1. Differences of Gaussian filters

New York Journal of Mathematics

New York J. Math. 14(2008)577–599.

R´ enyi dimension and Gaussian filtering.

II

Terry A. Loring

Abstract. We consider convolving a Gaussian of a varying scale against a Borel measureμ on Euclidean δ-dimensional space. The L^q norm of the result is differentiable in . We calculate this derivative and show how the upper order of its growth relates to the lower R´enyi dimension ofμ. We assumeqis strictly between 1 and∞and thatμis finite with compact support.

Consider choosing a sequencenof scales for the Gaussians g(x) =^−δe^−(|x|/)2.

Letfq denote theL^q norm for Lebesgue measure. The differences

˛˛˛‚‚g_n+1∗μ‚‚

q− g_n∗μ_q˛˛˛

between the norms at adjacent scalesnandn−1 can be made to grow more slowly than any positive power ofnby setting thenby a power rule. The correct exponent in the power rule is determined by the lower R´enyi dimension.

We calculate and find bounds on the derivative of the Gaussian kernel versions of the correlation integral. We show that a Gaussian kernel version of the R´enyi entropy sum is continuous.

Contents

1. Differences of Gaussian filters 578

2. Differentiating the norm of a filtered measure 580

3. Asymptotic indices 584

4. Gaussian kernel correlation integrals 589

5. Gaussian kernel R´enyi entropy sums 596

References 598

Received August 1, 2007 and in revised form October 11, 2008.

Mathematics Subject Classification. 28A80, 28A78.

Key words and phrases. Asymptotic indices, R´enyi dimension, generalized fractal di- mension, regular variation, Laplacian pyramid, correlation dimension, gaussian kernel.

This work was supported in part by DARPA Contract N00014-03-1-0900.

ISSN 1076-9803/08

577

1. Differences of Gaussian filters

Suppose μ is a finite Borel measure on R^δ with compact support. If δ = 2 we can think of μ as the abstraction of an image. In the context of image processing, it is common to look at the difference of convolutions of two Gaussians against μ. This is the case in the standard construction of a Laplacian pyramid ([3]). Traditionally, the scales of the kernels vary geometrically. For some purposes, it might be better to set the scales of the convolution kernels in a different pattern.

Given a functiongonR^δ, thought of as acting as a canonical filter kernel on measures, we rescale it as

g(x) =^−δg ⁻¹x

.

The most important cases we have in mind are where g is a Gaussian or a function of compact support that approximates a Gaussian. We now consider the problem of selecting some _n0 so that the differences

g_n∗μ−g_n−1 ∗μ behave nicely.

We will use the notation m for Lebesgue measure, fq=

R^δf(x)^qdm(x) ¹

q

=

R

R· · ·

Rf(x)^qdx₁dx₂· · ·dx_δ ¹

q

. One criterion for the selection of the scales _n is to keep

g_n∗μ−g_n−1 ∗μ

q

approximately constant for some choice of 1< q <∞. We don’t see a means of estimating this norm of differences, so we instead look at the difference of norms.

Ifμ is “fractal,” then we expect that setting_n in a geometric series will lead to exponential growth in

gn∗μ_q−g_n−1 ∗μ

q

.

If we set the_nvia a power law,_n=n^−t, we can reasonably hope that this difference is more or less constant.

Recall, say from [1] or [14], that the upper and lower R´enyi dimensions of μfor indexq are defined by

D^±_q(μ) = lim

→0 supinf

1 q−1

ln (S_μ^q()) ln() ,

where the standard partition functionS_μ^q() is taken to be

(1) S_μ^q() =

k∈Z^δ

μ(k+I)^q, and whereI is aδ-fold product of the unit interval [0,1).

The connection with Gaussian convolution is the formula, due to Gu´erin, lim sup

x→∞

ln

g_x−1 ∗μ_q

ln(x) =q−1 q

δ−D_q⁻(μ) .

See [8] or [12, Lemma 2.3]. For this formula to be valid, we need some restriction ong. For simplicity, we consider the case wheregis nonnegative and rapidly decreasing. By rapidly decreasing we mean that any order derivative g^(α) of g exists and decays at infinity more rapidly than any negative power of x. Certainly, less is needed. See [8].

