Math 191 lecture notes

1. L^{1} Fourier transform If f ∈ L^{1}(ℝ^{n}) then the Fourier transform is f̂: ℝ^{n} → ℂ defined by

f̂(ξ) = ∫ e^{−2πix·ξ}f(x)dx

More generally let M(ℝ^{n}) be the finite complex measures on ℝ^{n} with the norm

∥μ∥ = |μ|(ℝ^{n})

where ∥μ∥ is the total variation. Thus L^{1}(ℝ^{n}) is contained in M(ℝ^{n}) via the identification f → μ, dμ = fdx. We can generalize the definition of Fourier transform via

μ̂(ξ) = ∫ e^{−2πix·ξ}dμ(x)

\underline{Example 1} Let a ∈ ℝ^{n} and let δ_{a} be the Dirac measure at a, δ_{a}(E) = 1 if a ∈ E and δ_{a}(E) = 0 if a ∉ E. Then δ̂_{a}(ξ) = e^{−2πia·ξ}.

\underline{Example 2} Let Γ(x) = e^{−π|x|^{2}}. Then

Γ̂(ξ) = e^{−π|ξ|^{2}}  (1)

\underline{Proof} The integral in question is

Γ̂(ξ) = ∫ e^{−2πix·ξ}e^{−π|x|^{2}}dx

Notice that this factors as a product of one variable integrals. So it suffices to prove (1) when n = 1. For this we use the formula for the integral of a Gaussian: ∫_{−∞}^{∞} e^{−πx^{2}}dx = 1. It follows that

∫_{−∞}^{∞} e^{−2πixξ}e^{−πx^{2}}dx = ∫_{−∞}^{∞} e^{−π(x−iξ)^{2}}dx · e^{−πξ^{2}}
= ∫_{−∞−iξ}^{∞−iξ} e^{−πx^{2}}dx · e^{−πξ^{2}}
= ∫_{−∞}^{∞} e^{−πx^{2}}dx · e^{−πξ^{2}}
= e^{−πξ^{2}}

where we used contour integration at the next to last line. □

There are some basic estimates for the L^{1} Fourier transform, Propositions 1 and 2 below. Consideration of example 1 above shows that in complete generality not that much more can be said.

\underline{Proposition 1.1} If μ ∈ M(ℝ^{n}) then μ̂ is a bounded function, indeed

∥μ̂∥_{∞} ≤ ∥μ∥_{M(ℝ^{n})}  (2)

\underline{Proof} For any ξ

|μ̂(ξ)| = | ∫ e^{−2πix·ξ}dμ(x)|
≤ ∫ |e^{−2πix·ξ}|d|μ|(x)
= ∥μ∥
□

\underline{Proposition 1.2} If μ ∈ M(ℝ^{n}) then μ̂ is a continuous function.

\underline{Proof}Fix ξ and consider

μ̂(ξ + h) = ∫ e^{−2πix·(ξ+h)}dμ(x)

As h → 0 the integrands converge poitwise to e^{−2πix·ξ}. The convergence is dominated since all the integrands have absolute value 1 and |μ|(ℝ^{n}) < ∞. The result therefore follows from the DCT. □

We now list some basic formulas for the Fourier transform; the ones listed here are roughly speaking the ones that do not involve any differentiations. They can all be proved by using the formula e^{a+b} = e^{a}e^{b} and appropriate changes of variables. Let f ∈ L^{1}, τ ∈ ℝ^{n}, and let T be an invertible linear map from ℝ^{n} to ℝ^{n}. Then

1. Let f_{τ}(x) = f(x − τ). Then

f̂_{τ}(ξ) = e^{−2πiτ·ξ}f̂(ξ)  (3)

2. Let e_{τ}(x) = e^{2πix·τ}. Then

ê_{τ}f(ξ) = f(ξ − τ)  (4)

3. Let T^{−t} be the inverse transpose of T. Then

{ \atop f ∘ T} = |det(T)|^{−1}f̂∘ T^{−t}  (5)

4. Define f̃(x) = \overline{f(−x)}. Then

ˆf̃ = \overline{f̂}  (6)

We note some special cases of 3. If T is an orthogonal transformation (i.e. T T^{t} is the identity map) then { \atop f ∘ T} = f̂∘ T, since det(T) = ±1. In particular, this implies that if f is radial then so is f̂, since orthogonal trasnformations act transitively on spheres. If T is a dilation, i.e. T x = r · x for some r > 0, then 3. says that the Fourier transform of the function f(rx) is the function r^{−n}f̂(r^{−1}ξ). Replacing r with r^{−1} and multiplying through by r^{−n} we see that the reverse formula also holds: the Fourier transform of the function r^{−n}f(r^{−1}x) is the function f̂(rξ).

There is a general principle that if f is localized in space, then f̂ should be smooth, and conversely if f is smooth then f̂ should be localized. We now discuss some simple manifestations of this. Let D(x, r) = {y ∈ ℝ^{n} : |y − x| < r}

\underline{Proposition 1.3} Suppose that μ ∈ M(ℝ^{n}) and supp(μ) is compact. Then μ̂ is C^{∞} and

D^{α}μ̂ = { \atop (−2πix)^{α}μ}  (7)

Further, if supp(μ) ⊂ D(0, R) then

∥D^{α}μ̂∥_{∞} ≤ (2πR)^{|α|}∥μ∥  (8)

We are using multiindex notation here and will do so below as well. Namely, a multiindex is a vector α ∈ ℝ^{n} whose components are nonnegative integers. If α is a multiindex then by definition

D^{α} = \frac{∂^{α_{1}}}{∂x_{1}^{α_{1}}} … \frac{∂^{α_{n}}}{∂x_{n}^{α_{n}}}

x^{α} = Π_{j} = 1^{n}x_{j}^{α_{j}}

The length of α, denoted |α|, is ∑_{j} α_{j}. One defines a partial order on multiindices via

α ≤ β ⇔ α_{i} ≤ β_{i} for each i
α < β ⇔ α ≤ β and α ≠ β

\underline{Proof of Proposition 1.3} Notice that (8) follows from (7) and Proposition 1 since the norm of the measure (2πix)^{α}μ is ≤ (2πR)^{|α|}∥μ∥.

Furthermore, for any α the measure (2πix)^{α}μ is again a finite measure with compact support. Accordingly, if we can prove that μ̂ is C^{1} and that (7) holds when |α| = 1, then the lemma will follow by a straightforward induction.

Fix then a value j ∈ {1, …, n}, let e_{j} be the jth standard basis vector. Also fix ξ ∈ ℝ^{n} and consider the difference quotient

Δ(h) = \frac{μ̂(ξ + he_{j}) − μ̂(ξ)}{h}  (9)

This is equal to

∫ \frac{e^{−2πihx_{j}} − 1}{h}e^{−2πiξ·x}dμ(x)  (10)

As h → 0 the quantity

\frac{e^{−2πihx_{j}} − 1}{h}

converges pointwise to −2πix_{j}. Further |\frac{e^{−2πihx_{j}} −1}{h}| ≤ 2π|x_{j}| for each h. Accordingly the integrands in (10) are dominated by |2πx_{j}|, which is a bounded function on the support of μ. It follows by the DCT that

lim_{h→0} Δ(h) = ∫ lim_{h→0} \frac{e^{−2πihx_{j}} − 1}{h}e^{−2πiξ·x}dμ(x)

which is equal to

∫ −2πix_{j}e^{−2πiξ·x}dμ(x)

This proves the formula (7) when |α| = 1. Formula (7) and Proposition 2 imply that μ̂ is C^{1}. □

\underline{Remark} The estimate (8) is tied to the support of μ. However, the fact that μ̂ is C^{∞} and the formula (7) are valid whenever μ has enough decay to justify the differentiaions under the itegral sign. For example, they are valid if μ has moments of all orders, i.e. ∫ |X|^{N}d|μ|(x) < ∞ for all N.

Estimate (2) can be seen as justification of the idea that if μ is localized then μ̂ should be smooth. We now consider the reverse statement, μ smooth implies μ̂ localized.

\underline{Proposition 1.4} Suppose that f is C^{N} and furthermore that D^{α}f ∈ L^{1} for all α with 0 ≤ |α| ≤ N. Then

D̂^{α}f(ξ) = (2πiξ)^{α}f̂(ξ)  (11)

when |α| ≤ N and furthermore

|f̂(ξ)| ≤ C(1 + |ξ|)^{−N}  (12)

for a suitable constant C.

The proof is based on an integration by parts which is most easily justified when f has compact support. Accordingly we include the following lemma before giving the proof.

Let ϕ : ℝ^{n} → ℝ be a C^{∞} function with the following properties (Property 4. is actually irrelevant for present purposes):

1. ϕ(x) = 1 if |x| ≤ 1

2. ϕ(x) = 0 if |x| ≥ 2

3. 0 ≤ ϕ ≤ 1.

4. ϕ is radial.

Define ϕ_{k}(x) = ϕ(\frac{x}{k}); thus ϕ_{k} is similar to ϕ but lives on scale k instead of 1. If α is a multiindex, then there is a constant C_{α} such that |D^{α}ϕ_{k}| ≤ \frac{C_{α}}{k^{|α|}} uniformly in k and furthermore, if α ≠ 0 then the support of D^{α}ϕ is contained in the region k ≤ |x| ≤ 2k.

\underline{Lemma 1.5} If f is C^{N}, D^{α}f ∈ L^{1} for all α with |α| ≤ N and if we let f_{k} = ϕ_{k}f then ∥ lim_{k→∞} D^{α}f_{k} − D^{α}f∥_{1} = 0 for all α with |α| ≤ N.

\underline{Proof} It is obvious that

lim_{k→∞} ∥ϕ_{k}D^{α}f − D^{α}f∥_{1} = 0

so it suffices to show that

lim_{k→∞} ∥D^{α}(ϕ_{k}f) − ϕ_{k}D^{α}f∥_{1} = 0  (13)

However, by the Leibniz rule,

D^{α}(ϕ_{k}f) − ϕ_{k}D^{α}f = ∑_{β<α} c_{β}D^{α−β}fD^{β}ϕ_{k}

where the c_{β}’s are certain constants. Thus

∥D^{α}(ϕ_{k}f) − ϕ_{k}D^{α}f∥_{1} ≤ C ∑_{β<α} ∥D^{β}ϕ_{k}∥_{∞}∥D^{α−β}∥_{L^{1}({x:|x|≥k})}
≤ Ck^{−1} ∑_{β<α} ∥D^{α−β}∥_{L^{1}({x:|x|≥k})}

The last line clearly goes to zero as k → ∞. There are two reasons for this (either would suffice): the factor k^{−1}, and the fact that the L^{1} norms are taken only over the region |x| ≥ k. □

\underline{Proof of Proposition 4} If f is C^{1} with compact support then by integration by parts we have

∫ \frac{∂f}{∂x_{j}} (x)e^{−2πix·ξ}dx = 2πiξ_{j} ∫ e^{−2πix·ξ}f(x)dx

i.e. (11) holds when |α| = 1. An easy induction then proves (11) for all α provided f is C^{N} with compact support.

To remove the compact support assumption let f_{k} be as in Lemma 5. Then (11) holds for f_{k}. Now we pass to the limit as k → ∞. On the one hand { \atop D^{α}f_{k}} converges uniformly to D̂^{α}f as k → ∞ by Lemma 5 and Proposition 1. On the other hand f̂_{k} converges uniformly to f̂, so (2πiξ)^{α}f̂_{k} converges to (2πiξ)^{α}f̂ pointwise. This proves (11) in general.

To prove (12), observe that (11) and Proposition 1 imply that ξ^{α}f̂ ∈ L^{∞} if |α| ≤ N. On the other hand it is easy to estimate

C_{N}^{−1}(1 + |ξ|)^{N} ≤ ∑_{|α|≤N} |ξ^{α}| ≤ C_{N}(1 + |ξ|)^{N}  (14)

so (12) follows. □

Together with the inequality (14) let us note the inequality

1 + |x| ≤ (1 + |y|)(1 + |x − y|), x, y ∈ ℝ^{n}  (15)

which will be used several times below.

2. Schwartz space

The Schwartz space S is the functions f : ℝ^{n} → ℂ such that

1. f is C^{∞}.

2. x^{α}D^{β}f is a bounded function for each pair of multiindices α and β.

If f ∈ S then we define

∥f∥_{αβ} = ∥x^{α}D^{β}f∥_{∞}

It is possible to see that S with the family of norms ∥ · ∥_{αβ} is a Frechet space, but we don’t discuss such questions here. (see [19]) However, we define a notion of sequential convergence in S:

A sequence {f_{k}} ⊂ S converges in S to f ∈ S if lim_{k→∞} ∥f_{k} − f∥_{αβ} = 0 for each pair of multiindices α and β.

\underline{Examples} 1. Let C_{0}^{∞} be the C^{∞} functions with compact support. Then C_{0}^{∞} ⊂ S. Namely, to prove that x^{α}D^{β}f is bounded, just note that if f ∈ C_{0}^{∞} then D^{β}f is a

continuous function with compact support, hence bounded, and that x^{α} is a bounded function on the support of D^{β}f.

2. Let f(x) = e^{−π|x|^{2}}. Then f ∈ S.

For the proof, notice that if p(x) is a polynomial, then any first partial derivative \frac{∂}{∂x_{j}} (p(x)e^{−π|x|^{2}}) is again of the form q(x)e^{−π|x|^{2}} for some polynomial q. It follows by induction that for each D^{β}f is a polynomial times f for each β. Hence x^{α}D^{β}f is a polynomial times f for each α and β. This implies using L’Hospital’s rule that x^{α}D^{β}f is bounded for each α and β.

3. The following functions are not in S: f_{N}(x) = (1 + |x|^{2})^{−N} for any given N, and g(x) = e^{−π|x|^{2}} sin(e^{π|x|^{2}}). Roughly, although f_{N} decays rapidly at ∞ it does not decay rapidly enough, whereas g decays rapidly enough but its derivatives do not decay. Detailed verification left to reader.

We now discuss some simple properties of S, then some which are slightly less simple.

I. S is closed under differentiations and under multiplication by polynomials. Further, these operations are continuous on S in the sense that they preserve sequential convergence. Also f, g ∈ S implies fg ∈ S.

\underline{Proof} Let f ∈ S. If γ is a multiindex, then x^{α}D^{β}(D^{γ})f = x^{α}D^{β+γ}f, which is bounded since f ∈ S. So D^{γ}f ∈ S.

Further, by the Leibniz rule, x^{α}D^{β}(x^{γ}f) is a finite sum of terms each of which is a constant multiple of

x^{α}D^{δ}(x^{γ})D^{β−δ}f

for some δ ≤ β. Furthermore D^{δ}(x^{γ}) is a constant multiple of x^{γ−δ} if δ ≤ γ and otherwise is zero. Thus x^{α}D^{β}(x^{γ}f) is a linear combination of monomials times derivatives of f, and is therefore bounded. So x^{γ}f ∈ S.

The continuity statements follow from the proofs of the closure statements; we will normally omit these arguments. As an indication of how they are done, let us show that if γ is a multiindex then f → D^{γ}f is continuous. Suppose that f_{n} → f in S. Fix a pair of multiindices α and β.Applying the definition of convergence with the multiindices α and β + γ, we have

lim_{n→∞} ∥x^{α}D^{β+γ}(f_{n} − f)∥_{∞} = 0

Equivalently,

lim_{n→∞} ∥x^{α}D^{β}(D^{γ}f_{n} − D^{γ}f)∥_{∞} = 0

whcih says that D^{γ}f_{n} converges to D^{γ}f.

The last statement (that S is an algebra) follows readily from the product rule and the definitions. □

II. The following alternate definitions of S are often useful.

f ∈ S ⇔ (1 + |x|)^{N}D^{β}f is bounded for each N and α  (16)
f ∈ S ⇔ lim_{x→∞} x^{α}D^{β}f = 0 for each α and β  (17)

Indeed, (16) follows from the definition and (14). The backward implication in (17) is trivial, while the forward implication follows by applying the definition with α replaced by appropriate larger multiindices, e.g. α + e_{j} for arbitrary j ∈ {1, …, n}.

\underline{Proposition 2.1} C_{0}^{∞} is dense in S, i.e. for any f ∈ S there is a sequence \{f_{k}\} ⊂ C_{0}^{∞} with f_{k} → f in S.

\underline{Proof} This is almost the same as the proof of Lemma 1.5. Namely, define ϕ_{k} as there and consider f_{k} = ϕ_{k}f, which is evidently in C_{0}^{∞}. We must show that

x^{α}D^{β}(ϕ_{k}f) → x^{α}D^{β}f

uniformly as k → ∞. For this, we estimate

∥x^{α}D^{β}(ϕ_{k}f) − x^{α}D^{β}f∥_{∞} ≤ ∥ϕ_{k}x^{α}D^{β}f − x^{α}D^{β}f∥_{∞} + ∥x^{α}(D^{β}(ϕ_{k}f) − ϕ_{k}D^{β}f)∥_{∞}

The first term is bounded by sup_{|x|≥k} |x^{α}D^{β}f| and therefore goes to zero as k → ∞ by (17). The second term is estimated using the Leibniz rule by

C ∑_{γ<β} ∥x^{α}D^{γ}f∥_{∞}∥D^{β−γ}ϕ_{k}∥_{∞}  (18)

Since f ∈ S and ∥D^{β−γ}ϕ_{k}∥ ≤ \frac{C}{k} the expression (18) goes to zero as k → ∞. □

There is a stronger density statement which is sometimes needed. Define a C_{0}^{∞} tensor function to be a function f : ℝ^{n} → ℂ which is such that

f(x) = Π_{j}ϕ_{j}(x_{j})

where each ϕ_{j} ∈ C_{0}^{∞}(ℝ).

\underline{Proposition 2.1′} Linear combinations of C_{0}^{∞} tensor functions are dense in S.

\underline{Proof} In view of Proposition 2.1 it suffices to show that if f ∈ C_{0}^{∞} then there is a sequence {g_{k}} such that

1. Each g_{k} is a C_{0}^{∞} tensor function.

2. The supports of the g_{k} are contained in a fixed compact set E which is independent of k.

3. D^{α}g_{k} converges uniformly to D^{α}f for each α.

To construct {g_{k}}, we use the fact (a basic fact about Fourier series) that if f is a C^{∞} function in ℝ^{n} which is 2π-periodic in each variable then f can be expanded in a series

f(θ) = ∑_{ν∈ℤ^{n}} a_{ν}e^{iν·θ}

where the {a_{ν}} satisfy

∑_{ν}(1 + |ν|)^{N}|a_{ν}| < ∞

for each N. Considering partial sums of the Fourier series we therefore obtain a sequence of trigonometric polynomials p_{k} such that D^{α}p_{k} converges uniformly to D^{α}f for each α.

In constructing {g_{k}} we can assume that x ∈ suppf implies |x_{j}| ≤ 1, say, for each j - otherwise we work with f(Rx) for suitable fixed R instead and undo the rescaling at the end. Let ϕ be a C_{0}^{∞} function of one variable which is equal to 1 on [−1, 1] and vanishes outside [−2, 2]. Let f̃ be the function which is equal to f on [−π, π] × … × [−π, π] and is 2π-periodic in each variable. Then we have a sequence of trigonometric polynomials p_{k} such that D^{α}p_{k} converges uniformly to D^{α}f̃ for each α. Let g_{k}(x) = Π_{j=1}^{n}ϕ(x_{j}) · p_{k}(x). Then g_{k} clearly satisfies 1. and 2, and an argument with the product rule as in Lemma 1.5 and Proposition 2.2 will show that {g_{k}} satisfies 3. The proof is complete. □

The next proposition is an alternate definition of S using L^{1} instead of L^{∞} norms.

\underline{Proposition 2.2} A C^{∞} function f is in S iff the norms

∥x^{α}D^{β}f∥_{1}

are finite for each α and β. Furthermore a sequence {f_{k}} ⊂ S converges in S to f ∈ S iff

lim_{k→∞} ∥x^{α}D^{β}(f_{k} − f)∥_{1} = 0

for each α and β.

\underline{Proof} We only prove the first part; the equivalence of the two notions of convergence follows from the proof and is left to the reader.

First suppose that f ∈ S. Fix α and β. Let N = |α| + n + 1. Then we know that (1 + |x|)^{N}D^{β}f is bounded. Accordingly

∥x^{α}D^{β}f∥_{1} ≤ ∥(1 + |x|)^{N}D^{β}f∥_{∞}∥x^{α}(1 + |x|)^{−N}∥_{1}
< ∞

using that the function (1 + |x|)^{−n} is integrable.

For the converse, we first make a definition and state a lemma. If f : ℝ^{n} → ∞ is C^{k} and if x ∈ ℝ^{n} then

Δ_{k}^{f}(x) =^{def} ∑_{|α|=k} |D^{α}f(x)|

Let ω be the volume of the unit ball. We also now start to use the notation x ≲ y to mean that x ≤ Cy where C is a fixed but unspecified constant.

