Analysis for PhD students (2013)

Assignment 1

Problem 1. Let 0 < ω1 ≤ ω2 ≤ · · · be an increasing sequence ofpositive numbers satisfying

#{n ∈ N : ωn ≤ T} = cT 2 +O(T ) ∀T > 0,(1)

where c > 0 is some constant. Let α ≤ 2. Determine an asymptoticformula for

∑ωn<T ω−α

n as T → ∞.(10p)

Problem 2. Let n ∈ N and κ ∈ R. A vector x ∈ Rn is said to be ofDiophantine type κ if there exists some c > 0 such that for all k ∈ Zn

and q ∈ N we have |x− q−1k| > cq−κ. Prove that if κ > 1 + n−1 thenalmost every x ∈ Rn (w.r.t. Lebesgue measure) is of Diophantine typeκ.

(15p)

Problem 3. Let f be a bounded real-valued function on [a, b]. Provethat f is Riemann integrable iff {x ∈ [a, b] : f is discontinuous at x}has Lebesgue measure zero. (Hint: This is Folland’s exercise 2:23;note that Folland gives an outline of a proof in his formulation of theexercise.)

(15p)

Problem 4. Let µ be the Borel measure on R which is given by µ = δ+m1, where δ is the Dirac measure at 0 and m1 is the Lebesgue measurerestricted to [0, 1] (viz., δ(E) = I(0 ∈ E) and m1(E) = m(E ∩ [0, 1])for any Borel subset E ⊂ R, where I(·) is the indicator function.) Letµ2 be the product measure µ× µ on R2.(a). Find the Lebesgue decomposition of µ2 with respect to Lebesguemeasure on R2.(b). Give a formula for µ̂2(ξ), ξ ∈ R2.(c). Describe µ ∗ µ (which is a Borel measure on R) explicitly.

(15p)

Problem 5. For the the following two sequences {µN} inM(R2), provethat there is a measure µ ∈ M(R2) such that µN → µ in the weak*topology on M(R2) = C0(R

2)∗, and describe µ explicitly.

(a). µN given by∫R2 f dµN = N−1

∑Nk=1 f(

kN, 0), ∀f ∈ C0(R

2).

(b). µN given by∫R2 f dµN = N−2

∑N−1k=0

∑N−1m=0 f(

kN, kmN2 ), ∀f ∈ C0(R

2).(10p)

1

2

The next few exercises concern the rate of decay of χ̂E for varioussets E ⊂ Rn.

Problem 6. Prove that if E = [a1, b1]× · · · × [an, bn] is any box in Rn

then along certain rays χ̂E(ξ) does not decay faster than |ξ|−1, whilealong other rays χ̂E(ξ) decays as fast as |ξ|−n.(In more precise terms: Prove that there exist some ξ ∈ Rn \ {0},c > 0 and a sequence 0 < u1 < u2 < . . . such that limm→∞ um = ∞and |χ̂E(umξ)| > cu−1

m for all m. Also prove that there exist someξ ∈ Rn \ {0} and C > 0 such that |χ̂E(uξ)| < Cu−n for all u ≥ 1.)

(10p)

Problem 7. For E ⊂ Rn and δ > 0 we define ∂δE as the (open) setof all points in Rn which have distance < δ to some point in ∂E. Nowassume that there exist C > 0 and 0 < a ≤ 1 such that m(∂δE) < Cδa

for all 0 < δ ≤ 1. Then prove that E is Lebesgue measurable, and iffurthermore E is bounded then there is a constant K > 0 such that|χ̂E(ξ)| ≤ K|ξ|−a for all ξ ∈ Rn \ {0}. [Hint: Consider the Fouriertransform of χE − χη+E , where η is a suitably chosen vector in Rn.]

(10p)

Problem 8. Let B be a fixed ball in Rn. Prove that there is a constantC > 0 such that |χ̂B(ξ)| ≤ C(1 + |ξ|)− 1

2(n+1) for all ξ ∈ Rn.

[Hint: We here give an outline of a possible proof. Partial credit willbe given for carrying out one or some of these steps. (1) Using in-variance under rotations and Fubini’s theorem, prove that χ̂B(ξ) =∫ 1

−1f(x)e−2πi|ξ|x dx, where f(x) is the volume of a ball of radius

√1− x2

in Rn−1. Explicitly, f(x) = Vn−1(1−x2)12(n−1) where Vn−1 is the volume

of the (n − 1)-dimensional unit ball. (2) Prove that for k < 12(n + 1)

the kth derivative of f(x) equals∑k

j=1 Pj,k,n(x)(1− x2)12(n−1)−j , where

Pj,k,n(x) is a polynomial of degree ≤ j. (3) Now integrate by partsrepeatedly in the formula for χ̂B(ξ). For n odd there are no problemsto integrate by parts 1

2(n+1) times and this leads to the desired bound.

(4) For n even we integrate by parts 12n times; then split the range of

integration into (−1,−1+h), (−1+h, 1−h), (1−h, 1) for an appropri-ate h ∈ (0, 1), and apply integration by parts once more for the middleregion; the desired bound can now be deduced.]

(15p)

3

1. Set

A(T ) = #{ωn ≤ T} for T > 0.

Then for any T > ω1 we have

∑

ωn<T

ω−αn =

∫ T−

ω1/2

x−α dA(x) =[x−αA(x)

]x=T−

x=ω1/2+ α

∫ T−

ω1/2

x−α−1A(x) dx

= T−αA(T−) + α

∫ T

ω1/2

x−α−1A(x) dx

= T−α(cT 2 +O(T )

)+ α

∫ T

ω1/2

x−α−1(cx2 +O(x)

)dx.