Here is the main result.

Theorem 1.1. Suppose 1 < q < ∞. Suppose g :R^δ → R is nonnegative, nontrivial, rapidly decreasing and is radially nonincreasing. Suppose μ is a finite Borel measure on R^δ with compact support. Let I_q(μ) denote the set of positive t for which

∀α >0, lim

n→∞

g_n−t∗μ_q−g_(n−1)−t ∗μ

q

n^α = 0.

If D⁻_q(μ)< δ then

I_q(μ) =

0, q q−1

δ−D_q⁻(μ)₋₁ , while if D_q⁻(μ) =δ then I_q(μ) = (0,∞).

This is in sharp contrast with what happens if the scales of the kernels are set to geometric growth.

Theorem 1.2. Suppose q, g and μ are as in Theorem 1.1. If D⁻_q(μ) < δ then

lim sup

n→∞

g₂−n∗μ_q− g₂−n+1∗μ_q

n^α =∞ (∀α >0).

The proofs require an estimate on the derivative d

dλln

g_eλ∗μ_q ,

which we derive in Section2. Section3contains lemmas on the upper order of positive function of a positive variable and completes the proofs of the main theorems.

Let us use the notation fμ,q =

R^δf(x)^qdμ(x) ¹

q

.

The calculations from Section 2can be adjusted to estimate d

dλln

g_eλ∗μ_μ,q−1 . Equivalently, we find bounds on the derivative of

ln

R^δ

R^δg

x−y e^λ

dμ(y)

_q dμ(x)

.

This is the content of Section 4, which does not depend on Section 3. This should be of interest as it relates to computing the correlation dimension by probabilistic methods.

In Section 5 we consider adjusting the standard partition sum S_μ^q() by allowing soft cut-offs between the cells (bins). We cannot determine the derivative of the these Gaussian-kernel sums, but do demonstrate they are continuous in.

2. Differentiating the norm of a filtered measure

Recall that given a functiong on R^δ we rescale it as g(x) =^−δg

⁻¹x . This section’s goals are to compute

d

dg∗μ_q and to show that

d dλln

g_eλ∗μ_q

is bounded.

Lemma 2.1. Suppose g : R^δ → R is differentiable, bounded, and has bounded radial derivative. Let h : R^δ → R be the negative of the radial derivative of g,

h(x) =−

δ

j=1

x_j ∂g

∂x_j. If μ is a finite Borel measure on R^δ then

∂

∂[(g∗μ)(x)] =⁻¹((h∗μ)(x)−δ(g∗μ)(x)).

Proof. For any w,

∂

∂[g(w)] =

δ

j=1

∂g

∂x_j

w

∂

∂(w_j)

=

δ

j=1

∂g

∂x_j

w

w_j

=−⁻¹h(w).

Supposex is fixed. Assume 0< a≤≤b and y∈R^δ. Then

∂

∂[g(x−y)]

= ∂

∂

^−δg(⁻¹(x−y)

=^−δ(−h(⁻¹(x−y))(−⁻²) + (−δ^−δ−1)g(⁻¹(x−y))

=⁻¹(h(x−y)−δg(x−y)).

and so

∂

∂[g(x−y)]

≤a^−δ−1H+δa^−δ−1G,

where G and H are bounds on g and h. Since μ is finite, g(x−y) is integrable iny. The Dominated Convergence Theorem gives us

∂

∂[(g∗μ)(x)] = ∂

∂

R^δg(x−y)dμ(y)

=

R^δ−1(h(x−y)−δg(x−y)) dμ(y)

=⁻¹((h∗μ)(x)−δ(g∗μ)(x)). Notation 2.2. We shall usex∧y to denote the minimum of two numbers and x∨y for their maximum.

Theorem 2.3. Suppose 1 < q < ∞. Suppose g : R^δ → R is rapidly decreasing and let h : R^δ → R denote the negative of the radial derivative of g. Suppose g ≥0 and h ≥ 0. If μ is a finite Borel measure on R^δ with compact support then

d

dg∗μ_q =

R^δ(g∗μ)^q−1(h∗μ)dm

g∗μ^q−1_q −δg∗μ_q and

d dλln

g_eλ∗μ_q

=

R^δ(g_eλ∗μ)^q−1(h_eλ∗μ) dm

R^δ(g_eλ∗μ)^q dm

−δ.