\underline{Lemma 2.2’} Suppose f is a C^{∞} function. Then for any x

|f(x)| ≲ ∑_{0≤j≤n+1} ∥Δ_{j}^{f}∥_{L^{1}(D(x,1)}

This is contained in Lemma A2 which is stated and proved at the end of the section. To finish the proof of Proposition 2.2, we apply the preceding lemma to D^{β}f. This

gives

|D^{β}f(x)| ≲ ∑_{|γ|≤|β|+n+1} ∫_{D(x,1)} |D^{γ}f(y)|dy

therefore

(1 + |x|)^{N}|D^{β}f(x)| ≲ (1 + |x|)^{N} ∑_{|γ|≤|β|+n+1} ∫_{D(x,1)} |D^{γ}f(y)|dy

≲ ∑_{|γ|≤|β|+n+1} ∫_{D(x,1)}(1 + |y|)^{N}|D^{γ}f(y)|dy

where we used the elementary inequality

(1 + |x|) ≤ 2 min_{y∈D(x,1)}(1 + |y|)

It follows that

∥(1 + |x|)^{N}|D^{β}f∥_{∞} ≲ ∑_{|γ|≤|β|+n+1} ∥(1 + |x|)^{N}D^{γ}f∥_{1}

and then Proposition 2.2 follows from (14) □

\underline{Theorem 2.3} If f ∈ S then f̂ ∈ S. Furthermore the map f → f̂ is continuous from S to S.

\underline{Proof} As usual we explicitly prove only the first statement.

If f ∈ S then f ∈ L^{1}, so f̂ is bounded.

Thus if f ∈ S then { \atop D^{α}x^{β}f} is bounded for any given α and β, since D^{α}x^{β}f is again in S. However Propositions 1.3 and 1.4 imply that

{ \atop D^{α}x^{β}f}(ξ) = (2πi)^{|α|}(−2πi)^{−|β|}ξ^{α}D^{β}f̂(ξ)

so ξ^{α}D^{β}f̂ is again bounded, which means f̂ ∈ S. □

Appendix: pointwise Poincare inequalities

This is a little more technical than the preceding and we will omit some details. We prove a frequently used pointwise estimate for a function in terms of integrals of its gradient, which plays a similar role that the mean value inequality plays in calculus. Then we prove a generalization involving higher derivatives which includes Lemma 2.3. Let ω be the volume of the unit ball.

\underline{Lemma A1} Suppose that f is C^{1}. Then

|f(x) − \frac{1}{ω} ∫_{D(x,1)} f(y)dy| ≲ ∫_{D(x,1)} \frac{|∇f(y)|}{|x − y|^{n−1}} dy

\underline{Proof} Applying the fundamental theorem of calculus to the function

t → f(x + t(y − x))

shows that

|f(y) − f(x)| ≤ |x − y| ∫_{0}^{1} |∇f(x + t(y − x))|dt

Integrate this with respect to y over D(x, 1) and divide by ω. Thus

|f(x) − \frac{1}{ω} ∫_{D(x,1)} f(y)dy| ≤ \frac{1}{ω} ∫_{D(x,1)} |f(x) − f(y)|dy
≲ ∫_{D(x,1)} |x − y| ∫_{0}^{1} |∇f(x + t(y − x))|dtdy
= ∫_{0}^{1} ∫_{D(x,1)} |x − y||∇f(x + t(y − x))|dydt  (19)

Make the change of variables z = x + t(y − x), and then reverse the order of integration again. This leads to

(19) = ∫_{t=0}^{1} ∫_{D(x,t)} t^{−1}|z − x||∇f(z)|\frac{dz}{t^{n}} dt
= ∫_{D(x,1)} |z − x||∇f(z)| ∫_{t=|z−x|}^{1} t^{−(n+1)}dtdz
≲ |x − z|^{−(n−1)}|∇f(z)|dz

as claimed.

\underline{Lemma A.2} Suppose that f is C^{k}. Then

|f(x)| ≲ ∑_{0≤j<k} ∥Δ_{j}^{f}∥_{L^{1}(D(x,1))} + \{ \begin{matrix}∫_{D(x,1)} |x − y|^{−(n−k)}Δ_{k}^{f}(y)dy & if 1 ≤ k ≤ n − 1 \\ ∫_{D(x,1)} log \frac{1}{|x−y|}Δ_{n}^{f}(y)dy & if k = n \\ ∥Δ_{n+1}^{f}∥_{L^{1}(D(x,1))} & if k = n + 1 \end{matrix} (20)

The case k = n + 1 is Lemma 2.2’. Further, the case k = 1 follows immediately from Lemma A.1, so it has already been proved. To pass to general k we use an induction based on the following inequality (here a > 0, b > 0, |z − x| ≤ constant)

∫_{y∈D(x,C)} |x − y|^{−(n−a)}|z − y|^{−(n−b)}dy ≤ \{ \begin{matrix}C|x − z|^{−(n−a−b)} & if a + b < n \\ log \frac{C}{|z−x|} & if a + b = n \\ 1 & if a + b > n \end{matrix}  (21)

and furthermore

∫_{y∈D(x,C)} |x − y|^{−(n−1)} log \frac{1}{|z − y|}dy ≤ C  (22)

In fact, (21) may be proved by subdividing the region of integration in the three regimes |y − x| ≤ \frac{1}{2}|z − x|, |y − z| ≤ \frac{1}{2}|z − x| and “the rest” and noting that on the third regime the integrand is comparable to |y − x|^{−(2n−a−b)}, and (22) may be proved similarly.

We now prove (20) by induction on k. We have done the case k = 1. Suppose that 2 ≤ k ≤ n − 1 and that the cases up to and including k − 1 have been done. Then

|f(x)| ≲ ∑_{j≤k−2} ∥Δ_{j}^{f}∥_{L^{1}(D(x,1))} + ∫_{D(x,1)} |x − y|^{−(n−k+1)}Δ_{k−1}^{f}(y)dy

≲ ∑_{j≤k−2} ∥Δ_{k−1}^{f}∥_{L^{1}(D(x,1))} + ∫_{D(x,1)} |x − y|^{−(n−k)} ∫_{D(y,1)} Δ_{k−1}^{f}(z)dzdy

+ ∫_{D(x,1)} |x − y|^{−(n−k+1)} ∫_{D(y,1)} |y − z|^{−1}Δ_{k}^{f}(z)dzdy
≲ ∑_{j≤k−1} ∥Δ_{j}^{f}∥_{L^{1}(D(x,2))} + ∫_{D(x,2)} |x − z|^{−(n−k)}Δ_{k}^{f}(z)dy

For the first two inequalities we used (20) with k replaced by k − 1 and 1 respectively, and for the last inequality we reversed the order of integration and used (21). The disc D(x, 2) can be replaced by D(x, 1) using rescaling, so we have proved (20) for k ≤ n − 1.

To pass from k = n − 1 to k = n we argue similarly using the second case of (21), and to pass from k = n to k = n + 1 we argue similarly using (22). □

3. Fourier inversion and Plancherel

Convolution of ϕ and f is defined as follows:

ϕ ∗ f(x) = ∫ ϕ(y)f(x − y)dy  (23)

We assume the reader has seen this definition before but will summarize some facts mostly without giving the proofs. There is an issue of the appropriate conditions on ϕ and f under which the integral (23) makes sense. We recall

1. If ϕ ∈ L^{1} and f ∈ L^{p}, 1 ≤ p ≤ ∞, then the integral (23) is an absolutely convergent Lebesgue integral for a.e. x and

∥ϕ ∗ f∥_{p} ≤ ∥ϕ∥_{1}∥f∥_{p}  (24)

2. If ϕ is a continuous function with compact support and f ∈ L_{loc}^{1} then the integral (23) is an absolutely convergent Lebesgue integral for every x and ϕ ∗ f is continuous.

3. If ϕ ∈ L^{p} and f ∈ L^{p′}, \frac{1}{p} + \frac{1}{p′} = 1, then the integral (23) is an absolutely convegent Lebesgue integral for every x and ϕ ∗ f is continuous. Furthermore

ϕ ∗ f∥_{∞} ≤ ∥ϕ∥_{p}∥f∥_{p′}  (25)

Furthermore in any of the above situations convolution is commutative: the integral defining f ∗ ϕ is again convergent for the same values of x and f ∗ ϕ = ϕ ∗ f. This follows by making the change of variables y → x − y. Notice also the following fact:

supp(ϕ ∗ f) ⊂ suppϕ + suppf

where the sum E + F means {x + y : x ∈ E, y ∈ F }.

In many applications the function ϕ is fixed and very “nice” and one considers convolution as an operator

f → ϕ ∗ f

\underline{Lemma 3.1} If ϕ ∈ C_{0}^{∞} and f ∈ L_{loc}^{1} then ϕ ∗ f is C^{∞} and

D^{α}(ϕ ∗ f) = (D^{α}ϕ) ∗ f  (26)

\underline{Proof} It is enough to prove ϕ ∗ f is C^{1} and (26) holds for multiindices of length 1 since one can then use induction. Fix j and consider difference quotients

d(h) = \frac{1}{h}(ϕ ∗ f(x + he_{j}) − ϕ ∗ f(x))

Using commutativity of convolution we can rewrite this as

d(h) = ∫ \frac{1}{h}(ϕ(x + he_{j} − y) − ϕ(x − y))f(y)dy

The quotients

A_{h}(y) = \frac{1}{h}(ϕ(x + he_{j} − y) − ϕ(x − y))

are bounded by ∥\frac{∂ϕ}{∂x_{j}} ∥_{∞} by the mean value theorem. For fixed x and for |h| ≤ 1, the

support of A_{h} is contained in the fixed compact set E = \overline{D(x, 1)}+suppϕ. Thus the integrands A_{h}f are dominated by the L^{1} function ∥\frac{∂ϕ}{∂x_{j}} ∥_{∞}χ_{E}|f|. The DCT implies

lim_{h→0} d(h) = ∫ f(y) lim_{h→0} A_{h}(y)dy = ∫ f(y)\frac{∂ϕ}{∂x_{j}} (x − y)dy

This proves the formula (26) (when |α| = 1). The continuity of the partials then follows from 2. above. □

\underline{Corollary 3.1’} If f, g ∈ S then f ∗ g ∈ S.

\underline{Proof} By Lemma 3.1 it suffices to show that if (1 + |x|)^{N}f(x) and (1 + |x|)^{N}g(x) are bounded for every N then so is (1 + |x|)^{N}f ∗ g(x). This follows by writing out the definitions and using (15); details left to reader. □

Convolution interacts with the Fourier transform as follows: Fourier transform converts convolution to ordinary pointwise multiplication. Thus we have the following formulas:

{ \atop f ∗ g} = f̂ĝ, f, g ∈ L^{1}  (27)
f̂g = f̂∗ ĝ, f, g ∈ S  (28)

Formula (27) follows from Fubini’s theorem and is in many textbooks; proof left to reader. Formula (28) then follows easily from the inversion theorem, so we defer the proof until after Theorem 3.4.

Let ϕ ∈ S, and assume that ∫ ϕ = 1. Define ϕϵ(x) = ϵ^{−n}ϕ(ϵ^{−1}x). The family of functions {ϕϵ} is called an approximate identity. Notice that ∫ ϕϵ = 1 for all ϵ. Thus one can regard the ϕϵ as roughly convergent to the Dirac mass δ_{0} as ϵ → 0. Indeed the following fact is basic but quite standard; see any reasonable book on real analysis for the proof.

\underline{Lemma 3.2} Let ϕ ∈ S and ∫ ϕ = 1. Then:

1. If f is a continuous function which goes to zero at ∞ then ϕϵ ∗ f → f uniformly as ϵ → 0.

2. If f ∈ L^{p}, 1 ≤ p < ∞ then ϕϵ ∗ f → f in L^{p} as ϵ → 0.

Let us note the following corollary:

\underline{Lemma 3.3} Suppose f ∈ L_{loc}^{1}. Then there is a fixed sequence \{g_{k}\} ⊂ C_{0}^{∞} such that if p ∈ [1, ∞) and f ∈ L^{p} then g_{k} → f in L^{p}. If f is continuous and goes to zero at ∞ then g_{k} → f uniformly

The reason for stating the lemma in this way is that one sometimes has to deal with several notions of convergence simultaneously, e.g. L^{1} and L^{2} convergence, and it is convenient to be able to approximate f in both norms simulataneously.

\underline{Proof} Let ψ ∈ C_{0}^{∞}, ∫ ψ = 1, ψ ≥ 0, and let ϕ be as in Lemma 1.5. Let g_{k}(x) = ϕ(\frac{x}{k})ψ^{ϵ^{k}} ∗ f.

If f ∈ L^{p} then for large k the quantity ∥ψ^{ϵ^{k}} ∗ f∥_{L^{p}(|x|≥k} is bounded by ∥f∥_{L^{p}(|x|≥k−1)} using (24) and that suppψ^{ϵ^{k}} will be contained in D(0, 1). Accordingly, ∥g_{k} −ψ^{ϵ^{k}} ∗f∥_{p} → 0 as k → ∞. On the other hand, ∥ψ^{ϵ^{k}} ∗ f − f∥_{p} → 0 by Lemma 3.2. If f is continuous and goes to zero at ∞ then one can argue the same way using the first part of Lemma 3.2. Smoothness of g_{k} follows from Lemma 3.1 so the proof is complete. □

\underline{Theorem 3.4} (Fourier inversion) Suppose that f ∈ L^{1}, and assume that f̂ is also in L^{1}. Then for a.e. x

f(x) = ∫ f̂(ξ)e^{2πiξ·x}dξ  (29)

Otherwise stated,

ˆf̂(x) = f(−x) for a.e. x  (30)

The proof uses Lemma 3.2 and also the following facts:

A. The gaussian Γ(x) = e^{−ϕ|x|^{2}} satisfies Γ̂ = Γ, and therefore also satisfies (30). So at any rate there is one function f for which Theorem 3.4 is true. In fact this implies there are many such functions since if we form the functions

Γϵ(x) = e^{−πϵ^{2}|x|^{2}}

then we have

Γ̂ϵ(ξ) = ϵ^{−n}e^{−π \frac{|ξ|^{2}}{ϵ^{2}}}  (31)

and by applying this again with ϵ replaced by ϵ^{−1} one can verify that Γϵ satisfies (30). See the discussion after formula (5) on page 2.

B. The duality relation for the Fourier transform, i.e. the following

‘

\underline{Lemma 3.5} Suppose that μ ∈ M(ℝ^{n}) and ν ∈ M(ℝ^{n}). Then

∫ μ̂dν = ∫ ν̂dμ  (32)

In particular if f, g ∈ L^{1} then

∫ f̂(x)g(x)dx = ∫ f(x)ĝ(x)dx

\underline{Proof} This follows from Fubini’s theorem:

∫ μ̂dν = ∫ ∫ e^{−2πiξ·x}dμ(x)dν(ξ)
= ∫ ∫ e^{−2πiξ·x}dν(ξ)dμ(x)
= ∫ ν̂dμ

□

\underline{Proof of Theorem 3.4} Consider the integral in (29) with a damping factor included:

Iϵ(x) = ∫ f̂(ξ)e^{−πϵ^{2}|ξ|^{2}}e^{2πiξ·x}dξ  (33)

We evaluate the limit as ϵ → 0 in two different ways.

1. As ϵ → 0, Iϵ(x) → f̂(ξ)e^{2πiξ·x}dξ for each fixed x. This follows from the DCT, since f̂ ∈ L^{1}.

2. With x and ϵ fixed, define g(ξ) = e^{−πϵ^{2}|ξ|^{2}}e^{2πiξ·x}. Thus

Iϵ(x) = ∫ f(y)ĝ(y)dy

by Lemma 3.5. On the other hand we can evaluate ĝ using (31) and the fact that g(ξ) = e_{x}(ξ)Γϵ(ξ). Thus

ĝ(y) = Γ̂ϵ(y − x) = Γϵ(x − y)

where Γϵ(y) = ϵ^{−n}Γ(\frac{y}{ϵ}) is an approximate identity and we have used that Γ is even.

Accordingly,

Iϵ = f ∗ Γϵ

and we conclude by Lemma 3.2 that

Iϵ → f

in L^{1} as ϵ → 0.

Summing up, we have seen that the functions Iϵ converge pointwise to ∫ f̂(ξ)e^{2πix·ξ}dξ, and converge in L^{1} to f. This is only possible when (29) holds. □

Corollaries of the inversion theorem

The first corollary below is not really a corollary but a reformulation of the proof without the assumption that f̂ ∈ L^{1}. This is the form the inversion theorem takes for general f. Notice that the integrals Iϵ are well defined for any f ∈ L^{1}, since the Gaussian e^{−πϵ^{2}|x|^{2}} is integrable for each fixed ϵ. Corollary 3.6 is often stated as “the Fourier transform of f is Gauss-Weierstrass summable to f” and can be compared to the theorem on Cesaro summability for Fourier series.

\underline{Corollary 3.6} Suppose f ∈ L^{1} and define Iϵ(x) via (33). Then Iϵ → f in L^{1} as ϵ → 0. If 1 < p < ∞ and additionally f ∈ L^{p}, then Iϵ → f in L^{p} as ϵ → 0. If instead f is

continuous and goes to zero at ∞ then Iϵ → f uniformly.

‘

\underline{Proof} This follows from the preceding argument showing that Iϵ = Γϵ ∗ f, together with Proposition 3.2. □

\underline{Corollary 3.7} If f ∈ L^{1} and f̂ = 0, then f = 0.

This is immediate from Theorem 3.4. □

\underline{Theorem 3.8} The Fourier transform maps S \underline{onto} S.

\underline{Proof} Given f ∈ S, let F (x) = f(−x) and let g = F̂. Then g ∈ S by Theorem 2.3, and (30) implies

ĝ(x) = ˆ F̂(x) = F (−x) = f(x)
□

Let us also prove formula (28). Let f, g ∈ S. Then

{ \atop f̂∗ ĝ}(−x) = ˆ f̂(−x)ˆĝ(−x)
= f(x)g(x)

by (27) and Theorem 3.4. Using Theorem 3.4 again, it follows that f̂∗ ĝ = f̂g. □ \underline{Theorem 3.9} (Plancherel theorem, first version) If u, v ∈ S then

∫ û\overline{v̂} = ∫ u\overline{v}  (34)

‘ \underline{Proof} By the inversion theorem

∫ u(x)\overline{v}(x)dx = ∫ ˆ û(−x)\overline{v}(x)dx = ∫ ˆ û(x)\overline{v(−x)}dx

i.e.

∫ û\overline{v̂} = ∫ ˆ ûṽ

Apply the duality relation to the right member obtaining

∫ u\overline{v} = ∫ ûˆṽ

and now (34) follows from (6). □

Theorem 3.9 says that the Fourier transform restricted to Schwartz functions is an isometry in the L^{2} norm. Since S is dense in L^{2} (e.g. by Corollary 3.3) this suggests a way of extending the Fourier transform to L^{2}.

\underline{Theorem 3.10} (Plancherel theorem, second version) There is a unique bounded operator F : L^{2} → L^{2} such that Ff = f̂ when f ∈ S. F has the following additional properties:

1. F is a unitary operator.

2. Ff = f̂ if f ∈ L^{1} ∩ L^{2}.

\underline{Proof} The existence/uniqueness statement is immediate from Theorem 3.9, as is the fact that ∥Ff∥_{2} = ∥f∥_{2}. In view of this isometry property, the range of F must be closed and unitarity of F will follow if we show that the range is dense. However, the latter statement is immediate from Theorem 3.8 and Lemma 3.3.

It remains to prove 2., and of course 2. is true by definition if f ∈ S. Suppose f ∈ L^{1} ∩ L^{2}. By Lemma 3.3 there is a sequence {g_{k}} ⊂ S which converges to f both in L^{1} and in L^{2}. Then ĝ_{k} converges to f̂ uniformly by Proposition 1.1, whereas ĝ_{k} converges to Ff in L^{2} by boundedness of the operator F. It follows that Ff = f̂. □

Statement 2. allows us to use the notation f̂ for Ff if f ∈ L^{2} without any possible ambiguity. We may therefore extend the definition of the Fourier transform to L^{1} + L^{2} (in fact to σ-finite measures of the form μ + fdx, μ ∈ M(ℝ^{n}), f ∈ L^{2}) via { \atop f + g} = f̂+ ĝ.

\underline{Corollary 3.12} The following form of the duality relation is valid:

∫ ν̂ψ = ∫ ψ̂dν

if dν = dμ + fdx, μ ∈ M(ℝ^{n}), f ∈ L^{2}.

\underline{Proof} We have already proved this in Lemma 4.5 if f = 0, so it suffices to prove it when μ = 0, i.e. to show that

∫ f̂ψ = ∫ fψ̂

if f ∈ L^{2}, ψ ∈ S. This is true by Lemma 4.5 if f ∈ L^{1} ∩ L^{2}, and on the other hand for fixed ψ both sides depend continuously on f; in the case of the left side this follows from the Plancherel theorem. □

\underline{Theorem 3.13} If μ ∈ M(ℝ^{n}), f ∈ L^{2} and

f̂+ μ̂ = 0

then μ = −fdx. In particular, if μ ∈ M(ℝ^{n}) and μ̂ ∈ L^{2} then μ is absolutely continuous to Lebesgue measure with an L^{2} density.

\underline{Proof} By the Riesz representation theorem for measures on compact sets, the measure dμ + fdx will be zero provided

∫ ϕdμ + ϕfdx = 0  (35)

for continuous ϕ with compact support.

If ϕ ∈ C_{0}^{∞} then (35) follows from Corollary 3.12. In general, we choose(e.g. by Proposition 3.2) a sequence ϕ_{k} in C_{0}^{∞} which converges to ϕ uniformly and in L^{2}. We write down (35) for the ϕ_{k}’s and pass to the limit. This proves (35).