If α < 2 then this is

= cT 2−α +O(T 1−α) +cα

2− αT 2−α +

O(1) if α > 1

O(log T ) if α = 1

O(T 1−α) if α < 1

=2c

2− αT 2−α +

O(1) if α > 1

O(log T ) if α = 1

O(T 1−α) if α < 1.

(We here accept that the implied constant depends on “everythingexcept T”! Note that the error in the last term incorporates both thecontribution from the “O(x)” inside the integrand and the contribution

from the lower endpoint of integration;∫ T

ω1/2x1−α dx = (2−α)−1(T 2−α−

(ω1/2)2−α) = (2− α)−1T 2−α +O(1).)

On the other hand if α = 2 then we get

= c+O(T−1) + 2c log T +O(1) = 2c log T +O(1).

Answer: We have for all T > ω1 (or equivalently, for all T > 1):

∑

ωn<T

ω−αn =

{2c

2−αT 2−α if α < 2

2c log T if α = 2

}+

O(1) if α > 1

O(log T ) if α = 1

O(T 1−α) if α < 1.

The implied constant depends on α and {ωn} (but of course not on T ).

4

2. Assume κ > 1 + n−1. Let E be the set of all x ∈ Rn whichare not of Diophantine type κ. We wish to prove that E (is Lebesguemeasurable and) has Lebesgue measure zero, i.e. m(E) = 0. Note thatfor any ℓ ∈ Zn, x has Diophantine type κ iff x+ℓ has Diophantine typeκ. Hence

E =⋃

ℓ∈Zn

(ℓ+ (E ∩ [0, 1]n)

),

and since this is a countable union and

m(ℓ+ (E ∩ [0, 1]n)) = m(E ∩ [0, 1]n), ∀ℓ ∈ Zn,

it now suffices to prove that (E ∩ [0, 1]n is Lebesgue measurable and)m(E ∩ [0, 1]n) = 0.Next note that by definition of “Diophantine type κ” we have

E ∩ [0, 1]n =⋂

c>0

Dc(2)

where

Dc = {x ∈ [0, 1]n : there exist k ∈ Zn, q ∈ N such that |x− q−1k| ≤ cq−κ}.Note also that Dc ⊂ Dc′ whenever 0 < c < c′; hence (2) may equiva-lently be expressed as the countable intersection

E ∩ [0, 1]n =∞⋂

j=1

D1/j .(3)

Hence it suffices to prove that each set D1/j is Lebesgue measurable,and limj→∞m(D1/j) = 0. We will in fact prove that Dc is Lebesguemeasurable for each c > 0, and limc→0m(Dc) = 0.Next note that

Dc =⋃

q∈N

⋃

k∈Zn

Dc,q,k,(4)

where

Dc,q,k = {x ∈ [0, 1]n : |x− q−1k| ≤ cq−κ} = [0, 1]n ∩ Bq−1k(cq−κ).(5)

(Here we write By(r) ⊂ Rn for the closed ball of radius r centered aty ∈ Rn.) The formula (4) together with (5) proves that Dc is a Borelset (in fact a countable union of closed sets); hence also E is a Borelset.It now only remains to prove limc→0m(Dc) = 0. Let us assume

0 < c < 1. In (4), note that Dc,q,k = ∅ whenever q−1k has distance> cq−κ to [0, 1]n. Thus for any q ∈ N and k ∈ Zn we certainly haveDc,q,k = ∅ whenever q−1k lies outside the open cube (−1, 2)n (since

5

then the distance from q−1k to [0, 1]n is ≥ 1 > cq−κ). Hence we canrewrite the above formula for Dc as follows:

Dc =⋃

q∈N

⋃

k∈Zn∩(−q,2q)n

Dc,q,k,

Hence by Folland’s Theorem 1.8(b),

m(Dc) ≤∑

q∈N

∑

k∈Zn∩(−q,2q)n

m(Dc,q,k) ≤∑

q∈N

∑

k∈Zn∩(−q,2q)n

m(Bq−1k(cq−κ))

=∑

q∈N

#(Zn ∩ (−q, 2q)n

)· (cq−κ)nVn,

where Vn = m(B0(1)), the volume of the unit ball in Rn. (In the laststep we used Folland’s Theorem 2.44(b), with T =the diagonal matrixwith all entries= cq−κ.) We may continue:

≤ Vncn∑

q∈N

(3q)nq−κn = Vn(3c)n∑

q∈N

q(1−κ)n.

Recall that we are assuming κ > 1+n−1; hence (1−κ)n < −1 and thisimplies that the sum

∑q∈N q

(1−κ)n converges. We have thus proved:

m(Dc) ≤ Kncn,

where Kn is a positive constant which only depends on n. This implieslimc→0m(Dc) = 0, and we are done. �

6

3. We follow Folland’s hint. Thus we define the functions h : [a, b] →R and H : [a, b] → R by

H(x) = limδ→0+

sup|y−x|≤δ

f(y) ∈ R, h(x) = limδ→0+

inf|y−x|≤δ

f(y) ∈ R.

Note that for each x both these limits really exist in R, since f isa bounded function, and sup|y−x|≤δ f(y) is decreasing w.r.t. δ, andinf |y−x|≤δ f(y) is increasing w.r.t. δ. We also note:

h(x) ≤ f(x) ≤ H(x), ∀x ∈ [a, b].(6)

Lemma 1. For each x ∈ [a, b], we have H(x) = h(x) iff f is continuousat x.

