LECTURE NOTES ON OSCILLATORY INTEGRALSvitturi/... · 2 OSCILLATORY INTEGRALS...

LECTURE NOTES ON OSCILLATORY INTEGRALSAtelier d’Analyse Harmonique 2019

A large part of modern harmonic analysis is concerned with understanding can-cellation phenomena happening between different contributions to a sum or inte-gral. Loosely speaking, we want to know how much better we can do than if we hadtaken absolute values everywhere. A prototypical example of this is the oscillatoryintegral of the form ˆ

eiφλ(x)ψ(x)dx

(solutions of dispersive PDEs often take this form). Here ψ, called the amplitude,is usually understood to be “slowly varying” with respect to the real-valued φλ,called the phase, and thus the oscillatory behaviour is given mainly by the complexexponential eiφλ(x), where λ denotes a parameter or list of parameters and generallyφ′λ gets larger as λ grows, for example φλ(x) = λφ(x). Expressions of this formarise quite naturally in several problems, as we will see in Section 1, and typicallyone seeks to provide an upperbound on the absolute value of the integral abovein terms of the parameters λ. Intuitively, as λ gets larger the phase φλ changesfaster and therefore eiφλ oscillates faster, producing more cancellation between thecontributions of different intervals to the integral. We expect then the integral todecay as λ grows larger, and usually seek upperbounds of the form |λ|−α. Noticethat if you take absolute values inside the integral above you just obtain ‖ψ‖L1 , abound that does not decay in λ at all.The main tool we will use is simply integration by parts. In the exercises you willalso use a little basic complex analysis to obtain more precise information on certainspecial oscillatory integrals.Notation: We will use A . B or A = O(B) to denote the estimate A ≤ CB where C > 0 is someabsolute constant, and A ∼ B to denote the fact that A . B . A. If the constant C depends ona list of parameters L we will write A .L B or A = OL(B).We will further denote by Ckc the space of k-times differentiable functions with compact support.

1. Motivation

In this section we shall showcase the appearance of oscillatory integrals in analysiswith a couple of examples. You can find other interesting examples in the exercises,see particularly Exercise 8, Exercise 14 part viii), Exercise 15 and Exercise21.

1.1. Fourier transform of radial functions. Let f : Rd → C be a radiallysymmetric function, that is there exists a function f0 : R → C such that f(x) =f0(|x|) for every x ∈ Rd. Let’s suppose for simplicity that f ∈ L1(Rd) (equivalently,that f0 ∈ L1(R, rd−1dr)), so that it has a well-defined Fourier transform. It is easyto see1 that f must also be radially symmetric, that is there must exist g : R→ Csuch that f(ξ) = g(|ξ|); we want to understand its relationship with f0. Therefore

1By composing f with a rotation and using a change of variable in the integral defining f .1

2 OSCILLATORY INTEGRALS

we write using polar coordinates

f(ξ) =

ˆRdf(x)e−2πix·ξdx

=

ˆ ∞0

ˆSd−1

f0(r)e−2πirω·ξrd−1dσd−1(ω)dr,

=

ˆ ∞0

f0(r)rd−1(ˆ

Sd−1

e−2πirω·ξdσd−1(ω))dr

where dσd−1 denotes the surface measure on the unit (d − 1)-dimensional sphereSd−1 induced by the Lebesgue measure on the ambient space Rd. By inspection,we see that the integral in brackets above is radially symmetric in ξ, and so if wedefine

J(t) :=

ˆSd−1

e−2πitω·e1dσd−1(ω),

with e1 = (1, 0, . . . , 0), we have

f(ξ) = g(|ξ|) =

ˆ ∞0

f0(r)rd−1J(r|ξ|)dr. (1.1)

This is the relationship we were looking for: it allows one to calculate the Fouriertransform of f directly from the radial information f0. You might also notice thatJ is simply the Fourier transform of the spherical measure dσd−1.Now we claim that the function J is an example of oscillatory integral of the typementioned at the beginning. Indeed, observe that the inner product ω · e1 dependsonly on the first component of ω; thus we write ω = (cos θ, ω′ sin θ), with θ theangle between ω and e1 and ω′ ∈ Sd−2. By factorising the spherical measure dσd−1

along the e1 axis, we can use this change of variables to write

J(t) =

ˆSd−2

ˆ π

0

e−2πit cos θ(sin θ)d−2dθdσd−2(ω′)

=cd−2

ˆ π

0

e−2πit cos θ(sin θ)d−2dθ,

because the integrand does not depend on ω′ (here cd−2 =´Sd−2 dσd−2(ω′) =

σd−2(Sd−2)). It is now trivial to match the last expression to the one for an oscil-latory integral where the parameter is t. Later on we will see how to estimate it interms of |t| and show that J(t) is bounded by a constant multiple of (1+|t|)−(d−1)/2.

The information we get from how fast J(t) decays in terms of |t| tells us ofsomething interesting. Indeed, recall that if f ∈ L1(Rd) then its Fourier transformf is continuous. In general, when 1 < p ≤ 2 and f ∈ Lp(Rd), we know by Hausdorff-Young inequality that f is in Lp

′(Rd), but we don’t know whether it is continuous

or not. You can easily see this is generally not the case when p = 2. However, forradial functions one can show continuity of the Fourier transform if the exponentp is sufficiently close to 1! In particular

Proposition 1.1. If 1 ≤ p < 2dd+1 then the Fourier transform f of any radial

function f ∈ Lp(Rd) is continuous away from 0.

This follows purely from the decay of the oscillatory integral J and you will proveit in Exercise 1.

1.2. Counting the number of ways in which a number can be writtenas the sum of two squares. Consider the following diophantine-type problem:when can we write an integer n as the the sum of two squares, n = x2 + y2? If you

OSCILLATORY INTEGRALS 3

fiddle around with examples a bit you might find that certain numbers have sucha representation, say

5641 = 752 + 42,

while others, such as the successor 5642, have no such representation. Studyingthis problem more in depth you might discover the Fibonacci identity

(a2 + b2)(c2 + d2) = (ac− bd)2 + (ad+ bc)2,

and deduce as a consequence that it suffices to study the case where n is prime. Ifyou went further down this road, you might learn that there is a way to write pprime as x2 + y2 if and only if p ≡ 1 mod 4, and this gives the full answer for theproblem: n = x2 + y2 has a solution if and only if in the prime factorisation of nthe prime factors that are congruent to 3 mod 4 appear an even number of timeseach.However, after this, you might notice that some numbers have even more than onerepresentation as a sum of two squares: for example,

5645 = 742 + 132 = 672 + 342.

The next step would then be to try and count how many distinct solutions thereare. It turns out that this can be done in terms of the exponents in the primefactorisation of n (and the answer involves the Gaussian integers Z[i]); however,this is not too convenient, because factorising large integers is knowingly hard. Letus take a different, less algebraic approach instead.

We can generalise the question a little by considering sums of more than twosquares. Letting k ∈ N be fixed and given an integer n ∈ N, we define the k-thsum-of-squares function to be

rk(n) := #(x1, . . . , xk) ∈ Zk such that x21 + · · ·+ x2

k = n;in words, rk(n) is the number of ways in which n can be expressed as the sum ofk squares. Then we want to study the behaviour of the function rk(n) as n grows.In the following we will consider only k = 2.If you plot the graph of r2 you will quickly realise it is not a very regular function2.A natural approach to deal with this irregularity is to study the averaged sequenceinstead, hoping it will behave better. This should help us answer the question “inhow many ways can we expect to be able to write an arbitrary number n as asum of two squares on average?”. In formulas, we want to study the behaviour of∑Nn=1 r2(n)/N as N tends to ∞. The limit of the expression above is easy to find:

the key is to notice that n = x2 + y2 means that the point (x, y) ∈ Z2 belongs tothe circle of radius n1/2 centred at (0, 0), and therefore to the ball of radius N1/2

with the same centre. ThereforeN∑n=1

r2(n) = #(x, y) ∈ Z2 such that x2 + y2 ≤ N,

and a simple geometric argument shows that the limit of 1N

∑Nn=1 r2(n) is π (see

Exercise 2). The geometric argument actually says something more: namely, wecan also give an upperbound on the error, that is

N∑n=1

r2(n) = πN +O(N1/2).

The upperbound on the error is significant because it is of smaller order than themain term (which grows like O(N)). The limit of the averaged sequence was veryeasy to find, so now we ask a more sophisticated question: what is the behaviour

2Sequence A004018 in OEIS, https://oeis.org/A004018/graph.

https://oeis.org/A004018/graph


of the error term? How much can the sum deviate from πN? O(N1/2) is what onewould expect from a uniformly random behaviour, but it turns out that it is farfrom optimal, and we can do better. However, we do not know yet how much betterwe can do! It is still an open problem3 to establish the true asymptotic behaviourof the error term E(N) :=

∑Nn=1 r2(n)− πN . Using oscillatory integral techniques,

we can show the improved estimate that follows.

Proposition 1.2. For N large enough, we have the error term estimate

|E(N)| = O(N1/3).

If you stop for a moment to think about it, it is almost absurd that techniquesdeveloped to study oscillatory integrals can say something highly non-trivial aboutquestions concerning the integers.

Proof. The idea is to use the Poisson Summation formula to reveal some interestingoscillation hidden in the problem. In the end, the proof will again rely on a gooddecay estimate for a certain oscillatory integral related to J .Recall that, for f ∈ S(R2), that is a Schwartz function, the Poisson Summationformula says that ∑

n∈Z2

f(n) =∑n∈Z2

f(n).

The quantity we are interested in is Θ(N) :=∑n∈Z2 1B(0,N1/2)(n), but 1B(0,N1/2)

(the characteristic function of a ball of radius N1/2) is not in S(R2) - thus we haveto regularise it. Let δ > 0 be a small parameter to be chosen later and let ϕ be abump function in C∞, supported in the unit ball and such that

´R2 ϕ(x)dx = 1; we

let ϕδ denote the rescaled function δ−2ϕ(δ−1x) (so that´ϕδ = 1 too). Then we

defineχN,δ := 1B(0,N1/2) ∗ ϕδ,

which is in S(R2) and is an approximation to 1B(0,N1/2). Applying the PoissonSummation formula to this function, we obtain

Θδ(N) :=∑n∈Z2

χN,δ(n) = 1B(0,N1/2)(0)ϕ(0) +∑

n 6=0∈Z2

1B(0,N1/2)(n)ϕ(δn);

the first term is easily evaluated to be exactly πN , our main term! Thus the secondterm is an error we have to control. Observe that, since 1B(0,N1/2) is radial, wehave by (1.1) that

1B(0,N1/2)(n) =

ˆ N1/2

0

rJ(r|n|)dr.

The function J is not only an oscillatory integral, but it is itself oscillating, andwith the techniques of the next section you will be able to show in Exercise 9 thatˆ R

0

rJ(r)dr = O(R1/2). (1.2)

After a change of variables, estimate (1.2) shows that |1B(0,N1/2)(n)| . N1/4|n|−3/2.In the range |n| < δ−1 we have |ϕ(δn)| . 1 and therefore the above estimate givesus ∑

|n|<δ−1,n6=0

|1B(0,N1/2)(n)ϕ(δn)| . N1/4∑

|n|<δ−1,n6=0

|n|−3/2 ∼ N1/4δ−1/2.

3It is conjectured that for any ε > 0 the error E(N) is bounded by CεN1/4+ε for some constantCε > 0. The estimate is certainly false with ε = 0.


In the range |n| ≥ δ−1 things are even better, since ϕ decays fast, being in C∞; inparticular, we have |ϕ(δn)| . (δ|n|)−1 and thus, by the above same argument, thisrange also contributes O(N1/4δ−1/2). Summarising, we have proven that

Θδ(N) = πN +O(N1/4δ−1/2).

Now, observe that

Θδ((N1/2 − δ)2) ≤ Θ(N) ≤ Θδ((N

1/2 + δ)2); (1.3)

this is because, as you can easily see by expanding the convolution,

χ(N1/2−δ)2,δ ≤ 1B(0,N1/2) ≤ χ(N1/2+δ)2,δ.

Therefore we have, thanks to (1.3) and the fact that (N1/2 ± δ)2 ≈ N ± 2N1/2δ,

Θ(N)− πN = E(N) = O(N1/4δ−1/2) +O(N1/2δ),

and if we optimize by choosing δ = N−1/6 we see that both terms above are O(N1/3)and we are done.