Proof. Assume 0< a≤≤b.

Pick an integer k with k > ^δ+1_q . Since g is rapidly decreasing there is a C₁ so that

g(x)≤C₁(1∧ |x|^−k).

For allx,

g(x) =^−δg(⁻¹x)

≤a^−δC₁(1∧(b^k|x|^−k))

=a^−δC₁b^k(b^−k∧ |x|^−k).

If|y| ≤ ¹₂|x|and 2b≤ |x|then

g(x−y)≤a^−δC₁b^k(b^−k∧ |x−y|^−k)

≤a^−δC₁b^k

b^−k∧x 2

^−k

=a^−δC₁b^k2^k|x|^−k. Suppose

supp(μ)⊆

y∈R^δ|y| ≤R

. If|x| ≥(2b)∨(2R) then

(g∗μ)(x) =

|y|≤Rg(x−y)dμ(y)

≤μ R^δ

a^−δC₁b^k2^k|x|^−k. If|x| ≤(2b)∨(2R) then we have the estimate

(g∗μ)(x) =

g(x−y)dμ(y)

≤

a^−δC₁dμ(y)

=μ R^δ

a^−δC₁. For someC₂ andC₃,

(g∗μ)(x)≤C₂(C₃∧ |x|^−k) for all x.

We can repeat the previous argument for h. Possibly increasing C₂ and C₃ we can have

(h∗μ)(x)≤C₂(C₃∧ |x|^−k).

Therefore ∂

∂[(g∗μ)(x)]

=⁻¹((h∗μ)(x)−δ(g∗μ)(x))

≤a⁻¹((h∗μ)(x) +δ(g∗μ)(x))

≤a⁻¹(δ+ 1)C₂(C₃∧ |x|^−k)

and

∂

∂[((g∗μ)(x))^q]

=q((g∗μ)(x))^q−1 ∂

∂[(g∗μ)(x)]

≤qa⁻¹(δ+ 1)(C₂(C₃∧ |x|^−k))^q

=qa⁻¹(δ+ 1)C₂^q(C₃^q∧ |x|^−qk),

which is is integrable since k is larger than ^δ+1_q . We can use dominated convergence again. By Lemma 2.1,

d d

g∗μ^q_q

=

R^δq((g∗μ)(x))^q−1−1((h∗μ)(x)−δ(g∗μ)(x))dm(x)

= q

R^δ((g∗μ)(x))^q−1(h∗μ)(x)dm(x)−δ

R^δ((g∗μ)(x))^q dm(x)

. The two derivative formulas in the statement of the theorem now follow from

d

dg∗μ_q= 1

q g∗μ^1−q_p d d

g∗μ^q_q

and

d dλln

g_eλ∗μ_q

= d

d

=e^λg∗μ_q e^λ

g_eλ∗μ_q . Theorem 2.4. Suppose 1 < q < ∞. Suppose g : R^δ → R is rapidly decreasing and let h : R^δ → R denote the negative of the radial derivative of g. Suppose g ≥0 and h ≥ 0. If μ is a finite Borel measure on R^δ with compact support then

−δ⁻¹g∗μ_q≤ d

dg∗μ_q≤⁻¹h∗μq−δ⁻¹g∗μ_q and

−δ≤ d dλln

g_eλ∗μ_q

= h_eλ∗μq

g_eλ∗μq −δ.

Proof. The two lower bounds follow trivially from the last theorem and the fact that g andh are nonnegative.

H¨older’s inequality gives us

R^δ((g∗μ)(x))^q−1(h∗μ)(x)dm(x)

≤

R^δ((g∗μ)(x))^q dm(x) ^q−1

q

R^δ((h∗μ)(x))^q dm(x) ¹

q

=g∗μ^q−1_q h∗μq

and the upper bounds follow.

Corollary 2.5. Suppose 1< q < ∞. Suppose g :R^δ →R is nonnegative, nontrivial, rapidly decreasing and is radially nonincreasing. There is a finite constant C so that ifμis a finite Borel measure onR^δ with compact support then

−δ≤ d dλln

g_eλ∗μ_q

≤C for all λ. If g is a Gaussian, then we may take C = 0.