Proof. Assume that f is continuous at x. Let ε > 0 be given. Thenthere is some η > 0 such that f(x)−ε ≤ f(y) ≤ f(x)+ε for all y ∈ [a, b]with |y − x| ≤ η. It follows that for each δ with 0 < δ ≤ η we havesup|y−x|≤δ f(y) ≤ f(x) + ε and inf |y−x|≤δ f(y) ≥ f(x) − ε. ThereforeH(x) ≤ f(x) + ε and h(x) ≥ f(x) − ε. Using also (6) this implies|H(x)− h(x)| = H(x) − h(x) ≤ 2ε. Since this is true for every ε > 0we conclude H(x) = h(x).Conversely, assume H(x) = h(x). Let ε > 0 be given. By the

definition of H(x) and h(x), there is some δ > 0 such that |H(x) −sup|y−x|≤δ f(y)| ≤ ε and |h(x) − inf |y−x|≤δ f(y)| ≤ ε. Therefore, usingH(x) = h(x) and the triangle inequality, we have | sup|y−x|≤δ f(y) −inf |y−x|≤δ f(y)| ≤ 2ε. Now for every z ∈ [a, b] with |z − x| ≤ δ we have

f(z)− f(x) ≤(

sup|y−x|≤δ

f(y))− f(x) ≤ sup

|y−x|≤δ

f(y)− inf|y−x|≤δ

f(y) ≤ 2ε,

and similarly f(z)−f(x) ≥ −2ε. Since such a δ can be found for everyε, we conclude that f is continuous at x. �

Recall that we are only assuming that f is a bounded function[a, b] → R; we are not assuming that f is Riemann-integrable. Still,by an entirely similar argument as in the proof of Folland’s Theorem2.28(a) (see my notes, Section 3.7 for some more details regarding thatproof), we can choose a sequence P1, P2, P3, . . . of partitions of [a, b]such that Pk is a subsequence of Pk+1 for each k ∈ N, limk→∞mesh(Pk) =0, and

limk→∞

SPkf = I

b

a(f), limk→∞

sPkf = Iba(f).(7)

(Some more details on how to construct such a sequence {Pk}: First

find a sequence {P ′k} with limk→∞ SP ′

kf = I

b

a(f) and another sequence

{P ′′k } with limk→∞ sP ′′

kf = Iba(f). Then let Pk be the merge of P ′

k and

7

P ′′k , for each k. Since inserting more points in a partition decreases

SPf and increases sPf (equality allowed), the limit relations (7) nowhold. Next by inserting (if necessary) even more points in P2, then inP3, then in P4, etc, we can arrange so that also Pk is a subsequence ofPk+1 for each k ∈ N, and limk→∞mesh(Pk) = 0.) As in the proof ofTheorem 2.28(a), for any partition P of [a, b] we define the functionsGP : [a, b] → R and gP : [a, b] → R through

GP = M1χ[t0,t1] +

n∑

j=2

Mjχ(tj−1,tj ], gP = m1χ[t0,t1] +

n∑

j=2

mjχ(tj−1,tj ]

(8)

(where Mj = sup{f(t) : t ∈ [tj−1, tj ]}, mj = inf{f(t) : t ∈ [tj−1, tj]}).Then, as in the proof of Theorem 2.28(a), the limits

G(x) = limk→∞

GPk(x) and g(x) = lim

k→∞gPk

(x)

exist for every x ∈ [a, b]; and this definition make G and g into functionsin L1([a, b], m) satisfying

∫

[a,b]

Gdm = limk→∞

∫

[a,b]

GPkdm = lim

k→∞SPk

f = Ib

a(f)(9)

and∫

[a,b]

g dm = limk→∞

∫

[a,b]

gPkdm = lim

k→∞sPk

f = Iba(f).(10)

Lemma 2. H = G a.e. and h = g a.e.

Proof. We will prove that H(x) = G(x) and h(x) = g(x) hold for everyx ∈ [a, b] which does not belong to any of the sequences P1, P2, . . ..This implies that H(x) = G(x) for every x ∈ [a, b] outside a countableset, and this gives the claim of the lemma since every countable set hasLebesgue measure zero.Thus assume that x ∈ [a, b] and that x does not belong to any of the

sequences P1, P2, . . .. Let ε > 0 be given. Then by the definitions ofH(x) and h(x) there is some η > 0 such that

sup|y−x|≤η

f(y) ≤ H(x) + ε, inf|y−x|≤η

f(y) ≥ h(x)− ε.

Recall that limk→∞mesh(Pk) = 0; thus there is some k0 ∈ N such thatmesh(Pk) ≤ η for all k ≥ k0.Now consider any k ≥ k0. Writing Pk = {tj}nj=0 we choose j ∈

{1, 2, . . . , n} so that tj−1 < x < tj (such a j exists since by assumption

8

x /∈ {t0, t1, . . . , tn}). Then by the definitions of GPkand gPk

we have

GPk(x) = sup

y∈[tj−1,tj ]

f(y) and gPk(x) = inf

y∈[tj−1,tj ]f(y).

Hence sup|y−x|≤δ f(y) ≤ GPk(x) and inf |y−x|≤δ f(y) ≥ gPk

(x) for allδ < min(x − tj−1, tj − x), and therefore H(x) ≤ GPk

(x) and h(x) ≥gPk

(x). On the other hand |tj − tj−1| ≤ mesh(Pk) ≤ η and hence[tj−1, tj] ⊂ [x− η, x+ η] and

H(x) ≤ GPk(x) = sup

y∈[tj−1,tj ]

f(y) ≤ sup|y−x|≤η

f(y) ≤ H(x) + ε(11)

and

h(x) ≥ gPk(x) = inf

y∈[tj−1,tj ]f(y) ≥ inf

|y−x|≤ηf(y) ≥ h(x)− ε.(12)

The bounds (11) and (12) hold for all k ≥ k0, and such a k0 existsfor any ε > 0. Hence G(x) = limk→∞GPk

(x) = H(x) and g(x) =limk→∞ gPk

(x) = h(x). �

Since G is bounded and measurable and m is complete, Lemma 2implies that H ∈ L1([a, b], m) and

∫[a,b]

H dm =∫[a,b]

Gdm; similarly

h ∈ L1([a, b], m) and∫[a,b]

h dm =∫[a,b]

g dm. Hence in view of (9) and

(10) we have∫

[a,b]

H dm = Ib

a(f) and

∫

[a,b]

h dm = Iba(f).