To prove the last statement, suppose μ̂ ∈ L^{2} and choose (by Theorem 3.10) a function g ∈ L^{2} with ĝ = μ̂. Then dμ − gdx has Fourier trasnform zero so by the first part of the proof, dμ = gdx. □

All the basic formulas for the L^{1} Fourier transform extend to the L^{1} + L^{2}-Fourier transform by approximation arguments. This was done above in the case of the duality relation. Let us note in particular that the transformation formulas in section 1 extend. For example, in the case of (5), one has

{ \atop f ∘ T} = |det(T)|^{−1}f̂∘ T^{−t}  (36)

if f ∈ L^{1} + L^{2}. Since we already know this when f ∈ L^{1} it suffices to prove it when f ∈ L^{2}. Choose {f_{k}} ⊂ L^{1} ∩ L^{2}, f_{k} → f in L^{2}. Composition with T is continuous on L^{2}, as is Fourier transform, so we can write down (36) for the {f_{k}} and pass to the limit.

The L^{1} + L^{2} domain for the Fourier transform is is wide enough to include many natural examples. Note in particular that L^{p} ⊂ L^{1} +L^{2} if p ∈ (1, 2). Furthermore, certain homogeneous functions belong to L^{1} +L^{2} although none of them can belong to L^{p} for any fixed p. For example, |x|^{−a} belongs to L^{1} + L^{2} if \frac{n}{2} < a < n, since it belongs to L^{1} at the origin and to L^{2} at infinity. However, the L^{1} + L^{2} domain is not always sufficient. The most natural way to proceed would be to develop the idea of tempered distributions, but we don’t want to do this explicitly. Instead, we further broaden the definition of Fourier transform as follows: a tempered function is a function f ∈ L_{loc}^{1}(ℝ^{n}) such that

∫ (1 + |x|)^{−N}|f(x)|dx < ∞

for some constant N. Thus, roughly f has at most polynomial growth in the sense of L^{1} averages.

It is clear that if f is tempered and ϕ ∈ S then ∫ |ϕf| < ∞. Furthermore, the map ϕ → ∫ ϕf is continuous on S. It follows that ϕ ∗ f is well defined if ϕ ∈ S and f is tempered, and a simple estimation shows that ϕ ∗ f is again tempered.

If f and g are tempered functions then we say that g is the distributional Fourier transform of f if

∫ gϕ = ∫ fϕ̂  (37)

for all ϕ ∈ S. For given f, such a function g is unique using density properties of S as in several previous arguments. We denote g by f̂. Notice also that if f ∈ L^{1} + L^{2} then its L^{1} + L^{2}-Fourier transform coincides with its distributional Fourier transform by Proposition 3.12.

All the basic formulas, in particular (3), (4), (5), (6) extend to the case of distributional Fourier transforms, e.g. if g is the distributional Fourier transform of f then |detT |^{−1}f̂∘T^{−t} is the distributional Fourier transform of f ∘T. This may be seen by making approporiate changes of variable in the integrals in (37). Further, the inversion theorem is true in the sense that if f and g are tempered, g is the distributional Fourier transform of f then f(−x) is the distributional Fourier transform of g; this follows by substituting the inversion theorem for ϕ into (37) and making a change of variables x → −x.

We indicate how these arguments are carried out by proving the extended version of formula (27). Namely, if f is tempered, ψ ∈ S and f has a distributional Fourier transform then so does ψ ∗ f and

{ \atop ψ ∗ f} = ψ̂f̂  (38)

The proof is as follows. Let ϕ be another Schwartz function. Then

∫ (ψ̂f̂)ϕ = ∫ f̂(ψ̂ϕ)
= ∫ f ˆψ̂ϕ
= ∫ f(ˆψ̂ ∗ ϕ̂)
= ∫ f(x) ∫ ψ(−(x − y))ϕ̂(y)dydx
= ∫ ψ ∗ f(y)ϕ̂(y)

The second line followed from the definition of distributional Fourier transform, the third line from (28) and the next to last line used the inversion theorem for ψ. Comparing with the definition (37) we see that we have proved (38).

Let us note also that the inversion theorem is true for distributional Fourier transforms: if f is tempered and has a distrbutional Fourier transform f̂, then f̂ has the distributional Fourier transform f(−x). Here is the proof. If ϕ ∈ S then

∫ f(−x)ϕ(x)dx = ∫ f(x)ϕ(−x)dx
= ∫ f(x)ˆϕ̂(x)dx
= ∫ f̂(x)ϕ̂(x)dx

We used change of variables, the inversion theorem for ϕ and the definition (37) of f̂. Comparing again with (37) we have the stated result. □

4. Some specifics, and L^{p} for p < 2

We first discuss a couple of basic examples where the Fourier transform can be calculated, namely powers of the distance to the origin and complex Gaussians.

\underline{Proposition 4.1} Let h_{a}(x) = \frac{γ( \frac{a}{2} )}{π^{\frac{a}{2}}}|x|^{−a}. Then ĥ_{a} = h_{n−a} in the sense of L^{1} + L^{2} Fourier transforms if \frac{n}{2} <re(a) < n, and in the sense of distributional Fourier transforms if 0 <re(a) < n.

Here γ is the gamma function, i.e.

γ(s) = ∫_{0}^{∞} e^{−t}t^{s} \frac{dt}{t}

\underline{Proof} Suppose that a is real and \frac{n}{2} < a < n. Then h_{a} ∈ L^{1} + L^{2}. The functions of the form f(x) = c|x|^{−a} with c constant may be characterized by the following two transformation properties:

1. f is radial, i.e. f ∘ ρ = f for all linear ρ : ℝ^{n} → ℝ^{n} with ρ^{t}ρ =identity.

2. f is homogeneous of degree −a, i.e.,

f(ϵx) = ϵ^{−a}f(x)  (39)

for each ϵ > 0.

We will use the notation

fϵ(x) = f(ϵx)  (40)
fϵ(x) = ϵ^{−n}f(\frac{x}{ϵ})  (41)

Let f(x) = |x|^{−a}, \frac{n}{2} < a < n. Taking Fourier transforms we obtain from 1. and 2. the following (see the discussion on p.3 regarding special cases of (5)): f̂ is radial, and f̂ϵ = ϵ^{−a}f̂ which is equivalent to f̂ϵ = ϵ^{−(n−a)}f̂. It follows that f̂ = c|x|^{−(n−a)} so what remains is to evaluate the constant c. For this we use the duality relation taking the Schwartz function ψ to be the Gaussian Γ. Thus

∫ |x|^{−a}e^{−π|x|^{2}}dx = c ∫ |x|^{−(n−a)}e^{−π|x|^{2}}dx  (42)

To evaluate the left hand side, change to polar coordinates and then make the change of variable t = πr^{2}. Thus, if σ is the area of the unit sphere, we get

∫ |x|^{−a}e^{−π|x|^{2}}dx = σ ∫_{0}^{∞} e^{−πr^{2}}r^{n−a} \frac{dr}{r}

= σ ∫_{0}^{∞} e^{−t}(\frac{t}{π})^{\frac{n−a}{2}} \frac{dt}{2t}
= \frac{σ}{2} π^{−( \frac{n−a}{2} )}γ(\frac{n − a}{2})

and similarly the right hand side of (42) is c\frac{σ}{2} π^{−( \frac{a}{2} )}γ(\frac{a}{2}), so that

c = \frac{π^{\frac{a}{2}} γ(\frac{n−a}{2})}{π^{\frac{n−a}{2}} γ(\frac{a}{2})}

and the proposition is proved in the case \frac{n}{2} < a < n.

For the general case, fix ϕ ∈ S and consider the two integrals

A(z) = ∫ h_{z}ϕ̂

B(z) = ∫ h_{n−z}ϕ

Both A and B may be seen to be analytic in z in the indicated regime: since γ^{−1} is analytic this reduces to showing that

∫ |x|^{−z}ϕ(x)dx

is analytic when ϕ ∈ S, which may be done by using the DCT to justify complex differentiation under the integral sign.

Further, by Proposition 3.14 A and B agree when z is on the segment (\frac{n}{2}, n). So they agree everywhere by the uniqueness theorem. This proves that the distributional Fourier transform of h_{a} exists and is h_{n−a}. If rea > \frac{n}{2}, then h_{a} ∈ L^{1} + L^{2}, so its L^{1} + L^{2} and distributional Fourier transforms coincide. □

Let T be an invertible n×n real symmetric matrix. The signature of T is the quantity k_{+} − k_{−} where k_{+} and k_{−} are the numbers of positive and negaitive eigenvalues of T, counted with multiplicity. We also define

G_{T} (x) = e^{−πi⟨T x,x⟩}

and observe that G_{T} has absolute value 1 and is therefore tempered.

\underline{Proposition 4.2} Let T be an invertible n × n real symmetric matrix with signature σ. Then G_{T} has a distributional Fourier transform, which is equal to

e^{−πi \frac{σ}{4}} |detT |^{− \frac{1}{2}} G_{−T^{−1}}

\underline{Remark} This can easily be generalized to complex symmetric T with nonnegative imaginary part (the latter condition is needed, else G_{T} isn’t tempered). See [12], Theorem 7.6.1. If n = 1, we do this case in the course of the proof.

\underline{Proof} We need to show that

∫ e^{−πi⟨T x,x⟩}ϕ̂(x)dx = e^{^{\frac{−πi}{4}}σ}|detT |^{− \frac{1}{2}} ∫ e^{πi⟨T^{−1}x,x⟩}ϕ(x)dx  (43)

if ϕ ∈ S and T is invertible real symmetric.

First consider the n = 1 case. Let \sqrt{z} be the branch of the square root defined on the complement of the nonpositive real numbers and positive on the positive real axis. Thus \sqrt{±i} = e^{± \frac{πi}{4}} . Accordingly, (43) with n = 1 is equivalent with

∫ e^{−πzx^{2}} ϕ̂(x)dx = (\sqrt{z})^{−1} ∫ e^{−π \frac{x^{2}}{z}} ϕ(x)dx  (44)

if ϕ ∈ S and z is pure imaginary and not equal to zero. We prove this formula by analytic continuation from the real case.

Namely, if z = 1 then (44) is example 2 on p.1, and the case z real and positive then follows from scaling, i.e. the fact that the Fourier transform of fϵ is f̂ϵ. Both sides of (44) are easily seen to be analytic in z when rez > 0 and continuous in z when rez ≥ 0, z ≠ 0, so (44) is proved. □

Now consider the n ≥ 2 case. Observe that if (43) is true for a give T (and all ϕ) it is true also when A is replaced by UT U^{−1} for any U ∈ SO(n). This follows from the fact that { \atop f ∘ U} = f̂ ∘ U. However, since we did not give an explicit proof of the latter fact for distributional Fourier transforms we will now exhibit the necessary calculations. Let S = UT U^{−1}. Thus S and T have the same determinant and the same signature. Accordingly, if (43) holds for T then

∫ e^{−πi⟨Sx,x⟩}ϕ̂(x)dx = ∫ e^{−πi⟨T U^{−1}x,U^{−1}x⟩}ϕ̂(x)dx
= ∫ e^{−πi⟨T x,x⟩}ϕ̂(Ux)dx
= ∫ e^{−πi⟨T x,x⟩}{ \atop ϕ ∘ U}(x)dx
= e^{^{\frac{πi}{4}}σ}|detT |^{− \frac{1}{2}} e^{πi⟨T^{−1}x,x⟩}ϕ ∘ U(x)dx
= e^{^{\frac{πi}{4}}σ}|detT |^{− \frac{1}{2}} e^{πi⟨T^{−1}U^{−1}x,U^{−1}x⟩}ϕ(x)dx
= e^{^{\frac{πi}{4}}σ}|detS|^{− \frac{1}{2}} e^{πi⟨S^{−1}x,x⟩}ϕ(x)dx

We used { \atop ϕ ∘ U} = ϕ̂ ∘ U for Schwartz functions ϕ, see page 2.

It follows that it suffices to prove (43) when T is diagonal. If T is diagonal and ϕ is a tensor function, then the integrals in (43) factor as products of one variable integrals and (43) follows immediately from (44). The general case then follows using Proposition 2.2’ and the fact that integration against a tempered function defines a continuous linear functional on S.

We now briefly discuss the L^{p} Fourier transform, 1 < p < 2. The most basic result is the Hausdorff-Young theorem, which is a formal consequence of the Plancherel theorem and Proposition 1.1 via the following.

\underline{Riesz-Thorin interpolation theorem} Let T be a linear operator with domain L^{p_{0}} +L^{p_{1}}, 1 ≤ p_{0} < p_{1} ≤ ∞. Assume that f ∈ L^{p_{1}} implies

∥T f∥_{q_{0}} ≤ A_{0}∥f∥_{p_{0}}  (45)

and f ∈ L^{p_{1}} implies

∥T f∥_{q_{1}} ≤ B∥f∥_{p_{1}}  (46)

Suppose that for a certain θ ∈ (0, 1),

\frac{1}{p} = \frac{1 − θ}{p_{0}} + \frac{θ}{p_{1}}  (47)

and

\frac{1}{q} = \frac{1 − θ}{q_{0}} + \frac{θ}{q_{1}}  (48)

Then f ∈ L^{p} implies

∥T f∥_{q} ≤ A_{0}^{1−θ}A_{1}^{θ}∥f∥_{p}

For the proof see [13], [25], or numerous other textbooks.

We will adopt the convention that when indices p and p′ are used we must have

\frac{1}{p} + \frac{1}{p′} = 1

\underline{Proposition 4.3}(Hausdorff-Young) If 1 ≤ p ≤ 2 then

∥f̂∥_{p′} ≤ ∥f∥_{p}  (49)

\underline{Proof} We already know this if p = 1 or 2 and we therefore “interpolate” Namely, apply the Riesz-Thorin theorem with p_{0} = 1, q_{0} = ∞, p_{1} = q_{1} = 2, A_{0} = 1, A_{1} = ∞. The hypotheses (45) and (46) follow from Proposition 1.1 and Theorem 3.10 respectively. For given p, q, existence of θ ∈ (0, 1) making (47) and (48) true is equivalent with 1 < p < 2 and q = p′. The result follows. □

For later reference we insert here another basic result which follows from Riesz-Thorin, although this one (in contrast to Hausdorff-Young) could also be proved by elementary manipulation of inequalities.

\underline{Proposition 4.4}(Young’s inequality) Let ϕ ∈ L^{p}, ψ ∈ L^{r}, and let \frac{1}{q} = \frac{1}{p} − \frac{1}{r′}. Then the integral defining ϕ ∗ ψ is absolutely covergent for a.e. x and

∥ϕ ∗ ψ∥_{q} ≤ ∥ϕ∥_{p}∥ψ∥_{r}∥

\underline{Proof} View ϕ as fixed, i.e. define

T ψ = ϕ ∗ ψ

Inequalities (24) and (25) imply that

T : L^{1} + L^{p′} → L^{p} + L^{∞}

with

∥T ψ∥_{p} ≤ ∥ϕ∥_{p}∥ψ∥_{1}
∥T ψ∥_{∞} ≤ ∥ϕ∥_{p}∥ψ∥_{p′}

If \frac{1}{q} = \frac{1}{p} − \frac{1}{r′} then there is θ ∈ [0, 1] with

\frac{1}{r} = \frac{1 − θ}{1} + \frac{θ}{p′}

\frac{1}{q} = \frac{1 − θ}{p} + \frac{θ}{∞}

The result now follows from Riesz-Thorin. □

\underline{Remarks} 1. Unless p = 1 or 2 the constant 1 in the Hausdorff-Young inequality is not best possible; indeed the best constant is found by testing the Gaussian function Γ. This is much deeper and is due to Babenko when p′ is an even integer and to Beckner [1] in general. There are some related considerations in connection with Proposition 4.4, due also to Beckner.

2. Except in the case p = 2 the inequality (49) is not reversible, in the sense that there is no constant C such that ∥f∥_{p′} ≥ ∥f p when f ∈ S. Equivalently (in view of the inversion theorem) the result does not extend to the case p > 2. This is not at all difficult to show, but we discuss it at some length in order to illustrate a few different techniques used for constructing examples in connection with the L^{p} Fourier transform. Here is the most elementary argument.

\underline{Exercise} Using translation and muliplication by characters, construct a sequence of Schwartz functions {ϕ_{n}} so that

1. Each ϕ_{n} has the same L^{p} norm.

2. Each ϕ̂_{n} has the same L^{p′} norm.

3. The supports of the ϕ̂_{n} are disjoint.

4. The supports of the ϕ_{n} are “essentially disjoint meaning that

∥ ∑_{n=1}^{N} ϕ_{n}∥_{p}^{p} ≈ ∑_{n=1}^{N} ∥ϕ_{n}∥_{p}^{p} (≈ N)

uniformly in N and then use this to disprove the converse of Hausdorff-Young.

Here is a second argument based on Proposition 4.2. This argument can readily be adapted to show that there are functions f ∈ L^{p} for any p > 2 which do not have a distributional Fourier transform in our sense. See [12], Theorem 7.6.6.

Take n = 1 and fλ(x) = ϕ(x)e^{−πiλx^{2}}, ϕ ∈ C_{0}^{∞} fixed. Here λ is a large positive number. Then ∥fλ∥_{p} is independent of λ for any p. By the Plancherel theorem ∥f̂λ∥_{2} is also independent of λ. On the other hand f̂λ is the convolution of ϕ̂, which is in L^{1}, with

(\sqrt{iλ})^{−1}e^{πiλ^{−1}x^{2}}, which has L^{∞} norm λ^{− \frac{1}{2}} . Accordingly, if p < 2 then

∥f̂λ∥_{p′} ≤ ∥f̂λ∥2^{\frac{2}{p′}} ∥f̂λ∥_{∞}^{1− \frac{2}{p′}}
≲ λ^{−( \frac{1}{2} − \frac{1}{p′})}

Since ∥fλ∥_{p} is independent of λ this shows that when p < 2 there is no constant C such that C∥f∥_{p′} ≥ ∥f∥_{p} for all f ∈ S. □

Here now is another important technique (“randomization”) and a third disproof of the converse of Hausdorff-Young.

Let {ω_{n}}_{n=1}^{N} be independent random variables taking values ±1 with equal probability. Denote expectation (aka integral over the probability space in question) by 𝔼. and probability (aka measure) by Prob. Let {a_{n}}_{n=1}^{N} be complex numbers.

\underline{Proposition 4.5} (Khinchin’s inequality)

𝔼(| ∑_{n=1}^{N} a_{n}ω_{n}|^{p}) ≈ (∑_{n=1}^{N} |a_{n}|^{2})^{\frac{p}{2}}  (50)

for any 0 < p < ∞, where the implicit constant depends on p only.

Most books on probability and many analysis books give proofs. Here is the proof in the case p > 1. There are three steps.

(i)When p = 2 it is simple to see from independence that (50) is true with equality: expand out the left side and observe that the cross terms cancel.

(ii) The upper bound. This is best obtained as a consequence of a stronger (“subgaussian”)estimate. One can clearly assume the {a_{n}} are real and (52) below is for real {a_{n}}. Let t > 0. Start from the fact that

𝔼(e^{t ∑_{n} a_{n}ω_{n}}) = Π_{n}𝔼(e^{ta_{n}ω_{n}}) = Π_{n} \frac{1}{2}(e^{ta_{n}} + e^{−ta_{n}})

where the first equality follows from independence and the fact that e^{x+y} = e^{x}e^{y}. Use the numerical inequality

\frac{1}{2}(e^{x} + e^{−x}) ≤ e^{\frac{x^{2}}{2}}  (51)

to conclude that

𝔼(e^{t ∑_{n} a_{n}ω_{n}}) ≤ e^{\frac{t^{2}}{2} ∑_{n} a_{n}^{2}}

therefore

Prob(∑_{n} a_{n}ω_{n} ≥ λ) ≤ e^{−tλ+ \frac{t^{2}}{2} ∑_{n} a_{n}^{2}}

for any t > 0 and λ > 0. Taking t = \frac{λ}{∑_{n} a_{n}^{2}} gives

Prob(∑_{n} a_{n}ω_{n} ≥ λ) ≤ e^{− \frac{t^{2}}{2 ∑_{n} a^{2}_{n}}}  (52)

Prob(| ∑_{n} a_{n}ω_{n}| ≥ λ) ≤ 2e^{− \frac{λ^{2}}{2 ∑_{n} a^{2}_{n}}}

¿From this and the formula for the L^{p} norm in terms of the distribution function,

𝔼(|f|^{p}) = p ∫ λ^{p−1}Prob(|f| ≥ λ)dλ

one gets

𝔼(| ∑_{n} a_{n}ω_{n}|^{p}) ≤ 2p ∫ λ^{p−1}e^{− \frac{λ^{2}}{2 ∑_{n} a^{2}_{n}}} dλ = 2^{2+ \frac{p}{2}} pγ(\frac{p}{2})(∑_{n} a_{n}^{2})^{\frac{p}{2}}

This proves the upper bound.