Proof. We can apply [12, Lemma 2.1] to g. When S_μ^q() is the partition function (1), this tells us that there is aDso that

D⁻¹ ≤ ^δq−1q g∗μq

(S_μ^q())^1q ≤D.

From the proof of [12, Lemma 2.1] we see that D can be taken to depend only on q and g. This conclusion is valid for the negative radial derivative h as well. This is because h∗μq is invariant under translations ofh and clearlyh is bounded away from zero on some open set. Therefore

h∗μq

g∗μq

= ^δq−1q h∗μq

(S_μ^q())^1q

(S_μ^q())^{1q δq−1q} g∗μq

is bounded above and away from zero.

In the Gaussian case, we know that g∗μq is nonincreasing (cf. [12, Lemma 3.1]) and so the derivative is nonpositive.

3. Asymptotic indices

Given a functionf >0 on the positive reals, the quantities d(f) = lim sup

x→∞

ln (f(x)) ln(x) and

d(f) = lim inf

x→∞

ln (f(x)) ln(x)

are the upper and lower orders off. These provide a simple way to compare the asymptotic behavior of f(x) to x^c for various powers c. Equivalently, d(f) is the smallest extended real number so that

(2) c > d(f) =⇒ f(x)≤x^c for large x and d(f) is the largest extended real number so that (3) c < d(f) =⇒ f(x)≥x^c for large x.

The proof is not complicated. See [10]. For a look at how upper and lower order relate to regular variation, see [2].

In broad terms, if

c= lim

x→∞

ln (f(x)) ln(x)

exists, then f(x) behaves not so differently from x^c. There is no reason to think that if f(x) exists it must behave like x^c−1. However, if there are some bounds on the derivative of the log-log plot of f then we are able to deduce the upper order of |f|from the upper order off.

For the lower order on|f|, we have found no particularly interesting result that can be applied tog_x−1 ∗μq. The difficulty is that even if we assume g is a Gaussian we don’t know if the derivative of g_x−1 ∗μq is bounded away from zero.

We take the liberty of setting ln(0) =−∞, and indeed ^ln(0)_C =−∞. (This is to accommodate f(x) = 0 at some x and f(n) = f(n−1) at some n.) Both (2) and (3) remain valid.

Lemma 3.1. Suppose f : [1,∞) → (0,∞) is differentiable and that for some finite constant C,

(4)

d

dxln (f(e^x)) ≤C.

Given any nondecreasing sequence x_n with limit ∞, if

(5) lim

n→∞

ln(x_n+1) ln(x_n) = 1 then

lim sup

n→∞

ln (f(x_n))

ln(x_n) = lim sup

x→∞

ln (f(x)) ln(x) . Proof. The bound (4) implies

|ln (f(e^x))−ln (f(e^y))| ≤C|x−y| or

|ln (f(x))−ln (f(y))| ≤C|ln(x)−ln(y)|.

The rest of the proof mimics that of [12, Lemma 4.1], and is omitted.

Lemma 3.2. Suppose f : [1,∞) → (0,∞) is differentiable. If, for some finite constant C,

d

dxln (f(e^x)) ≤C for all x, then

lim sup

n→∞

ln|f(n)−f(n−1)|

ln(n) = lim sup

x→∞

ln|f(x)| ln(x)

= lim sup

x→∞

ln (f(x)) ln(x) −1.

Proof. Suppose f is a function with the bounds ±C on the slope of its log-log plot. For each n, the Mean Value Theorem gives us a numberx_n in the range

n−1≤x_n≤n for which

f(n)−f(n−1) =f(x_n).

From basic facts about nets we obtain lim sup

n→∞

ln|f(n)−f(n−1)|

ln(n) = lim sup

n→∞

ln|f(x_n)|

ln(n)

≤lim sup

n→∞

ln|f(x_n)| ln(x_n)

≤lim sup

x→∞

ln|f(x)| ln(x) . Letg(x) = ln (f(e^x)) so that|g(x)| ≤C and

g(x) = e^xf(e^x) f(e^x) . This can be rewritten as

f(x) =g(ln(x))x⁻¹f(x) and so we have

f(x)≤Cx⁻¹f(x).

