h (x) = 2φ( x 2 ) ’ φ( x + hed+1 2 ) ’ φ( x ’ hed+1 2 ), x ∈ Rd+1 ,

2

2 2

where ed+1 denotes the (d + 1)-th unit vector. By Proposition 8.7 we know that h is

conditionally positive semi-de¬nite of order m on Rd+1 and hence on Rd . But the restriction

to Rd is given by

= 2 φ( x 2 ) ’ φ( x + h) =: 2ψh ( x 2 ), x ∈ Rd .

2

h (x) 2 2 2

(m)

Thus, by the induction hypothesis, (’1)m ψh is completely monotone on (0, ∞) for every

h > 0, i.e.

(’1)m+ ψh ) = (’1)m+ φ (m+ ) (r ) ’ φ (m+ ) (r + h) ≥ 0

(m+

for r > 0, ∈ N0 . But this means in particular that

+1 φ (r + h) ’ φ (m+ ) (r )

(m+ )

≥0

(’1)m+

h

for all r > 0, ∈ N0 , and h > 0. Letting h tend to zero results in

(’1)m+1+ φ (m+1+ ) (r ) ≥ 0

for all r > 0, ∈ N0 , which ¬nishes the induction proof.

An argument similar to that in the proof of Theorem 7.14 yields:

Corollary 8.20 Suppose that the function φ of Theorem 8.19 is not a polynomial of degree

at most m; then φ( · 2 ) is conditionally positive de¬nite of order m on every Rd .

2

Micchelli™s result (Theorem 8.19) gives a very powerful tool for deciding whether a given

radial function is conditionally positive de¬nite on every Rd . To demonstrate its usefulness

we investigate inverse multiquadrics, power functions, and thin-plate splines again.

Example The multiquadrics φ(r ) = (’1) β (c2 + r 2 )β , c, β > 0, β ∈ N, are conditionally

positive de¬nite of order m = β on every Rd .

β

(c2 + r )β , we see that

Proof If we de¬ne f β (r ) = (’1)

β

β(β ’ 1) · · · (β ’ k + 1)(c2 + r )β’k .

(k)

f β (r ) = (’1)

(β)

Thus (’1) β f β (r ) = β(β ’ 1) · · · (β ’ β + 1)(c2 + r )β’ β is completely monotone

(m)

and m = β is the smallest possible choice that makes (’1)m f β completely mono-

tone.

Example The functions φ(r ) = (’1) β/2 r β , β > 0, β ∈ 2N, are conditionally positive

de¬nite of order m = β/2 on every Rd .

116 Conditionally positive de¬nite functions

β/2

r β/2 to see that

Proof De¬ne f β (r ) = (’1)

β β β

β/2

’ k + 1 r β/2’k .

(k)

f β (r ) = (’1) ’ 1 ...

2 2 2

( β/2 )

β/2

(r ) is completely monotone and m = β/2 is the smallest

Thus again (’1) fβ

possible choice.

Example The thin-plate or surface splines φ(r ) = (’1)k+1r 2k log(r ) are conditionally

positive de¬nite of order m = k + 1 on every Rd .

Proof Since 2φ(r ) = (’1)k+1r 2 log(r 2 ) we set f k (r ) = (’1)k+1r k log(r ). Then it is easy

to see that

f k( ) (r ) = (’1)k+1 k(k ’ 1) · · · (k ’ + 1)r k’ log(r ) + p (r ), 1¤ ¤ k,

where p is a polynomial of degree k ’ . This means in particular that f k(k) (r ) =

(’1)k+1 k! log(r ) + c and ¬nally that (’1)k+1 f k(k+1) (r ) = k! r ’1 , which is obviously com-

pletely monotone on (0, ∞).

8.5 Interpolation by conditionally positive de¬nite functions

The investigation of positive de¬nite functions was motivated by the interpolation problem

(6.2) and the Ansatz (6.1) for the interpolating function. The example mentioned at the

beginning of the present chapter showed that this de¬nition of the interpolating function

does not work in the case of conditionally positive de¬nite functions. But a slight change in

the de¬nition of the interpolation function ensures solvability of the interpolation matrix.

Instead of (6.1) we now de¬ne the interpolant to a function f at the centers X = {x1 , . . . , x N }

as

Q

N

s f,X (x) = ± j (x ’ x j ) + βk pk (x).

j=1 k=1

Here, Q denotes again the dimension of the polynomial space πm’1 (Rd ) and p1 , . . . , p Q

denote a basis of πm’1 (Rd ). To cope with the additional degrees of freedom, the interpolation

conditions

s f,X (x j ) = f (x j ), 1 ¤ j ¤ N,

are completed by the additional conditions

N

± j pk (x j ) = 0, 1 ¤ k ¤ Q.

j=1

Solvability of this system is therefore equivalent to solvability of the system

± f |X

A ,X P

= (8.10)

β

PT 0 0

8.6 Notes and comments 117

where A ,X = ( (x j ’ xk )) ∈ R N —N and P = ( pk (x j )) ∈ R N —Q . This last system is ob-

viously solvable if the matrix on the left-hand side, which we will denote by A ,X, is

invertible.

Theorem 8.21 Suppose that is conditionally positive de¬nite of order m and X is a

πm’1 (R )-unisolvent set of centers. Then the system (8.10) is uniquely solvable.

d

Proof Suppose that (±, β)T lies in the null space of the matrix A ,X . Then we have

,X ± + Pβ = 0,

A

P T ± = 0.

The second equation means that ± satis¬es condition (8.1) for all p ∈ πm’1 (Rd ). Multi-

plying the ¬rst equation by ± T gives 0 = ± T A ,X ± + (P T ±)T β = ± T A ,X ±. Since is

conditionally positive de¬nite of order m we can conclude that ± = 0 and thus Pβ = 0.

Finally, since X is πm’1 (Rd )-unisolvent, this means that β = 0.

For a solution of the system (8.10) it is not necessary to require X to be πm’1 (Rd )-

unisolvent. This is only needed for uniqueness. To see the solvability in the general case

let us set V := P(R Q ) ⊆ R N and A := A ,X . Then the orthogonal complement V ⊥ of V

is the null space of P T . The system (8.10) is solvable for every f |X if R N = AV ⊥ + V .

But this sum is a direct sum, because x ∈ AV ⊥ © V means that x = A± = Pβ with a

certain ± ∈ V ⊥ and a β ∈ R Q ; this implies that ± T A± = (P T ±)T β = 0 and hence ± = 0

and A± = 0. Knowing that the intersection of AV ⊥ and V contains only the zero vector

gives

N ≥ dim(AV ⊥ + V ) = dim AV ⊥ + dim V = dim V ⊥ + dim V = N ,

because A|V ⊥ : V ⊥ ’ AV ⊥ is bijective. But this means solvability.

It is obvious that the addition of polynomial terms of total degree at most m ’ 1 to the

expansion guarantees polynomial reproduction, i.e. if the data come from a polynomial of

total degree less than m then they are ¬tted by that polynomial.

The method described in this section can be generalized in a straightforward way by

using arbitrary linearly independent functions p1 , . . . , p Q on Rd instead of polynomials.

Moreover, the conditionally positive de¬nite function can be replaced by a conditionally