Hm={1,2,...,n}{i1(1),...,iα1(1),...,i1(m1),...,iαm1(m1)};\medskippm=n(α1+...+αm1);\medskipGm={(i1(m),...,iαm(m))Hmαm:i1(m)<...<iαm(m)}.

It is clear that for any kN we have GkCpk,αk(Hk) and consequently:

|Gk|=pk!αk!(pkαk)!=[n(α1++αk1)]!αk![n(α1++αk1+αk)]!.

Let us denote by An[α1,,αm] the set of all systems (G1,,Gm), where for any i{1,2,,m} the system of indexes Gi obtained in the aforementioned manner.

It is obvious that:

|An[α1,,αm]|=|G1|...|Gm|=k=1mpk!αk!(pkαk)!=1α1!...αm!k=1m[n(α1+...+αk1)]![n(α1+...+αk1+αk)]!=1α1!...αm!n!(n(α1+...+αm))!.

But (n(α1++αm))!=(nn)!=0!=1, so:

|An[α1,,αm]|=n!α1!...αm!.

Let us consider now the non-linear mappings fi:AY, for  i=1,m where AX. Using these mappings and the n-linear mapping Qm:YmY introduced by (???) consider the mapping:

F:AY,F(x)=Qm(f1(x),,fm(x)),
12

which represents a more general case of the mappings (???).

In this way we have the following extension of the Proposition 6.

Proposition 7

If the mappings fi:AY;i=1,m and AX admit the Fréchet differentials of the n order at the point xA, then the mapping F:AY defined by (???) admits as well the Fréchet differential of the n order at the same point x and for any h1,,hnX we have the equality:

F(n)(x)h1...hn=α1+...+αm=n (G1,,Gm)An[α1,,αm]TG1,...,Gm,
13

where for TG1,...,Gm we have denoted the expression:

(i1(k),,iαk(k))Gk;k=1,mQm(f1(α1)(x)hi1(1)...hiα1(1),,fm(αm)(x)hi1(m)...hiαm(m)).
14

For the case h1=...=hm=h we have:

F(n)(x)hn=α1++αm=nn!α1!...αm!Qm(f1(α1)(x)hα1,,fm(αm)(x)hαm).
15

III) Let us consider now for a fixed aX, the mapping:

Tm:XY,Tm(x)=Am(xa,,xa)m times=notAm(xa)m,
16

where AmLm(X,Y) is introduced by (???) and the mappings BL2(Y,Y) and UL(X,Y) verify the specified conditions.

For these mappings we have:

Proposition 8

The mappings defined by (???) admit Fréchet differentials of any order nN and for any h1,...,hnX we have:

\begin{equation} T_{m}^{\left( n\right) }h_{1}...h_{n}= \begin{cases} \theta _{Y}, & \text{for}\ n>m\ A_{m}\left( h_{1},\ldots ,h_{m}\right), & \text{for}\ m=n\ \tfrac {m!}{\left( m-n\right) !}A_{n}\left( x-a\right) ^{m-n}h_{1}...h_{n}, & \text{ for}\ m 17

IV) Taking into account the mappings (Am)mN, with AmLm(X,Y) introduced by (???), and if the numbers r1,,rmN we can consider the mapping:

F:XY,F(x)=Ar1++rm((xx1)r1,,(xxm)rm),
21

where x1,,xmX are arbitrary.

We have the following result:

Proposition 9

The mappings defined by (???) admit Fréchet differentials up to the order n included, where nr1+r2++rm at any point xX and:

F(n)(x)h1...hn==n!α1+...+αm=ni=1m(riαi)Ar1+...+rm((xx1)r1α1,,(xxm)rmαm,h1,,hn)

V) We will also consider another extension of Leibnitz’ formula concerning the derivative with the n order.

In this way let us consider X,Y,Z linear normed spaces and the mappings f:X(Y,Z) and g:XY.

Using the considered functions we consider the function:

F:XY,F(x)=[f(x)]g(x)
23

and for this function we have:

Proposition 10

If the mappings f:X(Y,Z) and g:XY admit Fréchet differentials of the order n, at the point xX, then the mapping defined by (???) also admits the Fréchet differential of the same order at the same point x and:

F(n)(x)h1...hn==k=0n1i1<...<ikn[f(k)(x)hi1...hik]g(nk)(x)hj1...hjnk

where {j1,,jnk}={1,,n}{i1,,ik} with j1<j2<...<jnk.