Recall that f is Riemann-integrable iff Ib

a(f) = Iba(f); the aboveshows that this holds iff

∫[a,b]

(H − h) dm = 0, and by (6) and Fol-

land’s Prop. 2.16 this holds iff H = h a.e. By Lemma 1 this holds iff{x ∈ [a, b] : f is discontinuous at x} has Lebesgue measure zero.Summing up, we have proved that f is Riemann-integrable if and

only if {x ∈ [a, b] : f is discontinuous at x} has Lebesgue measurezero. � �

9

4. (a). We have

µ2 = (δ +m1)× (δ +m1)

= (δ × δ) + (δ ×m1) + (m1 × δ) + (m1 ×m1) = λ+ ν,(13)

where

λ = (δ × δ) + (δ ×m1) + (m1 × δ)

and

ν = m1 ×m1.

[In the above computation we used the fact that products of measuressatisfy a certain distributive law; this is an immediate consequence of(e.g.) Folland’s Theorem 2.36; we give the details below in Lemma 3.]Let m2 be Lebesgue measure on R2. We claim that λ ⊥ m. Indeed,

set

E = ([0, 1]× {0}) ∪ ({0} × [0, 1]) ⊂ R2

Then m2(E) = 0 and λ(R2 \ E) = 0. This proves that λ ⊥ m2.Next note that for any Borel subset E ⊂ R2 we have

ν(E) = m2(E ∩R),(14)

where R is the closed unit square, R = [0, 1] × [0, 1]. (One wayto prove (14) is to use Folland’s Theorem 2.36, which gives ν(E) =∫Rm1(Ex) dm1(x) =

∫[0,1]

m1(Ex ∩ [0, 1]) dm1(x), where we now write

m1 for the Lebesgue measure on R. Set E ′ = E∩R; then E ′x = Ex∩[0, 1]

for x ∈ [0, 1] while E ′x = ∅ for x ∈ R \ [0, 1]; hence the last integral is

=∫Rm1(E ′

x) dm1(x) = m2(E ′) = m2(E ∩ R), as claimed.)

The formula (14) immediately implies that ν ≪ m2.Hence µ = λ+ν as above gives the Lebesgue decomposition of µ wrt

m2.

In the above proof we used the following distributive law:

Lemma 3. Let µ1 and µ2 be two σ-finite measures on a measurablespace (X,M). Let also (Y,N , ν) be a σ-finite measure space. Then(µ1+µ2)×ν = (µ1×ν)+(µ2×ν) (an identity of measures on M⊗N ).

Proof. This is immediate from the definition of product measures (cf.Folland Sec. 2.5). Indeed, clearly (µ1+ µ2)× ν and (µ1 × ν) + (µ2 × ν)agree on the algebra A of finite disjoint unions of rectangles. Hence byFolland’s Theorem 1.14 they agree on the σ-algebra generated by A;that is, they agree on M⊗N . (Alternatively, note that the lemma isalso an immediate consequence of Folland’s Theorem 2.36). �

10

(b). For any ξ = (ξ1, ξ2) ∈ R2,

µ̂2(ξ) =

∫

R2

e−2πiξ·x dµ2(x).

Note that µ2 = µ× µ is a positive measure, and µ2 is finite: µ2(R2) =

µ× µ(R×R) = µ(R)2 = 4. Furthermore the function x 7→ e−2πiξ·x hasabsolute value everywhere bounded by 1. Hence

∫R2 |e−2πiξ·x| dµ2(x) <

∞, and therefore we can apply Fubini’s theorem (Folland’s Theorem2.37(b)) as follows:

µ̂2(ξ) =

∫

R2

e−2πiξ·x dµ2(x)

=

∫

R

(∫

R

e−2πi(ξ1x1+ξ2x2) dµ(x2)

)dµ(x1)

=2∏

j=1

∫

R

e−2πiξjxj dµ(xj)

=2∏

j=1

(1 +

∫ 1

0

e−2πiξjxj dxj

)

=

2∏

j=1

(1 +

{e−2πiξj−1−2πiξj

if ξj 6= 0

1 if ξj = 0

})

=

(1 + 1−e−2πiξ1

2πiξ1)(1 + 1−e−2πiξ2

2πiξ2) if ξ1, ξ2 6= 0

2(1 + 1−e−2πiξ1

2πiξ1) if ξ1 6= 0, ξ2 = 0

2(1 + 1−e−2πiξ2

2πiξ2) if ξ1 = 0, ξ2 6= 0

4 if ξ1 = ξ2 = 0.

11

(c). Recall that the Dirac measure is an identity element for convo-lution (acting on arbitrary complex measures on Rn). Using this wehave µ ∗ µ = (δ +m1) ∗ (δ +m1) = δ + 2m1 +m1 ∗m1. Furthermorefor any Borel set E ⊂ R we have (by Folland’s (8.47))

m1 ∗m1(E) =

∫ 1

0

∫ 1

0

χE(x+ y) dx dy =

∫ 1

0

∫ y+1

y

χE(z) dz dy

=

∫ 2

0

∫ min(z,1)

max(z−1,0)

χE(z) dy dz

(We applied Fubini’s theorem, for the product set (y, z) ∈ [0, 1]× [0, 2](say) and integrand = I(y ≤ z ≤ y + 1)I(z ∈ E).)

=

∫

E

(min(z, 1)−max(z − 1, 0)

)dz =

∫

E

(1− |z − 1|

)dz.