2. Oscillatory integrals in one variable

In this section we present some techniques that allow one to estimate oscillatoryintegrals when one has a lowerbound for some of the derivatives of the phase. Wewill analyse objects of the form

I(λ) :=

ˆ b

a

eiλφ(x)dx

and more in general

Iψ(λ) :=

ˆ b

a

eiλφ(x)ψ(x)dx.

Before we start, let’s consider the following heuristic. Suppose that we have func-tions f, g and we know that g is “slowly varying” - that is, the derivative g′ is small.In order to find a good upperbound for an expression of the form |

´ baf(t)g(t)dt|,

we could take advantage of the fact that g′ is small by using integration by parts toestimate instead the expression −

´ baF (t)g′(t)dt+ (boundary terms), where F is a

primitive of f . What we gain is that now we are working with an integral for whichwe know one of the factors of the integrand, g′, is small. We will use this idea andvariations thereof over and over.

2.1. Principle of non-stationary phase. We begin by considering the followingcase. Let ψ ∈ C∞(R) have compact support in (a, b) (so that in particular ψ(a) =ψ(b) = 0) and suppose that the phase φ ∈ C∞ satisfies φ′(t) 6= 0 for all t ∈ [a, b].We claim that in this case the integral Iψ(λ) decreases very fast in λ, in particular

Proposition 2.1 (Principle of non-stationary phase). Let ψ, φ be as above, that isψ ∈ C∞c ((a, b)) and φ ∈ C∞ is such that φ′ 6= 0 on [a, b]. Then for every N > 0 wehave

|Iψ(λ)| .N,ψ,φ |λ|−N .

Remark 1. Notice that the bound given by the proposition above is only interest-ing when |λ| is large. Indeed, when |λ| ≤ 1 we can simply bound |Iψ(λ)| ≤ ‖ψ‖L1 .1 by taking the absolute value inside the integral.

Proof. The proof is a simple integration-by-parts argument.We want to use integration by parts a number of times. Notice that (eiλφ)′ =iλφ′eiλφ, so if we define the differential operator D by

Df(t) :=1

iφ′(t)

df

dt


we have λ−1D(eiλφ) = eiλφ. Notice D is well-defined because φ′ 6= 0. Usingintegration by parts we then have

ˆ b

a

eiλφ(t)ψ(t)dt =

ˆ b

a

λ−1D(eiλφ(t))ψ(t)dt

=eiλφψ

iλφ′

∣∣∣∣ba

+

ˆ b

a

eiλφ(t)λ−1Dtψ(t)dt

=λ−1

ˆ b

a

eiλφ(t)Dtψ(t)dt,

where the boundary term vanishes by the hypothesis on the support of ψ; hereDt denotes the transpose4 of the operator D, namely Dtf(t) = − 1

iddt

(fφ′

). By

repeating the argument N times we get

Iψ(λ) = λ−Nˆ b

a

eiλφ(Dt)N (ψ)dt,

and we conclude by taking absolute values (the resulting integral is finite).

We can interpret the principle of non-stationary phase as saying that, for ageneric phase φ, the behaviour of the oscillatory integrals Iψ(λ) is determined bythe points where φ′ = 0, because away from those points the integral contributesat most an error term ON (|λ|−N ). In Section 2.3 below we will show how thisheuristic can be made precise.

2.2. Van der Corput’s lemma. We will now consider more general situationsthan the non-stationary phase one.

We start by considering I(λ), that is the case where ψ ≡ 1, and we assume thatwe have |φ′(t)| > 1 for all t ∈ (a, b). We make a further important assumption, thatis we also assume that φ′ is monotonic5. Then we have

Proposition 2.2. If φ is such that φ′ is monotonic and |φ′| > 1 on [a, b], we have

|I(λ)| ≤ C|λ|−1

for an absolute constant C > 0.

Some observations are in order, before we proceed to the proof:i) First of all, the assumption that φ′ is monotonic is fundamental: indeed, the

statement is false otherwise! (prove this in Exercise 4; see also Exercise 5.)ii) Secondly, we cannot get a decay in λ better than |λ|−1, as the example φ(t) = t

shows (you are cordially invited to do the calculation).iii) Thirdly, by a simple rescaling of the phase, we can see that the proposition is

more general than it looks: if the lowerbound on φ′ becomes more generally|φ′| > µ, then the estimate becomes |I(λ)| ≤ C(µ|λ|)−1.

iv) Finally, notice that the constant C in the statement above depends neitheron the phase φ nor on the interval [a, b]! In particular, if we allowed theconstant to depend arbitrarily on the interval, we would trivially have that|I(λ)| ≤ |a− b|, which is rather uninteresting.

4Formally, the operator such that 〈Df, g〉 = 〈f,Dtg〉 for all f, g ∈ C∞c .

5Recall that a function f : R → R is monotonic non-decreasing if x < y ⇒ f(x) ≤ f(y),and monotonic non-increasing if x < y ⇒ f(x) ≥ f(y). A function is monotonic if it is eithermonotonic non-decreasing or monotonic non-increasing.


Observations ii)-iv) are actually related, thanks to the scaling behaviour of theinequality. Indeed, if we ask for which values of α we can have the inequality

|I(λ)| ≤ Cφ|λ|−α

hold with a constant Cφ that is independent6 of (a, b), then the answer is thatnecessarily it must be α = 1. Prove this in Exercise 6.Now we proceed with the proof.

Proof. We repeat the same integration-by-parts argument as before, except thistime the boundary terms do not vanish. Thus we have, with D the same differentialoperator as before, that

ˆ b

a

eiλφ(t)dt =( eiλφ(b)

iλφ′(b)− eiλφ(a)

iλφ′(a)

)+

ˆ b

a

eiλφ(t)λ−1Dt(1)dt.

Since |φ′| > 1, the boundary term is bounded by 2/|λ| simply by triangle inequality.As for the other term, λ−1Dt(1) = −(iλ)−1d/dt(1/φ′), and by taking absolutevalues inside the integral we have that it is bounded by

1

|λ|

ˆ b

a

∣∣∣ ddt

( 1

φ′

)∣∣∣dt.However, φ′ is monotonic and therefore so is 1/φ′; this means that in the lastintegral the derivative is single-signed, and therefore we can take the absolute valueoutside (something normally prohibited!) and obtain that it equals

1

|λ|

∣∣∣ˆ b

a

d

dt

( 1

φ′

)dt∣∣∣.

By the Fundamental Theorem of Calculus this is equal to |λ|−1∣∣ 1φ′(b)−

1φ′(a)

∣∣, whichis bounded by 1/|λ|, thanks to the monotonicity assumption. Thus we have proventhe theorem with C = 3.

The natural step after this is to investigate what happens when the condition|φ′| > 1 is violated but we still have lowerbounds for some higher derivatives.Indeed, we have

Theorem 2.3 (Van der Corput’s lemma). Let k ≥ 2 and let φ ∈ Ck be such that|φ(k)| > 1 on (a, b). Then

|I(λ)| ≤ Ck|λ|−1/k,

where Ck is an absolute constant depending only on k.

Notice that |λ|−1/k decays slower than |λ|−1 as λ→∞. This is to be expected,since the phase “slows down” near the zeros of φ′ and hence there is less overallcancellation. It is indeed sharp, as the example φ(t) = tk over [0, 1] shows (again,you are cordially invited to do the calculation - you can use complex integrationand regularize the integral by introducing a factor of e−εt

k

, then take ε→ 0 at theend; see also Exercise 13).Again, it goes without saying that if we have |φ(k)| > µ instead, then the inequalitybecomes |I(λ)| ≤ Ck(µ|λ|)−1/k. Moreover, the exponent 1/k is necessary for theinequality to hold with Ck independent of the interval (a, b) (see again Exercise6).

6As the subscript indicates, the constant might depend on φ in principle. However, this willnot be the case for us, at least for now.


Remark 2. You should have noticed that this time we are not making an explicitmonotonicity assumption such as in Proposition 2.2. However, we implicitely stillare! Indeed, the fact that |φ(k)| > 1 for some k ≥ 2 implies that φ(k−1) attainsvalue zero in at most one point of (a, b), and thus φ(k−2) attains value zero in atmost two points, and so on; iterating, we see that φ′ has at most k − 2 changes ofmonotonicity, or in other words we can partition (a, b) into at most k − 1 intervalssuch that φ′ is monotonic on each interval.

Proof. The proof proceeds by induction on k. The case k = 1 (with the additionalassumption that φ′ be monotonic) has been proven in Proposition 2.2. Notice thatif |φ′′| > 1 then φ′ is monotonic.Assume now that the statement is true for k − 1. Since |φ(k)| > 1, the (k − 1)-thderivative φ(k−1) can have at most one zero in (a, b). Denote this zero by t0 andsplit the integral as ˆ t0−δ

a

+

ˆ t0+δ

t0−δ+

ˆ b

t0+δ

for some δ > 0 to be chosen later. In the interval (a, t0 − δ) the function φ(k−1)

is never zero, but moreover we have by the assumption on its derivative φ(k) that|φ(k−1)| > δ; similarly on the interval (t0 + δ, b). By inductive hypothesis wetherefore have that∣∣∣ˆ t0−δ

a

eiλφ(t)dt∣∣∣+∣∣∣ˆ b

t0+δ

eiλφ(t)dt∣∣∣ ≤ 2Ck−1(δ|λ|)−1/(k−1).

As for the remaining integral, we just estimate trivially∣∣∣ˆ t0+δ

t0−δeiλφ(t)dt

∣∣∣ ≤ 2δ.

Putting everything together, we have shown that |I(λ)| ≤ 2Ck−1(δ|λ|)−1/(k−1) +2δ,and by choosing δ = |λ|−1/k this gives

|I(λ)| ≤ (2Ck−1 + 2)|λ|−1/k,

proving the induction (with Ck = 2Ck−1 + 2).

Although the constant Ck obtained with the above argument suffices for most (ifnot all) applications, it’s interesting to notice it is not optimal in k. See Exercise11 if you are interested in determining the correct behaviour of the optimal constantin k.

It is a simple matter of integrating by parts to extend Van der Corput’s lemmato the oscillatory integrals Iψ(λ) as well. We obtain

Corollary 2.4. Let ψ ∈ C1 and let the phase φ satisfy |φ(k)| > 1 in (a, b) for somek ≥ 1 (if k = 1, assume additionally that φ′ is monotonic). Then the inequality

|Iψ(λ)| ≤ C ′k[|ψ(b)|+

ˆ b

a

|ψ′(t)|dt]· |λ|−1/k

holds, with C ′k > 0 an absolute constant depending only on k.

Thus the overall constant will depend on a, b through the amplitude ψ (thoughnotice that if we assume ‖ψ‖L∞ + ‖ψ′‖L1 is finite, it is independent of a, b), but wehave isolated its dependence in the term in square brackets. The constant is stillindependent of the phase φ.

Proof. Let Φ(t) :=´ taeiλφ(s)ds, so that

Iψ(λ) =

ˆ b

a

Φ′(t)ψ(t)dt.


Integrating by parts and taking absolute values we have that |Iψ(λ)| is bounded by

|Φ(b)ψ(b)|+ˆ b

a

|Φ(t)||ψ′(t)|dt.

For the first term, by Theorem 2.3 we can estimate |Φ(b)| .k |λ|−1/k. Similarly,the second term can then be estimated by Ck|λ|−1/k

´ ba|ψ′(t)|dt. Summing these

contributions gives the stated bound.

Notice the integration-by-parts trick we used here is different from the one usedfor Propositions 2.1, 2.2.

2.3. Method of stationary phase. Now we go back to the observation made atthe end of Section 2.1. Recall that the portions of the integral Iψ(λ) where thephase satisfies φ′ 6= 0 contribute at most Oψ,N (λ−N ) for arbitrary N > 0 andthus can tipically be treated as an error term. The behaviour of Iψ(λ) is thereforedetermined by the zeros of φ′ and higher derivatives of the phase. In particular, onecan perform a full asymptotic expansion of the oscillatory integral Iψ(λ). This isknown as the method of stationary phase; it’s particularly useful in physics, where itprominently appears in the semi-classical approximation to Quantum Field Theory.We can state (one version of) the method as follows.

Theorem 2.5 (Method of stationary phase). Assume that φ ∈ C∞. Let k ≥ 2 andassume that

φ′(t0) = . . . = φ(k−1)(t0) = 0 and that φ(k)(t0) 6= 0.

