Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 22, 1997, 227–236

ESTIMATES OF AVERAGES OF FOURIER TRANSFORMS OF MEASURES WITH FINITE ENERGY

Dedicated to S. Igari

Per Sjölin

Royal Institute of Technology, Department of Mathematics S-100 44 Stockholm, Sweden; pers@math.kth.se

Abstract. Estimates of Fourier transforms of measures with finite energy are considered. In earlier papers spherical means of the Fourier transform have been considered. We shall here study more general means. We shall also in particular study the case when the measure is given by a radial function.

1. Introduction

Let ℳ denote the class of all finite positive Borel measures μ in R^{n} with compact support. The Fourier transform of μ ∈ ℳ is defined by

μ̂(ξ) = ∫ e^{−iξ·x} dμ(x)

and the α-energy of μ is given by

I_{α}(μ) = ∫∫ |x − y|^{−α} dμ(x) dμ(y) = c ∫ |x|^{α−n}|μ̂(x)|^{2} dx, 0 < α < n.

Let ℳ_{0} denote the class of all μ ∈ ℳ with diam(supp μ) ≤ 1. We also let θ denote the area measure on S^{n−1} and set

σ(μ)(r) = ∫_{S^{n−1}} |μ̂(rξ)|^{2} dθ(ξ), r > 0,

for n ≥ 2.

We shall here study questions of the following type. For which values of β does the estimate

(1)  σ(μ)(r) ≤ Cr^{−β}I_{α}(μ),  r > 1,

1991 Mathematics Subject Classification: Primary 42B10.

hold for all μ ∈ ℳ_{0} ? This problem has been studied by P. Mattila [2] and then by P. Sjölin [3]. To formulate the results we set

β(α) = sup{β; (1) holds for all μ ∈ ℳ_{0}}.

It is proved in the above papers that

(2)  β(α) ≥ α,  0 < α ≤ \frac{1}{2}(n − 1),
(3)  β(α) ≥ \frac{1}{2}(n − 1),  \frac{1}{2}(n − 1) < α ≤ \frac{1}{2}(n + 1),

and

(4)  β(α) ≥ α − 1,  \frac{1}{2}(n + 1) < α < n.

Upper bounds for β(α) are also known. It is easy to see that always β(α) ≤ α so it follows that β(α) = α for 0 < α ≤ \frac{1}{2}(n − 1). There are also known counterexamples, which for n = 2 show that

(5)  β(α) ≤ \frac{1}{2},  \frac{1}{2} < α ≤ 1,

and

(6)  β(α) ≤ \frac{1}{2}α,  1 < α < 2.

For n = 2 we therefore have β(α) = α, 0 < α ≤ \frac{1}{2} , and β(α) = \frac{1}{2} , \frac{1}{2} < α ≤ 1.

Estimates of the above type have been used in [2] and J. Bourgain [1] to study the Hausdorff dimension of distance sets. However, we shall not discuss this application here.

We also remark that to prove the estimate (1) for μ ∈ ℳ_{0} it is sufficient to prove (1) in the special case μ = f ∈ C_{0}^{∞}(R^{n}) where f ≥ 0 and supp f ⊂ B, where B denotes the open unit ball in R^{n}. This can be seen by approximating a general μ by smooth functions.

We shall here study some variants of the above problem. We shall first consider the case of radial functions. We shall consider the estimate

(7)  σ(f)(r) ≤ Cr^{−β}I_{α}(f),  r > 1,

assuming that f is measurable and radial in R^{n}, f ≥ 0 and f(x) = 0 for |x| > 1. We set

β_{1}(α) = sup{β; (7) holds for all f of the above type}.

We have determined β_{1}(α) and have the following theorem.

Theorem 1. One has β_{1}(α) = α, 0 < α ≤ n − 1, and β_{1}(α) = n − 1, n − 1 < α < n.

We shall then replace S^{n−1} in the definition of σ(μ)(r) by more general hypersurfaces. Let Q_{n−1} denote the open unit cube in R^{n−1} and assume that φ ∈ C^{∞}(Q_{n−1}) with φ real-valued. Set

S = {x ∈ R^{n}; x_{n} = φ(x′), x′ ∈ Q_{n−1}},

where we have set x = (x_{1}, …, x_{n}) and x′ = (x_{1}, …, x_{n−1}). We make the assumption that S has non-vanishing Gaussian curvature at every point. Also let ψ ∈ C_{0}^{∞}(R^{n}) and assume that the orthogonal projection of (supp ψ) ∩ S onto the hyperplane x_{n} = 0 has positive distance to ∂Q_{n−1} . Here we consider Q_{n−1} as a subset of the hyperplane x_{n} = 0. We let σ_{0} denote the area measure on S and now set

(8)  σ(μ)(r) = ∫_{S} |μ̂(rξ)|^{2}ψ(ξ) dσ_{0}(ξ),  r > 0.

