mat 636: lectures on fourier series
TRANSCRIPT
MAT 636: Lectures on Fourier series
Brad Rodgers
[Lecture 1]
1 Introduction
The subject of Fourier series concerns decomposing periodic functions into sumsof somewhat more basic functions that have nice algebraic and analytic proper-ties. In this course we will develop the basic theory of this subject.
In order to be concrete, in what follows we will work almost exclusively withfunctions f , defined on the real line, which are of period 1. The same theory,however, can be developed with only minor and obvious changes for functionsthat are of period L, for any L. Likewise, the same theory that we will develophere can be applied to functions that are defined on the unit circle, or, if youknow some group theory, to functions defined on T := R/Z with the groupoperation of addition. All such types of functions are, so to speak, differentembodiments of the same mathematical object.
For functions f of period 1, the more basic functions into which we willdecompose f(x) are {ei2πnx}n∈Z. There is an equivalent theory in which thesefunctions are replaced by {sin 2πnx, cos 2πnx}n∈N0
, but it is not as nice alge-braically.
Some knowledge of measure theory (from Analysis III at UZH) is assumed inthese notes, but with a little work, one could probably follow most of these noteswith only a superficial understanding of that subject. Please do not hesitate toemail me with any questions.
We begin with a definition.
Definition 1. For a function f of period 1 on the real line and such thatf ∈ L1[0, 1), the Fourier coefficients of f are defined by
f(n) :=
∫ 1
0
f(x)e−i2πnx dx.
We note the following fundamental facts, easily verified by calculus
Proposition 2. For j an integer∫ 1
0
ei2πjx dx =
{1, j = 0
0 j 6= 0.
1
Corollary 3. For j, k integers∫ 1
0
ei2πjxe−i2πkx dx = δj=k :=
{1, j = k
0 j 6= k.
Some texts at this point define the ‘formal Fourier series’
∞∑k=−∞
f(k)ei2πkx,
and write
f ∼∞∑
k=−∞
f(k)ei2πkx.
It is important to stress that this identification is formal, and we don’t knowat this point that this formalism has any literal meaning; indeed we don’t evenknow whether the sum on the right is guaranteed to converge in the traditionalsense. As we will later see, it sometimes doesn’t. A central question in thetheory of Fourier series is the extent to which f(x) actually does equal the sumon the right.
2 The Dirichlet kernel
In order to be more rigorous, we make the following definition.
Definition 4.
Sn(f, x) :=
n∑k=−n
f(k)ei2πkx.
By the formal Fourier series we’ve written above, one might have the some-what vague intuition that for large n, Sn(f, x) should look basically like f(x).This is certainly the case when f is a trigonometric polynomial, that is whenf(x) is a finite linear combination of the functions ei2πkx.
Proposition 5. If
f(x) =
N∑k=−M
ckei2πkx,
where M and N are finite numbers and ck are constants, then as n→∞,
Sn(f, x)→ f(x).
This is easy to verify. Indeed, in this case, as long as n ≥ N,M , we havethat Sn(f, x) = f(x).
Despite this proposition, it is not always the case, even for continuous f ,that Sn(f, x) → f(x) for all x (a fact due to Paul du Bois-Reymond). But wewill see very soon that statements very close to this are true.
For the moment we simply note some properties of these partial sums.
2
Proposition 6. For f of period 1 and in L1[0, 1),
Sn(f, x) =
∫ 1
0
f(t)Dn(x− t) dt where Dn(x) :=
n∑k=−n
ei2πkx.
Moreover,
Dn(x) =sin((n+ 1
2
)2πx
)sin(12 · 2πx
) .
Proof. The first of these claims is trivial to verify by using the definition of Dn
to evaluate the integral, and we did so in class. For the second, let ω = ei2πx.Then
Dn(x) = ω−n(1 + ω + · · ·+ ω2n)
= ω−nω2n+1 − 1
ω − 1
=ω(n+1/2) − ω−(n+1/2)
ω1/2 − ω−1/2.
This may be seen to be equal to the formula in the proposition, using the factthat (eix − e−ix)/2i = sinx.
Dn(x) is called the Dirichlet kernel, in honor of Lejeune Dirichlet. We aredeferring a proof that sometimes Sn(f, x) does not converge to f(x) until later.Nonetheless, if we take this claim on faith for the moment, it is natural to thenask the weaker question whether given a sequence of Fourier coefficients f(k)we can even recover the original function f(x). This is what we will turn to inthe next two sections.
3 Summability methods
In this section, we introduce a very first example of a much larger subjectknown as summability methods. It allows us to expand what is meant by asequence converging to a value, and this broadening of our understanding hasmany consequences.1
For us, the importance of summability methods is that they will allow us todefine analogues of the partial sums Sn(f, x) that converge to f(x) even whenthe partial sums do not.