If ψ ∈ C∞c is supported in a sufficiently small neighbourhood of t0, then there existcoefficients aj for j ∈ N (each depending only on finitely many derivatives of φ andψ at t0) such that

Iψ(λ) ' eiλφ(t0)λ−1/k∑j∈N

ajλ−j/k, (2.1)

where by ' we mean that for all n > 0 we have for the sum truncated at n− 1 that∣∣∣∣Iψ(λ)− eiλφ(t0)λ−1/kn−1∑j=0

ajλ−j/k

∣∣∣∣ .ψ,φ,n |λ|−n/kas λ→∞, and moreover for all n, ` > 0 we have∣∣∣∣( ddλ)`[Iψ(λ)− eiλφ(t0)λ−1/k

n−1∑j=0

ajλ−j/k

]∣∣∣∣ .ψ,φ,`,n |λ|−`−n/k.The coefficients aj can be determined explicitely - for example, when k = 2 an

explicit calculation of a0 shows that the main term in Iψ(λ) is

eiλφ(t0)( 2π

−iλφ′′(t0)

)1/2

ψ(t0).

This is excellent for a quick-and-dirty estimate of complicated integrals.We will not prove the above theorem here since it would make these notes unnec-essarily long, but if you fancy you can prove it yourself in Exercise 13 followingthe guidelines provided there.


3. Oscillatory integrals in several variables

In the previous section we have analysed the situation for single variable phases,that is for integrals over (intervals of) R. In this section, we will instead studythe higher dimensional situation, when the phase is a function of several variables.Things are unfortunately generally not as nice as in the single variable case, as youwill see.

In order to avoid having to worry about connected open sets of Rd (see Exercise18 for the sort of issues that arise in trying to deal with general open sets of Rd),in this section we will study mainly objects of the form

Iψ(λ) :=

ˆRdeiλu(x)ψ(x)dx,

where ψ has compact support. We have switched to u for the phase to remind thereader of the fact that it is a function of several variables now.

3.1. Principle of non-stationary phase - several variables. The principle ofnon-stationary phase we saw in Section 2.1 continues to hold in the several variablescase.Given a phase u, we say that x0 is a critical point of u if

∇u(x0) = (0, . . . , 0).

Proposition 3.1 (Principle of non-stationary phase - several variables). Let ψ ∈C∞c (Rd) (that is, smooth and compactly supported) and let the phase u ∈ C∞ besuch that u does not have critical points in the support of ψ. Then for any N > 0we have

|Iψ(λ)| .N,ψ,u |λ|−N .

Proof. The proof is simply a reduction to the single variable case of Proposition2.1. Indeed, by assumption there exists a c > 0 such that |∇u(x)| > 2c for x in thesupport of ψ, and therefore for any such x there exists a small ball Bx, centered at xand a unit vector ξx such that ξx ·∇u(y) > c for every y ∈ Bx. By compactness wecan find a finite collection of such balls, Bjj , that covers the support of ψ; witha partition of unity associated to this collection, we can therefore write ψ =

∑j ψj

with ψj supported in Bj . Since the number of balls is finite (depending only onψ, u), it suffices to prove the claimed bound for a single Iψj (λ). By a change ofcoordinates, we can assume ξj = (1, 0, . . . , 0) and sinceˆ

Rdeiλu(x)ψj(x)dx =

ˆRd−1

(ˆReiλu(x1,x2,...,xd)ψj(x1, x2, . . . , xd)dx1

)dx2 · · · dxd,

we can conclude by applying Proposition 2.1 to the inner integral and then inte-grating in the remaining variables (over the projection of Bj).

3.2. Van der Corput’s lemma in several variables. Proceeding by analogy towhat was done in Section 2, we now want to study what happens when we havea lowerbound on some (possibly mixed) derivative of the phase u. We look forinequalities that share the same scaling behaviour as the single variable ones inProposition 2.2 and Theorem 2.3.

We will denote by α = (α1, . . . , αd) a multi-index, that is an element of Nd, andby |α| the sum α1 + . . .+ αd. Then ∂α denotes the partial derivative of order |α|

∂α = ∂α1x1· · · ∂αdxd .

We have the following several-variables version of Van der Corput’s lemma (moreprecisely, of Corollary 2.4).


Theorem 3.2 (Van der Corput’s lemma in several variables). Let ψ ∈ C1c (B(0, 1))

(that is, compactly supported in the unit ball). Assume that for every x in thesupport of ψ the phase u satisfies

|∂αu(x)| > 1

for a certain fixed multi-index α, with |α| ≥ 1. Furthermore we assume u ∈ C |α|+1.Then we have

|Iψ(λ)| ≤ C|α|,u[‖ψ‖L∞ + ‖∇ψ‖L1

]· |λ|−1/|α|,

where the constant C|α|,u depends only on |α| and on the phase u (in particular, onthe derivatives of u up to order |α|+ 1).

One can appreciate the superficial similarity between Corollary 2.4 and Theorem3.2. We reiterate that if the hypothesis on the phase becomes |∂αu(x)| > µ thenthe inequality holds with |λ|−1/|α| replaced by (µ|λ|)−1/|α|.However, there are several key differences with Corollary 2.4 which make the abovetheorem weak:i) unfortunately, the constant now depends on the phase too, which can some-

times be an issue when one has to deal with multiple phases at once;ii) as you will see from the proof, the constant also depends on the size of the

support of ψ if one removes the extra hypothesis that ψ ∈ C1c (B(0, 1)), and

thus the inequality does not have the nice scaling properties that it had in thesingle variable case;

iii) the estimate is no longer tight: indeed, if we take a phase like u(x, y) = xy,so that |∂x∂yu| = 1, and an amplitude that is just a bump function supportednear the origin, the theorem gives us a bound of O(|λ|−1/2), but in reality onecan prove the better estimate7 O(|λ|−1) (see Section 3.3 or Exercise 19).

Proof. The proof will again work by reducing to the single variable case (and thisis the source of the inefficiencies mentioned above).Let k = |α|. In order to reduce to the single variable case, we need to show that theassumption |∂αu| > 1 implies a similar lowerbound for some derivative of the form(ξ ·∇)ku - that is, a k-th derivative in a fixed direction, (ξ ·∇)f(x) = d

dt

(f(x+ tξ)

),

so that we can apply single variable results. Indeed, the real vector space of partialderivatives of order k admits a basis of the above form: there exist unit vectorsξ(1), . . . , ξ(m) (where m = #β ∈ Nd s.t. |β| = k) such that any partial derivativeof order k can be expressed as a linear combination of derivatives (ξ(j) ·∇)k. Provethis in Exercise 16.Since ∂α can be expressed as a linear combination of (ξ(j) · ∇)k, for each x in thesupport of ψ there exists a unit vector ξx such that |(ξx · ∇)ku(x)| &k 1; moreover,since we have assumed that u ∈ Ck+1, we have that ‖u‖Ck+1 is bounded on thesupport of ψ, and therefore for a small ball Bx (of radius . ‖u‖−1

Ck+1) centered atx we have actually

|(ξx · ∇)ku(y)| &k 1 for all y ∈ Bx(with a worse constant than the one implicit in the previous lowerbound). As inthe proof of Proposition 3.1, we can find a finite covering of the support of ψ bysuch balls and use them to create a partition of unity that allows us to split ψ as∑j ψj with each ψj supported in one of such balls. It suffices to estimate Iψj (λ)

separately (ψ being supported in B(0, 1), the number of such balls only depends

7For a suggestive calculation: estimate´Ω exp(iλxy)dxdy where Ω = (x, y) : |x| + |y| ≤ 1

(hint: rotate coordinates and use single variable Van der Corput’s lemma).


on ‖u‖Ck+1). After a change of variables, we can assume that ξ = (1, 0, . . . , 0) forthe ball under examination, and therefore we write again (with x′ = (x2, . . . , xd))ˆ

B(0,1)

eiλu(x)ψj(x)dx =

ˆRd−1∩B(0,1)

(êiλu(x1,x

′)ψj(x1, x′)dx1

)dx′.

Applying Corollary 2.4 to the inner integral we get that it is bounded by

.k[‖ψ‖L∞ +

ˆ|∂x1

ψ(x1, x′)|dx1

]· |λ|−1/k;

integrating in the remaining variables gives us the estimate we want.

3.3. Method of stationary phase in several variables - non-degeneratephases. When |α| = 2, we can obtain better estimates than the ones given byTheorem 3.2, provided that we have some extra information about the phase beingwell-behaved. In particular, if x0 is a critical point for phase u, we say that it is anon-degenerate critical point if

det(Hess(u)(x0)) 6= 0,

where Hess(u) is the Hessian matrix(

∂2u∂xi∂xj

)ij.

If the phase has only non-degenerate critical points, we can assert an analogue ofTheorem 2.5 for phases of several variables.

Theorem 3.3. Assume that ψ ∈ C∞c (Rd) and that the phase u ∈ C∞(Rd) hasx0 as a non-degenerate critical point. If ψ is supported in a sufficiently smallneighbourhood of x0 (in particular, there are no other critical points of u in it),then there exist coefficients aj for j ∈ N (each depending only on finitely manyderivatives of u and ψ at x0) such that

Iψ(λ) ' eiλu(x0)λ−d/2∑j∈N

ajλ−j , (3.1)

where ' is in the same sense as in Theorem 2.5. In particular, |Iψ(λ)| .φ,ψ |λ|−d/2.

When u(x, y) = xy, we see that this gives Iψ(λ) = O(λ−1) as claimed before.We do not prove this theorem either (but if you have proven Theorem 2.5 inExercise 13, you can prove this one as well in Exercise 20 following the samestrategy with little extra effort), but rather make a remark.

Remark 3. The theorem above is limited to the case |α| = 2. Indeed, observethat in the single variable case the fact that φ(k)(t0) 6= 0 and φ(t0) = φ′(t0) =. . . = φ(k−1)(t0) = 0 implies that we can put the phase in the simple canonicalform φ(t) = tk (after a change of variable); however, in several variables there is nosimple analogue of this, except in the case |α| = 2, in which we can put the phaseu in a standard quadratic form - hence our limitation.

3.4. Estimates independent of the phase. As observed above, one of the down-sides of Van der Corput’s lemma in several variables is that the constant ends updepending on the phase u as well, which is sometimes a problem. It is also knownthat, if we consider the case ψ ≡ 1 and thus look at oscillatory integrals of the form´

Ωeiλu(x)dx with Ω a connected (bounded) open set, the constant cannot be made

independent of Ω (prove this in Exercise 18).It is reasonable to look for estimates with constants that are independent of as manyparameters as possible. The above observation forces us to give up independence inΩ, and therefore we concentrate on independence on the phase u. It turns out thatit is possible to obtain oscillatory integral estimates with constant independent ofthe phase, but it seems that one has to pay the price somehow (see the log in the


estimate below). Such estimates form a rich theory, but here we will only presentan example to give you a taste of what kind of results and techniques are involved.

Theorem 3.4. If u : [0, 1]2 → R is such that∣∣∣ ∂2u

∂x∂y(x, y)

∣∣∣ > 1 ∀(x, y) ∈ [0, 1]2

and furthermore∂3u

∂2x∂y(x, y) 6= 0 ∀(x, y) ∈ [0, 1]2,

we have ∣∣∣ˆ[0,1]2

eiλu(x,y)dxdy∣∣∣ . (log(2 + |λ|))1/2|λ|−1/2.

In particular, the constant is independent of u.

Obviously, there is an analogous statement for ∂x∂2yu in place of ∂2

x∂yu. No-tice that the assumption on this third mixed derivative is nothing more than amonotonicity assumption (which turns out to be necessary).

Proof. Assume λ > 1 for convenience. By Cauchy-Schwarz and Fubini, we have∣∣∣ˆ[0,1]2

eiλu(x,y)dxdy∣∣∣2 ≤ ˆ 1

0

∣∣∣ˆ 1

0

eiλu(x,y)dy∣∣∣2dx · ˆ 1

0

1dx

=

ˆ 1

0

ˆ 1

0

(ˆ 1

0

eiλ(u(x,y)−u(x,y′))dx)dydy′.

Letting φy,y′(x) := u(x, y)− u(x, y′) we have that

φ′y,y′(x) = ∂xu(x, y)− ∂xu(x, y′) =

ˆ y

y′∂x∂yu(x, t)dt,

which by the hypotheses on u implies that φ′y,y′ is monotonic (since φ′′y,y′ is single-signed) and that |φ′y,y′(x)| > |y − y′|. By Proposition 2.2 it follows that∣∣∣ˆ 1

0

eiλ(u(x,y)−u(x,y′))dx∣∣∣ . min1, (λ|y − y′|)−1.