For the case where h1=...hn=h we have:

F(n)(x)hn=k=0n(nk)[f(k)(x)hk]g(nk)(x)hnk.
25

VI) Let us consider now the mapping f:X(X,Y) supposing that for every xX the linear mapping f(x):XY has an inverse mapping [f(x)]1:YX.

Therefore we can consider the mapping:

g:X(Y,X),g(x)=[f(x)]1.
26

We obtain the following result:

Proposition 11

If the non-linear mapping f:X(X,Y) has a Fréchet differential at the point x, then the mapping g:X(Y,X) introduced by (???) also has a Fréchet differential at the same point x, and:

g(x)h=[f(x)]1f(x)h[f(x)]1,
27

for every hY.

3 The Fréchet differential of certain essential mappings that appear in the interpolation with multiple nodes

We will consider the sequence of mappings (An)nN where An:XnY are given by (???) and B(Y2,Y),U(X,Y) verify the assumptions specified in the first paragraph of the present paper.

For x1,,xmX and p1,,pmN we consider the mapping:

ω(x1p1,...,xmpm;x)(X,Y),ω(x1p1,...,xmpm;x)h=Ap1++pm+1((xx1)p1,,(xxm)pm,h).
28

It is clear that if xxiX0 for any i=1,m, from the imposed hypotheses it results that the mappings defined by (???) are bijections from X0 to Y0. Thus we can consider the inverse mapping defined on Y0, which can be extended first by linearity to sp(Y0) and then by continuity to cl(sp(Y0)).

We can thus consider the mapping:

g:X(cl(sp(Y0)),X),g(x)=[ω(x1p1,...,xmpm;x)]1.
29

Concerning this mapping we have:

Proposition 12

If for any k=1,m we have xxkX0, then for any nN, the mapping defined by (???) admits a Fréchet differential of the order n. For any h1,...,hnX and tcl(sp(Y0)) we have:

[g(n)(x)h1...hn]t==(1)nn!α1+...+αm=ni=1m(pi+αi1αi)An+1([ω(x1p1+α1,...,xmpm+αm;x)]1t,h1,...,hn)

Proof â–¼
Based on Proposition 11 we deduce that:

g(x)h=[P(x)]1P(x)h[P(x)]1,
31

where:

P(x)=ω(x1p1,...,xmpm;x).

Taking into account Propositions 7 and 9 we deduce that:

P(x)hu=ω(x1p1,...,xmpm;x)hu=k=1mpkCp1,...,pm(k)(x,h,u),
32

for any h,uX, where we denote by Cp1,...,pm(k)(x,h,u) the value of the mapping Ap1+...+pm+1 on the arguments:

(xx1)p1,...,(xxk1)pk1,(xxk)pk1,(xxk+1)pk+1,...,(xxm)pm,h,u.

If we choose tcl(sp(Y0)) and:

u=[ω(x1p1,...,xmpm;x)]1tX,

after that for k{1,2,...,m} it has been proved that:

Cp1,...,pm(k)(x,h,u)=\medskipB(Ep1,...,pk1,pk1,pk+1,...,pm(k)(x,h),U(u)),

where Ep1,...,pk1,pk1,pk+1,...,pm(k)(x,h) is the value of Ap1+...+pm at the arguments:

(xx1)p1,...,(xxk1)pk1,(xxk)pk1,(xxk+1)pk+1,...,(xxm)pm,h;

so:

Ep1,...,pk1,pk1,pk+1,...,pm(k)(x,h)=ω(x1p1,...,xk1pk1,xkpk1,xk+1pk+1,...,xmpm;x)h.

Thus:

ω(x1p1,...,xmpm;x)h[ω(x1p1,...,xmpm;x)]1t==k=1mpkB(ω(x1p1,...,xk1pk1,xkpk1,xk+1pk+1,...,xmpm;x)h,U(u))

We now show the equality:

U(u)=U[ω(x1p1,...,xmpm;x)]1t=A2(xxk,[ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)]1t).