With this definition of σ(μ)(r) we then define β_{2}(α) by

β_{2}(α) = sup{β; (1) holds for all μ ∈ ℳ_{0}}.

Then the following theorem holds.

Theorem 2. In the inequalities (2), (3) and (4) β(α) can be replaced by β_{2}(α).

Theorem 2 shows that the above estimates for the unit sphere S^{n−1} can be generalized to hypersurfaces with non-vanishing Gaussian curvature. We shall also consider a surface with vanishing Gaussian curvature and see that the situation then is different.

Let Q denote the unit cube in R^{n} and now set

σ(μ)(r) = ∫_{∂Q} |μ̂(rξ)|^{2} dσ_{0}(ξ), r > 0,

where σ_{0} denotes the area measure on ∂Q. Also set

β_{3}(α) = sup{β; (1) holds for all μ ∈ ℳ_{0}}.

In this case we have the following result.

Theorem 3. One has β_{3}(α) = 0, 0 < α ≤ 1, and β_{3}(α) = α−1, 1 < α < n.

2. Proofs

Proof of Theorem 1. Assume that f ∈ C_{0}^{∞}(R^{n}) and that f is radial, f ≥ 0 and supp f ⊂ B. Then f̂ is radial and we have

f̂(r) = c_{n}r^{1−n/2} ∫_{0}^{∞} f(s)J_{n/2−1}(rs)s^{n/2} ds, r > 0,

where J_{k} denotes the Bessel function of order k. Here we write f(s) = f(x) if s = |x| in the usual way.

Now assume that 0 < α ≤ n − 1. Also choose β so that

(9)  2 − n < β < 2 − n + α.

It follows that β < 1, that is β/2 < 1/2, and also β/2 > 1 − n/2, that is n/2 − 1 > −β/2. Using asymptotic estimates for Bessel functions (see E.M. Stein and G. Weiss [5, p. 158]) we then conclude that

(10)  |J_{n/2−1}(t)| ≤ C t^{−β/2},  t > 0.

Invoking this estimate we obtain

|f̂(r)| ≤ Cr^{1−n/2} ∫_{0}^{1}(rs)^{−β/2}s^{n/2}f(s) ds
= C r^{1−n/2−β/2} ∫_{0}^{1} s^{n/2−β/2}f(s) ds
= C r^{1−n/2−β/2} ∫_{R^{n}} |x|^{−β/2−n/2+1}f(x) dx
= C r^{1−n/2−β/2} ∫_{R^{n}} |ξ|^{−n/2+β/2−1}f̂(ξ) dξ.

Here we used the fact that the inequality 1 − n/2 < β/2 < 1 − n/2 + α/2 implies that 1 − n/2 − β/2 < 0 and 1 − n/2 − β/2 > −α/2 ≥ −(n − 1)/2 > −n. It follows that

\begin{aligned}|f̂(r)| & ≤ C r^{1−n/2−β/2}∥f∥_{1} ∫_{|ξ|≤1} |ξ|^{−n/2+β/2−1} dξ \\ & + C r^{1−n/2−β/2} ∫_{|ξ|≥1} |ξ|^{−n/2+β/2−1}|f̂(ξ)| dξ \end{aligned}

and hence

σ(f)(r) ≤ C r^{2−n−β}I_{α}(f) + C r^{2−n−β}(∫_{|ξ|≥1} |ξ|^{β/2−α/2−1}|ξ|^{α/2−n/2}|f̂(ξ)| dξ)^{2}.

The last integral is majorized by

(∫_{|ξ|≥1} |ξ|^{β−α−2} dξ)^{1/2}(∫ |ξ|^{α−n}|f̂(ξ)|^{2} dξ)^{1/2} = C I_{α}(f)^{1/2},

since β − α − 2 < −n.

We conclude that

σ(f)(r) ≤ C r^{2−n−β}I_{α}(f)

for all β satisfying (9), and choosing β close to 2 − n + α we see that β_{1}(α) ≥ α. On the other hand a well-known counter-example shows that β_{1}(α) ≤ α (see [3, p. 324]). One therefore concludes that β_{1}(α) = α for 0 < α ≤ n − 1.