This proves that m1 ∗ m1 is the measure on R which is given by the(“triangle shaped”) density z 7→ (1− |z − 1|)+ dz, i.e.:

d(m1 ∗m1) = (1− |x− 1|)+ dx.(15)

Hence an answer to the problem is: µ∗µ = δ+2m1+m1 ∗m1 wherem1 ∗m1 is the measure given by (15).Alternatively we could of course include all the non-singular part of

m1 ∗m1 in the density; then the answer is:

d(µ ∗ µ) = f(x) dx+ dδ,(16)

where f ∈ L1(R) is given by

f(x) =

0 if x ≤ 0 or x ≥ 2

2 + x if 0 < x ≤ 1

2− x if 1 < x < 2.

(Note that (16) provides the Lebesgue-Radon-Nikodym decompositionof µ ∗ µ.)

12

5. (a). Let f ∈ C0(R2) be given. Then the function [0, 1] ∋ x 7→

f(x, 0) is continuous and hence the Riemann integral∫ 1

0f(x, 0) dx ex-

ists (cf. Theorem 1.5 in my notes for the course). But∫

R2

f dµN = N−1N∑

k=1

f( kN, 0)

is a Riemann sum for this integral∫ 1

0f(x, 0) dx, corresponding to the

partition 0 ≤ 1N

≤ 2N

≤ · · · ≤ 1 of [0, 1]; cf. Def. 1.4 in my notes. Sincethe mesh of this partition tends to zero as N → ∞, it follows that

N−1N∑

k=1

f( kN, 0) →

∫ 1

0

f(x, 0) dx as N → ∞.

Also by Folland’s Theorem 2.28(a), the Riemann integral∫ 1

0f(x, 0) dx

equals the Lebesgue integral∫[0,1]

f(x, 0) dx. 1 (Of course the fact that

N−1∑N

k=1 f(kN, 0) tends to the Lebesgue integral

∫[0,1]

f(x, 0) dx can

also easily be proved directly; see Prop. 4 below.)

Answer: µ is the measure which is given by∫R2 f dµ =

∫ 1

0f(x, 0) dx

for any f ∈ C0(R2). (Then for any Borel set E ⊂ R2 we have µ(E) =

m1({x ∈ [0, 1] : (x, 0) ∈ E}), where m1 is Lebesgue measure on R.)

1Note that after the foundational results in Folland’s Chapter 2 are proved, we

use the notation∫ b

af dx also for the Lebesgue integral...

13

(b). Let us prove a somewhat more general result (which I have men-tioned in class; this sorts under “basic understanding of the Riemannintegral”):

Proposition 4. For any f ∈ C(Rn) we have

limN→∞

N−n∑

k∈{0,1,...,N−1}n

f(N−1k) =

∫

[0,1]nf(x) dx.

Proof. We have∫[0,1]n

f(x) dx =∫[0,1)n

f(x) dx since [0, 1]n \ [0, 1)n has

Lebesgue measure zero. For any N ∈ N, we can write [0, 1)n as adisjoint union:

[0, 1)n =⊔

k∈{0,1,...,N−1}n

(N−1k + [0, N−1)n

).

Hence∫

[0,1)nf(x) dx =

∑

k∈{0,1,...,N−1}n

∫

N−1k+[0,N−1)nf(x) dx,

and using the fact that each box N−1k + [0, N−1)n has Lebesgue mea-sure N−n we get

N−n∑

k∈{0,1,...,N−1}n

f(N−1k)−∫

[0,1)nf(x) dx

= N−n∑

k∈{0,1,...,N−1}n

∫

N−1k+[0,N−1)n

(f(N−1k)− f(x)

)dx,

and therefore, using the triangle inequality (for sums and for integrals;see Folland Prop. 2.22 for the latter),

∣∣∣∣∣∣N−n

∑

k∈{0,1,...,N−1}n

f(N−1k)−∫

[0,1)nf(x) dx

∣∣∣∣∣∣(17)

≤∑

k∈{0,1,...,N−1}n

∫

N−1k+[0,N−1)n

∣∣f(N−1k)− f(x)∣∣ dx.

Now let ε > 0 be given. Since f is continuous and [0, 1]n is compact,the restriction of f to [0, 1]n is uniformly continuous. Hence there issome N0 ∈ N such that |f(y) − f(x)| < ε for any two points x, y ∈[0, 1]n satisfying |y − x| ≤ n/N0. Then in particular for any N ≥ N0,any k ∈ {0, 1, . . . , N − 1}n and any x ∈ N−1k + [0, N−1)n we have

14

|x − N−1k| < n/N ≤ n/N0 and therefore |f(N−1k) − f(x)| < ε. Itfollows that for any N ≥ N0 the difference in (17) is

≤∑

k∈{0,1,...,N−1}n

∫

N−1k+[0,N−1)nε dx = ε.

We have proved that such an N0 exists for every ε > 0; hence thedifference in (17) tends to 0 as N → ∞. �

We next show how the above Prop. 4 immediately gives the solutionto problem (b). Let f ∈ C0(R

2) be given. Define f1 ∈ C(R2) through

f1(x, y) = f(x, xy).

Applying now our Prop. 4 to the function f1 gives:

N−2N−1∑

k=0

N−1∑

m=0

f( kN, kmN2 ) = N−2

N−1∑

k=0

N−1∑

m=0

f1(kN, mN)

→∫ 1

0

∫ 1

0

f1(x, y) dy dx as N → ∞.(18)

The limit in (18) equals:∫ 1

0

∫ 1

0

f(x, xy) dy dx =

∫ 1

0

x−1

∫ x

0

f(x, z) dz dx

=

∫

T

f(x, z) x−1 dy dx,

where T is the triangle T = {(x, y) ∈ R2 : 0 < y < x < 1}.Answer: µ is the measure x−1 dy dx restricted to the triangle

T = {(x, y) ∈ R2 : 0 < y < x < 1}.