Because the extension from tY0 to tcl(sp(Y0)) is evident it is enough to suppose that tY0.

From the bijectivity of the mappings (???) for tY0 we deduce the existence with a unique determination of the elements h,uX0 so that:

t=ω(x1p1,...,xmpm;x)h=ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)u.
35

The first member of this equality can be written as:

B(Ap1+...+pm((xx1)p1,...,(xxm)pm),U(h))

and the second is the value of the mapping Ap1+...+pm+2 at the arguments:

(xx1)p1,...,(xxk1)pk1,(xxk)pk+1,(xxk+1)pk+1,...,(xxm)pm,u

and this can be written as:

B(Ap1+...+pm((xx1)p1,...,(xxm)pm),A2(xxk,u)),

consequently the equality (???) becomes:

B(Ap1+...+pm((xx1)p1,...,(xxm)pm),U(h)A2(xxk,u))=θY.
36

Because for every k{1,2,...,m} we have xxkX0 we deduce that:

Ap1+...+pm((xx1)p1,...,(xxm)pm)Y0,

consequently the equality (???) will be possible only if:

U(h)=A2(xxk,u).
37

But from the same relation (???) it is clear that:

u=[ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)]1t

and:

h=[ω(x1p1,...,xmpm;x)]1t,

after that replacing these values in (???), we obtain the relation (???).

Because of the relation (???) we have the equality:

ω(x1p1,...,xmpm;x)h[ω(x1p1,...,xmpm;x)]1t==k=1mpkB(Ep1,...,pk1,pk1,pk+1,...,pm(k)(x,h),A2(xxk,qk1t))

where:

qk=ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x).

So:

B(Ep1,...,pk1,pk1,pk+1,...,pm(k)(x,h),A2(xxk,qk1t))==Ap1+...+pm+2((xx1)p1,...,(xxm)pm,qk1t,h)=B(Ap1+...+pm+1((xx1)p1,...,(xxm)pm,qk1t),U(h))=B(ω(x1p1,...,xmpm;x)qk1t,U(h)),

consequently:

ω(x1p1,...,xmpm;x)h[ω(x1p1,...,xmpm;x)]1t==k=1mpkB(ω(x1p1,...,xmpm;x)qk1t,U(h))

We remark that for any aY0 it exists a yY0 so that for any sY we have:

B(s,B(a,y))=s,
40

this fact being evident, the element yY0 is the symmetrical element of the element a in the group (Y0,B).

Because the fact that xxkX0 for any k=1,m we can deduce that:

a=Ap1+...+pm((xx1)p1,...,(xxm)pm)Y0

from where, using the relation (???), we deduce that for any sY we have:

B(s,B(Ap1+...+pm((xx1)p1,...,(xxm)pm),y))=s,

from where through the properties of the bilinear mapping B(Y2,Y) we have:

s=B(Ap1+...+pm((xx1)p1,...,(xxm)pm),B(s,y))=B(Ap1+...+pm((xx1)p1,...,(xxm)pm),U(U1B(s,y)))=Ap1+...+pm+1((xx1)p1,...,(xxm)pm,U1B(s,y));

consequently:

ω(x1p1,...,xmpm;x)U1B(s,y)=s,

from where:

B(s,y)=U[ω(x1p1,...,xmpm;x)]1s.

For the beginning let be s=B(u,v)Y with u,vY in the previous relation, so we have:

U[ω(x1p1,...,xmpm;x)]1B(u,v)=B(B(u,v),y)=B(B(yu),v)=B(U[ω(x1p1,...,xmpm;x)]1u,v).

Then in the previous relation we consider, with tY, the elements:

u=ω(x1p1,...,xmpm;x)[ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)]1t

and:

v=U(h),

consequently:

U[ω(x1p1,...,xmpm;x)]1B(u,U(h))==B(U[ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)]1t,h)=A2([ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)]1t,h),

thus:

[g(x)h]t==k=1mpkU1A2([ω(x1p1,...,xk1pk1,xkpk+1,xk+1pk+1,...,xmpm;x)]1t,h)=α1+...+αm=1i=1m(pi+αi1αi)U1A2([ω(x1p1+α1,...,xmpm+αm;x)]1t,h)

so the relation (???) is true for n=1.