Therefore

lim sup

x→∞

ln|f(x)|

ln(x) ≤lim sup

x→∞

lnC−ln(x) + ln(f(x)) ln(x)

= lim sup

x→∞

ln(f(x)) ln(x) −1.

To finish, we must show lim sup

x→∞

ln(f(x))

ln(x) −1≤lim sup

n→∞

ln|f(n)−f(n−1)|

ln(n) .

We can apply Lemma 3.1, because 1≤ ln(x_n+1)

ln(x_n) ≤ ln(n+ 1) ln(n−1) →1, and this tells us that it will suffice to show

lim sup

n→∞

ln (f(n))

ln(n) −1≤lim sup

n→∞

ln|f(n)−f(n−1)|

ln(n) .

Let

m= lim sup

n→∞

ln|f(n)−f(n−1)|

ln(n) .

Suppose we are givenδ >0. Then pickc=−1 with m < c < m+δ.

(If m = ∞ we have nothing to prove. If m = −∞ then modify this to picking c = −1 less then any given finite number C.) There is a natural number n₀ so that

n≥n₀ =⇒ |f(n)−f(n−1)| ≤n^c. For largen,

f(n) =f(n₀) +

n

k=n0+1

|f(k)−f(k−1)|

≤f(n₀) +

n

k=n0+1

k^c

≤f(n₀) + _n+1

n0

y^cdy

≤f(n₀) + 1

c+ 1(n+ 1)^c+1. For largen,

f(n)≤n^m+δ+1. Therefore

lim sup

n→∞

ln (f(n))

ln(n) ≤m+δ+ 1.

Since this is true for all δ >0, we are done. (If m = −∞ then we obtain this lim sup is less thanC, for all finite C, and so is also−∞.) Lemma 3.3. Suppose f : [1,∞) →(0,∞) is differentiable and that there is a finite constant C so that

d

dxln (f(e^x)) ≤C for all x. If

d(f) = lim sup

x→∞

ln(f(x)) ln(x) >0

then t >0

∀α >0, lim

n→∞

f(n^t)−f((n−1)^t)

n^α = 0

=

0, d(f)⁻¹ .

If d(f) = 0 then

t >0

∀α >0, lim

n→∞

f(n^t)−f((n−1)^t)

n^α = 0

= (0,∞).

Proof. For a sequence a_n>0, it is routine to show that

(6) lim sup

n

ln(a_n)

ln(n) ≤0 ⇐⇒ ∀α >0, lim

n→∞

a_n n^α = 0.

Let h(x) = ln(f(e^x)) so that |h(x)| ≤ C. With t > 0 to be specified below, defineg(x) =f(x^t). Then

d

dxln (g(e^x)) ≤tC and we may apply Lemma 3.2to g.

As to the upper order:

lim sup

x→∞

ln(g(x))

ln(x) = lim sup

x→∞

ln(f(x^t)) ln(x)

= lim sup

x→∞

ln(f(x)) ln(x^1t)

=td(f).

By Lemma3.2, lim sup

n→∞

lnf(n^t)−f((n−1)^t)

ln(n) = lim sup

n→∞

ln (|g(n)−g(n−1)|) ln(n)

= lim sup

x→∞

ln (g(x)) ln(x) −1

=td(f)−1.

Ift > d(f) then by (6) there exists α >0 so that f(n^t)−f((n−1)^t)

n^α →0.

Ift≤d(f) then

f(n^t)−f((n−1)^t)

n^α →0

for all positiveα.

Theorem1.1now follows, since if

f(x) =g_x−1 ∗μ_q then by [8], or [12, Lemma 2.3],

d(f) = q−1 q

δ−D⁻_q(μ) .

To prove Theorem1.2 requires only the following lemma.

Lemma 3.4. Suppose f : [1,∞) → (0,∞) is differentiable. If, for some finite constant C,

d

dxln (f(e^x)) ≤C for all x, and if

d(f) = lim sup

x→∞

ln(f(x)) ln(x) >0, then

lim sup

n→∞

lnf(2ⁿ)−f(2ⁿ⁻¹)

ln(n) =∞.