(iii) The lower bound. This follows from (i) and (ii) by duality. Namely

∑_{n} |a_{n}|^{2} = 𝔼(| ∑_{n} a_{n}ω_{n}|^{2})
≤ 𝔼(| ∑_{n} a_{n}ω_{n}|^{p})^{\frac{1}{p}} 𝔼(| ∑_{n} a_{n}ω_{n}|^{p′})^{\frac{1}{p′}} 𝔼(| ∑_{n} a_{n}ω_{n}|^{p})^{\frac{1}{p}}
≲ (∑_{n} |a_{n}|^{2})^{\frac{1}{2}} 𝔼(| ∑_{n} a_{n}ω_{n}|^{p})^{\frac{1}{p}}

so

𝔼(| ∑_{n} a_{n}ω_{n}|^{p})^{\frac{1}{p}} ≳ (∑_{n} |a_{n}|^{2})^{\frac{1}{2}}

as claimed. □.

To apply this in connection with the converse of Hausdorff-Young, let ϕ be a C_{0}^{∞} function and let {k_{j}}_{j=1}^{N} be such that the functions ϕ_{j} =^{def} ϕ(· − k_{j}) have disjoint support.

Thus ϕ̂_{n}(ξ) = e^{2πiξ·k_{n}} ϕ̂(ξ) The L^{p} norm of ∑_{n≤N} ω_{n}ϕ_{n} is independent of ω in view of the disjoint support, indeed

∥ ∑_{n≤N} ω_{n}ϕ_{n}∥_{p} = CN^{\frac{1}{p}}  (53)

where C = ∥ϕ∥_{p}.

Now consider the corresponding Fourier side norms, more precisely the expectation of their p′ powers, the quantity

𝔼(∥ ∑_{n≤N} ω_{n}ϕ̂_{n}∥_{p′}^{p′}  (54)

We have by Fubini’s theorem

(54) = 𝔼(∥ ∑_{n≤N} ω_{n}e^{2πiξ·k_{n}} ϕ̂(ξ))∥_{L^{p′}(dξ)}^{p′}

= ∫_{ℝ_{n}} |ϕ̂(ξ)|^{p′}𝔼(| ∑_{n≤N} ω_{n}e^{2πiξ·k_{n}}|^{p′})
≈ N^{\frac{p′}{2}}

where the last step used Khinchin.

It follows that we can make a choice of ω so that ∥ ∑_{n≤N} ω_{n}ϕ̂_{n}∥_{p′} ≲ N^{\frac{1}{2}} .

If p < 2 and if N is large then this is much smaller than the right hand side of (53), so we are done.

5. Uncertainty Principle A ellipsoid in ℝ^{n} is a set of the form

E = \{x ∈ ℝ^{n} : ∑_{j} \frac{|(x − a) · e_{j}|^{2}}{r_{j}^{2}} ≤ 1\}  (55)

for some a ∈ ℝ^{n} (called the center of R), some choice of orthonormal basis {e_{j}} (the axes) and some choice of positive numbers r_{j} (the axis lengths). If E and E^{∗} are two ellipsoids, then we say that R^{∗} is dual to E if E^{∗} has the same axes as R and reciprocal axis lengths, i.e. if E is given by (55) then R^{∗} should be of the form {x ∈ ℝ^{n} : ∑_{j} r_{j}^{2}(|(x − b) · e_{j}|)^{2} ≤ 1 for some choice of center point b.

The uncertainty principle is^{1} the heuristic statement that if a measure μ is supported on R, then for many purposes μ̂ may be regarded as being constant on any dual ellipsoid R^{∗}.

The simplest rigorous statement is

1This should be qualified by adding “as far as we are concerned”. There are various more sophisticated related statements which are also called uncertainty principle; see for example [10], [11] and references there.

\underline{Proposition 5.1}(L^{2} Bernstein inequality) Assume that f ∈ L^{2} and f̂ is supported in D(0, R). Then f is C^{∞} and there is an estimate

∥D^{α}f∥_{2} ≤ (2πR)^{|α|}∥f∥_{2}  (56)

\underline{Proof} Essentially this is an immediate consequence of the Plancherel theorem, see the calculation at the end of the proof. However there are some details to take care of if one wants to be rigorous.

The Fourier inversion formula

f(x) = ∫ f̂(ξ)e^{2πix·ξ}dξ  (57)

is valid (in the naive sense). Namely, note that the support assumption implies f̂ ∈ L^{1} and choose a sequence of Schwartz functions ψ_{k} which converges to f̂ both in L^{1} and in L^{2}. Let ϕ_{k} ∈ S satisfy ϕ̂_{k} = ψ_{k}. The formula (57) is valid for ϕ_{k}. As k → ∞ the left sides converge in L^{2} to f by Theorem 3.10 and the right sides converge uniformly to ∫ f̂(ξ)e^{2πix·ξ}dξ, which proves (57) for f.

Proposition 1.3 applied to f̂ now implies that f is C^{∞} and that D^{α}f is obtained by differentiation under the integral sign in (57). The estimate (56) holds since

∥D^{α}f∥_{2} = ∥D̂^{αf}∥_{2}∥(2πiξ)^{α}f̂∥_{2} ≤ (2πR)^{|α|}∥f̂∥_{2} = (2πR)^{|α|}∥f∥_{2}

A corresponding statement is also true in L^{p} norms, but proving this and other related results needs a different argument since there is no Plancherel theorem.

\underline{Lemma 5.2} There is a fixed Schwartz function ϕ such that if f ∈ L^{1} + L^{2} and f̂ is supported in D(0, R), then

f = ϕ^{R^{−1}} ∗ f

\underline{Proof} Take ϕ so that ϕ̂ is equal to 1 on D(0, 1). Thus { \atop ϕ^{R^{−1}}}(ξ) = ϕ̂(R^{−1}ξ) is equal to 1 on D(0, R), so { \atop ϕ^{R^{−1}} ∗ f − f} vanishes identically. Hence ϕ^{R^{−1}} ∗ f = f. □

\underline{Proposition 5.3}(Bernstein’s inequality for a disc) Suppose f ∈ L^{1} + L^{2}, and f̂ is supported in D(0, R). Then

1. For any α and p ∈ [1, ∞]

∥D^{α}f∥_{p} ≤ (CR)^{|α|}∥f∥_{p}

2. For any 1 ≤ p ≤ q ≤ ∞

∥f∥_{q} ≤ CR^{n( \frac{1}{p} − \frac{1}{q})}∥f∥_{p}

\underline{Proof} The function ψ = ϕ^{R^{−1}} satisfies

∥ψ∥_{r} = CR^{\frac{n}{r′}}  (58)

for any r ∈ [1, ∞] where C = ∥ϕ∥_{r}. Also, using the chain rule

∥∇ψ∥_{1} = R∥ϕ∥_{1}  (59)

We know that f = ψ ∗ f. In the case of first derivatives, 1. therefore follows from (58) and (24). The general case of 1. then follows by induction.

For 2., let r satisfy \frac{1}{q} = \frac{1}{p} − \frac{1}{r′}. Apply Young’s inequality obtaining

∥f∥_{q} = ∥ψ ∗ f∥_{q}
≤ ∥ψ∥_{r}∥f∥_{p}
≲ R^{\frac{n}{r′}} ∥f∥_{p}
= R^{n( \frac{1}{p} − \frac{1}{q})}∥f∥_{p}
□

We now extend the L^{p} → L^{q} bound to ellipsoids instead of balls, using change of variable. One could similarly extend the first part of Proposition 5.3 to ellipsoids centered at the origin, but the statement is awkward since one has to weight different directions differently, so we ignore this.

\underline{Proposition 5.4} (Bernstein’s inequality for an ellipsoid) Suppose f ∈ L^{1} + L^{2}, and f̂ is supported in an ellipsoid R. Then

∥f∥_{q} ≲ |R|^{−( \frac{1}{p} − \frac{1}{q})}∥f∥_{p}

if 1 ≤ p ≤ q ≤ q∞.

\underline{Proof} Let k be the center of R. Let T be a linear map taking the unit ball onto R − k. Let S = T^{−t}; thus T = S^{−t} also. Let f_{1}(x) = e^{−2πik·x}f(x) and g = f_{1} ∘ S, which has

ĝ(ξ) = |detS|^{−1}f̂_{1}(S^{−t}(ξ))
= |detS|^{−1}f̂(S^{−t}(ξ + k))
= |detT |f̂(T (ξ) + k))

Thus ĝ is supported in the unit ball, so

∥g∥_{q} ≲ ∥g∥_{p}

On the other hand,

∥g∥_{q} = |detS|^{− \frac{1}{q}} ∥f∥_{q} = |detT |^{\frac{1}{q}} ∥f∥_{q} = |R|^{\frac{1}{q}} ∥f∥_{q}

and likewise with q replaced by p. So

|R|^{\frac{1}{q}} ∥f∥_{q} ≲ |R|^{\frac{1}{p}} ∥f∥_{p}

as claimed. □

For some purposes one needs a related “pointwise statement”, roughly that if suppf̂ ⊂ R, then for each R^{∗}, the values on R^{∗} are controlled by the average over R^{∗}.

To formulate this precisely, let N be a large number, let ϕ(x) = (1 + |x|^{2})^{−N}. Suppose an ellipsoid is given, which we call R^{∗}. Define ϕ_{R^{∗}}(x) = ϕ(T (x − k)) where k is the center of R^{∗} and T is a selfadjoint linear map taking R^{∗} − k onto the unit ball. If T_{1} and T_{2} are two such maps then T_{1} ∘ T_{2}^{−1} is an orthogonal transformation, so ϕ_{R^{∗}} is well defined. Essentially, ϕ_{R^{∗}} is roughly equal to 1 on R^{∗} and decays rapidly as one moves away from R^{∗}. We could also write more explicitly

ϕ_{R∗}(x) = (1 + ∑_{j} \frac{|(x − k) · e_{j}|^{2}}{r_{j}^{2}} )^{−N}

\underline{Proposition 5.5} Suppose f ∈ L^{1} + L^{2} and f̂ is supported in an ellipsoid R. Then for any dual ellipsoid R^{∗} and any z ∈ R^{∗},

|f(z)| ≤ C_{N} \frac{1}{|R^{∗}|} ∫ |f(x)|ϕ_{R^{∗}}dx  (60)

\underline{Proof} Assume first that R is the unit ball, and R^{∗} is also the unit ball. Then f is the convolution of itself with a fixed Schwartz function ψ. Accordingly

|f(z)| ≤ ∫ |f(x)| |ψ(z − x)|dx
≤ C_{N} ∫ |f(x)|(1 + |z − x|^{2})^{−N}
≤ C_{N} ∫ |f(x)|(1 + |x|^{2})^{−N}

We used the Schwartz space bounds for ψ and that |1 + |z − x|^{2} ≳ 1 + |x|^{2} uniformly in x when |z| ≤ 1. This proves (60) when R = R^{∗} =unit ball.

Suppose next that R is centered at zero but R and R^{∗} are otherwise arbitrary. Let k and T be as above and consider

g(x) = f(T^{−1}x + k))

Its Fourier transform is supported on T^{−1}R, and if T maps R^{∗} onto the unit ball, then T^{−1} maps R onto the unit ball. Accordingly

|g(y)| ≤ ∫ ϕ(x)|g(x)|dx

if y ∈ D(0, 1).

f(T^{−1}z + k) ≤ ϕ(x)|f(T^{−1}x + k)|dx = |detT | ∫ ϕ_{R^{∗}}(x)|f(x)|dx

by changing variables. And |detT | = \frac{1}{|R^{∗}|}, so we get (60).

If R isn’t centered at zero, then we can apply the preceding with f replaced by e^{−2πik·x}f(x) where k is the center. □

\underline{Remarks} 1. Proposition 5.5 is an example of an estimate “with Schwartz tails”. It is not possible to make the stronger conclusion that, say, |f(x)| is bounded by the average of f over the double of R^{∗} when x ∈ R^{∗}, even in the one dimensional case with R = R^{∗} =unit interval. For this, consider a fixed Schwartz function g whose Fourier transform is supported in the unit interval [−1, 1]. Consider also the functions

f_{N} = (1 − \frac{x^{2}}{4} )^{N}g(x)

These have the same Fourier support as g and converge pointwise boundedly to zero on [−2, 2], except at the origin. It follows that there can be no estimate of the value of f_{N} at the origin by its average over [−2, 2].

2. All the estimates related to Bernstein’s inequality are sharp except for the values of the constants. For example, if E is an ellipsoid, E^{∗} a dual ellipsoid, N < ∞ then there is a function f with suppf̂ ⊂ E^{∗} and with

∥f∥_{1} ≥ |E|  (61)

,

|f(x)| ≤ Cϕ_{E}(x)  (62)

where ϕ_{E} = ϕ_{E}^{(N)} was defined above. In the case E = E^{∗} =unit ball this is obvious; take f to be any Schwartz function with Fourier support in the unit ball and with the appropriate L^{1} norm. The general case then follows as above by making changes of variable.

The estimates (61) and (62) imply that ∥f∥_{p} ≈ |E|^{\frac{1}{p}} for any p, so it follows that Proposition 5.4 is also sharp.

6. Stationary phase Let ϕ be a real valued C^{∞} function, let a be a C_{0}^{∞} function and define

I(λ) = ∫ e^{−πiλϕ(x)}a(x)dx

Here λ is a parameter, which we always assume to be positive. The issue is the behavior of the integral I(λ) as λ → +∞.

\underline{Some general remarks} 1. |I(λ)| is clearly bounded by a constant depending on a only. One may expect decay as λ → ∞ since when λ is large the integral will involve a lot of cancellation.

2. On the other hand, if ϕ is constant then |I(λ)| is independent of λ. So one needs to put nondegeneracy hypotheses on ϕ. As it turns out properties of a are less important. Note also that one can always cut up a with a partition of unity, which means that the question of how fast I(λ) decays can be “localized” to a small neighborhood of a point.

3. Suppose that ϕ_{1} = ϕ_{2} ∘ G where G is a diffeomorphism. Then

∫ e^{−πiλϕ_{2}(x)}a(x)dx = ∫ e^{−πiλϕ_{1}(G^{−1}x)}a(x)dx
= ∫ e^{−πiλϕ_{1}(y)}a(Gy)d(Gy)
= ∫ e^{−πiλϕ_{1}(y)}a(Gy)|J_{G}(y)|dy

where J_{G} is the Jacobian determinant. y → a(Gy)|J_{G}(y)| is again C_{0}^{∞} so we see that any bound for I(λ) which is independent of choice of a will be “diffeomorphism invariant”.

4. Recall from advanced calculus [15] the normal forms for function near a regular point or a nondegenerate critical point:

\underline{Straightening Lemma} Suppose Ω ⊂ ℝ^{n} is open, f : Ω → ℝ is C^{∞}, p ∈ Ω and ∇f(p) = 0. Theh there are neighborhoods U and V of 0 and p respectively and a C^{∞} diffeomorphism G : U → V with G(0) = p and

f ∘ G(x) = f(p) + x_{n}

\underline{Morse Lemma} Suppose Ω ⊂ ℝ^{n} is open, f : Ω → ℝ is C^{∞}, p ∈ Ω, ∇f(p) = 0, and suppose that the Hessian matrix H_{f}(p = (\frac{∂^{2}f}{∂x_{i}∂x_{j}} (p)) is invertible. Then, for a unique k (=number of positive eigenvalues of H_{f}; see Lemma 6.3 below) there are neighborhoods U and V of 0 and p respectively and a C^{∞} diffeomorphism G : U → V with G(0) = p and

f ∘ G(x) = f(p) + ∑_{j=1}^{k} x_{j}^{2} − ∑_{j=k+1}^{n} x_{j}^{2}

We consider now I(λ) first when a is supported near a regular point, and then when a is supported near a nondegenerate critical point. Degenerate critical points are easy to deal with if n = 1, see [24], chapter 8, but in higher dimensions they are much more complicated, and only the two dimensional case has been worked out, see [28].

\underline{Proposition 6.1} (Nonstationary phase) Suppose Ω ⊂ ℝ^{n} is open, ϕ : Ω → ℝ is C^{∞}, p ∈ Ω and ∇ϕ(p) ≠ 0. Suppose a ∈ C_{0}^{∞} has its support in a sufficiently small neighborhood of p. Then

∀N∃C_{N} : |I(λ)| ≤ C_{N}λ^{−N}

and furthermore C_{N} depends only on bounds for finitely many derivatives of ϕ and a and a lower bound for |∇ϕ(a)| (and on N).

\underline{Proof} The straightening lemma and the calculation in 3. above reduce this to the case ϕ(x) = x_{n} + c. In this case, letting e_{n} = (0, …, 0, 1)

I(λ) = e^{−πiλc}ϕ̂(\frac{λ}{2}e_{n})

and this has the requisite decay by Proposition 2.3. □

Now we consider the nondegenerate critical point case, and as in the preceding proof we first consider the normal form.

\underline{Proposition 6.2} Let T be a real symmetric invertible matrix with signature σ, let a be C_{0}^{∞} (or just in S) and define

I(λ) = ∫ e^{−πiλ⟨T x,x⟩}a(x)dx

Then, for any N,

I(λ) = e^{−πi \frac{σ}{4}} |detT |^{− \frac{1}{2}} λ^{− \frac{n}{2}} (a(0) + ∑_{j=1}^{N} λ^{−j}D_{j}a(0) + O(λ^{−(N+1)}))

Here Δ_{j} are certain explicit homogeneous constant coefficient differential operators of order 2j, depending on T only, and the implicit constant depends only on T and on bounds for finitely Schwartz space seminorms of a.

\underline{Proof} Essentially this is just another way of looking at the formula for the Fourier transform of an imaginary Gaussian. By Proposition 4.2, the definition of distibutional Fourier transform and the Fourier inversion theorem for ϕ we have

I(λ) = e^{−πi \frac{σ}{4}} λ^{− \frac{n}{2}} |detT |^{− \frac{1}{2}} ∫ ϕ̂(−ξ)e^{πiλ^{−1}⟨T^{−1}ξ,ξ⟩}dξ

We can replace ϕ̂(−ξ) with ϕ̂(−ξ) by making a change of variables, since the Gaussian is even. To understand the resulting integral, use that λ^{−1} → 0 as λ → 0 so the Gaussian term is approaching 1. To make this quantitative, use Taylor’s theorem for e^{ix}:

e^{πiλ^{−1}⟨T^{−1}ξ,ξ⟩} = ∑_{j=0}^{N} \frac{(πiλ^{−1}⟨T^{−1}ξ, ξ⟩)^{j}}{j!} + O(\frac{|ξ|^{2N+2}}{λ^{N+1}} )

uniformly in ξ and λ. Accordingly

∫ ϕ̂(ξ)e^{πiλ^{−1}⟨T^{−1}ξ,ξ⟩}dξ = ∫ ϕ̂(ξ)(1 + ∑_{j=1}^{N} \frac{(πiλ^{−1}⟨T^{−1}ξ, ξ⟩)^{j}}{j!} )dξ + O(∫ |ϕ̂(ξ)|\frac{|ξ|^{2N+2}}{λ^{N+1}} )

Now observe that ∫ ϕ̂(ξ)dξ = ϕ(0) by the inversion theorem and similarly

∫ ϕ̂(ξ)\frac{(πi⟨T^{−1}ξ, ξ⟩)^{j}}{j!} )dξ

is the value at zero of D_{j}ϕ for an appropriate differential operator D_{j}. This gives the result, since

∫ |ϕ̂(ξ)|ξ|^{2N+2}dξ

is bounded in terms of Schwartz space seminorms of ϕ̂, and therefore in terms of derivatives of ϕ. □

Now we consider the case of a general phase function with a nondegenerate critical point. It is clear that this should be reducible to the Gaussian case using the Morse lemma and remark 3. above. However, there is more calculation involved than in the proof of Proposition 5.1, since we need to obtain the correct form for the asymptotic expansion. We recall the following formula which follows from the chain rule:

\underline{Lemma 6.3} Suppose that ϕ is smooth, ∇ϕ(p) = 0 and G is a diffeomorphism, G(0) = p. Then

H_{ϕ∘G}(0) = DG(0)^{t}H_{ϕ}(p)DG(0)

Thus H_{ϕ}(p) and H_{ϕ∘G}(0) have the same signature and

det(H_{ϕ∘G}(0) = J_{G}(0)^{2}det(H_{ϕ}(p))

\underline{Proposition 6.4} Let ϕ be C^{∞} and assume that ∇ϕ(p) = 0 and H_{ϕ}(p) is invertible. Let σ be the signature of H_{ϕ}(p), and let Δ = 2^{−n}|det(H_{ϕ}(p)|. Let a be C_{0}^{∞} and supported in a sufficiently small neighborhood of p and define

I(λ) = ∫ e^{−πiλϕ(x)dx}a(x)dx

Then, for any N,

I(λ) = e^{−πiλϕ(p)}e^{−πi \frac{σ}{4}} Δ^{− \frac{1}{2}} λ^{− \frac{n}{2}} (a(p) + ∑_{j=1}^{N} λ^{−j}D_{j}a(p) + O(λ^{−(N+1)}))

Here D_{j} are certain differential operators of order^{2} ≤ 2j, with coefficients depending on ϕ, and the implicit constant depends on ϕ and on bounds for finitely many derivatives of a.