Let us suppose now that the relation (???) is true for n=k, namely:

[g(k)(x)h2...hk+1]t=(1)kk!××α1+...+αm=ki=1m(pi+αi1αi)U1Ak+1([ω(x1p1+α1,...,xmpm+αm;x)]1t,h2,...,hk+1)

and for h1X as well:

[g(k)(x+h1)h2...hk+1]t=(1)kk!××α1+...+αm=ki=1m(pi+αi1αi)U1Ak+1([ω(x1p1+α1,...,xmpm+αm;x+h1)]1t,h2,...,hk+1)

From the relations (???) and (???) through substraction, member by member, we obtain:

[(g(k)(x+h1)g(k)(x))h2...hk+1]t==(1)kk!α1+...+αm=ki=1m(pi+αi1αi)U1Ak+1(Z(x,h1)t,h2,...,hk+1),

where:

Z(x,h1)=[ω(x1p1+α1,...,xmpm+αm;x+h1)]1\medskip[ω(x1p1+α1,...,xmpm+αm;x)]1(Y,X).

At the same time:

Z(x,h1)={[ω(x1p1+α1,...,xmpm+αm;x)]1}h1+h1XR(x,h1),
45

where R(x,h1)(Y,X) and limh1θXR(x,h1)=0.

Because Ak+1:Xk+1Y is a k+1 linear mapping, from the already proved relation (???) we obtain:

[(g(k)(x+h1)g(k)(x))h2...hk+1]t==(1)k+1k!α1+...+αm=ki=1m(pi+αi1αi)β1+...+βm=1i=1m(pi+αi+βi1βi)××U1Ak+1(U1A2(Ωα1+β1,...,αm+βm(x)t,h1),h2,...,hm)++h1XR(x;h1,h2,...,hk+1)

where:

Ωα1+β1,...,αm+βm(x)=[ω(x1p1+α1+β1,...,xmpm+αm+βm;x)]1(Y,X)

and:

R(x;h1,h2,...,hk+1)==(1)kk!α1+...+αm=ki=1m(pi+αi1αi)U1Ak+1(R(x,h1)t,h2,...,hm).

Therefore:

R(x;h1,h2,...,hk+1)Yk!U1Ak+1tYh2X...hmX××α1+...+αm=ki=1m(pi+αi1αi)R(x,h)X,

from where:

limh1θXR(x;h1,h2,...,hk+1)Y=0.

If we denote γi=αi+βi for i=1,m it evidently results that:

γ1+...+γm=(α1+...+αm)+(β1+...+βm)=k+1

and:

(pi+αi1αi)(pi+αi+βi1βi)=(γiαi)(pi+γi1γi),

so if we denote:

Tα1+β1,...,αm+βm=U1Ak+1(U1A2(Ωα1+β1,...,αm+βm(x)t,h1),h2,...,hm),

we have:

α1+...+αm=ki=1m(pi+αi1αi)β1+...+βm=1i=1m(pi+αi+βi1βi)Tα1+β1,...,αm+βm==γ1+...+γm=k+1i=1m(pi+γi1γi)Tγ1,...,γmα1+...+αm=k  i=1m(γiαi)=(k+1)γ1+...+γm=k+1i=1m(pi+γi1γi)Tγ1,...,γm.

We have used the identity:

α1+...+αm=ki=1m(γiαi)=(γ1+...+γmα1+...+αm)=(k+1k)=k+1.

In this way, because k!(k+1)=(k+1)!, we are able to write the relation:

[(g(k)(x+h1)g(k)(x))h2...hk+1]t==(1)k+1(k+1)!γ1+...+γm=k+1  i=1m(pi+γi1γi)Tγ1,...,γm+h1XR(x;h1,h2,...,hk+1);

whence we deduce that there exists the mapping g(k+1)(x)(Yk+1,X).