Proof. Suppose for some α >0 there is an n₀ so that n≥n₀ =⇒ f(2ⁿ)−f(2ⁿ⁻¹)≤n^α. Then

n≥n₀ =⇒ f(2ⁿ)≤f(2ⁿ⁰) +n^α+1. Supposeβ >0. Then for some n₁≥n₀,

n≥n₁ =⇒ f(2ⁿ⁰) +n^α+1≤2^nβ. Therefore

lim sup

n→∞

ln (f(2ⁿ)) ln (2ⁿ) ≤β.

Lemma3.1 tells us lim sup

n→∞

ln (f(x))

ln (x) = lim sup

n→∞

ln (f(2ⁿ))

ln (2ⁿ) = 0.

4. Gaussian kernel correlation integrals

The probabilistic interpretations of the correlation integral

μ(x+B)^q−1dμ(x)

make it a common tool for determining the R´enyi dimensions of μ. Here B=

x∈R^δ|x| ≤1

and μ is a Borel probability measure on R^δ. When q = 2 the correlation integral is

μ(B(x))dμ(x) = Pr{|X₁−X₂| ≤},

where X₁ and X₂ are random locations in the probability space R^δ, μ

. Using a sharp cut-off for the allowed distance seems unwise in a numerical situation, as is discussed in [4,5, 6,7,8,11,13,16,17].

Consider the expectation of the scalar-valued random variable G

⁻¹|X₁−X₂|

for a function such as a GaussianG(x) =e^−x2, or anyG≥0 that is positive at 0 and rapidly decreasing. This expectation can be rewritten as follows.

Letg(x) =G(|x|) and

g(x) =^−δG(⁻¹|x|).

Then

E G

⁻¹X₁−X₂

=^δ g(x−y) dμ(y)dμ(x)

=^δg∗μ_μ,1. As is shown in [1],

D_q^±(μ) = lim

→0 supinf

1 q−1

ln

R^δ

^δg∗μ_q−1 dμ

ln()

=δ+ lim

→0 supinf

ln

g∗μ_μ,q−1

ln()

and more specifically there is a constantC = 0 so that C⁻¹ ≤

R^δδ(g∗μ)^q−1 dμ S_μ^p() ≤C for all .

The R´enyi dimensions of μcan be computed as

λ→−∞lim

supinf

ln(P_μ(e^λ)) λ

for at least the following six choices of partition function. (See [1,8, 9,15]

and Section 5.)

(7) P_μ^q() =

S_μ^q()_q−1¹

=

⎛

⎝

j∈Z^δ

μ(j+I)^q

⎞

⎠

q−11

;

(8) P_μ^q() =

R^δμ(x+B)^q−1dμ(x) ¹

q−1 ;

(9) P_μ^q() =

R^δμ(x+B)^q1

^δdm(x) ¹

q−1 ;

(10) P_μ^q() =

⎛

⎝

j∈Z^δ

R^δg

j−y

dμ(y) _q−1⎞

⎠

q−11

;

(11) P_μ^q() =

R^δ

R^δg

x−y

dμ(y)

_q−1 dμ(x)

¹

q−1

;

(12) P_μ^q() =

R^δ

R^δg

x−y

dμ(y)

_q 1

^δdm(x) _q−1¹

.

The functions (7) and (8) can be discontinuous in . (A sum of two point masses shows this.) Assuming μ has compact support, we find that (10) is continuous for 1 < q < ∞ (Theorem 5.2), that (11) is continuous for 1 < q <∞ and differentiable for 2< q < ∞, (Theorems 4.5and 4.2), and that (12) is differentiable for 1< q <∞ (Theorem2.3). Added smoothness should be an advantage in computational situations, as was pointed out in [17].

It is not clear if the function in (9) is continuous wheneverμis finite with compact support.

The bound on the last partition function that we found in Section2used a different normalizing constant. Recall there is a C so that

−δ ≤ d dλln

g_eλ∗μ_q

≤C.