Thus the above is bounded byˆ 1

0

ˆ 1

0

min1, (λ|y − y′|)−1dydy′

= λ−1

¨|y−y′|>λ−1

|y − y′|−1dydy′ +

¨|y−y′|<λ−1

1dydy′ . λ−1 log λ+ λ−1,

which gives the claim.

It is not known whether the estimate holds without the logarithmic term.


Exercises

This is a (long) list of exercises meant to improve your grasp on the topic andprovide you with useful ideas for the future - do the ones you like. Those that requirea bit more work are marked by F’s - they are not necessarily harder, just longermaybe. The ones unmarked are probably more important for a basic understandingthough. In any case, hints are given in the next section. Since the exercises aremeant to be a complement to the lectures, don’t be afraid to have a look at thehints - you are actually encouraged to do so.

Exercise 1. (F) Let p be such that 1 ≤ p < 2d/(d + 1). Show, by using thedecay estimate J(t) . (1 + |t|)−(d−1)/2, that if f is a radial function in Lp(Rd)and Rd 3 ξ 6= 0, then the Fourier transform f is continuous at ξ. (See hints for awalk-through)

Exercise 2. Show, using the geometric interpretation given in Section 1.2 of solu-tions to n = x2 + y2 as points in a disk, that there exists some constant C > 0 suchthat for all N > 1 ∣∣ N∑

n=1

r2(n)− πN∣∣ ≤ CN1/2

holds.

Exercise 3. If D is a (linear) differential operator, its transpose is given by thelinear operator Dt that satisfies

〈Df, g〉 = 〈f,Dtg〉

for all test functions f, g ∈ C∞c . While D satisfies Leibniz’s rule

D(fg) = gDf + fDg,

show that Dt in general does not.

Exercise 4. Show that Proposition 2.2 is false if the assumption that φ′ is mono-tonic is dropped. In other words, untangling the quantifiers, construct a familyof phases φλλ such that |φλ′| > 1 on (a, b) for each λ, but |λ||I(λ;φλ)| is notbounded as λ→ +∞.

[To destroy cancellation in I(λ), you should aim to make φλ′ oscillate at the scale|λ|−1. Indeed, consider the following: looking only at the real part of the integral,we are trying to make

´ 1

0cos(2πλφλ(t))dt large (and positive, say). The function

cos(2πλx) is positive for x in[− 1

4λ ,1

4λ

]+ 1

λZ, and negative otherwise. If you lookat the graph of φλ (and I encourage you to make a drawing of the argument thatfollows), you want it to spend as much time as possible in the horizontal bandsR ×

([− 1

4λ ,1

4λ

]+ 1

λZ)and as little time as possible in the complement, that is

in the bands R×([

14λ ,

34λ

]+ 1

λZ), so that the positive contribution outweighs the

negative one. To achieve this, φλ′ should be small when φλ is in the former bands(but not too small, since we still want |φλ′| > 1), and quite large when in the latterbands. In particular, φλ′ will not be monotone and it will oscillate between twobehaviours over an interval of length ∼ λ−1.]

Exercise 5. Show the following weaker version of Van der Corput’s lemma fork = 1: assume that |φ′| > µ but, instead of assuming that φ′ is monotonic, we onlyassume that |φ′′| < M on (a, b); prove that

|I(λ)| .( 1

µ+M(b− a)

µ2

)|λ|−1.


Exercise 6. Let φ be a phase that satisfies |φ(k)| > 1 on (a, b) for some k ≥ 1.Show that a necessary condition for inequality

|I(λ)| ≤ Cφ|λ|−α

to hold with Cφ independent of [a, b] is that α = 1/k.

Exercise 7. Show, using integration by parts (as many times as necessary) andVan der Corput’s lemma, that we have the estimate

|J(t)| .d (1 + |t|)−(d−1)/2,

where J is the function introduced in Section 1.

Exercise 8. The Airy equation is the dispersive PDE∂tu+ ∂3

xu = 0,

u(0, x) = f(x).

By taking a Fourier transform in the spatial variable x, it is not hard to see thatthe solution to the Airy equation can be written formally as the convolution

u(t, x) =1

2π

ˆf(y)

1

t1/3Ai(x− yt1/3

)dy,

where Ai(x) denotes the Airy function

Ai(x) :=

ˆRei(ξ

3+xξ)dξ,

which is clearly an oscillatory integral (provided the integral exists!). You will showthat it satisfies the following estimates:1) For x > 1 we have superpolynomial decay, that is for every N > 0 we have|Ai(x)| . |x|−N .

2) For −1 ≤ x ≤ 1 we have |Ai(x)| . 1.3) For x < −1 we have |Ai(x)| . |x|−1/4.Follow these steps:i) It will be useful to smoothly split R dyadically. Let ϕ be a C∞ bump function

supported in [−2, 2], with ϕ ≡ 1 on [−1, 1]; define ψ(ξ) := ϕ(ξ/2)− ϕ(ξ), andψj(ξ) := ψ(2−jξ). Show that ϕ(ξ) +

∑j∈N ψj(ξ) = 1 for every ξ and that ψj

is supported in 2j < |ξ| < 2j+2. We can make sense of the integral definingAi(x) as the limit as n→∞ of

êi(ξ

3+xξ)ϕ(ξ)dξ +

n∑j=0

êi(ξ

3+xξ)ψj(ξ)dξ.

ii) Splitting the integral using the decomposition above, show 1) by the principleof non-stationary phase. In doing this, you will also prove that the limit aboveindeed exists pointwise.

iii) Show 2) adapting the argument you just used for ii).iv) Show 3) by Corollary 2.4 (and the splitting).v) Show the simple dispersive estimate

‖u(·, t)‖L∞(R,dx) . |t|−1/3‖f‖L1(R).

[The integration in the definition of Ai(x) is over all of R instead of a finite intervalas in Section 2; that is why we introduced the splitting. More importantly, noticethat the phase xξ+ξ3 is not of the form λφ, or at least so it appears at first sight...]


Exercise 9. Show estimate (1.2) using integration by parts, Van der Corput’slemma and splitting the integrals where necessary; that is, when d = 2 (thus J(r) =´ π

0e−2πir cos θdθ) show that we have for large R∣∣∣ ˆ R

0

rJ(r)dr∣∣∣ . R1/2.

The point is that´ R

0rJ(r)dr is itself an oscillatory integral (expanding J and using

Fubini shows this is the case).[Notice this is a much better estimate than the one you would get from just pluggingin the estimate |J(r)| . |r|−1/2, which would give you the much larger O(R3/2).There is therefore additional cancellation to be exploited.]

Exercise 10. An estimate of the form

|t ∈ (a, b) s.t. |φ(t)| < λ| = O(λα)

for some α > 0 is called a sublevel-set estimate. Notice that it is only interesting forsmall λ. These estimates are in many ways related to oscillatory integral estimates,as this exercise will show:i) Show that Van der Corput’s lemma implies that, under the same hypotheses

on φ, for every λ > 0 we have

|t ∈ (a, b) s.t. |φ(t)| < λ| .k λ1/k. (♠)We stress that the constant only depends on k.

ii) Show conversely that if we assume that (♠) holds, we can prove Van derCorput’s lemma by splitting the interval (a, b) into t ∈ (a, b) s.t. |φ′(t)| < θand its complement, where θ is a parameter to be optimized (Remark 2 mightcome in handy).

iii) Now prove (♠) directly by induction in k (thus providing another proof of Vander Corput’s lemma, when combined with ii)).

iv) (Only if you know about p-adics, otherwise ignore) Let p be a prime and letZp be the ring of p-adic integers, with its non-archimedean valuation | · |p. IfP is a polynomial in Z[X], what does the sublevel-set

x ∈ Zp s.t. |P (x)|p < p−scorrespond to, in simpler terms? And what would an estimate like (♠) meanin this context?

Exercise 11. (F) The proof of Van der Corput’s lemma we have presented inthese notes gave us a constant that is exponential in k; in particular, you can easilysee that it gives Ck ∼ 2k (Theorem 2.3). This behaviour is not sharp. Indeed, ifBk denotes the smallest constant for which the inequality

|I(λ)| ≤ Bk|λ|−1/k

holds (under the hypotheses of Theorem 2.3), then the constant actually growslinearly in k, namely Bk ∼ k.In this exercise you will obtain the optimal behaviour of the constant, showing that|I(λ)| ≤ Ck|λ|−1/k. This will be achieved by using the same strategy as in ii)-iii)of Exercise 10, but with an improved dependency on k for estimate (♠).i) Show that, if E is a measurable subset of R with |E| > 0, for any k one can

find x0, . . . , xk ∈ E such that for any j ∈ 0, 1, . . . , k∏i : i 6=j

|xi − xj | ≥( |E|

2e

)k.

[hint: squeeze E into an interval of length |E|, then do the obvious thing.]


ii) Show the following generalisation of the mean-value theorem: let x0 < x1 <· · · < xk and f ∈ Ck([x0, xk]); then there exists y ∈ (x0, xk) such that

f (k)(y) = (−1)kk!

k∑j=0

f(xj)∏i : i 6=j(xi − xj)

.

[hint: use Lagrange interpolation at the xj ’s to get a polynomial approximant,then apply Rolle’s theorem to the difference k times.]

iii) Show that if φ is such that φ(k)(t) > 1 for all t ∈ R, then we have the sublevel-set estimate (see Exercise 10)

|t ∈ R s.t. |φ(t)| < λ| ≤ (2e)((k + 1)!)1/kλ1/k.

In particular, use i) and ii) above applied to E = t ∈ R s.t. |φ(t)| < λ andf = φ.

iv) Use Stirling’s approximation to show that ((k + 1)!)1/k ∼ k.v) Prove Van der Corput’s lemma using the same proof as in ii) of Exercise

10 (using the sublevel-set estimate above that has an improved constant) toconclude that

|I(λ)| ≤ Ck|λ|−1/k.

vi) Show that Bk cannot grow slower than k [check a canonical example anduse complex integration techniques; don’t expect a straightforward calculationthough.]

Exercise 12. You might have noticed that the statement of Theorem 2.5 (themethod of stationary phase) omits the case k = 1. Indeed, in this case there canbe no expansion in terms of powers of λ−1, since the integral is ON (λ−N ) for everyN > 0 (by non-stationary phase principle). However, in the case where the supportof ψ is not strictly contained in (a, b) (and thus non-stationary phase does notapply), a similar statement holds. Prove an asymptotic expansion of the form

λ−1∑j∈N

(aje

iλφ(a) + bjeiλφ(b)

)λ−j

for Iψ(λ), under the hypotheses that φ′ 6= 0 in (a, b) and that φ, ψ ∈ C∞ (inparticular, in general ψ(a), ψ(b) 6= 0). Calculate explicitely the first few coefficientsaj , bj .[hint: just use integration by parts repeatedly. ]

Exercise 13. (FF) In this exercise you will prove the method of stationary phaseas stated in Theorem 2.5. The proof will proceed in stages, and we can assumet0 = 0 and φ(0) = 0 for simplicity.i) Begin with the case k = 2, which will already contain all the main ideas.

Take φ(t) = t2 and ψ(t) = tme−t2

for some m > 0 (which is not compactlysupported, but still very concentrated around the origin); let a = −∞ andb = +∞. Show, using standard complex integration techniques, that Iψ(λ) =´ +∞−∞ eiλt

2

tme−t2

dt equals (fixing the principal branch of z−(m+1)/2)

(1− iλ)−(m+1)/2

ˆ +∞

−∞tme−t

2

dt,

and argue by a power series expansion in λ−1 that therefore it satisfies (2.1).ii) Now we keep the quadratic phase φ(t) = t2 but replace the gaussian factor

e−t2

directly with a compactly supported function η. Thus, let η ∈ C∞c andlet ψ(t) = tmη(t). Show, by splitting the region of integration smoothly into


one close to 0 and one away from 0 (according to a well chosen parameter),that in this case

|Itmη(λ)| .η,m |λ|−(m+1)/2.