We begin with an example sequence: αn = (−1)n. Here αn oscillates around0, taking the values of 1 and −1, and plainly doesn’t converge to anything.
1Indeed, several months ago there was a sort of controversy on the internet about theformula “1 + 2 + 3 + ... = −1/12.” Obviously such a formula can’t literally be true, but it canbe understood using summability methods (albeit summability methods that lie deeper thanwe will make use of in this course), and once understood it is seen to be an important identityin the theory of numbers.
3
Nonetheless, the average of this sequence tends to 0:
1
N + 1
N∑n=0
(−1)n =1− 1 + 1− · · · ± 1
N + 1=
0 or 1
N + 1→ 0.
Moreover, whenever a sequence converges to some value, so to will the averageof the sequence:
Theorem 7 (Cesaro). If αn → α, then
α0 + · · ·+ αNN + 1
→ α.
Proof. Define εn by αn = α+ εn. Then for any ε > 0 there exists some M suchthat |εn| ≤ ε for all n ≥M .
Nowα0 + · · ·+ αN
N + 1= α+
ε0 + · · ·+ εNN + 1
. (1)
Moreover, for any ε > 0 there is M as above, and
ε0 + · · ·+ εNN + 1
≤ |ε0|+ · · ·+ |εM |N + 1
+|εM+1|+ · · ·+ |εN |
N + 1
≤ |ε0|+ · · ·+ |εM |N + 1
+ε(N −M)
N + 1
→ 0 + ε,
as N →∞. Hence
lim supN→∞
∣∣∣∣ε0 + · · ·+ εNN + 1
∣∣∣∣ ≤ ε,and as ε was arbitrary
limN→∞
∣∣∣∣ε0 + · · ·+ εNN + 1
∣∣∣∣ = 0.
This and (1) prove the theorem.
When (α0 + · · · + αN )/(N + 1) → α, we say that α is the Cesaro limit ofαN . What we have shown is that anytime α is the limit of αN , it will also bethe Cesaro limit, and moreover, from the example, a sequence can sometimeshave a Cesaro limit when the sequence itself has no limit. This widened conceptof a converging sequence would have been familiar to the young mathematicianLipot Fejer a little over a century ago, and it ended up being exactly what heneeded to answer the question with which we ended section 1: can we recoverf(x) from the sequence f(k)?
4
4 The Fejer kernel
Theorem 8 (Fejer). For f of period 1 and continuous,
σN (f, x) :=S0(f, x) + S1(f, x) + · · ·+ SN (f, x)
N + 1→ f(x),
uniformly.
Before proving this theorem, we prove a lemma.
Lemma 9. For f of period 1 and in L1[0, 1),
σN (f, x) =
∫ 1
0
f(t)KN (x−t) dt where KN (x) :=1
N + 1
N∑j=−N
(N+1−|j|)ei2πjx.
Moreover,
KN (x) =1
N + 1
( sin(N+12 · 2πx
)sin(
12 · 2πx
) )2
.
Proof. Note that
σN (f, x) =1
N + 1
N∑n=0
∫ 1
0
f(t)Dn(x− t) dt
=
∫ 1
0
f(t)KN (x− t) dt,
where
KN (x) =1
N + 1
N∑n=0
Dn(x)
=1
N + 1
(1 + (ω−1 + 1 + ω) + · · ·+ (ω−N + ω−(N−1) + ·+ ωN )
)=
1
N + 1
N∑j=−N
(N + 1− |j|)ωj ,
where as before ω = ei2πx. This verifies the first claim.Yet one may check that
N∑j=−N
(N + 1− |j|)ωj =(ω−N/2(1 + ω + · · ·+ ωN )
)2=
(ω(N+1)/2 − ω−(N+1)/2
ω1/2 − ω−1/2
)2
=
( sin(N+12 · 2πx
)sin(
12 · 2πx
) )2
,
5
which verifies the second claim. Here the first line follows from a countingargument, and the second by summing the geometric series and some simplealgebra.
After Fejer, the function KN is called the Fejer kernel.The reader should take the time to graph both the functions SN and KN
for various values of N to get a feel for their behavior.
Lemma 10. Both the following are true:
i) KN (x) ≥ 0 for all x
ii)∫ 1
0KN (x) dx = 1
Proof. i) This is clear from Lemma 9, because the square of any real numberis non-negative.
ii) By Lemma 9 and then Proposition 2∫ 1
0
KN (x) dx =1
N + 1
∫ 1
0
N∑j=−N
(N + 1− |j|)ei2πjx dx = 1.