In the Gaussian case we hadC = 0. Since ln

R^δ

R^δg

x−y e^λ

dμ(y)

_q 1

e^δλ dm(x) ¹

q−1

= ln

e^δλg_eλ∗μq^q−1q

=δλ+ q q−1ln

g_e^λ∗μ_q

we have δ

1−q ≤ d dλln

R^δ

R^δg

x−y e^λ

dμ(y)

_q 1

e^δλ dm(x) ¹

q−1

≤C₁.

In the Gaussian case, we may take C₁=δ.

In this section we prove that for 2 ≤ q < ∞, there is a constant C depending on g and q so that for any finite Borel measure μ of compact

support, 0≤ d

dλln

⎛

⎝

R^δ

R^δg

x−y e^λ

dμ(y)

_q−1 dμ(x)

¹

q−1⎞

⎠≤C.

We will need a lower bound on

R^δδ(h∗μ)^q−1 dμ,

wherehis the negative of the radial derivative ofg. Sinceh(0) = 0 we need a small modification of a result in [1]. We are restricting our attention to the case 1< q <∞, which allows us to avoid the technicalities encountered in [1].

Here we use the notation from [12], so μ⁽⁾ is the sequence over Z^δ given by

μ⁽⁾_n =μ(n+I).

Lemma 4.1. Assume that g≥0 is rapidly decreasing and that 1< q <∞. There is a finite constant C so that for any finite Borel measure μ on R^δ,

^δg∗μ_μ,q−1 (S_μ^q())^q−11

≤C

for all >0. If also g(0)>0 then there is a c >0 so that c≤ ^δg∗μ_μ,q−1

(S_μ^q())^q−11 .

Proof. Define Γ over Z^δ by

Γ_n= sup{g(x) |x∈n+D}, where

D= (−1,1)×(−1,1)× · · · ×(−1,1).

Repeating an argument from [1], we find g∗μ^q−1_μ,q−1

=^−(q−1)δ g(⁻¹(x−y))dμ(y) _q−1

dμ(x)

=^−(q−1)δ

j∈Z^δ

j+I

⎛

⎝

k∈Z^δ

k+Ig(⁻¹(x−y))dμ(y)

⎞

⎠

q−1

dμ(x)

≤^−(q−1)δ

j∈Z^δ

⎛

⎝

k∈Z^δ

Γ_j−kμ⁽⁾_k

⎞

⎠

q−1

μ⁽⁾_j .

H¨older’s inequality and Young’s convolution inequality now tell us g∗μ^q−1_μ,q−1 ≤^−(q−1)δ

Γ∗μ⁽⁾

q

_q−1μ⁽⁾

q

≤^−(q−1)δΓ^q−1₁ μ^()q

q, i.e.,

^δg∗μ_μ,q−1

(S_μ^p())^q−11 ≤ Γ₁.

If g is positive at the origin, then since it is continuous we can rescale g usingg=g_η with the same properties asg, but with

inf{g(x)|x∈D}>0

and g=g_η. We can compareg∗μ_μ,q−1 and g_η∗μ_μ,q−1 as follows:

^δg∗μ_μ,q−1 S_μ^δ() ¹

q−1

= ^δg_η∗μ_μ,q−1 S_μ^δ() ¹

q−1

=

η^−δ

η^δδg_η∗μ_μ,q−1 (S_μ^q(η))^q−11

(S_μ^q(η))^q−11 (S_μ^q())^q−11

.

By [12, Theorem 3.4], there are constantsAand B so that e^{−A−B|ln(η)|}≤ S_μ^q(η)

S_μ^q() ≤e^A+B|ln(η)|

for all . Therefore ^δg∗μ_μ,q−1

(S_μ^q())^q−11 ≤

η^−δe^{A+B|q−1ln(η)|}

(η)^δg_η∗μ_μ,q−1 (S_μ^q(η))^q−11

,

and it suffices to prove the result in the case where inf{g(x)|x∈D}>0.

Let

γ_n = inf{g(x) |x∈n+D}. As above, we find

g∗μ^q−1_μ,q−1≥^−(q−1)δ

j∈Z^δ

⎛

⎝

k∈Z^δ

γ_j−kμ⁽⁾_k

⎞

⎠

q−1

μ⁽⁾_j

and so

g∗μ^q−1_μ,q−1 ≥^−(q−1)δ

j∈Z^δ

γ₀μ⁽⁾_j

_q−1 μ⁽⁾_j

=γ₀^−(q−1)δ

j∈Z^δ

μ⁽⁾_j

_q

and

^δg∗μ_μ,q−1 ≥γ

q−11

0

S_μ^q() ¹

q−1.