More precisely, let ϕ denote a smooth bump function supported in [−1, 1] anddecompose 1 = ϕ(t/δ)+(1−ϕ(t/δ)). For the region close to 0 you can take theabsolute value inside the integral and give a trivial estimate; for the region awayfrom 0, apply the integration by parts argument as in the proof of Proposition2.1 N times, where 2N is bigger than m+1, then take the absolute value insideand estimate the size of this integral. Finally, optimise in the parameter δ.

iii) Show, using the above same arguments, that if g ∈ S(Rd) (that is, g is aSchwartz function) is such that g ≡ 0 in a neighbourhood of 0, then for anyN > 0 one has |Ig(λ)| = ON,g(|λ|−N ). (The phase is still φ(t) = t2.)

iv) Now, still in the case φ(t) = t2, we tackle the general ψ case. Writeêiλt

2

ψ(t)dt =

êiλt

2

e−t2(et

2

ψ(t))η(t)dt,

where η ∈ C∞c is such that η(t) = 1 for t in the support of ψ. Perform a Taylorexpansion of et

2

ψ(t) to degree n and substitute into Iψ(λ). Then use i)-ii)-iii)to argue that (2.1) holds for each term you get out of this procedure.

v) Now consider the general k = 2 case. The idea is to deform the phase to turn itinto t2 again. Find a diffeomorphism from a sufficiently small neighbourhoodU of 0 that, by a change of variables, achieves this, and then conclude that(2.1) holds by iv).

vi) Generalise the above to k > 2. The substitute for i) will be the identityˆ ∞0

eiλtk

e−tk

tmdt = Ck,m(1− iλ)−(m+1)/k,

which is similarly proven by standard complex integration techniques.

Exercise 14. In these notes we have only considered amplitudes ψ that are ev-erywhere C∞. In applications however (see viii) below) more singular cases arise,and it is instructive to see that some of those cases can still be treated using thetechniques here developed. This exercise will moreover show you that sometimessome cancellation can also arise from the amplitude.Let P (t) =

∑dj=1 cjt

j be a polynomial of degree d. You will show that for arbitrary0 < ε < R it holds ∣∣∣ˆ

ε<|t|<ReiP (t) dt

t

∣∣∣ ≤ Cdwith a constant that depends only on d (and not on the coefficients of P !). Noticethat the amplitude 1/t is only C∞ away from 0, and has a bad singularity there.In the following, it will be really important that integration is over a set symmetricwith respect to the origin such as t ∈ R : ε < |t| < R = (−R,−ε)∪ (ε,R) above.i) Show by a change of variable that we can assume that cd = 1 (at the price of

changing ε,R; but these are arbitrary).ii) Show, using Corollary 2.4, that the portion of the integral over 1 < |t| < R

is bounded by Od(1), so that it suffices to concentrate on the remaining partε < |t| < 1. (Notice that even in the range |t| > 1 we cannot estimate theintegral by taking absolute values, since the resulting quantity grows like logR;we really need oscillatory integral estimates!).

iii) Let Pk(t) :=∑kj=1 cjt

j (thus Pd = P ). Using the trivial estimate |eiθ−1| . |θ|and taking absolute values, show that∣∣∣ˆ

ε<|t|<1

eiP (t) dt

t−ˆε<|t|<1

eiPd−1(t) dt

t

∣∣∣ .d 1.


iv) Use steps i)-iii) repeatedly to reduce to the case of integral´ε<|t|<1

eic1t dtt .v) It remains to show that the above integral is Od(1). Show that

á<|t|<b

dtt = 0

for any a, b > 0 (this is a form of cancellation!). Therefore, you can arbitrarilysubtract 1 from the exponential term eic1t above...

vi) Split the integral as´ε<|t|<1/c1

+´

1/c1<|t|<1and apply again Corollary 2.4 to

the second integral to show that it is Od(1).vii) It remains to estimate

´ε<|t|<1/c1

eic1t dtt . Using v), show that it is the sameas´ε<|t|<1/c1

(eic1t − 1)dtt and estimate the latter by taking the absolute valueand using the trivial estimate for |eiθ − 1|. This concludes the proof.

viii) Consider the operator

f 7→ limε→0,R→∞

ˆε<|t|<R

f(x− t, y − t2)dt

t;

this is known as the Hilbert transform along a parabola. Show that this oper-ator is L2(R2) → L2(R2) bounded. [hint: use Plancherel and take a Fouriertransform of the operator above; after some calculations, you should recognisea certain object... ]

Exercise 15. (F) The discrete counterparts to oscillatory integrals, which wehaven’t discussed in these notes, are exponential sums, that is sums of the form

N∑n=1

e2πif(n)

(sometimes they are normalised by dividing by N). Such sums are ubiquitous innumber theory, where they are used to encode all sorts of information, from thenumber of solutions to diophantine equations to the behaviour of the Riemann Zetafunction.It should not be too surprising that some of the techniques developed for oscillatoryintegrals transfer over to the treatment of exponential sums. Indeed, integration byparts has a very direct discrete counterpart in summation by parts. In this exerciseyou will prove some estimates for exponential sums in the spirit of the ones studiedin these lectures. We will use a notational convention from number theory, namelywe introduce the function e(x) := e2πix so that e2πif(n) = e(f(n)).You will prove an analogue of Proposition 2.2:i) Show the trigonometric identity

1

1− eiθ=

1

2+i

2cot(θ/2).

ii) Let g(n) = e(f(n))e(f(n))−e(f(n−1)) ; show by summation by parts that

N∑n=1

e(f(n)) =[e(f(N))g(N)− e(f(0))g(1)

]−N−1∑n=1

e(f(n))(g(n+ 1)− g(n)).

iii) Using i), show that

g(n+ 1)− g(n) = (i/2)[

cot(π(f(n+ 1)− f(n)))− cot(π(f(n)− f(n− 1)))].

iv) So far we just did a bunch of algebra; now we start doing analysis. Assumethat f ′ is monotonic in the interval [0, N ] and that for some 1/2 > δ > 0we have dist(f ′(x),Z) > δ for all x ∈ [0, N ] (that is, the derivative alwaysstays within two consecutive integers, never getting any closer than δ to any


of them). Show that the sequence n 7→ f(n)− f(n− 1) is also monotonic andthat for some fixed integer k we have for all n ∈ 1, . . . , N

k + δ ≤ f(n)− f(n− 1) ≤ k + 1− δ.

v) Since cot is also monotonic in intervals of the form [πk, π(k + 1)], show thatunder the hypotheses in iv) on f we have∣∣∣N−1∑

n=1

e(f(n))(g(n+ 1)− g(n))∣∣∣ ≤ |g(N)− g(0)|,

and deduce that the above is bounded by . δ−1. [hint: monotonicity allowsyou to take the absolute values from inside of a sum to outside of it; also,cot(x) ≤ 1/x for x ∈ (0, π/2].]

vi) Put everything together (check those remaining terms) and conclude that youhave shown: if f ′ is monotonic on interval I and dist(f ′(x),Z) > δ for allx ∈ I, then ∣∣∣ ∑

n∈I∩Ze(f(n))

∣∣∣ . δ−1. (♦1)

Compare this with Proposition 2.2 and appreciate the similarities betweenthe respective proofs.

Now we move on to the 2nd derivative. You will prove an analogue of Van derCorput’s lemma, that is: if f ∈ C2 is such that for all x ∈ I we have

λ < |f ′′(x)| < Cλ

(for some C > 1), then∣∣∣ ∑n∈I∩Z

e(f(n))∣∣∣ . C|I|λ1/2 + λ−1/2. (♦2)

The strategy will be to partition I into intervals where (♦1) above applies, andwhere it doesn’t to use trivial estimates, and finally optimize between the two.a) Assume for convenience that f ′′ is positive and let I = [a, b]. Show that the

range of f ′ is J := [f ′(a), f ′(b)] and show that |J | ≤ Cλ|I|.b) Partition J into intervals close to integers and intervals that avoid integers. More

precisely, let 1/2 > δ > 0 be a parameter to be chosen and define

Ak :=[k − δ, k + δ] ∩ J,Bk :=[k + δ, k + 1− δ] ∩ J.

Show that Ak and Bk are not empty for at most O(Cλ|I| + 1) values of k andthat they partition J .

c) Since f ′ is monotonic, we can partition I into a disjoint union of intervals(f ′)−1(Ak) and (f ′)−1(Bk). Show that on the first ones we have trivially∣∣∣ ∑

n∈(f ′)−1(Ak)∩Z

e(f(n))∣∣∣ . δ/λ

and on the second ones by (♦1) we have∣∣∣ ∑n∈(f ′)−1(Bk)∩Z

e(f(n))∣∣∣ . δ−1.

d) Now put everything together to show that we have∣∣∣ ∑n∈I∩Z

e(f(n))∣∣∣ . (Cλ|I|+ 1)

( δλ

+1

δ

),


and conclude estimate (♦2) by choosing an optimal δ (i.e. minimise δ/λ+ 1/δ).Again, you should appreciate the similarities between this proof and the proofof Van der Corput’s lemma we have given in the notes.

e) An estimate is only interesting if it beats the trivial one, which in this case is|∑n∈I∩Z e(f(n))| ≤ |I| (by triangle inequality; here |I| > 1). For which range

of values of λ is estimate (♦2) interesting?f) Consider truncations of the Riemann Zeta function along the critical line, that

is sums of the form∑Bn=A n

−1/2−it. Argue that in order to bound such sums,by summation by parts one can reduce to study sums of the form

∑B′

n=A′ n−it

instead. As n−it = e(−(t/2π) log n), verify that the estimates (♦1), (♦2) thatyou proved above are well suited to treat these last oscillatory sums. This ishow bounds on the growth of

∣∣ζ( 12 + it

)∣∣ as t→∞ are usually proven.

Exercise 16. Let V denote the real vector space of partial derivatives of order k,that is the space of differential operators of the form∑

α :|α|=k

cα∂α.

Show that V admits a basis of vectors all of the form (ξ · ∇)k, where the ξ’s areunit vectors.Equivalently, you can identify V with the real vector space of homogeneous poly-nomials of degree k in d variables.[hint: Using the identification with homogeneous polynomials, it suffices to find aninner product on V such that if a polynomial is orthogonal to all polynomials ofthe form (ξ ·X)k, with X = (X1, . . . , Xd), then it has to be the zero polynomial.]

Exercise 17. Show again the estimate

|J(t)| .d (1 + |t|)−(d−1)/2

as in Exercise 7, but this time using the method of stationary phase in severalvariables as in Theorem 3.3. Use the expression for J in terms of an integral overthe sphere, then express the half-sphere as the graph of a function.Relatedly, recall that if µ is a finite measure on Rd we can define its Fourier trans-form dµ as

dµ(ξ) :=

ˆRde−2πix·ξdµ(x).

Calculate the Fourier transform dσ(ξ) of the spherical measure dσ on the (d− 1)-dimensional sphere Sd−1 and deduce its decay properties from the above. Oscilla-tory integrals provide a great way to obtain such estimates.Finally, consider the flat surface in Rd given by [−1, 1]d−1 × 0 and let dµ be itssurface measure. Show that there is a direction in which dµ does not decay at all.This should convince you that the source of the extra cancellation that makes dσdecay is the curvature of the underlying surface.

Exercise 18. In this exercise you will show that in dimension greater than 1 it isimpossible to have estimates of the form∣∣∣ ˆ

Ω

eiλu(x)dx∣∣∣ ≤ C|λ|−1/|α|

with constant C independent of Ω (here we are assuming |∂αu| > 1).i) Repeat the proof you gave for i) of Exercise 10 to show that the inequality

above would imply

x ∈ Ω s.t. |u(x)| < λ ≤ C ′λ1/|α|


with C ′ independent of Ω. Thus it will suffice to disprove the sublevel-setestimates.

ii) First we show that if Ω is not bounded, there is no hope to have an estimatelike the above. Consider Ω = [−R,R]2 and u(x, y) = 1

2 (x + y)2. We have∂x∂yu = 1. Show that, however,

|(x, y) ∈ Ω s.t. |u(x, y)| < λ| ∼ Rλ1/2,

so there is an unavoidable dependence on diam(Ω).iii) Now we show that even if we take Ω contained in a bounded set (say [0, 1]2),

the constant will still depend on properties of Ω. Here the trick will be toconsider a set Ω that looks a bit like a comb with N teeth. In particular,define

Ωj := [1/2, 1]× [j/N, (j + 1/2)/N ]

for j = 0, . . . , N − 1 (this will be the j-th “tooth”) and

Ω := [0, 1/2]× [0, 1] ∪N−1⋃j=0

Ωj .