2Actually the order is exactly 2j but we have no need to know that

\underline{Proof} We can assume that ϕ(p) = 0; else we replace ϕ with ϕ − ϕ(p). Choose a diffeomorphism G by the Morse lemma and apply remark 3. Thus

Iλ = ∫ e^{−πiλ⟨T y,y⟩}a(Gy)|J_{G}(y)|dy

where T is a diagonal matrix with diagonal entries ±1 and with signature σ. Also |J_{G}(0)| = Δ^{− \frac{1}{2}} by Lemma 6.3 and an obvious calculation of the Hessian determinant of the function y → ⟨T y, y⟩. Let D be associated to this T as in Proposition 5.2 and let b(y) = a(Gy)|J_{G}(y)|. Then

I(λ) = e^{−πi \frac{σ}{4}} λ^{− \frac{n}{2}} (b(0)) + ∑_{j=1}^{N} λ^{−j}D_{j}b(0) + O(λ^{−(N+1)})

by Proposition 6.2. Now b(0) = |J_{G}(0)| = Δ^{− \frac{1}{2}} , so we can write this as

I(λ) = e^{−πi \frac{σ}{4}} Δ^{− \frac{1}{2}} λ^{− \frac{n}{2}} (a(p)) + ∑_{j=1}^{N} λ^{−j}Δ^{\frac{1}{2}} D_{j}b(0) + O(λ^{−(N+1)})

Further, it is clear for the chain rule and product rule that any 2jth order derivative of b at the origin can be expressed as a linear combination of derivatives of a at p of order ≤ 2j with coefficients depending on G, i.e. on ϕ. Otherwise stated, the term Δ^{\frac{1}{2}} D_{j}b(0) can be expressed in the form D̃_{j}a(p) where D̃_{j} is a new differential operator of order ≤ 2j with coefficients depending on ϕ. This gives the result. □

In practice, it is often more useful to have estimates for I(λ) instead of an asymptotic expansion. Clearly an estimate |I(λ)| ≲ λ^{− \frac{n}{2}} could be derived from Proposition 6.4, but one also sometimes needs estimates for the derivatives of I(λ) with respect to suitable parameters. For now we just consider the technically easiest case where the parameter is λ itself.

\underline{Proposition 6.5} (i) Assume that ∇ϕ(p) ≠ 0. Then, for a supported in a small neigh-

borhood of p, \frac{d^{j}I(λ)}{dλ^{j}} ≤ C_{jN}λ^{−N} for any N.

(ii)Assume that ∇ϕ(p) = 0, and H_{ϕ}(p) is invertible. Then, for a supported in a small

neighborhood of p, |\frac{d^{k}(e^{πiλϕ(p)}I(λ)}{dλ^{k}}| ≤ C_{k}λ^{−( \frac{n}{2} +k)}.

\underline{Proof} We only prove (ii), since (i) follows easily from Proposition 5.1 after differentiating under the integral sign as in the proof below. For (ii) we need the following.

\underline{Claim}. Let \{ϕ_{i}\}_{i=1}^{M} be real valued smooth functions and assume that ϕ_{i}(p) = 0, ∇ϕ_{i}(p) = 0. Let Φ = Π_{i=1}^{M}ϕ_{i}. Then all partial derivatives of Φ of order less than 2M also vanish at p.

\underline{Proof} By the product rule any partial D^{α}Φ is a linear combination of terms of the form

Π_{i=1}^{M}D^{β_{i}}ϕ_{i}

with ∑_{i} β_{i} = α. If |α| < 2M then some β_{i} must be less than 2 so by hypothesis all such terms vanish at p.

To prove the proposition, differentiate I(λ) under the integral sign obtaining

\frac{d^{k}(e^{πiλϕ(p)}I(λ)}{dλ^{k}} = ∫ (ϕ(x) − ϕ(p))^{k}a(x)e^{−πiλϕ(x)}dx

Let b(x) = (ϕ(x) − ϕ(p))^{k}a(x) We just saw that all partials of b of order less than 2k vanish at p. Now look at the expansion in Proposition 6.4 replacing a with b and setting N = k − 1. By the claim the terms D_{j}b(p) must vanish when j < k, as well as b(p) itself.

Hence Proposition 5.4 shows that \frac{d^{k}(e^{πiλϕ(p)}I(λ)}{dλ^{k}} = O(λ^{−( \frac{n}{2} +k)}) as claimed. □

As an application we estimate the Fourier transform of surface measure on the sphere S^{n−1} ⊂ ℝ^{n}. For this and for other similar calculations one wants to work with an integral over a submanifold instead of over ℝ^{n}. This is not significantly different since it is always possible to work in local coordinates. However, things are easier if one uses the local coordinates as economically as possible. Recall then that if ϕ : ℝ^{n} → ℝ is smooth and if M is a k-dimensional submanifold, p ∈ M, and if F : U → M is a local coordinate (more precisely the inverse map to one) near p, then ϕ ∘ F will have a critical point at F^{−1}p iff ∇ϕ(p) is orthogonal to the tangent space to M at p; in particular this is independent of the choice of F.

Notice that σ̂ is a radial function, because surface measure is rotation invariant (exercise: prove this rigorously), and is smooth by Proposition 1.3. It therefore suffices to consider σ̂(λe_{n}) where e_{n} = (0, …, 0, 1) and λ > 0.

Put local coordinates on the sphere as follows: the first “local coordinate” is the map

x → (x, \sqrt{1 − |x|^{2}})

ℝ^{n−1} ⊃ D(0, \frac{1}{2}) → S^{n−1}

The second is the map x → (x, −\sqrt{1 − |x|^{2}})

ℝ^{n−1} ⊃ D(0, \frac{1}{2}) → S^{n−1}

and the remaining ones map onto sets whose closures do not contain {±e_{n}}. Let {q_{k}} be a suitable partition of unity subordinate to this covering by charts. Define ϕ(x) = e_{n} · x, ϕ : ℝ^{n} → ℝ. Thus the gradient of ϕ is e_{n} and is normal to the sphere at ±e_{n} only.

Now

σ̂(λe_{n}) = ∫ e^{−2πiλe_{n}·x}dσ(x)
= ∑_{j=1}^{k} ∫ e^{−2πiλe_{n}·x}q_{k}(x)dσ(x)

= ∫_{D(0, \frac{1}{2} )} e^{−2πiλ\sqrt{1−|x|^{2}}}q_{1}(x)\sqrt{1 + |x|^{2}}dx + ∫_{D(0, \frac{1}{2} )} e^{2πiλ\sqrt{1−|x|^{2}}}q_{2}(x)\sqrt{1 + |x|^{2}}dx
+ ∑_{k≥3} ∫ e^{−2πiλϕ_{k}(x)}a_{k}(x)dx  (63)

where the dx integrals are in ℝ^{n−1}, and the phase functions ϕ_{k} for k ≥ 3 have no critical points in the support of a_{k}. The Hessian of 2\sqrt{1 − |x|^{2}} at the origin is −2 times the identity matrix, and in particular is invertible. It is also clear that the first and second terms are complex conjugates. We conclude from Proposition 6.5 that

σ̂(λe_{n}) = Re(a(λ)e^{2πiλ}) + y(λ)

with

\frac{d^{j}a(λ)}{dλ^{j}} ≲ λ^{− \frac{n−1}{2} −j}  (64)
\frac{d^{j}y(λ)}{dλ^{j}} ≲ λ^{−N}  (65)

for any N. In fact σ is real and even and therefore σ̂ must be real valued. Multiplying y by e^{−2πiλ} does not affect the estimate (65), so we can absorb y into a and rewrite this as

σ̂(λe_{n}) = Re(a(λ)e^{2πiλ})

where a satisfies (64). Since σ̂ is radial, we have proved the following.

\underline{Corollary 6.6} The function σ̂ (is a C^{∞} function and) satisfies

σ̂(x) = Re(a(|x|)e^{2πi|x|})  (66)

where for large r

|\frac{d^{j}a}{dr^{j}}| ≤ C_{j}r^{−( \frac{n−1}{2} +j)}  (67)

Furthermore, looking at the first term in (63), and using the expansion of Proposition 6.4, with N = 0, we can obtain the leading behavior at ∞. Namely, for the first term in (63) at its critical point x = 0 we have a phase function 2\sqrt{1 − |x|^{2}} with ϕ(0) = 2, Δ = 1 and signature −(n − 1), and an amplitude q_{2}(x)\sqrt{1 + |x|^{2}} which is 1 at the critical point. By Proposition 6.4 the integral is

e^{−2πiλ}e^{\frac{πi}{4} (n−1)}λ^{− \frac{n−1}{2}} + O(λ^{− \frac{n+1}{2}} )

The second term is the complex conjugate and the others are O(λ^{−N}) for any N. Whence the quantity (63) is

2λ^{− \frac{n−1}{2}} cos(2π(λ − \frac{n − 1}{8})) + O(λ^{− \frac{n+1}{2}} )

and we have proved

\underline{Corollary 6.7} For large |x|

σ̂(x) = 2|x|^{− \frac{n−1}{2}} cos(2π(|x| − \frac{n − 1}{8})) + O(|x|^{− \frac{n+1}{2}} )

\underline{Remarks} Of course it is possible to consider surfaces other than the sphere, see for example [12], Theorem 7.7.14. The main point in regard to the latter is that nondegeneracy of the critical points of the phase function which arises when calculating the Fourier transform is equivalent to nonzero Gaussian curvature, so a hypersurface with nonzero Gaussian curvature everywhere behaves essentially the same as the sphere, whereas if there are flat directions the decay becomes weaker. Obtaining derivative bounds like Corollary 6.6 in the above manner requires a somewhat more complicated version of Proposition 6.5 where ϕ and a depend on an auxiliary parameter z, which we now explain without giving the proofs.

Suppose that ϕ(x, z) is a C^{∞} function of x and z, where x ∈ ℝ^{n}, and z ∈ ℝ^{k} should be regarded a parameter. Assume that for a certain p and z_{0} we have ∇_{x}ϕ(p, z_{0}) = 0 and that the matrix of second x-partials of ϕ at (p, z_{0}) is invertible.

1. Prove the following: there are neighborhoods U of z_{0} and V of p and a smooth function κ : U → V with the following property: if z ∈ V, then ∇_{x}ϕ(x, z) = 0 if and only if x = κ(z).

2. Let a(x, z) be C_{0}^{∞} and supported in a small enough neighborhood of (p, z_{0}). Define

I(λ, z) = ∫ e^{−πiλϕ(x,z)}a(x, z)dx

Prove the following:

|\frac{d^{j+k}(e^{πiλϕ(κ(z),z)}I(λ, z)}{dλ^{j}dz^{k}})| ≤ C_{jk}λ^{−( \frac{n}{2} +j)}

This is the analogue of Proposition 6.5 for general parameters.

7. Restriction problem We now suppose given a function f : S^{n−1} → ℂ and consider the Fourier transform

f̂dσ(ξ) = ∫_{S^{n−1}} f(x)e^{−2πix·ξ}dσ(x)  (68)

If f is smooth, then one can use stationary phase to evaluate f̂dσ to any desired degree of precision, just as with Corollary 6.6. In particular this leads to the bound

|f̂dσ(ξ)| ≤ C∥f∥_{C^{2}}(1 + |ξ|)^{− \frac{n−1}{2}}

say, where ∥f∥_{C^{2}} = ∑_{0≤|α|≤2} ∥D^{α}f∥_{L^{∞}}.

On the other hand there can be no similar decay estimate for functions f which are just bounded. The reason for this is that then there is no distinguished reference point in Fouirer space. Thus, if we let f_{k}(x) = e^{2πik·x} and set ξ = k, we have

|{ \atop f_{k}dσ}(ξ)| = σ(S^{n−1}) ≈ 1

Taking a sum of the form f = ∑_{j} j^{−2}f_{k_{j}}, where |k_{j}| → ∞ sufficiently rapidly, we obtain a continuous function f such that there is no estimate

|f̂dσ(ξ)| ≤ C(1 + |ξ|)^{−ϵ}

for any ϵ > 0.

On the other hand if we consider instead L^{q} norms then the issue of a distinguished origin is no longer relevant. The following is a long standing open problem in the area.

\underline{Restriction conjecture} (Stein) Prove that if f ∈ L^{∞}(S^{n−1}) then

∥f̂dσ∥_{q} ≤ C_{q}∥f∥_{∞}  (69)

for any q > \frac{2n}{n−1}.

The example of a constant function shows that the regime q > \frac{2n}{n−1} would be best possible. Namely Corollary 6.7 implies that σ̂ ∈ L^{q} if and only if q · \frac{n−1}{2} > n.

The corresponding problem for L^{2} densities f was solved in the 1970’s:

\underline{Theorem 7.1} (P. Tomas-Stein) If f ∈ L^{2}(S^{n−1}) then

∥f̂dσ∥_{q} ≤ C∥f∥_{L^{2}(S^{n−1})}  (70)

for q ≥ \frac{2n+2}{n−1}, and this range of q is best possible.

\underline{Remarks} 1. Notice that the assumptions on q in (70) and (69) are of the form q > q_{0} or q ≥ q_{0}. The reason for this is that there is an obvious estimate

∥f̂dσ∥_{∞} ≤ ∥f∥_{1}

by Proposition 1.1, and it follows by the Riesz-Thorin that if (69) or (70) holds for a given q then it also holds for any larger q.

2. The restriction conjecture (69) is known to be true when n = 2; this is due to C. Fefferman and Stein, early 1970’s. See [9] and [24].

3. Of course there is a difference in the L^{q} exponent in (70) and the one which is conjectured for L^{∞} densities. Until fairly recently it was unknown (in three or more dimensions) whether the estimate (69) was true even for some q less than the Stein-Tomas exponent \frac{2n+2}{n−1}. This was first shown by Bourgain [2], a paper which has been the starting point for a lot of recent work.

4. The fact that q ≥ \frac{2n+2}{n−1} is best possible for (70) is due I believe to A. Knapp. We now discuss the construction. Notice that in order to distinguish between L^{2} and L^{∞} norms, one should use a function f which is highly localized. Next, in view of the nice behavior of rectangles under the Fourier transform discussed e.g. in our section 5, it is natural to take the support of f to be the intersection of S^{n−1} with a small rectangle. Now we set up the proof.

Let

Cδ = {x ∈ S^{n−1} : 1 − x · e_{n} ≤ δ^{2}}

where e_{n} = (0, …, 0, 1}. Since |x − e_{n}|^{2} = 2(1 − x · e_{n}), it is easy to show that

|x − e_{n}| ≤ C^{−1}δ ⇒ x ∈ Cδ ⇒ |x − e_{n}| ≤ Cδ  (71)

for an appropriate constant C. Now let f = fδ be the indicator function of Cδ. We calculate ∥f∥_{L^{2}(S^{n−1})} and ∥f̂dσ∥_{q}. Constants are of course independent of δ.

In the first place ∥f∥_{2} is the square root of the measure of Cδ so by (71) and the dimensionality of the sphere we have

∥f∥_{2} ≈ δ^{\frac{n−1}{2}}  (72)

The support of fdσ is contained in the rectangle centered at e_{n} with length about δ^{2} in the e_{n} direction and length about δ in the orthogonal directions. We look at f̂dσ on the dual rectangle centered at 0. Suppose then that |ξ_{n}| ≤ C_{1}^{−1}δ^{−2} and that |ξ_{j}| ≤ C_{1}^{−1}δ^{−1} when j < n; here C_{1} is a large constant. Then

|f̂dσ(ξ)| = | ∫_{Cδ} e^{−2πix·ξ}dσ(x)

= | ∫_{Cδ} e^{−2πi(x−e_{n})·ξ}dσ(x)

≥ ∫_{Cδ} cos(2π(x − e_{n}) · ξ)dσ(x)

Our conditions on ξ imply if C_{1} is large enough that |(x − e_{n}) · ξ| ≤ \frac{π}{3}, say, for all x ∈ Cδ. Accordingly

|f̂dσ(ξ)| ≥ \frac{1}{2}|Cδ| ≈ δ^{n−1}

Our set of ξ has volume about δ^{−(n+1)}, so we conclude that

∥f̂dσ∥_{q} ≳ δ^{n−1− \frac{n+1}{q}}

Comparing this estimate with (72) we find that if (70) holds then

δ^{n−1− \frac{n+1}{q}} ≲ δ^{\frac{n−1}{2}}

uniformly in δ ∈ (0, 1]. Hence n − 1 − \frac{n+1}{q} ≥ \frac{n−1}{2}, i.e. q ≥ \frac{2n+2}{n−1}.

Before giving the proof of Theorem 7.1 we need to discuss convolution of a Schwartz function with a measure, since this wasn’t previously considered. Let μ ∈ M(ℝ^{n}); assume μ has compact support for simplicity, although this assumption is not really needed. Define

ϕ ∗ μ(x) = ∫ ϕ(x − y)dμ(y)

Observe that ϕ ∗ μ is C^{∞}, since differentiation under the integral sign is justified as in Lemma 3.1.

It is convenient to use the notation μ̌ for μ̂(−x). We need to extend some of our formulas to the present context. In particular the following extends (28) since if μ ∈ S then the Fourier transform of μ̌ is μ by Theorem 3.4.

ϕ̂μ̌ = ϕ̂ ∗ μ when ϕ ∈ S  (73)
ϕ̂μ = ϕ̂ ∗ μ̂ when ϕ ∈ S  (74)

Notice that (73) can be interpreted naively: Proposition 1.3 and the product rule imply that ϕμ̌ is a Schwartz function. To prove (73), by uniqueness of distributional Fourier transforms it suffices to show that

∫ ψ̂ϕμ̌ = ∫ ψ(ϕ̂ ∗ μ)

if ψ is another Schwartz function which is done as follows; denote T x = −x.

∫ ψ̂(x)ϕ(x)μ̂(−x)dx = ∫ ψ̂(−x)ϕ(−x)μ̂(x)dx
= ∫ (ψ̂ϕ) ∘ T μ̂
= ∫ { \atop (ψ̂ϕ) ∘ T} dμ by the duality relation
= ∫ (ˆψ̂ ∘ T) ∗ { \atop ϕ ∘ T} dμ
= ∫ ψ ∗ (ϕ̂ ∘ T)dμ

= ∫ ∫ ψ(y)ϕ̂(−x + y)dydμ(x)
= ∫ ψ(y)ϕ̂ ∗ μ(y)dy

For (74), again let ψ be another Schwartz function. Then

∫ ϕ̂μψdx = ∫ ψ̂ϕdμ by the duality relation
= ∫ { \atop ϕ̌ ∗ ψ}dμ
= ∫ ϕ̌ ∗ ψμ̂dx by the duality relation
= ∫ ϕ̂ ∗ μ̂ψdx

The last line may be seen by writing out the definition of the convolution and using Fubini’s theorem. Since this worked for all ψ ∈ S we get (74).

‘ \underline{Lemma 7.2} Let f, g ∈ S, let μ be a (say) compactly supported measure, then

∫ f̂\overline{ĝ}dμ = ∫ μ̂ ∗ \overline{g} · fdx  (75)

\underline{Proof} Recall that

ˆg̃ = \overline{ĝ}

so that

ˆ\overline{ĝ} = ˆ ˆ g̃ = \overline{g}

by the inversion theorem. Now apply the duality relation and (74), obtaining

∫ f̂\overline{ĝ}dμ = ∫ f ˆ\overline{ĝ}μdx
= ∫ f · (\overline{g} ∗ μ̂)dx

as claimed. □

\underline{Lemma 7.3} Let μ be a finite positive measure. The following are equivalent for any q and any C.

1. ∥f̂dμ∥_{q} ≤ C∥f∥_{2}, f ∈ L^{2}(dμ)

2. ∥ĝ∥_{L^{2}(dμ)} ≤ C∥g∥_{q′}, g ∈ S

3. ∥μ̂ ∗ f∥_{q} ≤ C^{2}∥f∥_{q′}, f ∈ S \underline{Proof} Let g ∈ S, f ∈ L^{2}(dσ). By the duality relation

∫ ĝfdμ = ∫ f̂dμ · gdx  (76)

If 1. holds then the right side of (76) is ≤ ∥g∥_{q′}∥f̂dμ∥_{q} ≤ C∥g∥_{q′}∥f∥_{2} for any f ∈ L^{2}(μ). Hence so is the left side. This proves 2. by duality. If 2. holds then the left side is ≲ ∥ĝ∥_{L^{2}(dμ)}∥f∥_{L^{2}(dμ)} ≤ C∥g∥_{q′}∥f∥_{L^{2}(dμ)} for g ∈ S. Hence so is the right side. Since S is dense in L^{q′} this proves 1. by duality.

If 3. holds, then the right side of (75) is ≤ C^{2}∥f∥_{q′}^{2} when f = g ∈ S. Hence so is the left side which proves 2. If 2. holds then, for any f, g ∈ S, using also the Schwartz inequality the left side of (75) is ≤ C^{2}∥f∥_{q′}∥g∥_{q′}. Hence the right side of (75) is also ≤ C^{2}∥f∥_{q′}∥g∥_{q′}, which proves 3. by duality. □

\underline{Remark} One can fit lemma 7.2 into the abstract setup

T : L^{2} → L^{q} ⇔ T^{∗} : L^{q′} → L^{2} ⇔ T T^{∗} : L^{q′} → L^{q}

This is the standard way to think about the lemma, although it is technically a bit easier to present the proof in the above ad hoc manner. Namely, If T is the operator f → f̂dσ then T^{∗} is the operator f → f̂, where we regard f̂ as being defined on the measure space associated to μ; and T T^{∗} is convolution with μ̂.

\underline{Proof of Theorem 7.1} We will not give a complete proof; we only prove (70) when q > \frac{2n+2}{n−1} instead of ≥. For the endpoint, see for example [27], [6], [23], [24].

We will show that if q > \frac{2n+2}{n−1}, then

∥σ̂ ∗ f∥_{q} ≤ C_{q}∥f∥_{q′}  (77)

which suffices by Lemma 7.3.