We then assume that n − 1 < α < n. Choosing β = 1 we see that the inequality (10) still holds and we can argue as above. Observing that β − α − 2 = −α − 1 < −n implies that

∫_{|ξ|≥1} |ξ|^{β−α−2} dξ < ∞,

we conclude that

σ(f)(r) ≤ C r^{1−n}I_{α}(f).

It follows that β_{1}(α) ≥ n − 1.

To prove that β_{1}(α) ≤ n − 1 we shall use a counter-example. Assume that (7) holds for all f satisfying the conditions after the statement of (7). Then we also have

(11)  σ(f)(r) ≤ C r^{−β}I_{α}(|f|),  r > 1,

if f is radial, measurable, complex-valued and f(r) = 0 for r > 1.

We choose

f(s) = f_{R}(s) = e^{−iRs}φ(s), s > 0,

where R is large and φ ∈ C_{0}^{∞}(R), supp φ ⊂ (\frac{1}{2}, 1), φ ≥ 0 and φ (\frac{3}{4}) = 1. It is then clear that

(12)  I_{α}(|f_{R}|) ≤ C.

According to [5, p. 158], we have

J_{n/2−1}(t) = c_{1} \frac{e^{it}}{t^{1/2}} + c_{2} \frac{e^{−it}}{t^{1/2}} + 𝒪(t^{−3/2}), t → ∞,

232 Per Sjölin and hence

f̂(R) = cR^{1−n/2} ∫_{0}^{1}[c_{1} \frac{e^{iRs}}{(Rs)^{1/2}} + c_{2} \frac{e^{−iRs}}{(Rs)^{1/2}} + 𝒪((Rs)^{−3/2})]s^{n/2}f(s) ds
= cc_{1}R^{1/2−n/2} ∫_{0}^{1} e^{iRs}s^{n/2−1/2}f(s) ds
+ cc_{2}R^{1/2−n/2} ∫_{0}^{1} e^{−iRs}s^{n/2−1/2}f(s) ds + 𝒪(R^{−1/2−n/2})
= cc_{1}R^{1/2−n/2} ∫_{0}^{1} s^{n/2−1/2}φ(s) ds
+ cc_{2}R^{1/2−n/2} ∫_{0}^{1} e^{−i2Rs}s^{n/2−1/2}φ(s) ds + 𝒪(R^{−1/2−n/2}).

We conclude that

|f̂(R)| ≥ c_{0}R^{1/2−n/2}

where c_{0} > 0 and R is large. According to (11) we have

σ(f_{R})(R) ≤ C R^{−β}I_{α}(|f_{R}|)

and invoking the above estimates we obtain

R^{1−n} ≤ C R^{−β}. It follows that β ≤ n − 1 and hence β_{1}(α) ≤ n − 1. We have proved that

β_{1}(α) = n − 1 for n − 1 < α < n and hence the proof of Theorem 1 is complete.

Proof of Theorem 2. We let f ∈ C_{0}^{∞}(R^{n}), f ≥ 0, supp f ⊂ B and now set dθ = ψdσ_{0} . According to E.M. Stein [4, p. 348], we then have

(13)  |θ̂(ξ)| ≤ C|ξ|^{(1−n)/2}.

To prove (2) with β(α) replaced by β_{2}(α) we observe that

σ(f)(r) = ∫_{S} f̂(rξ) \overline{f̂(rξ)} ψ(ξ) dσ_{0}(ξ)
= ∫_{S}(∫_{R^{n}} e^{−irξ·x}f(x) dx) \overline{(∫_{R^{n}} e^{−irξ·y}f(y) dy)} ψ(ξ) dσ_{0}(ξ)
= ∫_{R^{n}} ∫_{R^{n}}(∫_{S} e^{−irξ·(x−y)} ψ(ξ) dσ_{0}(ξ))f(x)f(y) dx dy
= ∫∫ θ̂(r(x − y))f(x)f(y) dx dy.

Assuming 0 < α ≤ \frac{1}{2}(n − 1) and invoking (13) we then obtain
σ(f)(r) ≤ C ∫∫ (r|x − y|)^{−α}f(x)f(y) dx dy ≤ Cr^{−α}I_{α}(f)

and (2) follows.

