j=1

Then Lemma 6.7 leads to ± ≡ 0.

Now we are able to construct positive de¬nite functions just by choosing the measure.

This is even simpler if μ has a Lebesgue density.

Corollary 6.9 If f ∈ L 1 (Rd ) is continuous, nonnegative, and nonvanishing then

f (ω)e’i x ω dω,

T

x ∈ Rd ,

(x) :=

Rd

is positive de¬nite.

Proof We use the measure μ de¬ned for any Borel set B by

μ(B) = f (x)d x.

B

Then the carrier of μ is equal to the support of f . However, since f is nonnegative and

74 Positive de¬nite functions

Fig. 6.1 The Gaussian for ± = 1 and ± = 2 (on the left) and the inverse multiquadrics with c = 1 for

β = 1/2 and β = 3/2 (on the right).

nonvanishing, its support must contain an interior point, and hence the Fourier transform

of f is positive de¬nite by the preceding theorem.

It is time for our ¬rst example. The reader can probably see that the Gaussian is a good

candidate for a positive de¬nite function. We formulate the result more generally using an

additional scale parameter.

(x) = e’±

2

, ± > 0, is positive de¬nite on every Rd .

x

Theorem 6.10 The Gaussian 2

Proof We reduce the proof to that for the function G(x) := e’ x 2 /2 , which satis¬es G = G.

2

If we introduce G ± = G(·/±) then we have = G 1/√2± and G ± = ± d G(±·) by Theorem

5.16. Collecting these results we derive

(x) = 2’d (π ±)’d/2 e’ ω 2 /(4±) ’i x T ω

e dω,

2

Rd

which means that is positive de¬nite.

The left-hand part of Figure 6.1 shows the univariate function φ(r ) = e’±r for ± = 1

2

and ± = 2.

The Gaussian is positive de¬nite for every scaling parameter ± > 0. However, the correct

choice of ± in a particular interpolation problem is crucial. On the one hand, if the parameter

is too large then the Gaussian becomes a sharp peak, which immediately carries to the surface

and leads to a rather poor representation; the interpolation matrix, however, is then almost

diagonal and has a low condition number. On the other hand, a small value for ± leads to

a better surface reconstruction but corresponds to a very large condition number for the

interpolation matrix. We shall come back to this relation between condition number and

approximation property later on. Figure 6.2 demonstrates the effect of the scale parameter

on the reconstruction of a smooth function from a 6 — 6 grid. The picture on the left comes

closest to the original function.

Of course, Corollary 6.9 is useful when constructing positive de¬nite functions. But if a

function is given it would be useful to have a tool to check whether it is positive de¬nite.

Motivated by the Gaussian we formulate:

Theorem 6.11 Suppose that ∈ L 1 (Rd ) is a continuous function. Then is positive de¬-

nite if and only if is bounded and its Fourier transform is nonnegative and nonvanishing.

6.2 Bochner™s characterization 75

Fig. 6.2 The reconstruction of a smooth function using a Gaussian with scale parameter ± = 10, ± =

100, ± = 1000.

Proof Suppose that is bounded and has a nonnegative and nonvanishing Fourier trans-

form. It suf¬ces to show that ∈ L 1 (Rd ) is satis¬ed. Then the Fourier inversion formula

can be applied and Theorem 6.8 ¬nishes the proof.

As in the proof of Bochner™s theorem we choose gm from Theorem 5.20 and get

(2π )d/2 (0) = (2π)d/2 lim (x)gm (x)d x

m’∞ Rd

= (2π)d/2 lim (ω)gm (ω)dω

m’∞ Rd

= (ω)dω,

Rd

since is nonnegative. Thus is in L 1 (Rd ).

If, conversely, is positive de¬nite then we know from Theorem 6.2 that is bounded

and from Bochner™s theorem that is the inverse Fourier transform of a nonnegative ¬nite

Borel measure μ on Rd . Furthermore, using Theorem 5.16, Theorem 5.20 and Fubini™s

theorem we ¬nd that

(x) = lim (ω)gm (ω ’ x)dω

m’∞ Rd

(ω)gm (ω)e’i x ω dω

T

= lim

m’∞ Rd

= lim (2π )’d/2 e’iω · dμ(·)gm (ω)e’i x ω dω

T T

m’∞ Rd Rd

= lim (2π )’d/2 gm (ω)e’iω

T

(·+x)

dωdμ(·)

m’∞ Rd Rd

= lim gm (’· ’ x)dμ(·)

m’∞ Rd

≥ 0.

Thus is nonnegative. Now we can proceed as in the ¬rst part of the proof to show that

∈ L 1 (Rd ) and L 1 (Rd ) = (2π)

d/2

(0). Hence cannot vanish identically.

76 Positive de¬nite functions

The proof of the last theorem shows in particular that it suf¬ces to show that is

nonnegative when is not the zero function, because then cannot vanish.

The crucial point in the last proof was to establish that a bounded, continuous, and

integrable function with a nonnegative Fourier transform also has its Fourier transform in

L 1 (Rd ). The assumptions can be weakened. For an integrable Fourier transform, it suf¬ces

to assume that is integrable, continuous at zero, and has a nonnegative Fourier transform

(see Stein & Weiss [180]). But since the functions in which we are interested are always

continuous and bounded we will use these facts to shorten the proofs.

Corollary 6.12 If ∈ C(Rd ) © L 1 (Rd ) is positive de¬nite then its nonnegative Fourier