The relevant properties of σ will be

σ(D(x, r)) ≲ r^{n−1}  (78)

which reflects the n − 1-dimensionality of the sphere, and the bound

|σ̂(ξ)| ≲ (1 + |ξ|)^{− \frac{n−1}{2}}  (79)

from Corollary 6.6.

Let ϕ be a C^{∞} function with the following properties:

suppϕ ⊂ \{x : \frac{1}{4} ≤ x ≤ 1\}

If |x| ≥ 1 then ∑_{j≥0} ϕ(2^{−j}x) = 1

Such a function may be obtained as follows: let χ be a C^{∞} function which is equal to 1 when |x| ≥ 1 and to 0 when |x| ≤ \frac{1}{2}, and let ϕ(x) = χ(2x) − χ(x).

We now cut up σ̂ as follows

σ̂ = K_{−∞} + ∑_{j=0}^{∞} K_{j}

where

K_{j}(x) = ϕ(2^{−j}x)σ̂(x)

K_{−∞}(x) = (1 − ∑_{j=0}^{∞} ϕ(2^{−j}x))σ̂

Then K_{−∞} is a C_{0}^{∞} function, so

∥K_{−∞} ∗ f∥_{q} ≲ ∥f∥_{p}

by Young’s inequality, provided q ≥ p.

We now consider the terms in the sum. The logic will be that we estimate convolution with K_{j} as an operator from L^{1} to L^{∞} and from L^{2} to L^{2}, and then we use Riesz-Thorin. We have

∥K_{j}∥_{∞} ≲ 2^{−j \frac{n−1}{2}}

by (78). Using the trivial ∥K_{j} ∗ f∥_{∞} ≤ ∥K_{j}∥_{∞}∥f∥_{1} we conclude our L^{1} → L^{∞} bound,

∥K_{j} ∗ f∥_{∞} ≲ 2^{−j \frac{n−1}{2}} ∥f∥_{1}  (80)

On the other hand, we can use (79) to estimate K̂_{j}. Namely, let ψ = ϕ̂. Note also that σ̂ = σ̌, since σ and therefore σ̂ are invariant under x → −x. Accordingly, we have

K̂_{j} = ψ^{2^{−j}} ∗ σ

using (73) and the fact that ϕ̂ϵ = ϕ̂ϵ. Since ψ ∈ S it follows that

|K̂_{j}(ξ)| ≤ C_{N}2^{jn} ∫ (1 + 2^{j}|ξ − η|)^{−N}dσ(η)

for any fixed N < ∞. Therefore

|K̂_{j}(ξ)| ≤ C_{N}2^{jn} (∫_{D(ξ,2^{−j})}(1 + 2^{j}|ξ − η|)^{−N}dσ(η)

+ ∑_{k≥0} ∫_{D(ξ,2^{k+1−j})\D(ξ,2^{k−j})}(1 + 2^{j}|ξ − η|)^{−N}dσ(η))

≤ C_{N}2^{jn} (σ(D(ξ, 2^{−j})) + ∑_{k≥0} 2^{−N(k−j)}σ(D(ξ, 2^{k+1−j}))

≤ C_{N}2^{jn} (2^{−j(n−1)} + ∑_{k≥0} 2^{−N(k−j)}2^{(n−1)(k−j)})
≲ 2^{j}

where we used (79) at the next to last line, and at the last line we fixed N to be, say, equal to n, and summed a geometric series. Thus

∥K̂_{j}∥_{∞} ≲ 2^{j}  (81)

Now we mention the trivial but important fact that

∥K ∗ f∥_{2} ≤ ∥K̂∥_{∞}∥f∥_{2}  (82)

if say K and f are in S. This follows since

∥K ∗ f∥_{2} = ∥{ \atop K ∗ f}∥_{2}
= ∥K̂f̂∥_{2}
≤ ∥K̂∥_{∞}∥f̂∥_{2}
= ∥K̂∥_{∞}∥f∥_{2}

Combining (81) and (82) we conclude

∥K_{j} ∗ f∥_{2} ≲ 2^{j}∥f∥_{2}  (83)

Accordingly, by (80), (83) and Riesz-Thorin, we have

∥K_{j} ∗ f∥_{q} ≲ 2^{jθ}2^{−j \frac{n−1}{2} (1−θ)}∥f∥_{q′}

if \frac{θ}{2} + \frac{1−θ}{∞} = \frac{1}{q}. This works out to

∥K_{j} ∗ f∥_{q} ≲ 2^{j( \frac{n+1}{q} − \frac{n−1}{2} )}∥f∥_{q′}  (84)

for any q ∈ [2, ∞]. If q > \frac{2n+2}{n−1} then the exponent \frac{n+1}{q} − \frac{n−1}{2} is negative, so we conclude that

∑_{j} ∥K_{j} ∗ f∥_{q} ≲ ∥f∥_{q′}

Since f is a Schwartz function the sum

K_{−∞} ∗ f + ∑_{j} K_{j} ∗ f

may easily seen to converge pointwise to σ̂ ∗ f. We conclude using Fatou’s lemma that ∥σ̂ ∗ f∥_{q} ≲ ∥f∥_{q′}, as claimed. □

\underline{Further remarks} 1.Notice that the L^{2} estimate in the preceding argument was based only on dimensionality considerations. This suggests that there should be an L^{2} bound for f̂dσ valid under very general conditions.

\underline{Theorem 7.4} Let μ be a positive finite measure satisfying the estimate

ν(D(x, r)) ≤ Cr^{α}  (85)

Then there is a bound

∥f̂dν∥_{L^{2}(D(0,R)} ≤ CR^{\frac{n−α}{2}} ∥f∥_{L^{2}(dν)}  (86)

The proof uses the following “generic” test for L^{2} boundedness.

\underline{Lemma 7.5} (Schur’s test) Let (X, μ) and (Y, ν) be measure spaces and K(x, y) a measurable function on X × Y with

∫_{X} |K(x, y)|dμ(x) ≤ A for each y  (87)
∫_{X} |K(x, y)|dν(y) ≤ B for each x  (88)

Define T_{K}f(x) = ∫ K(x, y)f(y)dν(y). Then, for f ∈ L^{2}(dν), the integral defining T_{K}f converges a.e. (dμ(x)) and there is an estimate

∥T_{K}f∥_{L^{2}(dμ)} ≤ \sqrt{AB}∥f∥_{L^{2}(dν)}  (89)

\underline{Proof} It is possible to use Riesz-Thorin here, since (88) implies ∥T_{K}f∥_{∞} ≤ B∥f∥_{∞} and (87) implies ∥T_{K}f∥_{1} ≤ B∥f∥_{1}.

A more “elementary” argument goes as follows. If a and b are positive numbers then we have

\sqrt{ab} = min_{ϵ∈(0,∞)} \frac{1}{2}(ϵa + ϵ^{−1}b)  (90)

since ≤ is the arithmetic-geometric mean inequality and ≥ follows by taking ϵ = \sqrt{\frac{b}{a}}. To prove (89) it suffices to show that if ∥f∥_{L^{2}(dμ)} ≤ 1, ∥g∥_{L^{2}(dν)} ≤ 1, then

∫ ∫ |K(x, y)||f(x)||g(y)|dμ(x)dν(y) ≤ \sqrt{AB}  (91)

To show (91), we estimate

∫ ∫ |K(x, y)||f(x)||g(y)|dμ(x)dν(y) = \frac{1}{2} {min \atop ϵ} (ϵ ∫ ∫ |K(x, y)||g(y)|^{2}dμ(x)dν(y)

+ϵ^{−1} ∫ ∫ |K(x, y)||f(x)|^{2}dν(y)dμ(x))

≤ \frac{1}{2} {min \atop ϵ} (ϵA ∫ |g(y)|^{2}dν(y) + ϵ^{−1}B ∫ |f(x)|^{2}dμ(x))
≤ \frac{1}{2} {min \atop ϵ}(ϵA + ϵ^{−1}B)
= \sqrt{AB}

□

To prove Theorem 7.4 let ϕ be an even Schwartz function which is ≥ 1 on the unit disc and whose Fourier transform has compact support. (exercise: show such a function exists) In the usual way define ϕ_{R^{−1}}(x) = ϕ(R^{−1}x). Then

∥f̂dν∥_{L^{2}(D(0,R))} ≤ ∥ϕ^{R^{−1}}(x)f̂dν(−x)∥_{L^{2}(dx)}
= ∥{ \atop ϕ^{R^{−1}}} ∗ (fdν)∥_{2}

by the duality relation.

The last line is the L^{2}(dx) norm of the function

∫ R^{n}ϕ̂(R(x − y))f(y)dν(y)

We have estimates

∫ |R^{n}ϕ̂(R(x − y))|dx = ∥ϕ̂∥_{1} < ∞

for each fixed y, by change of variables, and

∫ |R^{n}ϕ̂(R(x − y))|dν(y) ≲ R^{n−α}

for each fixed x, by (85) and the compact support of ϕ̂. By Lemma 7.5

∥ ∫ R^{n}ϕ̂(R(x − y))f(y)dν(y)∥_{L^{2}(dx)} ≲ R^{\frac{n−α}{2}} ∥f∥_{L^{2}(dν)}

and the proof is complete. □

2. Another remark is that it is possible to base the whole proof of Theorem 7.1 on the stationary phase asymptotics in section 6, instead of explicitly using the dimensionality of σ. This sort of argument has the obvious advantage that it is more flexible, since it works also in other situations where the “convolution kernel” σ̂(x − y) is replaced by a kernel K(x, y) satisfing appropriate conditions. See for example [21], [23], [24]. On the other hand, it is more complicated and is not as relevant in connection with more delicate questions such as the restriction conjecture, which is known to be false in most of the more general situations (see [4], [18], [22]). We give a brief sketch omitting details. The basic result is the so-called variable coefficient Plancherel theorem, due to Hormander.

Let ϕ be a real valued C^{∞} function defined on ℝ^{n} × ℝ^{n}, let a ∈ C_{0}^{∞}(ℝ^{n} × ℝ^{n}) and consider the “oscillatory integral operators”

Tλf(x) = ∫ e^{−πiλϕ(x,y)}a(x, y)f(y)dy  (92)

Since a has compact support it is obvious that these map L^{2}(ℝ^{n}) to L^{2}(ℝ^{n}) with a norm bound independent of λ, but we want to show that the norm decays in a suitable way as λ → ∞. As with the oscillatory integrals of section 6 this will not be the case if ϕ is too degenerate. In the present situation, note that if ϕ depends on x only, then the factor e^{−πiλϕ} in (92) may be taken outside the integral sign so the norm is independent of λ. Similarly, if ϕ depends on y only then the factor e^{−πiλϕ} may be incorporated into f. We conclude in fact that if ϕ(x, y) = a(x) + b(y) then ∥Tλ∥_{L^{2}→L^{2}} is independent of λ. This strongly suggests that the appropriate nondegneracy condition should involve the “mixed Hessian”

H̃_{ϕ} = (\frac{∂^{2}ϕ}{∂x_{i}∂y_{j}} )_{i,j=1}^{n}

since the mixed Hessian vanishes identically if ϕ(x, y) = a(x) + b(y).

\underline{Theorem A} (Hormander) Assume that

det(H̃_{ϕ}(x, y)) ≠ 0

at all points (x, y) ∈suppa. Then

∥Tλ∥_{L^{2}→L^{2}} ≤ Cλ^{− \frac{n}{2}}

\underline{Sketch of proof} This is evidently related to stationary phase, but one cannot apply stationary phase directly to the integral (92), since f isn’t smooth. Instead, one looks at TλTλ^{∗} which is an integral operator T_{K} with kernel

K(x, y) = ∫ e^{−πiλ(ϕ(x,z)−ϕ(y,z))}a(x, z)\overline{a(y, z)}dz  (93)

The assumption about the mixed Hessian guarantees that the phase function in (93) has no critical points if x and y are close together. Using a version^{3} of “nonstationary phase” one can obtain the estimate

∀N∃C_{N} : |K(x, y)| ≤ C_{N}(1 + λ||x − y|)^{−N}

provided |x − y| is less than a suitable constant. It follows that if a has small support then

∫ |K(x, y)|dy ≲ λ^{−n}

for each fixed x and similarly

∫ |K(x, y)|dx ≲ λ^{−n}

for each y. Then Schur’s test shows that ∥TλTλ^{∗}∥_{L^{2}→L^{2}} ≲ λ^{−n}, so ∥Tλ∥_{L^{2}→L^{2}} ≲ λ^{− \frac{n}{2}} . The small support assumption on a can then be removed using a partition of unity. □.

3One needs something a bit more quantitative than our Proposition 6.1; the necessary lemma is best proved by integration by parts. See for example [23].

It is possible to generalize this to the case where the rank of H̃_{ϕ} is ≥ k, where k ∈ \{1, …, n\}; just replace the exponent \frac{n}{2} by \frac{k}{2}. Furthermore the compact support assumption on a may be replaced by “proper support” (see the statement below), and finally one can obtain L^{q′} → L^{q} estimates by interpolating with the trivial ∥Tλ∥_{L^{1}→L^{∞}} ≤ 1. Here then is the variable coefficient Plancherel, souped up in a manner which makes it applicable in connection with Stein-Tomas. See the references mentioned above.

\underline{Theorem B}(Hormander+??) Assume that a is a C^{∞} function supported on the set {(x, y) ∈ ℝ^{n} × ℝ^{n} : |x − y| ≤ C} and all of whose partial derivatives are bounded. Let ϕ be a real valued C^{∞} function defined on a neighborhood of suppa, all of whose partial derivatives are bounded, and such that the rank of H̃_{ϕ}(x, y) is at least k everywhere, and furthermore the sum of the absolute values of the determinants of the k by k minors of H̃_{ϕ} is bounded away from zero. Then there is a bound

∥Tλ∥_{L^{q′}→L^{q}} ≤ Cλ^{− \frac{k}{q}}

when 2 ≤ q ≤ ∞.

Now look back at the proof we gave for Theorem 7.1. The main point was to obtain the bound (84). Now, K_{j}(x) is the real part of

K̃_{j}(x) =^{def} ϕ(2^{−j}x)a(|x|)e^{−2πi|x|}

where a satisfies the estimates in Corollary 6.6. Accordingly it suffices to prove (84) with K_{j} replaced by K̃_{j}. Let T_{j} be convolution with K̃_{j}, and rescale by 2^{j}; thus we consider the operator

f → T_{j}(f_{2^{j}})_{2^{−j}}

This is an integral operator S_{j} whose kernel is

2^{nj}ϕ(x − y)a(2^{j}|x − y|)e^{−2πi2^{j}|x−y|}

We want to apply Theorem B to S_{j}; toward this end we make the following observations.

(i) From the estimates in Corollary 6.6, we see that the functions

2^{\frac{n−1}{2} j}ϕ(x − y)a(2^{j}|x − y|)

have derivative bounds which are independent of j, and clearly they are supported in \frac{1}{4} ≤ |x − y| ≤ 1.

(ii) The mixed Hessian of the function |x − y| has rank n − 1. This is a calculation which we leave to the reader, just noting that the exceptional direction corresponds to the direction along the line segment \overline{xy}.

It follows that the operators 2^{− \frac{n+1}{2} j}S_{j} satisfy the hypotheses of Theorem B with k = n − 1, uniformly in j, if we take λ = 2^{j+1}. Accordingly

∥S_{j}f∥_{q} ≲ 2^{( \frac{n+1}{2} − \frac{n−1}{q})j}∥f∥_{q′}

and therefore using change of variables

2^{− \frac{n}{q} j}∥T_{j}f∥_{q} ≲ 2^{( \frac{n+1}{2} − \frac{n−1}{q})j}2^{− \frac{n}{q′} j}∥f∥_{q′}

i.e.

∥T_{j}f∥_{q} ≲ 2^{( \frac{n+1}{q} − \frac{n−1}{2} )j}∥f∥_{q′}

which is (84). □

\underline{Exercise}: Use Theorem A for an appropriate phase function, and a rescaling argument of the preceding type to prove the bound

∥f̂∥_{2} ≤ C∥f∥_{2}

This explains the name “variable coefficient Plancherel theorem”.

\underline{8. Hausdorff measures} Fix α > 0, and let E ⊂ ℝ^{n}. If ϵ > 0, then one defines

Hϵ_{α}(E) = inf(∑_{j=1}^{∞} r_{j}^{α})

where the infimum is taken over all countable coverings of E by discs D(x_{j}, r_{j}) with r_{j} < ϵ. It is clear that Hϵ_{α}(E) increases as ϵ decreases, and we define

H_{α}(E) = lim_{ϵ→0} Hϵ_{α}(E)

It is also clear that Hϵ_{α}(E) ≤ Hϵ_{β}(E) if α > β and ϵ ≤ 1; thus

H_{α}(E) is a nonincreasing function of α  (94)

\underline{Remarks} 1. If H_{α}^{1}(E) = 0, then H_{α}(E) = 0. This follows readily from the definition since a covering showing that H_{α}^{1}(E) < δ will necessarily consist of discs of radius < δ^{\frac{1}{α}}.

2. It is also clear that H_{α}(E) = 0 for all E if α > n, since one can then cover ℝ^{n} by discs D(x_{j}, r_{j}) with ∑_{j} r_{j}^{α} arbitrarily small.

\underline{Lemma 8.1} There is a unique number α_{0}, called the Hausdorff dimension of E or dimE, such that H_{α}(E) = ∞ if α < α_{0} and H_{α}(E) = 0 if α > α_{0}.

\underline{Proof} Define α_{0} to be the supremum of all α such that H_{α}(E) = ∞. Thus H_{α}(E) = ∞ if α < α_{0}, by (94). Suppose α > α_{0}. Let β ∈ (α_{0}, α). Define M = 1 + H_{β}(E) < ∞. If ϵ > 0, then we have a covering by discs with ∑_{j} r_{j}^{β} ≤ M and r_{j} < ϵ. So

∑_{j} r_{j}^{α} ≤ ϵ^{α−β} ∑_{j} r_{j}^{β} ≤ ϵ^{α−β}M

which goes to 0 as ϵ → 0. Thus H_{α}(E) = 0. □.

\underline{Further remarks} 1. The set function H_{α} may be seen to be countably additive on Borel sets, i.e. defines a Borel measure. See standard references in the area like [5], [7], [17]. This is part of the reason one considers H_{α} instead of, say, H_{α}^{1}. Notice in this connection that if E and F are disjoint compact sets, then evidently H_{α}(E ∪ F) = H_{α}(E) + H_{α}(F) This statement is already false for H_{α}^{1}.

2. The Borel measure H_{n} coincides with \frac{1}{ω} times Lebesgue measure, where ω is the volume of the unit ball. If α < n, then H_{α} is non-sigma finite; this follows e.g. by Lemma 8.1, which implies that any set with nonzero Lebesgue measure will have infinite H_{α}-measure.

\underline{Examples} The canonical example is the usual \frac{1}{3}-Cantor set on the line. This has a covering by 2^{n} intervals of length 3^{−n}, so it has finite H_{\frac{log 2}{log 3}} -measure. It is not difficult to show that in fact its H_{\frac{log 2}{log 3}} -measure is nonzero; this can be done geometrically, or one can apply Proposition 8.2 below to the Cantor measure. In particular, the dimension of the Cantor set is \frac{log 2}{log 3}.

Now consider instead a Cantor set with variable dissection ratios {ϵ_{n}} i.e. one starts with the interval [0, 1], removes the middle ϵ_{1} proportion, then removes the middle ϵ_{2} proportion of each of the resulting intervals and so forth. If we assume that ϵ_{n+1} ≤ ϵ_{n}, and let ϵ = lim_{n→∞} ϵ_{n}, then it is not hard to show that the dimension of the resulting set E will be \frac{log 2}{log( \frac{2}{1−ϵ})}. In particular, if ϵ_{n} → 0 then dimE = 1.

On the other hand, H_{1}(E) will be positive only if ∑_{n} ϵ_{n} < ∞, so this gives examples of sets with zero Lebesgue measure but “full” Hausdorff dimension.

There are numerous other notions of dimension. We mention only one of them, the Minkowski dimension, which is only defined for compact sets. Namely, if E is compact then let Eδ = {x ∈ ℝ^{n} : dist(x, E) < δ}. Let α_{0} be the supremum of all numbers α such that, for some constant C,

|Eδ| ≥ Cδ^{n−α}

for all δ ∈ (0, 1]. Then α_{0} is called the lower Minkowski dimension and denoted d_{L}(E). Let α_{1} be the supremum of all numbers α such that, for some constant C,

|Eδ| ≥ Cδ^{n−α}

for a sequence of δ’s which converges to zero. Then α_{1} is called the upper Minkowski dimension and denoted D_{U}(E).

It would also be possible to define these like Hausdorff dimension but restricting to coverings by discs all the same size, namely: define a set S to be δ-separated if any two distict points x, y ∈ S satisfy |x − y| > δ. Let Eδ(E) (“δ-entropy on E) be the maximal possible cardinality for a δ-separated subset^{4} of E. Then it is not hard to show that

d_{L}(E) = lim inf_{δ→0} Eδ(E)

d_{U}(E) = lim sup_{δ→0} \frac{log Eδ(E)}{log \frac{1}{δ}}

Notice that a countable set may have positive lower Minkowski dimension; for example, the set \{\frac{1}{n}\}_{n=1}^{∞} ∪ \{0\} has upper and lower Minkowski dimension \frac{1}{2}.