It is clear that:

Tγ1,...,γm=U1Ak+1(U1A2([ω(x1p1+γ1,,xmpm+γm;x)]1t,h1),h2,,hk+1)=U1Ak+1(U1A2(Wγ1,...γm(t),h1),h2,...,hk+1)=U1B(UU1A2(Wγ1,...γm(t),h1),Ak(h2,...,hk+1))=U1Ak+1(Wγ1,...γm(t),h1,h2,...,hk+1),

where we have denoted:

Wγ1,...γm=[ω(x1p1+γ1,...,xmpm+γm;x)]1(Y,X).

Therefore:

[g(k+1)(x)h1...hk+1](t)=(1)k+1(k+1)!××γ1+...+γm=k+1i=1m(pi+γi1γi)U1Ak+1(Wγ1,...γm(t),h1,h2,...,hk+1),

and so the equality (???) is true for n=k+1.

Based on the principle of mathematical induction the equality (???) is true for every nN.

Proposition 12 is proved.

4 The construction of an abstract interpolation polynomial   with multiple nodes

Let us consider the linear normed spaces X and Y, the set DX and the function f:DY.

The general interpolation problem has the following setting.

Being given the distinct elements x0,x1,...,xmD, the numbers r0,r1,..., rmN and the values f(j)(xi)(Xj,Y) with j=0,ri1; i=0,m, determine a (UB) polynomial P:XZ with the minimum degree so that for any i=0,m and j=0,ri1 we have P(j)(xi)=f(j)(xi), the equality being understood as one between the elements of (Xj,Y).

To provide an answer to this problem let us suppose that the hypotheses from the preliminaries are fulfilled. Then let us consider the mappings that we have introduced by (???).

To simplify the writing we will introduce:

Xi=(x1,...,xi1,xi+1,...,xm)Xm1,Ri=(r1,...,ri1,ri+1,...,rm)Nm1,Ji=(α1,...,αi1,αi+1,...,αm)Nm1,|Ji|=α1+...+αi1+αi+1+...+αmN.

together with the mapping from (X,Y):

Wm,i(Xi,Ri,Ji)=ω(x1r1+α1,...,xi1ri1+αi1,xi+1ri+1+αi+1,...,xmrm+αm;xi).
47

If xixjX0 for any i,j{1,2,...,m} then, based on what we have established in the previous paragraph, the mapping (???) is invertible on Y0, and it is possible to prolong the inverted mapping to cl(sp(Y0)).

We denote this inverted mapping as [Wm,i(Xi,Ri,Ji)]1(Y,X).

As an answer to the aforementioned interpolation problem, we have the following theorem:

Theorem 13

If for every i,j{1,2,...,m} with ij we have xixjX0 and f(j)(xi)(X,cl(sp(Y0))), there exists a (UB) polynomial with the degree n=r1+...+rm1 that fulfills the conditions of the general interpolation problem. The expression of this (UB) polynomial is:

Hn(x1r1,...,xmrm;f)(x)=i=1mj=0ri1lij(f)(x),
48

where lij(f):XY has the form:

lij(f)(x)=1j!k=0rij1(1)kQn,i,j,k(f;x)
49

where Qn,i,j,k(f;x) is the value of the mapping Anri+k+2(Xnri+k+2,Y) at the arguments:

(xx1)r1,...,(xxi1)ri1,(xxi)k,(xxi+1)ri+1,...,(xxm)rm; Zn,i,j,k(x)

with:

Zn,i,j,k(x)=|Ji|=k  j=1,jim(rj+αj1αj)[Wm,i(Xi,Ri,Ji)]1f(j)(xi)(xxi)j.
50

Proof â–¼
To simplify the writing we will use the notation:
Hn(x1r1,...,xmrm;f)=Hn:XY.

The theorem is proved, if in the form (???) the element lij(f)(x) is under the form (???) with the specification (???).

As Hn:XY is an abstract interpolation polynomial it is necessary that:

[lij(f)](t)(xs)=δisδtjf(j)(xi)
51

where δpq is Kronecker’s symbol. In the equality (???) we have the values i,s{1,2,...,m} and for a fixed i the indices j,t{0,1,...,ri1}.