Theorem 4.2. Suppose 2 ≤ q < ∞. Suppose g : R^δ → R is rapidly decreasing and let h:R^δ→Rdenote the negative of the radial derivative of g. Suppose g≥0 andh≥0. If μis a finite Borel measure on R^δ then

d

dg∗μ_μ,q−1 =

R^δ(g∗μ)^q−2h∗μ dμ

g∗μ^q−2_μ,q−1 −δg∗μ_μ,q−1 and

d dλln

g_e^λ∗μ_μ,q−1

=

R^δ(g_eλ∗μ)^q−2h_eλ∗μ dμ

R^δ(g_eλ∗μ)^q−1 dμ −δ.

Proof. For restricted to some interval [a, b], it follows from Lemma 2.1 that

∂

∂((g∗μ)(x))^q−1

≤(q−1)a−(q−1)(δ+1)μ^q−1₁ g^q−2_∞ (h_∞+δg_∞). Dominated convergence yields

d d

R^δ(g∗μ)^q−1 dμ

= q−1

R^δ(g∗μ)^q−2h∗μ dμ−δg∗μ^q−1_μ,q−1

and so d

dg∗μ_μ,q−1

= 1

g∗μ^2−q_μ,q−1

R^d(g∗μ)^q−2h∗μ dμ−δg∗μ^q−1_μ,q−1

=

R^δ(g∗μ)^q−2h∗μ dμ

g∗μ^q−2_μ,q−1 −δg∗μ_μ,q−1

.

We use

d dλln

g_eλ∗μ_μ,q−1

= d

d

=e^λg∗μ_μ,q−1 e^λ g_eλ∗μ_μ,q−1

and find d dλln

g_eλ∗μ_μ,q−1

=

R^δ(g_eλ∗μ)^q−2h_eλ∗μ dμ

g_eλ∗μ^q−1_μ,q−1 −δ.

Theorem 4.3. Suppose 2 ≤ q < ∞. Suppose g : R^δ → R is rapidly decreasing and let h : R^δ → R denote the negative of the radial derivative of g. Suppose g ≥0 and h ≥ 0. If μ is a finite Borel measure of compact support on R^d then

−δ⁻¹g∗μ_μ,q−1≤ d d

g∗μ_μ,q−1

≤⁻¹h∗μ_μ,q−1−δ⁻¹g∗μ_μ,q−1 and

−δ ≤ d dλln

g_eλ∗μ_μ,q−1

≤ h_eλ∗μ_μ,q−1 g_eλ∗μ_μ,q−1 −δ.

Proof. If 2< q <∞, we can apply H¨older’s inequality and we find

R^δ(g∗μ)^q−2(h∗μ)dμ

≤

R^δ(g∗μ)^q−1 dμ ^q−2

q−1

R^δ(h∗μ)^q−1 dμ ¹

q−1

. This is trivially true as well whenq = 2. We can rewrite this as

R^δ(g∗μ)^q−2(h∗μ) dμ≤ g∗μ^q−2_μ,q−1h∗μ_μ,q−1

and the inequalities follow from the last result and the fact that g∗μ and

h∗μare nonnegative.

Corollary 4.4. Suppose 2≤ q < ∞. Suppose g :R^δ →R is nonnegative, nontrivial, rapidly decreasing and is radially nonincreasing. There is a finite constant C so that if μ is a finite Borel measure of compact support on R^δ then

0≤ d dλln

⎛

⎝

R^δ

R^δg

x−y e^λ

dμ(y)

_q−1 dμ(x)

_q−1¹ ⎞

⎠≤C.

Proof. By Lemma 4.1, we have an upper bound on h_eλ∗μ_μ,q−1 and a lower bound on g_eλ∗μ_μ,q−1 that depends only on q and g. Therefore Theorem4.3gives us a C₁ so that

−δ ≤ d dλln

g_eλ∗μ_μ,q−1

≤C₁