If E is a compact set then let P (E) be the probability measures supported on E.

\underline{Proposition 8.2} Suppose E ⊂ ℝ^{n} is compact. Assume there is μ ∈ P (E) with

μ(D(x, r)) ≤ Cr^{α}  (95)

for a suitable constant C and all x ∈ ℝ^{n}, r > 0. Then H_{α}(E) > 0. Conversely if H_{α}(E) > 0 there is μ ∈ P (E) such that (95) holds.

\underline{Proof} The first part is easy: let \{D(x_{j}, r_{j})\} be any covering of E by discs. Then

1 = μ(E)
≤ ∑_{j} μ(D(x_{j}, r_{j}))
≤ C ∑_{j} r_{j}^{α}

which shows that H_{α}(E) ≥ C^{−1}.

The proof of the converse involves constructing a suitable measure, which is most easily done using dyadic cubes. Thus we let Q_{k} be all cubes of side length ℓ(Q) = 2^{−k} whose vertices are at points of 2^{−k}ℤ^{n}. We can take these to be closed cubes, for definiteness. It is standard to work with these in such contexts, because of their nice combinatorics: if Q ∈ Q_{k} then there is a unique Q̃ ∈ Q_{k−1} with Q ⊂ Q̃; and furthermore if we fix Q_{1} ∈ Q_{k−1} then Q_{1} is the union of those Q ∈ Q_{k} with Q̃ = Q_{1}, and the union is disjoint except for edges. A dyadic cube is a cube which is in Q_{k} for some k.

If Q is a dyadic cube, then clearly there is a disc D(x, r) with Q ⊂ D(x, r) and r ≤ Cℓ(Q). Likewise, if we fix D(x, r) then there are a bounded number of dyadic cubes Q_{1} … Q_{C} with ℓ(Q_{j}) ≤ Cr and whose union contains D(x, r). From these properties, it is easy to see that the definition of Hausdorff measure and also property (95) could equally well be given in terms of dyadic cubes. Thus, except for the values of the constants,

4Exercise: show that Eδ(E) is comparable to the minimum number of δ-discs required to cover E

μ satisfies (95) ⇔ μ(Q) ≤ Cℓ(Q)^{α} for all dyadic cubes Q

Furthermore, if we define

hϵ_{α}(E) = inf( ∑_{Q∈F} ℓ(Q)^{α} : E ⊂ ∪_{Q∈F} Q

where F runs over all coverings of E by dyadic cubes of side ℓ(Q) < ϵ, and

h_{α}(E) = lim_{ϵ→0} hϵ_{α}(E))

then we have

C^{−1}Hϵ_{α}(E) ≤ hϵ_{α}(E) ≤ CHϵ_{α}(E)

therefore

h_{α}(E) > 0 ⇔ H_{α}(E) > 0

We return now to the proof of Proposition 8.2. We may assume that E is contained in the unit cube [0, 1] × … × [0, 1]. By the preceding remarks and Remark 1. above we may assume that h_{α}^{1}(E) > 0, and it suffices to find μ ∈ P (E) so that μ(Q) ≤ Cℓ(Q)^{α} for all dyadic cubes Q with ℓ(Q) ≤ 1. We now make a further reduction.

\underline{Claim} It suffices to find, for each fixed m ∈ ℤ^{+}, a positive measure μ with the following properties

μ is supported on the union of the cubes Q ∈ Q_{m} which intersect E  (96)
∥μ∥ ≥ C^{−1}  (97)
μ(Q) ≤ ℓ(Q)^{α} for all dyadic cubes with ℓ(Q) ≥ 2^{−m}  (98)

Here C is independent of m.

Namely, if this can be done, then denote the measures satisfying (96), (97), (98) by μ_{m}. (98) implies a bound on ∥μ_{m}∥, so there is a weak* limit point μ. (96) then shows that μ is supported on E, (98) shows that μ(Q) ≤ ℓ(Q)^{α} for all dyadic cubes, and (97) shows that ∥μ∥ ≥ C^{−1}. Accordingly, a suitable scalar multiple of μ gives us the necessary probability measure.

There are a number of ways of constructing the measures satisfying (96), (97), (98). Roughly, the issue is that (97) and (98) are competing conditions and one has to find a measure μ with the appropriate support, and with total mass roughly as large as possible given that (97) holds. This can be done for example by using finite dimensional convexity theory (exercise!). We present a different (more constructive) argument taken from [5], chapter 2.

We fix m, and will construct a finite sequence of measures ν_{m}, …, ν_{0}, in that order; ν_{0} will be the measure we want.

Start by defining ν_{m} to be the unique measure with the following properties

1. On each Q ∈ Q_{m}, ν_{m} is a scalar multiple of Lebesgue measure.

2. If Q ∈ Q_{m} and Q ∩ E = ∅, then ν_{m}(Q) = 0.

3. If Q ∈ Q_{m} and Q ∩ E ≠ ∅, then ν_{m}(Q) = 2^{−mα}.

If we set k = m, then ν_{k} has the following properties: it is absolutely continuous to Lebesgue measure and

ν_{k}(Q) ≤ ℓ(Q)^{α} if Q is a dyadic cube with side 2^{−j}, k ≤ j ≤ m  (99)

If Q_{1} is a dyadic cube of side 2^{−k}, then there is a covering F_{Q_{1}} of Q_{1} ∩ E by

dyadic cubes contained in Q_{1} such that ν_{k}(Q_{1}) ≥ ∑_{Q∈F_{Q_{1}}} ℓ(Q)^{α} (100)

Assume now that 1 ≤ k ≤ m and we have constructed an absolutely continuous measure ν_{k} with properties (96), (99) and (100). We will construct ν_{k−1} having these same properties, where in (99) and (100) k is replaced by k − 1. Namely, to define ν_{k−1} it suffices to define ν_{k−1}(Y) when Y is contained in a cube Q ∈ Q_{k−1}. Fix Q ∈ Q_{k−1}. Consider two cases

(i) ν_{k}(Q) ≤ ℓ(Q)^{α}. In this case we let ν_{k−1} agree with ν_{k} on subsets of Q.

(ii) ν_{k}(Q) ≤ ℓ(Q)^{α}. In this case we let ν_{k−1} agree with cν_{k} on subsets of Q, where c is the scalar \frac{2^{−(k−1)α}}{ν_{k}(Q)} .

Notice that ν_{k−1}(Y) ≤ ν_{k}(Y) for any set Y, and furthermore ν_{k−1}(Q) ≤ ℓ(Q)^{α} if Q ∈ Q_{k−1}. These properties and (99) for ν_{k} give (99) for ν_{k−1}, and (96) for ν_{k−1} follows trivially from (96) for ν_{k}. To see (100) for ν_{k−1}, fix Q ∈ Q_{k−1}. If Q was in case (ii), then ν_{k−1}(Q) = ℓ(Q)^{α}, so we can use the covering by the singleton {Q}. If Q was in case (i), then for each of the cubes Q_{j} ∈ Q_{k} whose union is Q we have the covering of Q_{j} ∩ E associated with (100) for ν_{k}. Since ν_{k} and ν_{k−1} agree on subsets of Q, we can simply put these coverings together to obtain a suitable covering of Q ∩ E. This concludes the inductive step from ν_{k} to ν_{k−1}.

We therefore have constructed ν_{0}. It has properties (96), (97) (since for ν_{0} this is equivalent to (99)), and by (100) and the definition of h_{α}^{1} it has property (98). □

Let us now define the α-dimensional energy of a (positive) measure μ with compact support^{5} by the formula

I_{α}(μ) = ∫ ∫ |x − y|^{−α}dμ(x)dμ(y)

We always assume that 0 < α < n.

We also define

V_{μ}^{α}(y) = ∫ |x − y|^{−α}dμ(x)

Thus

I_{α}(μ) = ∫ V_{μ}dμ  (101)

The “potential” V_{μ}^{α} is very important in other contexts (namely elliptic theory, since it is harmonic away from suppμ when α = n − 2) but less important than the energy here. Nevertheless we will use it in a technical way below. Notice that it is actually the convolution of the function |x|^{−α} with the measure μ.

Roughly, one expects a measure to have I_{α}(μ) < ∞ if and only if it satisfies (95); this precise statement is false, but we see below that nevertheless the Hausdorff dimension of a compact set can be defined in terms of the energies of measures in P (E).

\underline{Lemma 8.3} (i) If μ is a probability measure with compact support satisfying (95), then I_{β}(μ) < ∞ for all β < α.

(ii) Conversely, if μ is a probability measure with compact support and with I_{α}(μ) < ∞, then there is another probability measure ν such that ν(X) ≤ 2μ(X) for all sets X and such that ν satisfies (95).

\underline{Proof} (i) We can assume that the diameter of the support of μ is ≤ 1 then

V_{μ}^{β}(x) ≲ ∑_{j=0}^{∞} 2^{jβ}μ(D(x, 2^{−j}))

Accordingly, if μ satisfies (95), and β < α, then

V_{μ}^{β}(x) ≲ ∑_{j=0}^{∞} 2^{jβ}2^{−jα}
≲ 1

It follows by (101) that I_{α}(μ) < ∞.

(ii) Let F be the set of points x such that V_{μ}^{α}(x) ≤ 2I_{α}(μ). Then μ(F) ≥ \frac{1}{2} by (101). Let χ_{F} be the indicator function of F and let dν(x) = \frac{1}{μ(F)}χ_{F} (x)dμ(x). We need to show that ν satisfies (95). Suppose first that x ∈ F. If r > 0 then

r^{−α}ν(D(x, r)) ≤ V_{ν}(x) ≤ 2V_{μ}(x) ≤ 4I_{α}(μ)

5The compact support assumption is not needed; it is included to simplify the presentation.

which verifies (95) when x ∈ F. For general x, consider two cases. If r is such that D(x, r)∩F = ∅ then evidently ν(D(x, r)) = 0. If D(x, r)∩F ≠ ∅, then let y ∈ D(x, r)∩F. Then ν(D(x, r) ≤ ν(D(y, 2r)) ≲ r^{α} by the first part of the proof. □

\underline{Proposition 8.4} If E is compact then the Hausdorff dimension of E coincides with the number

sup(α : ∃μ ∈ P (E) with I_{α}(μ) < ∞)

\underline{Proof} Denote the above supremum by s. If β < s then by (ii) of Lemma 8.3 E supports a measure with μ(D(x, r)) ≤ Cr^{β}. Then by Proposition 8.2, H_{β}(E) > 0, so β ≤dimE. So s ≤dimE. Conversely, if β <dimE then by Proposition 8.2 E supports a measure with μ(D(x, r)) ≤ Cr^{β+ϵ} for some ϵ > 0. Then I_{β}(μ) < ∞, so β ≤ s, which shows that dimE ≤ s. □

The energy is a quadratic expression in μ and is therefore susceptible to Fourier transform arguments. Indeed, the following formula is essentially just Lemma 7.2 combined with the formula for the Fourier transform of |x|^{−α}.

\underline{Proposition 8.5} Let μ be a positive measure with compact support and 0 < α < n. Then

∫ ∫ |x − y|^{−α}dμ(x)dμ(y) = c_{α} ∫ |μ̂(ξ)|^{2}|ξ|^{−(n−α)}dξ  (102)

where c_{α} = \frac{γ( \frac{n−a}{2} )π^{a− \frac{n}{2}}}{γ( \frac{a}{2} )} .

\underline{Proof} Suppose first that f ∈ L^{1} is real and even and that dμ(x) = ϕ(x)dx with ϕ ∈ S. Then we have

∫ f(x − y)dμ(x)dμ(y) = ∫ |μ̂(ξ)|^{2}f̂(ξ)dξ  (103)

This is proved like Lemma 7.1 using (73) instead of (74). Now fix ϕ. Then both sides of (103) are easily seen to define continuous linear maps from f ∈ L^{2} to ℝ. Accordingly,

(103) remains valid when f ∈ L^{1} + L^{2}, ϕ ∈ S. Applying Proposition 4.1, we conclude (102). The analytic continuation argument in section 4 can then be used to remove the assumption a > \frac{n}{2}, so we have proved (102) if dμ(x) = ϕ(x)dx, ϕ ∈ S. To pass to general measures, we use the following fact

\underline{Lemma 8.6} Let ϕ be any radial decreasing Schwartz function with L^{1} norm 1. Then

∫ |x − y|^{−α}ϕ(y)dy ≲ |x|^{−α}

where the implicit constant depends only on α, not on the choice of ϕ.

We sketch the proof as follows: one can easily reduce to the case where ϕ = \frac{1}{|D(0,R)|}χ_{D(0,R)}, and this case can be done by explicit calculation.

Now let ϕ(x) = e^{−π|x|^{2}}. We have then ϕϵ ∗ μ ∈ S, so

∫ ∫ (∫ ∫ |x − y|^{−α}ϕϵ(x − z)ϕϵ(y − w)dxdy) dμ(z)dμ(w) = c_{α} ∫ |μ̂(ξ)|^{2}|ϕ̂(ϵξ)|^{2}|ξ|^{−(n−α)}dξ

(104)

Now let ϵ → 0. On the left side of (104), the expression inside the parentheses converges pointwise to |z − w|^{−α} using a minor variant on Lemma 3.2. If I_{α}(μ) < ∞ then the convergence is dominated in view of the preceding lemma, so the integrals on the left side converge to I_{α}(μ). If I_{α}(μ) = ∞, then this remains true, using Fatou’s lemma. On the right hand side of (104) we can argue similarly: the integrands converge pointwise to μ̂(ξ)|^{2}|ξ|^{−(n−α)}. If ∫ μ̂(ξ)|^{2}|ξ|^{−(n−α)}dξ < ∞ then the convergence is dominated since the factors ϕ̂(ϵξ) are bounded by 1, so the integrals converge to ∫ μ̂(ξ)|^{2}|ξ|^{−(n−α)}dξ. If ∫ μ̂(ξ)|^{2}|ξ|^{−(n−α)}dξ = ∞ then this remains true by Fatou’s lemma. Accordingly, we can pass to the limit from (104) to obtain the lemma. □

\underline{Corollary 8.7} Suppose μ is a measure with

|μ̂(ξ)| ≤ C|ξ|^{−β}  (105)

or more generally that (105) is true in the sense of L^{2} means;

∫_{D(0,N)} |μ̂(ξ)|^{2}dξ ≤ CN^{n−2β}  (106)

Then the dimension of the support of μ is at least 2β.

\underline{Proof} It suffices by Proposition 8.4 to show that if (106) holds then I_{α}(μ) < ∞ for all α < 2β. However,

∫_{|ξ|≥1} |μ̂(ξ)|^{2}|ξ|^{−(n−α)}dξ ≲ ∑_{j=0}^{∞} 2^{−j(n−α)} ∫_{2^{j}≤|ξ|≤2^{j+1}} |μ̂(ξ)|^{2}dξ

≲ ∑_{j=0}^{∞} 2^{−j(n−α)}2^{j(n−2β)}
< ∞

if α < 2β and (106) holds. This completes the proof in view of Proposition 8.5. □

One can ask the converse question, whether a compact set with dimension α must support a measure μ with

|μ̂(ξ)| ≤ Cϵ(1 + |ξ|)^{− \frac{α}{2} +ϵ}  (107)

for all ϵ > 0. The answer is (emphatically) no^{6} Indeed, there are many sets with positive dimension which do not support any measure whose Fourier transform goes to zero as |ξ| → ∞. The easiest way to see this is to consider the line segment E = [0, 1]×{0} ⊂ ℝ^{2}. E has dimension 1, but if μ is a measure supported on E then μ̂(ξ) depends on ξ_{1} only, so cannot go to zero at ∞. If one considers only the case n = 1 this question is related to the classical question of “sets of uniqueness”. See e.g. [20], [30]. One can show for example that the standard \frac{1}{3} Cantor set does not support any measure such that μ̂ vanishes at ∞.

6On the other hand, if one interprets decay in an L^{2} averaged sense the answer becomes yes, because the calculation in the proof of the above corollary is reversible.

Indeed, it is nontrivial to show that a “noncounterxample” exists, i.e. a set E with given dimension α which supports a measure satisfying (107). We describe a construction of such a set due to R. Kaufman in the next section.

As a typical application (which is also important in its own right) we now discuss a special case of Marstrand’s projection theorem. Let e be a unit vector in ℝ^{n} and E ⊂ ℝ^{n} a compact set. The projection P_{e}(E) is the set {x · e : x ∈ ℝ^{n}}. We want to relate the dimensions of E and of its projections. Notice first of all that dimP_{e}E ≤dimE; this follows from the definition of dimension and the fact that the projection P_{e} is a Lipschitz function.

A reasonable example, although not very typical, is a smooth curve in ℝ^{2}. This is one dimensional, and most of its projection will be also one dimensional. However, if the curve is a line, then one of its projections will be just a point.

\underline{Theorem 8.8} (Marstrand’s projection theorem for one dimensional projections) Assume that E ⊂ ℝ^{n} is compact and dimE = α. Then

(i) If α ≤ 1 then for a.e. e ∈ S^{n−1} we have dimP_{e}E = α.

(ii) If α > 1 then for a.e. e ∈ S^{n−1} the projection P_{e}E has positive one dimensional Lebesgue measure.

\underline{Proof} If μ is a measure supported on E, e ∈ S^{n−1} then the projected measure μ_{e} is defined by

∫ fdμ_{e} = ∫ f(x · e)dμ(x)

for continuous f. Notice that μ̂_{e} may readily be calculated from this definition:

μ̂_{e}(k) = ∫ e^{−2πikx·e}dμ(x)
= μ̂(ke)

Let α <dimE, and let μ be a measure supported on E with I_{α}(μ) < ∞. We have then

∫ |μ̂(ke)|^{2}|k|^{−1+α}dkdσ(e)  (108)

by Proposition 8.5 and polar coordinates. Thus, for a.e. e we have

∫ |μ̂(ke)|^{2}|k|^{−1+α}dk < ∞

It follows by Proposition 8.5 with n = 1 that for a.e. e the projected measure μ_{e} has finite α-dimensional energy. This and Proposition 8.4 give part (i), since μ_{e} is supported on the projected set P_{e}E. For part (ii), we note that if dimE > 1 we can take α = 1 in (108). Thus for a.e. e μ̂_{e} belongs to L^{2}. By Theorem 3.13, this condition implies that μ_{e} is absolutely continuous to Lebesgue measure. Accordingly P_{e}E must have positive Lebesgue measure. □

\underline{Remark} Theorem 8.8 has a natural generalization to k-dimensional instead of 1dimensional projections, which is proved in the same way. See [7].

9. Sets with maximal Fourier dimension, and distance sets

A. Sets with maximal Fourier dimension Jarnik’s theorem is the following Proposition 9A1. Fix a number α > 0, and let

E_{α} = \{x ∈ [0, 1] : ∃ infinitely many rationals \frac{a}{q} such that |x − \frac{p}{q}| ≤ q^{−(2+α)}\}

\underline{Proposition 9A1} The Hausdorff dimension of E_{α} is equal to \frac{2}{2+α}.

\underline{Proof} We show only that dimE_{α} ≤ \frac{2}{2+α}. The converse inequality is not much harder (see [7]) but we have no need to give a proof of it since it follows from Theorem 9A2 below using Corollary 8.7.

For the upper bound, consider the set of intervals I_{aq} = (\frac{a}{q} − q^{−(2+α)}, \frac{a}{q} − q^{−(2+α)}), where 0 ≤ a ≤ q are integers. Then

∑_{a,q} |I_{aq}|^{β} ≈ ∑_{q} q · q^{−β(2+α)}

which is finite if β > \frac{2}{2+α}. For any given q_{0} the set \{I_{aq} : q > q_{0}\} covers E_{α}, which therefore has H_{β}(E_{α}) = 0 when β > \frac{2}{2+α}, as claimed. □.

\underline{Theorem 9A2}(Kaufman [14]). For any α > 0 there is a positive measure μ supported on a compact subset of E_{α} such that

|μ̂(ξ)| ≤ Cϵ|ξ|^{− \frac{1}{2+α} +ϵ}  (109)

for all ϵ > 0.

This shows then that Corollary 8.7 is best possible of its type.

The proof is most naturally done using periodic functions so we start with the following general remarks concerning “periodization” and “deperiodization”. Let 𝕋^{n} be the n-torus which we regard as [0, 1] × … × [0, 1] with edges identified; thus a function on 𝕋^{n} is the same as a function on ℝ^{n} periodic for the lattice ℤ^{n}.

If f ∈ L^{1}(𝕋^{n}) then one defines its Fourier coefficients by

f̂(k) = ∫_{𝕋^{n}} f(x)e^{−2πik·x}dx k ∈ ℤ^{n}

and one also makes the analogous definition for measures. If f is smooth then one has |f̂(k)| ≤ C_{N}|k|^{−N} for all N and ∑_{j} f̂(j)e^{2πij·x} = f(x).

Also, if f ∈ L^{1}(ℝ^{n}) one defines its periodization by

f_{per}(x) = ∑_{ν∈ℤ^{n}} f(x − ν)

Then f_{per} ∈ L^{1}(𝕋^{n}), and we have

\underline{Lemma 8A3} If k ∈ ℤ^{n} then f̂_{per}(k) = f̂(k).