In order to fulfill the conditions (???) we search the mappings lij(f):XY under the form:

lij(f)(x)=B(gi(x),Sij(x));j=0,ri1;i=1,m;
52

where for any i=1,m we have:

gi(x)=Anri+1((xx1)r1,...|i...,(xxm)rm)
53

here, (xx1)r1,...|i...,(xxm)rm is an abbreviation for:

(xx1)r1,...,(xxi1)ri1,(xxi+1)ri+1,...,(xxm)rm

and:

Sij(x)=B(Aj(xxi)j,hij(x));j=0,ri1;i=1,m;
54

where hij:XY is a mapping to be determined.

From the relations (???),(???) and (???) it is clear that:

lij(t)(f)(xs)=0

for si, but also in the situation where s=i and t<j.

We will determine the mappings hij:XY from (???) so that:

lij(j)(f)(xi)=f(j)(xi), and lij(t)(f)(xi)=0\ for t>j.
55

The relations (???)(???) indicate that for the abstract interpolation polynomial to be determined, it is necessary to choose for hij a (UB) abstract polynomial with the degree rij1.

We will now consider Taylor’s formula for the case of non-linear mappings between linear normed spaces, formula which for a function F:ΩY, where Ω is an open and convex set of the linear normed space X, that admits Fréchet differentials up to the n+1 Fréchet differential, included, is expressed through:

F(x)k=0nF(k)(x0)k!(xx0)kYxx0Xn+1(n+1)!supθ[0,1]F(n+1)(x0+θ(xx0)).
56

If F:ΩY is a (UB) abstract polynomial with the degree n then using the proposition 8 we deduce that F(n+1)(y)=Θn+1, therefore from the inequality (???) we deduce that in this case we will have:

F(x)=k=0nF(k)(x0)k!(xx0)k.
57

Because the mapping hij:XY from the equality (???) is a (UB) abstract polynomial with the degree rij1, we have:

hij(x)=k=0rij11k!hij(k)(xi)(xxi)k.
58

From (???),(???) and (???) we deduce that:

lij(f)(x)=k=0rij11k!B(gi(x),B(Aij(xxi)j,hij(k)(xi)(xxi)k))
59

and this relation indicates that the problem is solved if the elements
hij(k)(xi)(Xk,Y) are determined and the equalities (???) are fulfilled.

Let us define for every xX the mapping g~i(x)(Y,Y) by g~i(x)t=B(gi(x),t) for tY.

From this equality we deduce that:

B(Aj(xxi)j,hij(x))=[g~i(x)]1lij(f)(x).
60

Considering the equality of the Fréchet differentials of the order j+k of the mappings from the first and the second member of the equality (???) and using the relations (???) and (???) we obtain:

s=0j+k(j+ks)B([Aj(xxi)j](j+ks)tj+ks,hij(s)(x)ts)==s=0j+k(j+ks){([g~i(x)]1)(j+ks)tj+ks}lij(s)(x)ts.

On the account of Proposition 8 and of the fact that:

lij(s)(f)(xi)=δsjf(j)(xi),

if in the equality (???) we replace x=xi we obtain:

j!B(Ajtj,hij(k)(xi)tk)={([g~i(x)]1)x=xi(k)tk}f(j)(xi)tj.
62

If we introduce:

g~~i(x)=ω(x1r1,...,xi1ri1,xi+1ri+1,...,xmrm;x)(X,Y)

it is obvious that g~i(x)t=g~~i(x)U1(t).

On account of the Proposition 12 we deduce that:

{([g~~i(x)]1)(k)tk}u==(1)kk!|Ji|=ks=1,sim(rs+αs1αs)U1Ak+1([Wm,i(Xi,Ri,Ji)]1u,tk).

From g~i(x)=g~~i(x)U1 we have [g~i(x)]1=U[g~~i(x)]1 therefore from (???) we deduce that:

{([g~i(x)]1)(k)tk}u==(1)kk!|Ji|=ks=1,sim(rs+αs1αs)Ak+1([Wm,i(Xi,Ri,Ji)]1u,tk)

From (???) and (???) we immediately deduce that:

B(Ajtj,hij(k)(xi)tk)=(1)kk!j!|Ji|=ks=1,sim(rs+αs1αs)××Ak+1([Wm,i(Xi,Ri,Ji)]1f(j)(xi)tj,tk)

Replacing now in (???) t=xxi we obtain:

B(Ajj(xxi),hij(k)(xi)(xxi)k)==(1)kk!j!|Ji|=ks=1,sim(rs+αs1αs)××Ak+1([Wm,i(Xi,Ri,Ji)]1f(j)(xi)(xxi)j,(xxi)k)

and from the relations (???) and (???) results the equality:

lij(f)(x)=1j!k=0rij1(1)k|Ji|=ks=1,sim(rs+αs1αs)××Δn,i,j,k(r1,...,ri1,ri+1,...,rm)(f),

where Δn,i,j,k(r1,...,ri1,ri+1,...,rm)(f) is the value of the mapping Anri+k+2 at the arguments:

(xx1)r1,...,(xxi1)ri1,(xxi)k,(xxi+1)ri+1,......,(xxm)rm,[Wm,i(Xi,Ri,Ji)]1f(j)(xi)(xxi)j.

Because:

|Ji|=ks=1,sim(rs+αs1αs)[Wm,i(Xi,Ri,Ji)]1f(j)(xi)(xxi)j=\medskipZn,i,j,k(x)

evidently:

|Ji|=ks=1,sim(rs+αs1αs)Δn,i,j,k(r1,...,ri1,ri+1,...,rm)(f)=Qn,i,j,k(f;x).

This last equality indicates that the theorem is proved.

Bibliography

1

I. K. Argyros, Polynomial Operator Equation in Abstract Spaces and Applications, CRC Press Boca Raton Boston London New York Washington D.C., (1998).

2

A. Diaconu, Interpolation dans les espaces abstraits. Méthodes itératives pour la resolution des équation opérationnelles obtenues par l’interpolation inverse (I), Babeş-Bolyai University, Faculty of Mathematics, Research Seminars, Preprint No. 4, 1981, Seminar of Functional Analysis and Numerical Methods, pp. 1–52.

3

A. Diaconu, Interpolation dans les espaces abstraits. Méthodes itératives pour la resolution des équation opérationnelles obtenues par l’interpolation inverse (II), Babeş-Bolyai University, Faculty of Mathematics, Research Seminars, Preprint No. 1, 1984, Seminar of Functional Analysis and Numerical Methods, pp. 41–97.

4

A. Diaconu, Interpolation dans les espaces abstraits. Méthodes itératives pour la resolution des équation opérationnelles obtenues par l’interpolation inverse (III), Babeş-Bolyai University, Faculty of Mathematics, Research Seminars, Preprint No. 1, 1985, Seminar of Functional Analysis and Numerical Methods, pp. 21–71.

5

A. Diaconu, Sur quelques propriétés des dérivées de type Fréchet d’ordre supérieur, Babeş-Bolyai University, Faculty of Mathematics, Research Seminaries, Seminar of Functional Analysis and Numerical Methods, Preprint No. 1, (1983), pp. 13–26.

6

A. Diaconu, Remarks on Interpolation in Certain Linear Spaces (I), Studii ın metode de analiză numerică şi optimizare, Chişinău: USM-UCCM., 2, 2(1), (2000), pp. 3–14.

7

A. Diaconu, Remarks on Interpolation in Certain Linear Spaces (II), Studii ın metode de analiză numerică şi optimizare, Chişinău: USM-UCCM., 2, 2 (4), (2000), pp. 143–161.

8

V. L. Makarov and V. V. Hlobistov, Osnovı teorii polinomialnogo operatornogo interpolirovania, Institut Mathematiki H.A.H. Ukrain, Kiev, (1998) ( in Russian ).

9

I. Păvăloiu, Interpolation dans des espaces linéaires normés et application, Mathematica, Cluj, 12(35), 1, (1970), pp. 149–158.

10

I. Păvăloiu, Consideraţii asupra metodelor iterative obţinute prin interpolare inversă, Studii şi cercetări matematice, 23, 10, (1971), pp. 1545–1549 (in Romanian).

11

I. Păvăloiu, Introducere ın teoria aproximării soluţiilor ecuaţiilor, Editura Dacia, Cluj-Napoca, (1976), ( in Romanian ).

12

P. M. Prenter, Lagrange and Hermite interpolation in Banach spaces, J. Approx. Theory, 4 (1971), pp. 419–432. \includegraphics[scale=0.1]{ext-link.png}