= r ’1 J3 (r ) ’ ( j ’ 4)r ’ j t j’2 J3 (t)dt.

0

√

Since Jν (r ) decays as O(1/ r ) we can bound the integral term in the last expression:

r r

r’j t j’2 J3 (t)dt ¤ c1r ’ j + c2r ’ j t j’2’1/2 dt

0 r0

’j ’3/2

¤ cr ’1

¤ c3r + c4 r

for every j ≥ 1. Knowing that the sum (9.4) is zero leads us ¬nally to

n

’2

0= c j ( j + 2 ’ 2) · · · j = (’1) D [r 2 p(r )](1) = γ j p ( j) (1)

j=1 j=0

9.4 Compactly supported functions of minimal degree 127

with certain constants γ j , 0 ¤ j ¤ , γ = 1. But as the ¬rst ’ 1 derivatives of p vanish

at 1, we derive p ( ) (1) = 0.

Obviously, the result of Theorem 9.9 remains true for odd space dimensions and general

functions φ with t ’ φ(t)t d’1 ∈ L 1 [0, ∞) and is simply a consequence of the radiality. For

even space dimensions the result is in this generality wrong; see Gneiting [70]. Hence, for

arbitrary space dimensions and functions the generalization becomes φ ∈ C (d’1)/2 (0, ∞).

Theorem 9.10 Suppose that φ is a continuous and even function of the form (9.1) that

is positive de¬nite on Rd . Then there exist integers k, ∈ N0 such that φ possesses 2k

continuous derivatives around 0 and 2k + + d/2 continuous derivatives around 1.

Proof For brevity we will use the following abbreviations. If φ is positive de¬nite on Rd

we will denote this by φ ∈ PDd . Furthermore φ ∈ C (x0 ) will mean that there is an open

neighborhood U of x0 with φ ∈ C (U ).

For k = 0 this is simply Theorem 9.9. The case where the space dimension d = 1 gives

a special version of Theorem 6.14.

Hence, it remains to prove this theorem for d ≥ 2. Assume that the function φ satis¬es

φ ∈ C 2k (0) © C m (1) © PDd ,

with k and m chosen maximal. Then we have m ≥ d/2 by Theorem 9.9. We will discuss

the cases m ≥ k and m < k separately. If m ≥ k is satis¬ed then we have

ψ := Dk φ ∈ C 0 (0) © C m’k (1) © PDd+2k,

which implies by Theorem 9.9 that ψ possesses at least d/2 + k continuous derivatives

around 1, i.e.

d/2 +k

ψ ∈C (1).

This means that m ’ k ≥ d/2 + k, i.e. m ≥ d/2 + 2k.

We are ¬nished as soon as we have shown that 0 ¤ m < k is impossible. So let us take

m < k; this implies that

ψ := Dm φ ∈ C 2(k’m) (0) © C 0 (1) © PDd+2m ;

but this means that 0 ≥ d/2 + m by the same arguments as before. The last inequality

cannot be satis¬ed because we have d ≥ 2. Hence m < k is impossible and this concludes

the proof.

9.4 Compactly supported functions of minimal degree

Knowing that functions of the form (9.1) must necessarily have an even number of con-

tinuous derivatives, it is our aim in this section to ¬nd those functions that are of minimal

degree with respect to a given space dimension d and a given smoothness 2k. Of course,

we mean by the degree of the function φ the degree of the polynomial p.

128 Compactly supported functions

De¬nition 9.11 With φ (r ) = (1 ’ r )+ we de¬ne

φd,k = I k φ d/2 +k+1 . (9.5)

As the operator I respects a compact support and maps polynomials to polynomials, the

functions φd,k are of the form (9.1). A possible iterative scheme to compute the polynomials

is stated in the next theorem. A proof can be easily achieved by induction.

Theorem 9.12 Within its support [0, 1] the function φd,k has the representation

+2k

d (j,k r j

)

pd,k (r ) =

j=0

with = d/2 + k + 1. The coef¬cients can be computed recursively for 0 ¤ s ¤ k ’ 1:

d (j,0 = (’1) j

)

, 0¤ j ¤ ,

j

+2s d (j,s

)

() ()

= , d1,s+1 = 0, s ≥ 0,

d0,s+1

j +2

j=0

d (j’2,s

)

d (j,s+1

)

=’ , s ≥ 0, 2 ¤ j ¤ + 2s + 2.

j

Furthermore, precisely the ¬rst k odd coef¬cients d (j,k vanish.

)

These functions are not only of the form (9.1) but also of minimal degree and hence the

answer to our initial question.

Theorem 9.13 The functions φd,k are positive de¬nite on Rd and are of the form

0 ¤ r ¤ 1,

pd,k (r ),

φd,k (r ) =

r > 1,

0,

with a univariate polynomial pd,k of degree d/2 + 3k + 1. They possess continuous

derivatives up to order 2k. They are of minimal degree for given space dimension d and

smoothness 2k and are up to a constant factor uniquely determined by this setting.

Proof We already know that these functions are of the speci¬c form. They are posi-

tive de¬nite on Rd , because we have for the Fourier transform Fd φd,k = Fd+2k φ d/2 +k+1

by Theorem 9.6. Moreover, the function φ d/2 +k+1 is positive de¬nite on Rd+2k by

Theorem 6.20.

From Lemma 9.8 we know that φd,k possesses 2k continuous derivatives, which is the

stated smoothness. The degree is given by Theorem 9.12.

Finally, suppose that there is a function ψ which is positive de¬nite on Rd , is of the

form (9.1), and possesses 2k continuous derivatives. Assume further that ψ is of minimal

degree. Then we can form the function ψ := Dk φ, which is still of the form (9.1) and

˜

9.4 Compactly supported functions of minimal degree 129

Table 9.1 Compactly supported functions of minimal degree

Space dimension Function Smoothness

φ1,0 (r ) = (1 ’ r )+ C0

d=1 φ1,1 (r ) = (1 ’ r )3 (3r + 1) C2

™ +

φ1,2 (r ) = (1 ’ r )5 (8r 2 + 5r + 1) C4

™ +

φ3,0 (r ) = (1 ’ r )2 C0

+

φ3,1 (r ) = (1 ’ r )4 (4r + 1) C2

™ +

d¤3

φ3,2 (r ) = (1 ’ r )6 (35r 2 + 18r + 3) C4

™ +

φ3,3 (r ) = (1 ’ r )8 (32r 3 + 25r 2 + 8r + 1) C6

™ +

φ5,0 (r ) = (1 ’ r )3 C0

+

d¤5 φ5,1 (r ) = (1 ’ r )5 (5r + 1) C2

™ +

φ5,2 (r ) = (1 ’ r )7 (16r 2 + 7r + 1) C4