The inequality (3) for β_{2}(α) then follows trivially and it remains to prove (4) for β_{2}(α). We choose f as above and assume (n + 1)/2 < α < n and r > 1. We also choose φ_{0} ∈ C_{0}^{∞}(R^{n}) so that φ_{0}(x) = 1 for |x| ≤ 2. One then has

σ(f)(r) = ∫∫ θ̂(r(x − y))f(x)f(y) dx dy
= ∫∫ θ̂(r(x − y))φ_{0}(x − y)f(x)f(y) dx dy
= ∫∫ K_{r}(x − y)f(x)f(y) dx dy = ∫ K_{r} ∗ f(x)f(x) dx,

where we have set K_{r}(x) = θ̂(rx)φ_{0}(x). It follows that

σ(f)(r) = c ∫ K̂_{r}(ξ)f̂(ξ) \overline{f̂(ξ)} dξ

and we claim that

(14)  |K̂_{r}(ξ)| ≤ C \frac{r^{1−α}}{|ξ|^{n−α}} .

If (14) holds then

σ(f)(r) ≤ Cr^{1−α} ∫ |ξ|^{α−n}|f̂(ξ)|^{2} dξ = Cr^{−(α−1)}I_{α}(f)

and we conclude that β_{2}(α) ≥ α − 1. To complete the proof of the theorem it therefore remains to prove (14). We have

K̂_{r}(ξ) = ∫_{R^{n}} e^{−iξ·x} θ̂(rx)φ_{0}(x) dx = ∫_{R^{n}} e^{−iξ·x}(∫_{S} e^{−rx·y} dθ(y))φ_{0}(x) dx
= ∫_{S}(∫_{R^{n}} e^{−ix·(ξ+ry)}φ_{0}(x) dx) dθ(y) = ∫_{S} φ̂_{0}(ξ + ry)ψ(y) dσ_{0}(y)
= ∫_{Q_{n−1}} φ̂_{0}(ξ′ + ry′, ξ_{n} + rφ(y′))F (y′) dy′,

where F denotes a bounded function on Q_{n−1} .

Letting N denote a large integer and performing a change of variable we then obtain

(15)  \begin{aligned}|K̂_{r}(ξ)| & ≤ C ∫_{Q_{n−1}} \frac{1}{1 + |ξ′ + ry′|^{N}} dy′ \\ & = C ∫_{rQ_{n−1}} \frac{1}{1 + |ξ′ + x′|^{N}} dx′r^{1−n} ≤ Cr^{1−n}. \end{aligned}

To prove (14) we then consider two cases. If |ξ| ≤ Ar, then

r^{1−n} ≤ C \frac{r^{1−α}}{|ξ|^{n−α}}

and (14) follows from (15). On the other hand, if |ξ| > Ar (and A is large enough) then |ξ + ry| ≥ c|ξ| for y ∈ S ∩ supp ψ, and since

K̂_{r}(ξ) = ∫_{S} φ̂_{0}(ξ + ry) dθ(y)

we directly obtain

|K̂_{r}(ξ)| ≤ C|ξ|^{−N}.

Here we have used the fact that φ̂_{0} ∈ 𝒮 . Therefore (14) follows also in this case. The proof of Theorem 2 is complete.

Proof of Theorem 3. We may assume that

σ(μ)(r) = ∫ |ξ_{2}| ≤ 1 ; ⋮ ; |ξ_{n}| ≤ 1 |μ̂(r, rξ_{2}, …, rξ_{n})|^{2}dξ′

where ξ′ = (ξ_{2}, …, ξ_{n}).

It is obvious that β_{3}(α) ≥ 0, 0 < α ≤ 1, and the fact that β_{3}(α) ≥ α − 1, 1 < α < n, follows from the argument in the second part of the proof of Theorem 2. In fact, in this part of the proof we never used the assumption about non-vanishing Gaussian curvature. It therefore remains to prove that

(16)  β_{3}(α) ≤ 0,  0 < α ≤ 1,

and

(17)  β_{3}(α) ≤ α − 1,  1 < α < n.

To prove this we shall use a counter-example. We first observe that if (1) holds for all μ ∈ ℳ_{0} then also

(18)  σ(μ)(r) ≤ Cr^{−β}I_{α}(|μ|),  r > 1,

for all complex Borel measures μ with diam(supp μ) < 1.