We define the phase u piecewise. Let ϕ be a smooth positive function suchthat ϕ ≡ 0 on [0, 5/8] and ϕ ≡ 1 on [3/4, 1]. Then on each tooth we prescribeu to behave the same way: for (x, y) ∈ Ωj , u(x, y) = y− jϕ(x)/N . On the restof the comb, we prescribe: for (x, y) ∈ [0, 1/2]× [0, 1], u(x, y) = y. Show that|∂yu| = 1 on Ω and show that for λ sufficiently smaller than N−1 we have

|(x, y) ∈ Ω s.t. |u(x, y)| < λ| & Nλ,

so that the constant necessarily depends on Ω.

Exercise 19. (F) In this exercise you will prove a result that is weaker than the fullasymptotics provided in Theorem 3.3 but which is sufficient for most applications.In particular, under the same hypotheses (in particular, the non-degeneracy of thecritical point of u), you will prove the estimate

|Iψ(λ)| .u,ψ,d |λ|−d/2.

The proof is quite similar to that of Theorem 3.4. For starters, let u, ψ satisfy thehypotheses of Theorem 3.3. We expand |Iψ(λ)|2 into a double integral in dxdy andchange variables to (y, z) by introducing z = x− y. We obtain

|Iψ(λ)|2 =

ˆRd

ˆRdeiλ(u(y+z)−u(y))ψ(y + z)ψ(y)dydz.

i) Verify the calculations above.ii) The expression we have is of the form

´Rd Jλ(z)dz with Jλ an oscillatory

integral. You will show that for any N > 0 it is |Jλ(z)| .N (1 + λ|z|)−N .Show that this will imply that |Iψ(λ)|2 . |λ|−d, hence concluding the estimatewe want.

iii) Now you will prove the claimed |Jλ(z)| .N (1 + λ|z|)−N using essentiallythe same proof given for Proposition 2.1. The oscillatory term of Jλ(z) iseiλ(u(y+z)−u(y)); calculate its ∇y gradient and use this information to craft adifferential operator D such that D(eiλ(u(y+z)−u(y))) = eiλ(u(y+z)−u(y)).

iv) Your differential operator should have the form D = (iλ)−1θz · ∇y for someexplicit Rd-valued function θz(y). Show by simple Taylor expansion argumentsthat for any multi-index α one has

‖∂αy θz(y)‖ .α |z|−1.


v) Now use integration by parts N times to show that

Jλ(z) =

êiλ(u(y+z)−u(y))(Dt)N (ψ(y + z)ψ(y))dy

and, using iv) above, conclude the estimate (Dt being the transpose of D, whichyou will have to work out explicitely).

Exercise 20. (FF) In this exercise you will prove the method of stationaryphase in several variables, that is Theorem 3.3, following the same strategy usedin Exercise 13 (which you should attempt before attempting this one). Assumex0 = 0.Let x = (x1, . . . , xd), and if α ∈ Nd is a multi-index let xα := xα2

1 · . . . · xαdd .

i) Assume the phase is purely a non-degenerate quadratic form Q(x) =∑`j=1 x

2j−∑d

j=`+1 x2j . Take ψ(x) = e−‖x‖

2

xα and show that (3.1) holds, that isˆRdeiλQ(x)e−‖x‖

2

xαdx ' λ−d/2−|α|/2∑j∈N

a`,αj λ−j .

[hint: the integral factorises, then you can repeat what you did in Exercise13.]

ii) Now we need to prove a result for a compactly supported ψ. Indeed, let η(x) asmooth function compactly supported around the origin and let ψ(x) = xαη(x).You will show that, with the same purely quadratic phase Q as above, we have

|Iψ(λ)| .α,η,Q |λ|−(d+|α|)/2.

To do so, we will need to use integration by parts at some point with respectto some ∂xk , and thus it will be convenient to localise in that direction. Weachieve this as follows: we roughly divide Rd into the cones

Γk := x ∈ Rd s.t. ‖x‖2/d ≤ |xk|2

for k ∈ 1, . . . , d; you can see that Γk is a cone with axis the xk-axis. Showthat

Rd =

d⋃k=1

Γk.

Using slightly larger cones

Γk := x ∈ Rd s.t. ‖x‖2/(2d) ≤ |xk|2,

argue by a partition of unity that it suffices to reduce to the case where theamplitude is ψ(x) · Gk(x) with Gk a function homogeneous of degree 0 (thatis, Gk(x) = Gk(x/‖x‖)), smooth away from the origin, supported in Γk andidentically 1 in a conical neighbourhood of the xk-axis (say, points x such that(9/10)‖x‖2 ≤ |xk|2).

iii) This step will be the exact analogue of step ii) in Exercise 13. With phaseQ and amplitude xαη(x)Gk(x), split the region of integration smoothly into‖x‖ . δ and the complement. For the first part use a trivial estimate, and forthe second part use repeated integration by parts with respect to the differentialoperator

Dkf(x) := ± 1

2iλxk∂xk .

Optimizing in δ, conclude the estimate in ii). [hint: notice that Dk applied toeiλQ returns exactly eiλQ.]


iv) Using the same arguments as above, show that if g ∈ S(Rd) is identically 0near the origin, then for any N > 0

|Ig(λ)| .N,g |λ|−N .v) For a generic amplitude ψ and phase Q as above, writeˆ

eiλQ(x)ψ(x)dx =

êiλQ(x)e−‖x‖

2(e‖x‖

2

ψ(x))η(x)dx

with η compactly supported and identically 1 on the support of ψ. Perform aTaylor expansion of e‖x‖

2

ψ(x) to some finite degree, then use i)-iv) above toargue that (3.1) holds for this oscillatory integral.

vi) Finally, if u is a generic phase satisfying the hypotheses of Theorem 3.3, argueby Morse’s Lemma that there is a smooth change of variables that lets us putu in the form Q above (provided ψ has small enough support). Therefore,conclude by v). [If you do not know about Morse’s lemma, it says more or lessprecisely what we claimed; look it up, it is a very convenient tool.]

Exercise 21. (FFF) Oscillatory integrals are prototype solutions for many par-tial differential equations:1) Show that the oscillatory integral

u(x, t) :=

êi(t|ξ|

2+x·ξ)ψ(ξ)dξ

is a solution for the Schrödinger equation i∂tu−∆u = 0.2) Show that the oscillatory integral

u(x, t) :=

êi(t|ξ|+x·ξ)ψ(ξ)dξ

is a solution for the wave equation ∂2t u−∆u = 0.

3) Show that the oscillatory integral

u(x, t) :=

êi(tm〈ξ/m〉+x·ξ)ψ(ξ)dξ,

where 〈ξ〉 (the so-called japanese bracket) denotes (1 + |ξ|2)1/2, is a solution forthe Klein-Gordon equation ∂2

t u − ∆u + m2u = 0 (m > 0 is a constant - themass).

Now consider the wave equation in Rd above. You will prove some decay estimatesfor a packet of waves of frequency approximately 2k.Let ψ(ξ) be a C∞ function supported in the annulus 1 < |ξ| < 2 and radiallysymmetric, for simplicity. Fix an integer k ∈ Z and consider the solution to thewave equation

Φk(x, t) :=

êi(x·ξ+t|ξ|)ψ

( ξ2k

)dξ,

which is frequency localised in 2k < |ξ| < 2k+1. You will show that for any N > 0the above solution obeys the estimate

|Φk(x, t)| .N 2kd(1 + 2k||x| − |t||)−N (1 + 2k|t|)−(d−1)/2. (♣)The proof will boil down to a number of applications of the non-stationary phaseprinciple, which will reduce matters to estimating the contribution of the criticalregion of integration where the phase t|ξ|+ x · ξ is stationary.i) As a warm-up, give a physical interpretation of (♣) when t is large. For the

sake of the interpretation, replace the fast-decaying term (1 + 2k||x| − |t||)−Nby the characteristic function 1[−1,1](2

k||x| − |t||) and draw its support in aspace-time diagram. Show that the estimate is compatible with the energyconservation law for waves.


ii) Show by rescaling that it suffices to prove (♣) for the case k = 0.iii) Show that the symmetries of the problem allow one to assume that t > 0 and

x = (x1, 0, . . . , 0) with x1 ≥ 0.iv) Now you have an oscillatory integral with phase φ(ξ) = t|ξ| + x1ξ1. Show

by using the non-stationary phase principle in ξ1 that when t . 1 one has|Φ0(x1, t)| .N (1 + x1)−N for any N > 0. This proves (♣) in the smalltime regime. Notice that the phase is not of the form λφ and therefore onecannot use the non-stationary phase principle (Proposition 2.1) out of the box;however, the proof adapts easily.

v) Assume then that t & 1. We concentrate on proving the localization in |x|−|t|.Show, again by non-stationary phase principle in ξ1, that if t > 4x1 or 1

4x1 > t

then one has |Φ0(x1, t)| .N (1 + |x1 − t|)−N for any N > 0. Show that thisproves (♣) in this regime.

vi) Assume therefore that t & 1 and that 4x1 > t > 14x1. Show that the phase

φ(ξ) = t|ξ|+ x1ξ1 has critical points only if t = x1 and the set of such criticalpoints is contained in (−∞, 0]e1. We expect that the main contribution tothe integral defining Φ0 comes from these critical points; therefore, smoothlyexcise a thin cone with axis Re1, say Γ := ξ = (ξ1, ξ

′) ∈ R× Rd−1 s.t. |ξ1| >(99/100)|ξ|, from the support of ψ and show by non-stationary phase principlein ξ′ that the remaining integral (outside the cone) is dominated by ON ((1 +t)−N ) (which is acceptable since it is dominated by (♣) in this regime).

vii) Show by non-stationary phase principle in ξ1 that one can do away with theportion of the cone Γ where ξ1 > 0. Indeed, in that region |∇φ| & x1; showthat this contribution is bounded by ON ((1+x1)−N ), which is again acceptablein this regime.

Critical region Γ

Supp(ψ)

viii) Finally, we come to the critical region (the portion of the cone where |ξ| ∼ 1and ξ1 < 0). You have shown that if x1 = t and ξ = (−ξ1, 0) then∇φ = 0. Nowshow that, even when x1 6= t, if |ξ′| . t−1/2 then |φ(−ξ1, ξ′)−φ(−ξ1, 0)| ≤ 1/10(this means that the phase is barely changing). Taking advantage of thisinformation, split (smoothly) the critical region dyadically into a sub-regionwhere |ξ′| . t−1/2 and sub-regions where |ξ′| ∼ 2jt−1/2, where j ∈ N (and2jt−1/2 . 1 because |ξ′| . 1 on the critical region). For the former, showthat ∂ξ1φ(ξ) = x1 − t + O(1). This is good when |x1 − t| & t, but if x1 − t issmall then the integration by parts argument behind the non-stationary phaseprinciple accumulates some unfortunate powers of t (coming from the fact that|∂jξ1φ| ∼ t for j > 2). To compensate for this, observe that there is oscillationin the ξ′ directions as well, and thus you should apply a non-stationary phaseargument both in ξ1 and ξ′. Show therefore that this contribution is boundedexactly by (♣). [If it helps, consider first the totally critical case x1 = t. ]

ix) For the remaining sub-regions, show that for each j the quantity ∂ξ1φ(ξ) −(x1− t) is positive and has size ∼ 22j (recall 22j . t). Use once again the non-stationary phase principle in ξ1, ξ′ to estimate the contribution of each region.Finally, sum up in j and show that the result is controlled by (♣) again.


Hints

Hints for Exercise 1: Take the estimate

|J(t)| . (1 + |t|)−(d−1)/2 (♥)for granted, and follow these steps:(i) Split the function f ∈ Lp(Rd) as f(x) = f1(x) + f2(x) := f(x)1B(0,1)(x) +

f(x)(1 − 1B(0,1)(x)), where 1B(0,1) denotes the characteristic function of theball of radius 1 centered at 0. Argue that f1 is in L1(Rd) and therefore thecontribution to f given by f1 is continuous at ξ 6= 0.

(ii) Let ρ = |ξ| and write the radial Fourier transform g(ρ) of f2 according toequation (1.1); then in the integral use a change of variable and argue that itwill suffice to show the continuity of the function Φ(ρ) := ρdg(ρ).