\underline{Proof}

f̂(k) = ∫_{ℝ_{n}} f(x)e^{−2πik·x}dx
= ∑_{ν∈ℤ^{n}} ∫_{[0,1]×...×[0,1]−ν} f(x)e^{−2πik·x}dx

= ∑_{ν∈ℤ^{n}} ∫_{[0,1]×...×[0,1]} f(x − ν)e^{−2πik·(x−ν)}dx

= ∫_{[0,1]×...×[0,1]} ∑_{ν∈ℤ^{n}} f(x − ν)e^{−2πik·(x−ν)}dx

= ∫_{[0,1]×...×[0,1]} f_{per}(x)e^{−2πik·x}dx

At the last line we used that e^{−2πik·ν} = 1. □

Suppose now that f is a smooth function on 𝕋^{n}; regard it as a periodic function on ℝ^{n}. Let ϕ ∈ S and consider the function F (x) = ϕ(x)f(x). We have then

F̂(ξ) = ∑_{ν∈ℤ^{n}} f̂(ν) ∫ e^{−2πi(ξ−ν)·x}ϕ(x)dx
= ∑_{ν∈ℤ^{n}} f̂(ν)ϕ̂(ξ − ν)

This formula extends by a limiting argument to the case where the smooth function f is replaced by a measure; we omit this argument^{7}. Thus we have the following: let μ be a measure on 𝕋^{n}, let ϕ ∈ S, and define a measure ν on ℝ^{n} by

7It is based on the fact that every measure on 𝕋^{n} is the weak∗ limit of a sequence of absolutely continuous measures with smooth densities, which is a corollary e.g. of the Stone-Weierstrass theorem.

dν(x) = ϕ(x)dμ({x})  (110)

where {x} is the fractional part of x. Then for ξ ∈ ℝ^{n}

ν̂(ξ) = ∑_{k∈ℤ^{n}} f̂(k)ϕ̂(ξ − k)  (111)

A corollary of this formula by simple estimates with absolutely convergent sums, using the Schwartz decay of ϕ̂, is the following

\underline{Lemma 9A4} If μ is a measure on 𝕋^{n}, satisfying

|μ̂(k)| ≤ C(1 + |k|)^{−α}

for a certain α > 0, and if ν ∈ M(ℝ^{n}) is defined by (110), then

|μ̂(ξ)| ≤ C′(1 + |ξ|)^{−α}

The detailed proof is left to the reader.

We now start to construct a measure on the 1-torus 𝕋, which will be used to prove Theorem 9A2.

Let ϕ be a nonnegative C_{0}^{∞} function on ℝ supported in [−1, 1] and with ∫ ϕ = 1. Define ϕϵ(x) = ϵ^{−1}ϕ(ϵ^{−1}x and let Φϵ be the periodization of ϕϵ.

Let P(M) be the set of prime numbers which lie in the interval (\frac{M}{2} , M]. By the prime number theorem |P(M)| ≈ \frac{M}{log M} for large M. If p ∈ P(M) then the function

Φϵ_{p}(x) =^{def} Φϵ(px) is again 1-periodic^{8} and we have

Φ̂ϵ_{p}(k) = \{ \begin{matrix}ϕ̂(ϵ\frac{k}{p}) & if p | k \\ 0 & otherwise \end{matrix}  (112)

To see this, start from the formula

Φ̂ϵ(k) = ϕ̂ϵ(k) = ϕ̂(ϵk)

which follows from Lemma 9A1. Thus

Φϵ(x) = ∑_{k} ϕ̂(ϵk)e^{2πik·x}

Φϵ_{p}(x) = ∑_{k} ϕ̂(ϵk)e^{2πikp·x}

8Of course for fixed p it is p^{−1}-periodic

which is equivalent to (112).

Now define

F = \frac{1}{|P(M)|} ∑_{p∈P(M)} Φϵ_{p}

F depends of course on ϵ and M but we suppress this dependence.

\underline{Lemma} The Fourier coefficients of F behave as follows:

F̂(0) = 1  (113)
F̂(k) = 0 if 0 < |k| ≤ \frac{M}{2}  (114)

For any N there is C_{N} such that

|F̂(k)| ≤ C_{N} \frac{log |k|}{M}(1 + \frac{ϵ|k|}{M})^{−N} for all k ≠ 0  (115)

\underline{Proof} Both (113) and (114) are selfevident from (112). For (115) we use that a given integer k > 0 has at most \frac{log k}{log M} different prime divisors in the interval [M, 2M). Hence, by (112) and the Schwartz decay of ϕ̂,

|F̂(k)| ≤ \frac{log M}{M} · \frac{log |k|}{log M} · C_{N}(1 + \frac{ϵ|k|}{M})^{−N}

as claimed. □

We now make up our mind to choose ϵ = M^{−(1+α)}, and denote the function F by F_{M}. Thus we have the following

suppF_{M} ⊂ \{x : ∃p ∈ P_{M}, a ∈ [0, p] : |x − \frac{a}{p}| ≤ p^{−(2+α)}  (116)

|F̂_{M}(k)| ≤ C_{N} \frac{log |k|}{M}(1 + \frac{|k|}{M^{2+α}})^{−N}  (117)

and furthermore F_{M} is nonnegative and satisfies (113) and (114).

Notice that a corollary of (109) is that |F̂_{M}(k)| ≲ |k|^{− \frac{1}{2+|α|}} log |k|. In view of (113) we could now try to prove the theorem by taking a weak limit of the measures F_{M}dx and then using Lemma 9A4. However this would not be correct since the set E_{α} is not closed, and ((116) notwithstanding) there is no reason why the weak limit should be supported on E_{α}. Indeed, (114) and (113) imply easily that the weak limit is Lebesgue measure. The following is the standard way of getting around this kind of problem. It has something in common with the classical “Riesz product” construction; wee [?]. We omit details; it is a nice exercise to work them out.

I. First consider a fixed smooth function ψ. Using (mainly) properties (114) and (113), show that if M has been chosen large enough then

|{ \atop ψF_{M}}(k) − ψ̂(k)| ≲ \{ \begin{matrix}\frac{log |k|}{M}(1 + \frac{|k|}{M^{2+α}})^{−100} & when |k| ≥ \frac{M}{4} \\ M^{−100} & when |k| ≤ \frac{M}{4} \end{matrix}

II. Let {M_{k}} be a sequence which goes to infinity extremely rapidly. Consider the functions

G_{m} = Π_{k≤m}F_{M_{k}}

Using step I, and a suitable induction on m, show that if M_{k+1} has been chosen large enough compared with M_{k} then

\frac{1}{2} ≤ Ĝ_{m}(0) ≤ \frac{3}{2}

for each m and that the estimate for F̂_{M} holds also for the G’s, i.e.

|Ĝ_{m}(k)| ≲ |k|^{− \frac{1}{2+α}} log |k|

III. Now let μ be a weak∗ limit point of the sequence {G_{m}dx}. The support of μ will be contained in the intersection of the supports of the {F_{M_{k}}}, so will be a compact subset of E_{α}. From step II, we will have |μ̂(k)| ≲ |k|^{− \frac{1}{2+α}} log |k|. The theorem now follows by Lemma 9A4. □

B. Distance sets If E is a compact set in ℝ^{2} (or in ℝ^{n}) the distance set Δ(E) is defined as follows.

Δ(E) = {|x − y| : x, y ∈ E

One version of Falconer’s distance set problem is the conjecture that

dimE > 1 ⇒ |Δ(E)| > 0

One can think of this as a version of the Marstrand projection theorem where the nonlinear projection (x, y) → |x − y| replaces the linear ones. In fact, it is also possible to make the stronger conjecture that the “pinned” distance sets

{|x − y| : y ∈ E}

should have positive measure for some x ∈ E, or for a set of x ∈ E with large Hausdorff dimension, and this would be analogous to Theorem 8.8 with the nonlinear maps x → (z − y| replacing the projections P_{e}.

Alternately, one can consider this problem as a continuous analogue of a well known open problem in discrete geometry (Erdos’ distance set problem): prove that for finite sets F ⊂ ℝ^{2} there is a bound |Δ(F)| ≥ Cϵ^{−1}|F |^{1−ϵ}, ϵ > 0. The example F = ℤ^{2} ∩D(0, N), N → ∞ can be used to show that in Erdos’ problem one cannot take ϵ = 0, and a related example [8] shows that in Falconer’s problem it does not suffice to assume that H_{1}(E) > 0. The current best result on Erdos’ problem is ϵ = \frac{1}{5} due to Szekely [26] (there were many previous contributions) and on Falconer’s problem the current best result is dimE > \frac{4}{3} due to myself [29] using previous work of Mattila [16] and Bourgain [3].

The strongest results on Falconer’s problem have been proved using Fourier transforms in a manner analogous to the proof of Theorem 8.8. We describe the basic strategy, which is due to Mattila [16]. Namely, observe that if μ is a measure on E, then there is a natural way to put a measure on Δ(E), namely push forward the measure μ × μ by the map Δ : (x, y) → |x − y|. If one can show that the pushforward measure has an L^{2} Fourier transform, then Δ(E) must have positive measure by Theorem 3.13.

In fact one proceeds slightly differently for technical reasons. Namely, if μ is a measure in ℝ^{2} then [16] one associates to it the measure ν defined as follows. Let ν_{0} = Δ(μ × μ), i.e.

∫ fdν_{0} = ∫ f(|x − y|)dμ(x)dμ(y)

Observe that

∫ t^{− \frac{1}{2}} dν(t) = I_{\frac{1}{2}} (μ)

Thus if I_{\frac{1}{2}} (μ) < ∞, as we will always assume, then the measure we now define will be in M(ℝ). Namely, let

dν(t) = e^{i \frac{π}{4}} t^{− \frac{1}{2}} dν_{0}(t) + e^{−i \frac{π}{4}} |t|^{− \frac{1}{2}} dν_{0}(−t)  (118)

\underline{Proposition 9B1} (Mattila [16]) Assume that I_{α}(μ) < ∞ for some α > 1. Then the following are equivalent: (i) ν̂ ∈ L^{2}(ℝ), and (ii) the estimate

∫_{R=1}^{∞}(∫ |μ̂(Re^{iθ})|^{2}dθ)^{2}RdR < ∞  (119)

\underline{Corollary 9B2} [16] Suppose that α > 1 is a number with the following property: if μ is a positive compactly supported measure with I_{α}(μ) < ∞ then

∫ |μ̂(Re^{iθ})|^{2}dθ ≤ C_{μ}R^{−(2−α)}  (120)

Then it follows that any compact set with dimension > α must have a positive measure distance set.

\underline{Proof of the corollary} Assume dimE > α. Then E supports a measure with I_{α}(μ) < ∞. Then

∫_{R=1}^{∞} R(∫ |μ̂(Re^{iθ})|^{2}dθ)^{2}dR ≲ ∫_{R=1}^{∞} R(∫ |μ̂(Re^{iθ})|^{2}dθ)R^{−(2−α)}RdR
< ∞

On the first line, we used (120) to estimate one of the two angular integrals, and the last line then follows by recognizing that the previous line corresponds to the Fourier representation of the energy in Proposition 8.5. It follows by Proposition 9B1 that Δ(E)∪ −Δ(E) supports a measure whose Fourier transform is in L^{2}, and then the corollary follows using Theorem 8.13. □.

At the end of the section we will prove (120) in the easy case α = \frac{3}{2} where it follows from the uncertainty principle; we believe this is due to P. Sjolin. It is known [29] that (120) holds when α > \frac{4}{3}, and this is essentially sharp; (120) fails when α < \frac{4}{3}. The negative result follows from a variant on the Knapp argument (Remark 4. in section 7); this is due to [16], and is presented also in several other places, e.g. [29], but the positive result requires more sophisticated L^{p} type arguments.

Before proving the proposition we recording a few more formulas. Let σ_{R} be angular measure on the circle of radius R centered at zero; thus we are normalizing arclength measure on this circle to have total mass 2π. Let μ be any measure with compact support; then

∫ |μ̂(Re^{iθ})|^{2}dθ = ∫ σ̂_{R} ∗ μdμ  (121)

This is just one more instance of the formula which first appeared in Lemma 7.2 and was used in the proof of Proposition 8.5. This version is contained in Lemma 7.2 if μ has a Schwartz space density, and a limiting argument like the one in the proof of Proposition 8.4 shows it holds for general μ. We also record the asymptotics for σ̂_{R} which may of course be obtained for those for σ_{1} using dilations; notice that the passage from σ_{1} to σ_{R} preserves the total mass, i.e. essentially σ_{R} = (σ_{1})^{R}. We conclude

σ̂_{R}(x) = 2(R|x|)^{− \frac{1}{2}} cos(2π(R|x| − \frac{1}{8})) + O((R|x|)^{− \frac{3}{2}} )  (122)

when R|x| ≥ 1, and |σ̂_{R}| is clearly also bounded independently of R.

\underline{Proof of the proposition} From the definition of ν we have

ν̂(k) = e^{i \frac{π}{4}} ∫ |x − y|^{− \frac{1}{2}} e^{−2πik|x−y|}dμ(x)dμ(y) + e^{−i \frac{π}{4}} ∫ |x − y|^{− \frac{1}{2}} e^{2πik|x−y|}dμ(x)dμ(y)
= 2 ∫ |x − y|^{− \frac{1}{2}} cos(2π(R|x − y| − \frac{1}{8}))dμ(x)dμ(y)

On the other hand, by (121) and (122) we have

∫ |μ̂(ke^{iθ})|^{2}dθ = |k|^{− \frac{1}{2}} ∫ 2|x − y|^{− \frac{1}{2}} cos(2π(R|x − y| − \frac{1}{8}))dμ(x)dμ(y)

+O(∫_{|x−y|≥|k|^{−1}}(|k||x − y|)^{− \frac{3}{2}} dμ(x)dμ(y)) + O(∫_{|x−y|≤|k|^{−1}}(|k||x − y|)^{− \frac{1}{2}} The last error term arises by comparing σ̂_{R}, which is bounded, to the main term on the righthand side of (122), which is O((R|x|)^{− \frac{1}{2}} , in the regime R|x| < 1. We may combine the two error terms to obtain

∫ |μ̂(ke^{iθ})|^{2}dθ = |k|^{− \frac{1}{2}} ∫ 2|x − y|^{− \frac{1}{2}} cos(2π(R|x − y| − \frac{1}{8}))dμ(x)dμ(y)
+O(∫ (|k||x − y|)^{−α}dμ(x)dμ(y))

for any α ∈ [\frac{1}{2}, \frac{3}{2}]. Therefore

ν̂(k) = |k|^{\frac{1}{2}} ∫ |μ̂(ke^{iθ})|^{2}dθ + O(|k|^{\frac{1}{2} −α}I_{α}(μ))

The error term here is evidently bounded by |k|^{\frac{1}{2} −α}I_{α}(μ) for any α ∈ (1, \frac{3}{2}), and therefore belongs to L^{2}(|k| ≥ 1). We conclude then that ν̂ belongs to L^{2} on |k| ≥ 1 if and only if |k|^{\frac{1}{2}} ∫ |μ̂(ke^{iθ})|^{2}dθ ∈ L^{2}(dk) does. This gives the proposition, since ν̂ (being the Fourier transform of a measure) clearly belongs to L^{2} on [−1, 1]. □

\underline{Proposition 9B3} If α ≥ 1 and if μ is a positive measure with compact support^{9} then

∫ |μ̂(Re^{iθ})|^{2}dθ ≲ I_{α}(μ)R^{−(α−1)}

where the implicit constant depends on a bound for the radius of a disc centered at 0 which contains the support of μ. In particular, (120) holds if α = \frac{3}{2}.

\underline{Corollary 9B4} (originally due to Falconer [8] with a different proof) If dimE > \frac{3}{2} then the distance set of E has positive measure.

\underline{Proofs} The corollary is immediate from the proposition and Corollary 9B2. The proof of the proposition is very similar to the proofs of Bernstein’s inequality and of Theorem 7.4. We can evidentlya ssume R is large. Let ϕ be a C_{0}^{∞} function whose Fourier transform is ≥ 1 on the support of μ. Let dν(x) = \frac{1}{ψ(x)}dμ(x). Then it is obvious (from the definition, not the Fourier representation) that I_{α}(ν) ≤ I_{α}(μ). Also μ̂ = ϕ ∗ ν̂. Accordingly

∫ |μ̂(Re^{iθ}|^{2}dθ = ∫ ϕ ∗ ν̂(Re^{iθ})|^{2}dθ
≲ ∫ |ϕ(Re^{iθ} − x)|ν̂(x)|^{2}dxdθ
= ∫ |ν̂(x)|^{2} ∫ |ϕ(x − Re^{iθ})|dθdx
≲ R^{−1} ∫_{| |x|−R|≤C} |ν̂(x)|^{2}dx

9Here, as opposed to in some previous situations, the compact support is important.

≲ R^{−1+2−α} ∫ |x|^{−(2−α)}ν̂(x)|^{2}dx
≈ R^{1−α}I_{α}(μ)

Here the second line follows by writing

|ϕ ∗ ν̂(Re^{iθ})| ≤ ∫ \sqrt{|ϕ(Re^{iθ} − x)|} · \sqrt{|ϕ(Re^{iθ} − x)|}|ν̂(x)|dx

and applying the Schwartz inequality. The fourth line follows since for fixed x the set of θ where ϕ(x − Re^{iθ}) ≠ 0 has measure ≲ R^{−1}, and is empty if |x| − R is large. The proof is complete. □

\underline{Remark} The exponent α − 1 is of course far from sharp; the sharp exponent is \frac{α}{2} if α > 1, \frac{1}{2} if α ∈ [\frac{1}{2}, 1] and α if α < \frac{1}{2}.

\underline{Exercise} Prove this in the case α ≤ 1. (This is a fairly hard exercise.)

\underline{Exercise} Carry out Mattila’s construction (formula (118) and the preceding discussion) to the case where μ is a measure in ℝ^{n} instead of ℝ^{2}, and prove analogues of Proposition 9A1, Corollary 9A2, Proposition 9A3. Conclude that a set in ℝ^{n} with dimension greater than \frac{n+1}{2} has a positive measure distance set. (See /citeMattila. The dimension result is also due originally to Falconer. The conjectured sharp exponent is \frac{n}{2})

References

[1] W. Beckner, Inequalities in Fourier analysis, Ann. of Math. 102 (1975), 159-182.

[2] J. Bourgain, Besicovitch type maximal operators and applications to Fourier analysis, Geometric and Functional Analysis 1 (1991), 147-187.

[3] J. Bourgain, Hausdorff dimension and distance sets, Israel J. Math 87(1994), 193-201.

[4] J. Bourgain, \underline{L^{p} estimates for oscillatory integrals in several variables}, Geometric and Functional Analysis 1(1991), 321-374.

[5] L. Carleson, \underline{Selected Problems on Exceptional Sets}

[6] K.M. Davis, Y.C. Chang, \underline{Lectures on Bochner-Riesz Means}

[7] K. J. Falconer, \underline{The geometry of fractal sets}, Cambridge University Press, 1985.

[8] K. J. Falconer, On the Hausdorff dimension of distance sets, Mathematika 32(1985), 206-212.

[9] C. Fefferman, Inequalities for strongly singular convolution operators, Acta Math. 124(1970), 9-36.

[10] G. Folland, \underline{Harmonic Analysis in Phase Space}

[11] V. Havin, B. Joricke, \underline{The Uncertainty Principle in Harmonic Analysis}, Springer-Verlag, 1994.

[12] L. Hormander, \underline{The Analysis of Linear Partial Differential Operators}, volume 1, 2nd edition, Springer Verlag 1990.

[13] Y. Katznelson, \underline{An Introduction to Harmonic Analysis}

[14] R. Kaufman, On the theorem of Jarnik and Besicovitch, Acta Arithmetica 39(1981), 265-267.

[15] J. E. Marsden, M. J. Hoffman, \underline{Elementary Classical Analysis}

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

[17] P. Mattila, \underline{Geometry of sets and measures in Euclidean spaces}, Cambridge University Press, 1995.

[18] W. Minicozzi, C. Sogge, Negative results for Nikodym maximal functions and related oscillatory integrals in curved space, . Math. Res. Lett. 4 (1997), 221–237.

[19] W. Rudin, \underline{Functional Analysis}

[20] R. Salem, \underline{Algebraic Numbers and Fourier Analysis}

[21] C. Sogge, \underline{Fourier integrals in Classical Analysis}, Cambridge University Press.

[22] C. Sogge, \underline{Concerning Nikodym-type sets in 3-dimensional curved space}, Journal Amer. Math. Soc., 1999.

[23] E.M. Stein, in \underline{Beijing Lectures in Harmonic Analysis}, edited by E. M. Stein.

[24] E. M. Stein, \underline{Harmonic Analysis}, Princeton University Press 1993.

[25] E. M. Stein, G. Weiss, \underline{An Introduction to Fourier Analysis on Euclidean Spaces} [26] L. Szekely, Crossing numbers and hard Erdos problems in discrete geometry, Comb.

Prob. Comput. 6(1997), 353-358.

[27] P. A. Tomas, A restriction theorem for the Fourier transform, Bull. Amer. Math. Soc. 81(1975), 477-478.

[28] A. Varchenko, Newton polyhedra and estimations of oscillatory integrals, Funct. Anal. Appl. 18(1976), 175-196.

[29] T. Wolff, Decay of circular means of Fourier transforms of measures, International Mathematics Research Notices 10(1999), 547-567.

[30] A. Zygmund, \underline{Trigonometric Series} 70