We choose φ ∈ C_{0}^{∞}(R) such that supp φ ⊂ (−\frac{1}{4}, \frac{1}{4}), φ ≥ 0 and φ̂(0) > 0. Also choose ψ ∈ C_{0}^{∞}(R^{n−1}) such that supp ψ ⊂ B(0; \frac{1}{2}), ψ ≥ 0 and ψ̂(0) > 0. We set

f(x) = f(x_{1}, x′) = e^{iRx_{1}}φ(x_{1})R^{n−1}ψ(Rx′),

where x′ = (x_{2}, …, x_{n}) and R is large. Also set g = |f| so that g(x_{1}, x′) = φ(x_{1})R^{n−1}ψ(Rx′).

We assume that (1) holds and by use of (18) it then follows that

(19)  σ(f)(r) ≤ Cr^{−β}I_{α}(g).

We have

f̂(ξ_{1}, ξ′) = φ̂(ξ_{1} − R)ψ̂(ξ′/R)

and hence

(20)  σ(f)(R) ≥ c > 0.

One also has

ĝ(ξ_{1}, ξ′) = φ̂(ξ_{1}) ψ̂(ξ′/R)

and we shall estimate

I_{α}(g) = c ∫_{R^{n}} |ξ|^{α−n}|ĝ(ξ)|^{2} dξ.

We set D_{1} = {ξ = (ξ_{1}, ξ′) ∈ R^{n}; |ξ_{1}| ≤ R^{ε}, |ξ′| ≤ R^{1+ε}} and D_{2} = R^{n} \ D_{1} , where ε denotes a small positive number, and write

\begin{aligned}I_{α}(g) & = c ∫∫_{D_{1}} |ξ|^{α−n}|φ̂(ξ_{1})|^{2}|ψ̂(ξ′/R)|^{2} dξ_{1} dξ′ \\ & + c ∫∫_{D_{2}} |ξ|^{α−n}|φ̂(ξ_{1})|^{2}|ψ̂(ξ′/R)|^{2} dξ_{1} dξ′ = A_{1} + A_{2}. \end{aligned}

Since φ̂ and ψ̂ ∈ 𝒮 it is easy to prove that

∫_{R^{ε}}^{∞} |φ̂(t)|^{2} dt ≤ CR^{−N}  and  ∫_{|ξ′|>R^{1+ε}} |ψ̂(ξ′/R)|^{2} dξ′ ≤ CR^{−N},

where N denotes a large positive integer, and hence A_{2} ≤ CR^{−N}.

We also have

A_{1} ≤ C ∫_{{|ξ_{1}|≤R^{ε} \atop |ξ′|≤R^{ε}}} |ξ|^{α−n} dξ + CR^{ε} ∫_{R^{ε}}^{R^{1+ε}} t^{α−2} dt = A_{11} + A_{12}.

It is clear that A_{11} ≤ CR^{εα} and to estimate A_{12} we first consider the case 0 < α < 1. Then A_{12} ≤ CR^{ε} and hence

(21)  I_{α}(g) ≤ CR^{ε}.

Combining (19), (20) and (21) we then obtain

c ≤ CR^{−β+ε}

and it follows that β ≤ ε. We conclude that β_{3}(α) ≤ 0 for 0 < α < 1.

We then have to consider the case 1 ≤ α < n. For α = 1 one obtains A_{12} ≤ CR^{ε} log R. Hence I_{α}(g) ≤ CR^{ε} for every ε > 0 and we can argue as above to obtain β_{3}(1) ≤ 0.

In the case 1 < α < n we get A_{12} ≤ CR^{ε+α−1} and (19) yields

c ≤ CR^{−β+ε+α−1}.

In this case we conclude that β ≤ α − 1 and hence β_{3}(α) ≤ α − 1. Hence we have proved (16) and (17) and the proof of the theorem is complete.

References

[1] Bourgain, J.: Hausdorff dimension and distance sets. - Israel J. Math. 87, 1994, 193–201.

[2] Mattila, P.: Spherical averages of Fourier transforms of measures with finite energy; dimension of intersections and distance sets. - Mathematika 34, 1987, 207–228.

[3] Sjölin, P.: Estimates of spherical averages of Fourier transforms and dimensions of sets. - Mathematika 40, 1993, 322–330.

[4] Stein, E.M.: Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals. - Princeton Univ. Press, 1993.

[5] Stein, E.M., and G. Weiss: Introduction to Fourier Analysis on Euclidean Spaces. - Princeton Univ. Press, 1971.

Received 23 November 1995