15

6. Assume E = [a1, b1]×· · ·× [an, bn], where aj < bj for j = 1, . . . , n.Then

χ̂E(ξ) =n∏

j=1

∫ bj

aj

e−2πiξjxj dxj =n∏

j=1

{e−2πiξjaj−e−2πiξjbj

2πiξjif ξj 6= 0

bj − aj if ξj = 0

}

=n∏

j=1

{e(−ξjaj)(1−e(ξj (aj−bj)))

2πiξjif ξj 6= 0

bj − aj if ξj = 0

},

where we use the notation

e(x) = e2πix.

Now let e1 = (1, 0, . . . , 0). Then for any u ∈ R \ {0} we have

χ̂E(ue1) =

( n∏

j=2

(bj − aj)

)e(−ua1)(1− e(u(a1 − b1)))

2πiu

and thus

|χ̂E(ue1)| =( n∏

j=2

(bj − aj)

) |1− e(u(a1 − b1))|2π|u|

Now if we take u = (b1−a1)−1(k+ 1

2) with k = 0, 1, 2, . . . then e(u(a1−

b1)) = −1 and therefore

|χ̂E(ue1)| =( n∏

j=2

(bj − aj)

)π−1|u|−1 =

( n∏

j=2

(bj − aj)

)π−1|ue1|−1.

This proves that along the ray ξ ∈ R+e1 the function χ̂E(ξ) does notdecay faster than |ξ|−1. (The same fact also holds along any other rayR+ej with ej being the standard unit vector ej = (0, . . . , 0, 1, 0, . . . , 0).)Next let us fix any α = (α1, . . . , αn) ∈ Rn with αj 6= 0 for all j. Then

for every u ∈ R \ {0} we have

|χ̂E(uα)| =n∏

j=1

∣∣∣∣e(−uαjaj)(1− e(uαj(aj − bj)))

2πiuαj

∣∣∣∣ ≤n∏

j=1

2

2π|uαj|

=

( |α|nπn∏n

1 |αj|

)|uα|−n.

This proves that along the ray ξ ∈ R+α the function χ̂E(ξ) decaysas fast as |ξ|−n.

16

7. We first prove that E is Lebesgue measurable: For each δ > 0we have ∂E ⊂ ∂δE and thus m(∂E) ≤ m(∂δE). (Note that ∂E isclosed and ∂δE is open so that both these sets are indeed Borel setsand afortiori Lebesgue measurable.) Applying this for δ → 0 and usingthe assumption we conclude that m(∂E) = 0. Let E◦ be the interiorof E. Now E \ E◦ is a subset of ∂E and hence Lebesgue measurablewith measure zero (since the Lebesgue measure is complete), and Eequals the union of E◦ (which is open, hence Lebesgue measurable)and E \ E◦. Hence E is Lebesgue measurable.Next we prove the assertion about the decay of χ̂E . (Thus we are

now assuming both that E is bounded and that m(∂δE) < Cδa forall 0 < δ ≤ 1.) For any η ∈ Rn we have χη+E = τηχE, and thus byFolland’s Theorem 8.22(a), χ̂η+E(ξ) = e−2πiξ·ηχ̂E(ξ). Thus if we set

fη = χE − χη+E ∈ L1(Rn)

(note that fη ∈ L1(Rn) holds since both χE and χη+E are in L1(Rn)and this holds since E is bounded) then

f̂η(ξ) = (1− e−2πiξ·η)χ̂E(ξ).(19)

Now note that

fη(x) =

1 if x ∈ E \ (η + E)

−1 if x ∈ (η + E) \ E0 otherwise,

and

(E \ (η + E)) ∪ ((η + E) \ E) ⊂ ∂δ(E), ∀η ∈ Rn, δ > |η|.Hence if η ∈ Rn and |η| < 1 then for all δ ∈ (|η|, 1] we have |f̂η(ξ)| ≤m(∂δ(E)) < Cδa. We have thus proved:

|f̂η(ξ)| ≤ C|η|a, ∀ξ ∈ Rn and η ∈ Rn with |η| < 1.(20)

Given now any ξ ∈ Rn with |ξ| ≥ 1, let us take η = 12|ξ|−2ξ; then

|η| = 12|ξ|−1 ≤ 1

2and η · ξ = 1

2. Hence (19) and (20) together give

|χ̂η(ξ)| ≤C|η|a

|1− e−2πi 12 |

= 12C|η|a = 2−1−aC|ξ|−a.

That is, if we setK = 2−1−aC then |χ̂E(ξ)| ≤ K|ξ|−a holds for all ξ with|ξ| ≥ 1. But we also know that χ̂E is bounded, since it is the Fouriertransform of an L1-function. Hence by increasing K if necessary, wecan make sure that also |χ̂E(ξ)| ≤ K for all ξ ∈ Rn with |ξ| < 1. Itthen follows, afortiori, that |χ̂E(ξ)| ≤ K|ξ|−a for all ξ ∈ Rn \ {0}. �

17

8. After a translation we may assume that the center of B is theorigin; note that by Folland’s Theorem 8.22(a), translating B does notaffect the absolute value of χ̂B(ξ). Next by Folland’s Theorem 8.22(b)(applied with T given by Tx = t−1x, t > 0), we may assume that B isthe unit ball (open, say).[Some details for the last step: If B is the open unit ball centered

at the origin, then an arbitrary open ball centered at the origin can beexpressed as tB for some t > 0. Now Folland’s Theorem 8.22(b) givesχ̂tB(ξ) = tnχ̂B(tξ). Therefore, if we assume that we have solved theproblem for B, then there is a constant C > 0 such that |χ̂tB(ξ)| =tn|χ̂B(tξ)| ≤ Ctn(1+|tξ|)− 1

