are dealt with in the same way. Finally, we have already discussed the fact that an even

φ ∈ C 2 (R) leads to a continuous function Dφ.

It is time for another example, which straight away proves the beauty of the whole concept.

It comes from Wu [203]. De¬ne f (r ) = (1 ’ r 2 )+ for ∈ N. Then g := f — f (2 ·) is

positive de¬nite on R, because its Fourier transform is the squared Fourier transform of f .

Moreover, it is of the form (9.1) with a polynomial of degree 4 + 1 and it is in C 2 . Thus

gk, := Dk g is positive de¬nite on R2k+1 , is of the form (9.1) with a polynomial of degree

4 ’ 2k + 1 and is in C 2 ’2k if ≥ k. Later on we will see that these functions do not have

minimal degree for a given smoothness.

The ¬rst step in characterizing functions of the form (9.1) with minimal degree by their

smoothness and positive de¬niteness (i.e. by the space dimension) is to show that none of

them has an odd number of continuous derivatives. To do this we ¬rst take a closer look at

the smoothness of φ in a neighborhood of 1.

Theorem 9.9 Let φ be an even function of the form (9.1) that is positive de¬nite on Rd .

Then φ satis¬es φ ∈ C d/2 (0, ∞).

Proof On account of the form (9.1), the function = φ( · 2 ) is in L 1 (Rd ). Since φ is

also positive de¬nite on Rd we know from Corollary 6.12 that the Fourier transform of

also belongs to L 1 (Rd ). This means in other words that

∞

|Fd φ(t)|t d’1 dt < ∞.

0

Since both and are in L 1 (Rd ) we can recover from its Fourier transform . But as

both functions are radial this can be written in the form

∞

φ(r ) = r ’(d’2)/2 Fd φ(t)t d/2 J(d’2)/2 (r t)dt. (9.3)

0

Using Leibniz™ formula and the fact that the Bessel functions and their derivatives possess the

9.3 Piecewise polynomial functions with local support 125

asymptotic behavior Jν(m) (r ) = O(r ’1/2 ) for r ’ ∞ (see Proposition 5.6), we can conclude

for r > 0 that

d

r ’(d’2)/2 J(d’2)/2 (r t) ¤ c(r )t ’1/2

for t ’ ∞.

dr

Thus we can differentiate times under the integral in (9.3) as long as ¤ (d ’ 1)/2 . In

this case the integral in the representation of φ ( ) can be bounded by a constant factor times

∞

’1/2

Fd φ(t)t d/2+ dt.

0

This means that we can form up to (d ’ 1)/2 derivatives of φ. This ¬nishes the proof for

odd d because then d/2 = (d ’ 1)/2 .

For even space dimension d = 2 it remains to show that the th derivative of p vanishes

at 1. Therefore, suppose that p(r ) = n c j r j ; then, for r > 0 the d-variate Fourier

j=0

transform of φ is given by

r

t

Fd φ(r ) = r ’d t d/2 J(d’2)/2 (t)dt

p

r

0

r

n

= r ’d cjr’j t j+ J ’1 (t)dt

j=0

0

’d

=: r I (r ).

is positive de¬nite on Rd we have I ≥ 0, ≡ 0. Now, making use of

Since

(d/dt)[t ’ν Jν (t)] = ’t ’ν Jν+1 (t) we obtain via integration by parts

n r

cjr’j

I (r ) = t j+ J ’1 (t)dt

0

j=0

n r

cjr’j ’2 ’ +2

= t j+2 t J ’1 (t)dt

0

j=0

n r

0

cjr’j ’1

= ’2 (t) r + ( j + 2 ’ 2)

t j+ J t j+ J ’2 (t)dt

0

j=0

n n r

c j r ’ j ( j + 2 ’ 2) ’1

=’ ’2 (r ) + t j+

cjr J J ’2 (t)dt

0

j=0 j=0

n r

c j r ’ j ( j + 2 ’ 2) ’1

= t j+ J ’2 (t)dt

0

j=0

’1)

because p(1) = c j = 0. Knowing that p(1) = p (1) = · · · = p ( (1) = 0, we can

126 Compactly supported functions

iterate this process to derive ¬nally

n

I (r ) = ’ c j ( j + 2 ’ 2)( j + 2 ’ 4) . . . j J’2 (r )

j=1

r

n

c j ( j + 2 ’ 2)( j + 2 ’ 4) . . . j( j ’ 2)r ’ j

+ t j’1 J’2 (t)dt.

j=1

0

The asymptotic behavior of the ¬rst summand is determined by

νπ π

2

+ O(r ’3/2 )

Jν (r ) =

cos r ’ ’

πr 2 4

√

(see Proposition 5.6), which means that it decays as O(1/ r ) for r ’ ∞, with varying

sign unless the sum

n

c j ( j + 2 ’ 2)( j + 2 ’ 4) · · · j (9.4)

j=1

is equal to zero. We are now going to prove that the second summand in the last represen-

tation of I (r ) decays as O(1/r ) for r ’ ∞. This means that the second summand cannot

compensate the change in sign of the ¬rst summand for large r and thus the sum (9.4) must

equal zero.

Using Jn = (’1)n J’n , n ∈ N, and (d/dt)[t ν Jν (t)] = t ν Jν’1 (t) we obtain, via integration

by parts,

r r

’j ’j

J’2 (t)dt = r

j’1

t j’1 J2 (t)dt

r t

0 0

r

= r’j t j’4 t 3 J2 (t)dt

0

r

r

= r ’ j t j’1 J3 (t) 0 ’ ( j ’ 4)r ’ j t j’2 J3 (t)dt

0