From this we have the very important
Proposition 11. All of the following are true:
(1)∫ 1
0KN (x) dx = 1
(2)∫ 1
0|KN (x)| dx is bounded as N →∞
(3) For δ > 0,limN→∞
KN (x) = 0
uniformly for δ ≤ |x| ≤ 1/2.
Proof. (1) and (2) follow from Lemma 10. For (3), note that for |x| ∈ [δ, 1/2],
KN (x) ≤ 1
N + 1
1
(sinπδ)2→ 0
as N →∞.
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We can now outline the idea behind Fejer’s proof. We have that for small δand large N ,
σN (f, x) =
∫ 1
0
f(t)KN (x− t) dt
=
∫ 1/2
−1/2f(x− τ)KN (τ) dτ
≈∫ δ
−δf(x− τ)KN (τ) dτ by (3) of Proposition 11
≈ f(x)
∫ δ
−δKN (τ) dτ by the continuity of f and (2) of Prop. 11
≈ f(x)
∫ 1/2
−1/2KN (τ) dτ by (3) of Proposition 11
= f(x) by (1) of Proposition 11.
It is not too difficult to make this argument rigorous once this outline is inplace.
[Lecture 2]
Proof of Theorem 8. We define κ := supN ‖KN‖L1[−1/2,1/2]. By (2) of Propo-sition 11, we know κ < +∞. Let ε > 0 be an arbitrary positive number anddefine ε′ := ε/κ.
We proceed backwards through the outlined steps above. Note that for anyδ > 0,
limN→∞
∫δ≤|τ |≤1/2
KN (τ) dτ = 0,
by (3) of Proposition 11. Note also, by property (1) of this proposition, that∫ δ
−δKN (τ) dτ =
∫ 1/2
−1/2KN (τ) dτ −
∫δ≤|τ |≤1/2
KN (τ) dτ
= 1−∫δ≤|τ |≤1/2
KN (τ) dτ
Hence,
supx
∣∣∣∣f(x)
∫ δ
−δKN (τ) dτ − f(x)
∣∣∣∣ = supx|f(x)|
∣∣∣∣ ∫δ≤|τ |≤1/2
KN (τ) dτ
∣∣∣∣→ 0 (2)
as N →∞.Yet because f is continuous and periodic, f must be uniformly continuous2.
So there must exist some δ > 0 such that
|f(x)− f(x− τ)| ≤ ε′
2To see this, note that f is uniformly continuous on [−1, 2] as this interval is compact, andthe periodicity of f implies that this uniform continuity may be extended to all of R
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for all x ∈ R and all |τ | < δ. But then for all N and x,∣∣∣∣ ∫ δ
−δf(x− τ)KN (τ) dτ − f(x)
∫ δ
−δKN (τ) dτ
∣∣∣∣ ≤ ε′ ∫ δ
−δ|KN (τ)| dτ
≤ ε. (3)
Moreover, note that
supx
∣∣∣∣ ∫ δ
−δf(x− τ)KN (τ) dτ −
∫ 1/2
−1/2f(x− τ)KN (τ) dτ
∣∣∣∣≤ sup
x|f(x)|
∫δ≤|τ |≤1/2
|KN (τ)| dτ
→ 0. (4)
Finally, it is indeed clear that
σN (f, x) =
∫ 1/2
−1/2f(x− τ)KN (τ) dτ,
by the periodicity of f and KN . Yet
supx|σN (f, x)− f(x)| = sup
x
∣∣∣∣ ∫ 1/2
−1/2f(x− τ)KN (τ) dτ −
∫ δ
−δf(x− τ)KN (τ) dτ
+
∫ δ
−δf(x− τ)KN (τ) dτ − f(x)
∫ δ
−δKN (τ) dτ
+ f(x)
∫ δ
−δKN (τ) dτ − f(x)
∣∣∣∣≤ sup
x
∣∣∣∣ ∫ 1/2
−1/2f(x− τ)KN (τ) dτ −
∫ δ
−δf(x− τ)KN (τ) dτ
∣∣∣∣+ sup
x
∣∣∣∣ ∫ δ
−δf(x− τ)KN (τ) dτ − f(x)
∫ δ
−δKN (τ) dτ
∣∣∣∣+ sup
x
∣∣∣∣f(x)
∫ δ
−δKN (τ) dτ − f(x)
∣∣∣∣,by the triangle inequality. Each of these terms as N → ∞ is controlled by theequations (4), (3), and (2) respectively, so we see that
lim supN→∞
(supx|σN (f, x)− f(x)|
)≤ 0 + ε+ 0 = ε.
As ε was arbitrary, this proves the claim.