2(n+1), and this is2 ≤ C ′(1+|ξ|)− 1

2(n+1) where

C ′ is a constant which only depends on C, t, n. Hence if we can solvethe problem for the unit ball B then the problem is also solved for allother balls.]Next note that by the same Theorem 8.22(b), we have χ̂B(Tξ) =

χ̂B(ξ) for every rotation T of Rn, and every ξ ∈ Rn. In particular wehave χ̂B(ξ) = χ̂B(|ξ|e1) where e1 = (1, 0, . . . , 0) ∈ Rn. Now

χ̂B(ξ) = χ̂B(|ξ|e1) =∫

Rn

χB(x)e−2πi|ξ|e1·x dx

=

∫

R

(∫

Rn−1

χB((x1, y)) dy

)e−2πi|ξ|x1 dx1,

where in the last step we used Fubini’s theorem, writing Rn = R ×Rn−1 (using the notation x = (x1, y) ∈ Rn, with x1 ∈ R, y ∈ Rn−1).Continuing,

χ̂B(ξ) =

∫ 1

−1

m({y ∈ Rn−1 : (x1, y) ∈ B}

)e−2πi|ξ|x1 dx1,

where m is the (n − 1)-dimensional Lebesgue measure on Rn−1. Now

note that for given x1 ∈ (−1, 1), (x1, y) ∈ B holds iff |y| <√

1− x21,

i.e. the set {y ∈ Rn−1 : (x1, y) ∈ B} is the open ball in Rn−1 of

radius√

1− x21 and center 0; thus its volume is (1 − x2

1)12(n−1)Vn−1

where Vn−1 is the volume of the (n − 1)-dimensional unit ball. (Thus

2The statement here is obvious if one notes that the task is only to prove that

the function “decays like |ξ|− 1

2(n+1) as |ξ| → ∞”; on top of this we only have to

use the fact that the function is bounded on any given compact set, which is clearsince it is continuous. To be a bit more precise: Clearly the statement holds with

C′ = Ctn sup{( 1+|ξ|1+|tξ| )

1

2(n+1) : ξ ∈ Rn}, and the fact that this supremum is finite

follows from the fact that the function ξ → 1+|ξ|1+|tξ| is bounded.

18

Vn−1 =π

12 (n−1)

Γ( 12(n+1))

, cf. Folland’s Cor. 2.55.) Therefore

χ̂B(ξ) = Vn−1

∫ 1

−1

(1− x21)

12(n−1) e−2πi|ξ|x1 dx1.(21)

We will bound the size of this function (for large |ξ|) using repeatedintegration by parts. Assume first that n is odd. Then

P (x1) = (1− x21)

12(n−1)

is a polynomial in x1 of degree n − 1. Hence by integrating by partsn−1 times we could obtain an explicit formula for χ̂B(ξ) as |ξ|−n timesa polynomial in e±2πi|ξ| and |ξ|. However the crucial question when wewish to determine the rate of decay as |ξ| → ∞ is: How many times canwe integrate by parts before we pick up non-zero ’boundary terms’ from

the integration limits x1 = ±1? Clearly if n = 1 then χ̂B(ξ) =sin(2π|ξ|)

π|ξ|

for ξ 6= 0, χ̂B(ξ) = 2, and this clearly satisfies the desired bound. Nowassume n ≥ 3, odd. Integrating by parts k times in (21), assumingP (j)(±1) = 0 for j = 0, 1, . . . , k − 1, we get

χ̂B(ξ) = Vn−1

∫ 1

−1

P (k)(x1)e−2πi|ξ|x1

(2πi|ξ|)k dx1.

However if we take k so that P (j)(±1) = 0 for j = 0, 1, . . . , k − 2 butP (k−1)(±1) may be non-zero, then we get:

χ̂B(ξ) = Vn−1P (k−1)(1) e−2πi|ξ| − P (k−1)(−1) e2πi|ξ|

(2πi|ξ|)k

+Vn−1

∫ 1

−1

P (k)(x1)e−2πi|ξ|x1

(2πi|ξ|)k dx1.

Using P (x1) = ((1 − x1)(1 + x1))12(n−1) we see that the zeros of P (x1)

at x1 = 1 and x1 = −1 have order exactly = 12(n − 1). Hence

if we take k = 12(n + 1) then we are in the situation of the last

formula; P (j)(±1) = 0 for j = 0, 1, . . . , k − 2 but both P (k−1)(1)and P (k−1)(−1) are non-zero. Now this formula (using the fact that

|∫ 1

−1P (k)(x1)e

−2πi|ξ|x1 dx1| ≤∫ 1

−1|P (k)(x1)| dx1, where the right hand

side is a finite positive number independent of ξ) shows that χ̂B(ξ)

decays like |ξ|−k = |ξ|− 12(n+1) as |ξ| → ∞, i.e. |χ̂B(ξ)| ≪ |ξ|− 1

2(n+1)

for |ξ| ≥ 1 (say). But we also know that χ̂B is a bounded continuousfunction, since it is the Fourier transform of an L1-function. Hence|χ̂B(ξ)| ≪ (1 + |ξ|)− 1

2(n+1) for all ξ ∈ Rn, as desired.

19

It remains to treat the case of n being even. Let us set

v = 12(n− 1) ∈ {1

2, 32, 52, . . .}

and

fv(x) = (1− x2)v, x ∈ [−1, 1].