(iii) It suffices to show continuity of Φ from above and from below, so take ρ′ > ρand show that you can write (abusing notation a little, we write f2 for theradial part of f2(x) as well)

Φ(ρ)− Φ(ρ′) =

ˆ ρ′

ρ

f2

( sρ

)sd−1J(s)ds (3.2)

+

ˆ +∞

ρ

(f2

( sρ

)− f2

( sρ′

))sd−1J(s)ds. (3.3)

(iv) Fixing ρ, use Hölder’s inequality and (♥) (and the hypothesis on p) to showthat the integral

´ +∞ρ|f2

(sρ

)|sd−1|J(s)|ds is finite; hence deduce by standard

arguments that term (3.2) tends to 0 as ρ′ → ρ+.(v) Using again Hölder’s inequality and (♥), argue that term (3.3) also tends to 0

as ρ′ → ρ+ (you might want to do a change of variables s = ρt and then argueby the continuity the Lp norm with respect to dilations). Putting everythingtogether, the proof is concluded.

Hints for Exercise 2: Prove upper- and lower-bounds for∑Nn=1 r2(n); tile the

plane into squares with centers in the Z2 lattice.Hints for Exercise 3: Try with D = a(x) d

dx .Hints for Exercise 4: Let a = 0, b = 1 for simplicity. Construct phases φλ

with oscillatory first derivatives where the oscillations happen at scale λ−1. A goodchoice would be φλ(t) = 2πt+λ−1θ(λt) with θ a smooth periodic function of period1. Show that

λI(λ) = λ

ˆ 1

0

ei(2πt+θ(t))dt+O(1),

and furthermore show that you can choose θ such that the integral above is 6= 0and such that |φλ′| > 1.

Hints for Exercise 5: Just adapt the proof of Proposition 2.2.Hints for Exercise 6: Do a change of variable t = θs in I(λ), with θ a

parameter. If the inequality is to be invariant with respect to this transformation,a certain condition involving α and k will have to be met.

Hints for Exercise 7: Let Jd be the function

Jd(t) :=

ˆ π

0

e−2πit cos θ(sin θ)d−2dθ

and show by integration by parts that we have the recurrence relation

(2πit)2Jd(t) = (d− 3)(d− 5)Jd−4(t)− (d− 3)(d− 4)Jd−2(t);

iterate this to reduce to the case d = 3, 2. For J3, compute the integral directly andget a bound of O(|t|−1); for J2, show by Van der Corput’s lemma that we have a


bound of O(|t|−1/2) (besides the trivial bound of O(1)). Notice that you will needto split the integral in 3 parts (say,

´ π/40

+´ 3π/4

π/4+´ π

3π/4) in order to apply Van der

Corput, since there is not a uniform lowerbound on φ′ or φ′′ over the entire interval[0, π].

Hints for Exercise 8: Find a rescaling of the ξ variable that allows you torewrite the phase as |x|3/2φ(ξ) = |x|3/2(sgn(x)ξ + ξ3), which is of the form westudied.(ii) When x > 1, φ′ 6= 0. Show that this implies |

éxp(i|x|3/2φ(ξ))ϕ(ξ)dξ| .N

min1, |x|−3N/2 by non-stationary phase. For the ψj(ξ) terms instead, applythe proof of the non-stationary phase principle to each term, with D thedifferential operator such that D(eiφ(ξ)) = eiφ(ξ), to show that it contributes

|x|−3N/2

ˆ|(Dt)Nψj(ξ)|dξ.

To show that this is summable in j, show that one will have

(Dt)Nψj(ξ) =

N∑k=0

cφ,k,N (ξ)2−jkψ(k)(2−jξ)

for some (integrable) functions cφ,k,N depending on φ′ that one can computeexplicitely, and conclude using the fact that

∑j∈N 2−jk|ψ(k)(2−jξ)| .k 1.

If you cannot convince yourself that (Dt)Nψj can be put in that form, try thisother way: show by induction in N that every term that appears in the expan-sion of (Dt)Nψj is of the form (ψj)

(k)(ξ)ξ`(3ξ2 + 1)−m (up to some constant)and that moreover k − ` + 2m > 1. Then show that

´|(ψj)(k)(ξ)||ξ|`(3ξ2 +

1)−mdξ ∼ 2j2−(k−`+2m)j , which is summable in j.(iii) When −1 < x < 1, φ might have critical points (if x < 0), but only in the

support of ϕ. Bound the ϕ contribution by O(1) trivially, and the contribu-tion of the ψj by the same argument above (since the non-stationary phaseprinciple applies to them again).

(iv) When x < −1, φ certainly has two critical points (±1/√

3) inside the supportof ϕ. For the ψj terms, argue as above that they contribute ON (|x|−3N/2)and can thus be ignored. For ϕ, isolate the critical points by splitting ϕ(ξ) =ϕ(100ξ)+(ϕ(ξ)−ϕ(100ξ)) and similarly split the integral. Notice the supportof ϕ(100ξ) is [−1/50, 1/50] and thus does not contain any critical points. Forthe contribution

éxp(i|x|3/2φ(ξ))(ϕ(ξ) − ϕ(100ξ))dξ, notice that |φ′′| & 1

and use Corollary 2.4 with k = 2.Hints for Exercise 9: Here one needs to be a little clever: the point is to notice

that the phase 2πr cos θ has first derivative non-zero around the problematic pointθ = π/2. If ϕ denotes a smooth bump function supported in [−π/4, π/4], split J(r)as J1(r) + J2(r) where

J1(r) :=

ê−2πir cos θϕ(θ − π/2)dθ.

Argue by the non-stationary phase principle that |J1(r)| .N (1 + r)−N and thattherefore

´ R0rJ1(r)dr = O(1). For J2, use Fubini and integration by parts to

show thatˆ R

0

rJ2(r)dr = −ˆ π

0

(Re−2πiR cos θ

2πi cos θ+e−2πiR cos θ

(2πi cos θ)2− 1

(2πi cos θ)2

)(1−ϕ(θ−π/2))dθ.

Where 1−ϕ(θ−π/2) does not vanish, one has that the 2nd derivative of the phasecos θ is bounded from below; conclude therefore using Corollary 2.4 term by term.

Hints for Exercise 10:


i) You need to rewrite the measure of the sublevel set as a nice oscillatory integral.Show that |t ∈ (a, b) : |φ(t)| < λ| =

´ ba1(|φ(t)|/λ < 1)dt. Choose some

smooth non-negative bump function ϕ such that 1(|φ(t)|/λ < 1) ≤ ϕ(φ(t)/λ)for every t and consequently replace the latter in the integral. Apply Fourierinversion to the term ϕ(φ(t)/λ), followed by Fubini, and apply Van der Corput’slemma to the resulting oscillatory integral.

ii) Splitting (a, b) as suggested, (♠) takes care of |´|φ′|<θ exp(iλφ(t))dt| by tak-

ing the absolute value inside and |φ′| ≥ θ consists of a bounded number ofintervals (how many? See Remark 2), for each of which one can apply Propo-sition 2.2 to the associated oscillatory integral. Finally, choose θ in order tooptimize the resulting bound (that is, make it as small as possible).

iii) The base case k = 1 is easy. For the general k case, use the same strategy asin ii): split (a, b) into those points where |φ′(t)| < θ, to which the inductivehypothesis applies, and those where |φ′(t)| ≥ θ, which consist of boundedlymany intervals on each of which you can apply the base case.

iv) What does it mean for x ∈ Zp to have |x|p ≤ p−1? And what does it mean forit to have |x|p ≤ p−2? etc.

Hints for Exercise 11:i) Squeezing E into an interval can only make the left hand side of the inequality

smaller, so it is fine to do. Once E is an interval, just take uniformly distributedpoints inside it as your xj ’s and perform the simple calculation that results.

ii) By Lagrange interpolation, you can find a polynomial P (X) of degree k suchthat P (xj) = f(xj) for all j ∈ 0, . . . , k. Argue that P (k) − f (k) must havea zero inside (x0, xk). Then notice that P (k)(X) must be a constant, preciselyk! times the coefficient of Xk. Calculate that coefficient explicitely (the con-straints P (xj) = f(xj) give you a linear system of equations in the coefficientsof P , with a nice Vandermonde matrix appearing; solving the system withCramer’s rule is a breeze).

iii) Part i) gives you the points xj and for each j reversing the inequality one has(∏i : i 6=j |xi−xj |

)−1

≤ (2e)k|E|−k. Then use ii) together with this inequality,

the fact that |φ(k)| > 1 and that |φ(xj)| < λ for each j.v) Remember Remark 2: how many intervals is |φ′| > θ made of at most?vi) Use (a, b) = (0, 1) and φ(t) = tk. To do this you will need complex integration.

Use a regularisation: evaluate instead´ 1

0eit

k

e−εtk

dt and take ε → 0 at theend. To evaluate the integral, see it as the integral of holomorphic functionezk

along the path Γ0 := (i− ε)1/kt s.t. t ∈ [0, 1] (fixing a branch of the k-throot). Complete the path Γ0 to a well-chosen closed path Γ that includes partof the real line and conclude using Cauchy’s integral theorem and some simpleestimates on the decay of |ezk | along certain directions of the complex plane.

Hints for Exercise 12: The first step is as in the proof of Proposition 2.2, andone hasˆ b

a

eiλφ(t)ψ(t)dt =(eiλφ(b)ψ(b)

iλφ′(b)− eiλφ(a)ψ(a)

iλφ′(a)

)−ˆ b

a

eiλφ(t)( ψ

iλφ′

)′(t)dt.

Iterating this on the integral on the right hand side shows that it is of size O(λ−2)and therefore the above gives you a0, b0. Keep applying integration by parts untilnecessary.

Hints for Exercise 13:i) You should see the integral as the complex integral of (a multiple of) the

holomorphic function zme−z2

along the path Γ0 := (1−iλ)1/2R ⊂ C. Truncate


to Γ+0 := (1 − iλ)1/2t : t ∈ [0, R] and complete this path to a closed

path Γ that includes the real interval [0, R]. Use Cauchy’s integral theoremand simple estimates on |zme−z2 | to conclude that as R → ∞ this (plus anequal contribution from a similarly defined Γ−0 ) gives precisely what is claimed.The power series expansion of (1 − iλ)−(m+1)/2 is standard, just write it asλ−(m+1)/2(λ−1 − i)−(m+1)/2 and expand the function (z − i)−(m+1)/2 in the zvariable.

ii) It is straightforward to show that´|t|m|η(t)|ϕ(t/δ)dt .m,η δm+1. The other

term looks a bit nasty, but luckily one does not have to make all the compu-tations implied. Let η(t)(1 − ϕ(t/δ)) =: ωδ(t) for convenience. Show that thesupport of ωδ(t) and all its derivatives is contained in δ . |t| . 1. The inte-gration by parts argument in the proof of Proposition 2.1 reduces matters toestimating λ−N

´|(Dt)N (tmωδ(t))|dt with Df(t) := (2it)−1f ′(t). (Dt)N is too

horrible to evaluate precisely (this is due to the fact that Dt is not a derivation,that is it does not satisfy Leibniz’s rule; see Exercise 3), but we do not needto. Show by induction on N that every term in the expansion of (Dt)Nf isof the form f (j)t−(2N−j) for j ∈ 0, . . . , N. Apply this to function tmωδ(t)together with the general Leibniz’s rule (the binomial expansion of (fg)(j)) toconclude that (Dt)N (tmωδ(t)) is of the form

m∑k=0

N∑j=k

cj,k,Nω(j−k)δ (t)t−(2N−m)+(j−k)

for some constants cj,k,N whose precise value does not concern us. Show that,by the support of ω(j−k)

δ and the fact that 2N > m + 1, the integral of eachterm is at most λ−Nδ−(2N−m)+1 (careful with ω

(j−k)δ , when expanded this

contains factors of δ−` for ` ≤ j − k). Therefore the quantity to be estimatedis controlled by δm+1 + λ−Nδ−(2N−m)+1, and it remains for you to choose agood δ that minimises this.

iii) Just repeat the second part of the above with δ ∼ 1.iv) Taylor expansion gives you e−t

2

ψ(t) = P (t) + tn+1Rn(t) for some polynomialP of degree n and a remainder Rn(t) that is well-behaved. Write the term´

exp(iλt2) exp(−t2)P (t)η(t)dt as

−êiλt

2

e−t2

P (t)dt−êiλt

2

e−t2

P (t)(1− η(t))dt;

use i) on each monomial of P (t) to deal with the first integral, and use iii) onthe second integral to show it is an error term (show that exp(−t2)P (t)(1−η(t))is a Schwartz function).For´

exp(iλt2)tn+1(exp(−t2)Rn(t)η(t))dt, use ii) to show that it is also anerror term. Choosing n large enough gives the result.

v) Simply show that under the given hypotheses s = |φ(t)|1/2 defines a diffeo-morphism. Finally, if t = θ(s) denotes the inverse diffeomorphism, show thatψ(θ(s))θ′(s) satisfies the same properties as ψ. Performing the change of vari-able t 7→ s concludes the argument.

vi) Repeat the proof you gave for i) basically word by word and adapt the rest.Hints for Exercise 14:i) The important thing is that dt/t is invariant with respect to changes of variables

of the form t 7→ αt.ii) Since P is a polynomial, we have P (d)(t) = d!. With ψ(t) = 1/t, the constant

in Corollary 2.4 is ∼d 1/R = Od(1).iii) eiP (t) − eiPd−1(t) = eiPd−1(t)

(eit

d − 1)and´ 1

0|t|d|t| dt = Od(1).


iv) Steps i)-ii)-iii) together have the effect of removing the top monomial from P .Repeat until there is only the last monomial c1t.

vi) Here λ−1 = |c1|−1, but with ψ(t) = 1/t the constant in Corollary 2.4 evaluatesto |c1|.

vii) We can add or subtract 0 for free to any quantity and bear in mind this is onlyrelevant when c1 > 1.

viii) You can consider ε,R fixed for the sake of the exercise. Calculate the Fouriertransform of the expression and show by Fubini that it equals

f(ξ, η)

ˆε<|t|<R

e−2πi(ξt+ηt2) dt

t.