This is a continuous function on [−1, 1], and C∞ in (−1, 1). We com-pute:

f ′v(x) = v(−2x)(1 − x2)v−1;

f ′′v (x) = 4v(v − 1)x2(1− x2)v−2 − 2v(1− x2)v−1,

and for general k ∈ Z≥0 we see that

f (k)v (x) =

k∑

j=1

Pj,k,v(x)(1− x2)v−j ,(22)

where Pj,k,v(x) is a polynomial of degree≤ j (and in particular Pk,k,v(x) =v(v−1) · · · (v−k+1)(−2x)k). This is proved using induction, by “im-mediate inspection”.

From (22) we see that f(k)v ∈ C([−1, 1]) and f

(k)v (−1) = f

(k)v (1) = 0

for every k ≤ 12n−1, and we may therefore integrate by parts k = 1

2n−1

times to obtain

χ̂B(ξ) = Vn−1

∫ 1

−1

fv(x) e−2πi|ξ|x dx =

Vn−1

(2πi|ξ|) 12n−1

∫ 1

−1

f( 12n−1)

v (x) e−2πi|ξ|x dx.

Turning now to k = 12n we see from (22) that f

(n/2)v has only “square

root singularities” at x = ±1 (coming from the j = k term in (22);

all the other terms are continuous on all [−1, 1]); that is, f(n/2)v is

continuous on the open interval (−1, 1) and |f (n/2)v (x)| ≪ (1− x)−

12 as

x → 1−, and |f (n/2)v (x)| ≪ (1 + x)−

12 as x → −1+. This means that∫ 1

−1|f (n/2)

v (x)| dx < ∞, and hence we may integrate by parts once more

in the above formula! (To do it carefully: Write∫ 1

−1as limh→0+

∫ 1−h

−1+h,

then integrate by parts, and finally take the limit.) This gives:

χ̂B(ξ) =Vn−1

(2πi|ξ|) 12n

∫ 1

−1

f( 12n)

v (x) e−2πi|ξ|x dx,

and from this we may conclude χ̂B(ξ) ≪ |ξ|− 12n as |ξ| → ∞. However

in order to get the (optimal) decay rate |ξ|− 12(n+1) we try to integrate

by parts once more. Now f( 12n+1)

v (x) blow up like (1− x)−32 as x → 1−

and similarly as x → −1+; thus we cannot integrate by parts all the

20

way out to the boundary points ±1. Instead we split the integral asfollows: For any 0 < h < 1 we have∫ 1

−1

f( 12n)

v (x) e−2πi|ξ|x dx

=

(∫ −1+h

−1

+

∫ 1

1−h

)f( 12n)

v (x) e−2πi|ξ|x dx+

∫ 1−h

−1+h

f( 12n)

v (x) e−2πi|ξ|x dx.

Applying integration by parts to the integral over [−1+h, 1−h] (usingthe fact that the integrand is smooth in an open interval containing[−1 + h, 1− h]) we get

=

(∫ −1+h

−1

+

∫ 1

1−h

)f( 12n)

v (x) e−2πi|ξ|x dx+

[f( 12n)

v (x)e−2πi|ξ|x

−2πi|ξ|

]x=1−h

x=−1+h

+

∫ 1−h

−1+h

f( 12n+1)

v (x)e−2πi|ξ|x

2πi|ξ| dx

The absolute value of this is (noticing that the bounds become sym-metric w.r.t x ↔ −x, and using “≪” with an implied constant whichmay depend on n but is independent of ξ and h):

≪∫ 1

1−h

(1− x)−12 dx+ h− 1

2 |ξ|−1 + |ξ|−1

∫ 1−h

0

(1− x)−32 dx

≪ h12 + h− 1

2 |ξ|−1.

We optimize this bound by choosing h = |ξ|−1 (this is ok only if |ξ| > 1,since we require that 0 < h < 1) and this leads to:

∣∣∣∣∫ 1

−1

f( 12n)

v (x) e−2πi|ξ|x dx

∣∣∣∣≪ |ξ|− 12 ,

for all ξ ∈ Rn with |ξ| > 1. Thus:

χ̂B(ξ) ≪ |ξ|− 12(n+1)

for all ξ ∈ Rn with |ξ| > 1. As before this implies that |χ̂B(ξ)| ≪(1 + |ξ|)− 1

2(n+1) for all ξ ∈ Rn, and we are done. �

Remark on the connection with Bessel functions. The inte-gral in (21) may be recognized as a Bessel function. (For example thiscan be found by using Maple: Try writingint((1-x1^2)^(1/2)*exp(-2*Pi*I*r*x1),x1=-1..1); andint((1-x1^2)^1*exp(-2*Pi*I*r*x1),x1=-1..1); and

21

int((1-x1^2)^(3/2)*exp(-2*Pi*I*r*x1),x1=-1..1);, etc.) Indeed,by [1, 8.411.10] we have

χ̂B(ξ) = Vn−1

∫ 1

−1

(1− x21)

12(n−1) e−2πi|ξ|x1 dx1

= Vn−1Γ(12)Γ(n+1

2)(π|ξ|)−n

2 Jn/2(2π|ξ|)= |ξ|− 1

2nJn/2(2π|ξ|).

But Jn/2(2π|ξ|) = O(|ξ|− 12 ) as |ξ| → ∞ (cf., e.g., [2, p. 199 (1)] or

wikipedia). This gives the desired bound, χ̂B(ξ) ≪ |ξ|− 12(n+1) as |ξ| →

∞.

References

1. I. S. Gradshteyn and I. M. Ryzhik, Table of integrals, series, and products, sev-enth ed., Elsevier/Academic Press, Amsterdam, 2007, Translated from the Rus-sian, Translation edited and with a preface by Alan Jeffrey and Daniel Zwillinger,With one CD-ROM (Windows, Macintosh and UNIX).

2. G. N. Watson, Theory of bessel functions, Cambridge, 1922.