The phase ξt + ηt2 is a polynomial, so apply what you just proved to get anL∞ bound for it. Conclude using Plancherel.

Hints for Exercise 15:i) Just a boring high-school trigonometry exercise.ii) Write

∑n e(f(n)) as

∑n g(n)(e(f(n))−e(f(n−1))) and check that every term

that appears in this sum appears exactly once in the right hand side (with thecorrect sign) and viceversa.

iv) Write f(n)− f(n− 1) =´ nn−1

f ′(s)ds.v) Apply triangle inequality to the left hand side, thus erasing the e(f(n)) fac-

tors. Then appeal to monotonicity to take the absolute values outside of thesummation and observe that the resulting expression is a telescopic sum, thusonly the first and last terms survive. Finally, use the trigonometric identity ini) to evaluate g(N), g(0), combine with periodicity of cot and iv).

vi) The remaining terms are bounded by |g(N)| + |g(1)| and are dealt with as inthe final part of v).

a) The bound on the length of J is an immediate consequence of the mean valuetheorem.

b) Ak ∪Bk has length 1 and they are all disjoint.c) The first sum runs over those n such that |f ′(n)− k| < δ. Since |f ′′| > λ, there

are at most O(δ/λ) such values of n.For the second sum, verify that the hypotheses of (♦1) as in vi) apply.

d) Summing over k gives the factor of Cλ|J |+ 1 by part b).e) It must be the case that both C|I|λ1/2 < |I| and λ−1/2 < |I|...

Hints for Exercise 16: Try with inner product 〈P,Q〉 := [P (∇)]Q, where ifP (X) =

∑α:|α|=k aαX

α then P (∇) =∑α:|α|=k aα∂

α. What does it mean that〈(ξ ·X)k, Q〉 = 0?

Hints for Exercise 17: Decompose the sphere smoothly in a bounded numberof pieces (see Exercise 20 part ii) for an idea how). Do a change of variables: apoint ω ∈ Sd−1 is given by ω = (x, φ(x)) where x ∈ Rd−1 and φ(x) =

(1−x2

1− . . .−x2d−1)1/2. Performing the change of variables, one has that

´Sd−1 exp(−2πitω1)dσ(ω)

is a sum of integrals of the formˆRd−1

e−2πitφ(x)S(x)ψ(x)dx,

where ψ is a smooth bump function supported around the origin and S(x)dx givesthe surface measure of Sd−1 in the x-coordinates. Justify all of the above, checkthe Hessian of φ and apply Theorem 3.3 directly.

Hints for Exercise 18:ii) just solve |u(x, y)| < λ for x, y.iii) Show that for x close to 1 the function u(x, y) is constant in x and periodic

in y with period 1/N . Thus concentrate on a single tooth and show that for


these x the set of y in this tooth such that |u(x, y)| < λ has measure boundedfrom below by & λ.

Hints for Exercise 19:ii) Just perform the integration

´Rd(1 + λ|z|)−Ndz with N large. If you prefer,

you can split Rd into dyadic annuli z ∈ Rd : 2j < |z| ≤ 2j+1, estimate theintegral over each annulus and sum in j ∈ Z.

iii) ∇yeiλ(u(y+z)−u(y)) = iλeiλ(u(y+z)−u(y))∇y(u(y + z) − u(y)), so take the innerproduct of this vector by vector (iλ)−1 ∇y(u(y+z)−u(y))

|∇y(u(y+z)−u(y))|2 to get the pure expo-nential factor back.

iv) From the previous point, θz(y) is the vector ∇y(u(y+z)−u(y))|∇y(u(y+z)−u(y))|2 . For the numer-

ator, observe that by hypothesis we can control |∂βy u| .β,u,ψ 1 for any multi-index β; show by the Fundamental Theorem of Calculus that ∂αy (u(y + z) −u(y)) =

´ 1

0∇y(∂αy u(y+tz))·zdt and deduce that |∂αy (u(y+z)−u(y))| .α,u,ψ |z|.

For the denominator, use the hypothesis that ψ has small support to argue thatby Taylor expansion ∇y(u(y + z) − u(y)) is approximately [Hess(u)](y)z, anduse the hypothesis on Hess(u) to argue that this is &u,ψ |z| uniformly in y.Finally, combine the above information and the generalized Leibniz’s rule toestimate ∂αy θz.

v) Show that Dtf = (iλ)−1∇ · (fθz) (that is, the divergence of the Rd-valuedfunction fθz). Let Ψ(y, z) := ψ(y + z)ψ(y) for simplicity. Show by inductionthat (Dt)NΨ contains only terms of the form ∂α0

j0Ψ · ∂α1

j1θz · . . . · ∂αNjN θz with

α0 + α1 + . . . + αN = N , and deduce by iv) that |(Dt)NΨ| is thus controlledby a sum of terms of the form |z|−N |∂α0

j0Ψ|.

Hints for Exercise 20:i) The integral factorises into a product of integrals that are exactly like the ones

in i) of Exercise 13.ii) The decomposition of Rd into cones follows from a trivial pigeonholing argu-

ment. Building the partition of unity 1 =∑dk=1Gk(x) is also a standard

exercise: simply take non-negative bump functions ρk on the sphere Sd−1

with support contained in Γk ∩ Sd−1 and identically 1 on Γk, then defineGk(x) := ρk(x/‖x‖)

(∑dj=1 ρj(x/‖x‖)

)−1 and show this is C∞ and satisfiesthe other conditions.

iii) See the hints for ii) of Exercise 13, for this is extremely similar. The trivialestimate near the origin will give you a contribution

´|xα|φ(x/δ)dx . δd+|α|.

Then you will end up having to estimate the integral of (Dtk)N (xαωδ(x)Gk(x)),

where ωδ(x) = η(x)(1 − ϕ(x/δ)). You can again appeal to induction to provethat the terms of this expression are of the form x

−(2N−j)k ∂jk(xαωδ(x)Gk(x)),

then apply Leibniz’s rule to expand the ∂jk derivative and conclude that sinceall the terms contain factors of ∂`kGk they are supported in Γk, where cruciallyit holds |xk| ∼ ‖x‖. Performing the integration will give a total contribution ofO(λ−Nδ−(2N−|α|)+d), which combined with the trivial estimate and optimisedin δ will give you the desired estimate |Ixαη(λ)| . |λ|−(d+|α|)/2.

iv) Again, repeat the second part of the proof you gave for step iii) above, butwith δ ∼ 1.

v) See the hints for part iv) of Exercise 13 and follow them closely.vi) Morse’s lemma says precisely what we want: if f ∈ C∞(Rd) is such that

f(0) = 0, ∇f(0) = 0 but det(Hess(f)) 6= 0 (where Hess(f) denotes the Hessianof f), then there exists a neighbourhood U of 0 and a C∞-diffeomorphismΦ : Rd → Rd such that Φ(0) = 0 and f Φ−1(y) = x2

1 + . . .+x2k−x2

k+1− . . .−x2d

for all y ∈ Φ(U). Moreover, (k, d− k) is the signature of Hess(f).


Use the change of variable provided by the diffeomorphism Φ to turn the phaseu(x) into a quadratic phase and conclude by appealing to the previous steps.

Hints for Exercise 21:i) At time t the wave solution is concentrated in a shell of width ∼ 1 and radius∼ t, and it has amplitude ∼ t−(d−1)/2. Multiply the square of the amplitudeby the volume occupied by the wave to get an estimate for the energy.

iv) The gradient of the phase is ∇φ(ξ) = t ξ|ξ| + x1e1, so if t . 1 and x1 & 1 onehas ∂ξ1φ is bounded from below by & x1. Show that for higher derivativesone has |∂kξ1φ| . 1. Apply the integration by parts argument of Proposi-tion 2.1 to the integral in ξ1 with respect to the differential operator D1 =(i∂ξ1φ(ξ))−1∂ξ1 . Show by induction in N that |(Dt

1)Nψ| .N,ψ x−N1 (by nowyou should know how to proceed: show that (Dt

1)Nψ consists of terms of theform (∂ξ1φ)−`∂β1

ξ1ψ∏mk=2(∂kξ1φ)βk where ` ≥ N ...). Finish by integrating in the

remaining variables ξ2, . . . , ξd.v) Show that one always has |∂ξ1φ(ξ)| & |x1 − t| & t in this regime. Then repeat

the analysis done in iv), but be more careful this time: now you can onlysay (show this) that |∂kξ1φ| . t. In order not to lose the decay, you will haveto show by induction that the terms of the form (∂ξ1φ)−`∂β1

ξ1ψ∏mk=2(∂kξ1φ)βk

appearing in the expansion of (Dt1)Nψ actually satisfy `−

∑mk=1 kβk = 0 and

β1 +∑mk=2(k − 1)βk = N , thus yielding the correct decay of |x1 − t|−N .

vi) For the smooth excision of a cone, refer to the hints for parts ii)-iii) of Exercise20. For the estimate, observe that ignoring the ξ1 component one has |∇φ(ξ)| &t|ξ′|/|ξ| and outside the cone one has |ξ′| & |ξ|. Upon a further smooth partitionof unity into cones with axes the coordinate directions ξ2, . . . , ξd, one can reduceto the case where |ξj | & |ξ| and thus |∂ξjφ(ξ)| & t. Then apply the sameargument as in v) above with differential operator Dj = (i∂ξjφ(ξ))−1∂ξj toconclude.

vii) Same reasoning as in vi).viii) You are allowed to use Taylor expansion of |ξ| = ξ1(1 + (|ξ′|2/ξ2

1))1/2 in termsof |ξ′|2/ξ2

1 , both for estimating the difference in the phases and for estimat-ing ∂ξ1φ. The non-stationary phase principle in both ξ1, ξj (after once againusing a smooth partition of unity to reduce to the case where |ξj | & |ξ|)will result in having to estimate (for example)

´|(Dt

jDt1)Nψ(ξ)|dξ. This is

quite excruciating, but you can again use an inductive argument to show thatthe integrand consists entirely of terms of the form (∂jφ)−`(∂1φ)−`

′ · ∂αj ∂β1 ψ ·∏

k+k′≥2(∂kj ∂k′

1 φ)γ(k,k′) where the various parameters involved satisfy certainfavourable contraints. Then one can see that |∂1φ| & |x1 − t|, |∂2

1φ| . 1 (thisbecause of the restriction to |ξ′| . t−1/2), |∂jφ| ∼ t and in general |∂kj ∂k

′

1 φ| . tfor all k + k′ ≥ 2. Using these facts and the induction one can verify that|(Dt

jDt1)Nψ(ξ)| . |x1 − t|−N . The factor of t−(d−1)/2 will come from the fact

that the region of integration is approximately a cylinder of length 1 and radiust−1/2.

ix) This part is a repetition of part viii) above essentially, except for the fact thatnow we can assume ξj ∼ 2jt−1/2 instead and consequently one has to updatethe upper- and lower-bounds on the derivatives of the phase accordingly.

If you find typos, mistakes or generally have questions please send an email [email protected]

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