This proof may seem somewhat hard-going at first; if so, it may help toreview the outline given above it once again.
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5 Consequences of Fejer’s theorem
We note some consequences of this remarkable result.
Corollary 12. If f and g are continuous functions of period 1, and f(k) = g(k)for all k ∈ Z, then f(x) = g(x) for all x.
Proof. By Fejer’s theorem, σN (f, x) → f(x) and σN (g, x) → g(x) uniformly.
But if f(k) = g(k) for all k, then σN (f, x) = σN (g, x).
We will later extend this corollary to functions that are only integrable,rather than continuous.
Closely related to Corollary 12, we also have that trigonometric polynomialsare dense in the space of continuous functions with supremum norm. Said in adifferent language,
Theorem 13. Let f be a continuous function of period 1. For any ε > 0, thereexists a trigonometric polynomial
P (x) =
N∑k=−M
ckei2πkx
such thatsupx|f(x)− P (x)| < ε.
Proof. We need only note that σN (f, x) is a trigonometric polynomial, andσN (f, x)→ f(x) uniformly.
We can ‘bootstrap’ this theorem to show that trigonometric polynomials aredense in L1 as well. In the more concrete language of the theorem above,
Theorem 14. Let φ ∈ L1[0, 1) also be of period 1 on the real line. For anyε > 0, there exists a trigonometric polynomial P (t) such that
‖φ− P‖L1[0,1) < ε.
Proof. We recall (from Analysis III or, for instance, Theorem 3.14 of Rudin’s“Real and Complex Analysis”) that for any ε > 0 there exists a continuousfunction f such that
‖φ− f‖L1[0,1) < ε/2.
But likewise from Theorem 13, there exists a trigonometric polynomial P suchthat
‖f − P‖L1[0,1) ≤ ‖f − P‖L∞[0,1) < ε/2.
Hence,‖φ− P‖L1[0,1) ≤ ‖φ− f‖L1[0,1) + ‖f − P‖L1[0,1) < ε.
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There is a simple estimate that lies very near the start of the theory ofFourier series:
Proposition 15. For f ∈ L1[0, 1) of period 1,
|f(k)| ≤ ‖f‖L1[0,1)
for all k.
Hence Fourier coefficients of a given function remain bounded. The proof ofthis Proposition is a straightforward exercise of taking absolute values inside anintegral (with the triangle inequality) and is left to the reader. In fact, we cannow show rather more than this.
Theorem 16 (Riemann-Lebesgue lemma). For f ∈ L1[0, 1) of period 1
lim|k|→∞
f(k) = 0.
Proof. Choose any ε > 0. There exists a trigonometric polynomial
P (x) =
N∑k=−M
ckei2πkx
such that‖f − P‖L1[0,1) < ε
by Theorem 14. Note that for |k| > N,M ,
P (k) = 0.
Yet, as f = (f − P ) + P , we have by the triangle inequality,
|f(k)| ≤∣∣∣ ∫ 1
0
(f(x)− P (x))e−i2πkx dx∣∣∣+∣∣∣ ∫ 1
0
P (x)e−i2πkx dx∣∣∣
≤ ‖f − P‖L1[0,1) + |P (k)|≤ ε
when |k| is larger than N and M . As ε has been chosen arbitrarily, this is thedefinition of the statement,
lim|k|→∞
f(k) = 0.
We will not prove so in this class, but not much more can be said about therate at which the Fourier coefficients of arbitrary integrable functions f decaythan that f(k) → 0. In fact, for any positive sequence αk → 0, there exists an
integrable function f such that f(k) ≥ αk for all k. (See Katznelson, Section
10
I.4, and exercise 4.1, if you’re curious.) Later, we will see that the Fouriercoefficients of functions in L2, rather than L1, can be described more exactly.
We do now have, however, enough information to show that when a func-tion’s Fourier coefficients are absolutely summable, then their partial sums con-verge to the original function.
Proposition 17. For f continuous and of period 1, if∑k∈Z|f(k)| < +∞,
thenSn(f, x)→ f(x)
for all x.
Proof. A proof very similar was covered in homework 7. We know that byFejer, limσn(f, x) = f(x) for all x. On the other hand as f(k) is absolutelysummable, limSn(f, x) exists for all x. Cesaro’s theorem thus implies thatlimσn(f, x) = limSn(f, x) so that limSn(f, x) = f(x) for all x.
We note in passing: it is easy to see that in the situation of Proposition 17,
‖f‖L∞[0,1) ≤ ‖f‖`1(Z),
as
|f(x)| =∣∣∣∣∑k
f(k)ei2πkx∣∣∣∣ ≤∑
k
|f(k)|
for all x. Note that Proposition 15 can be phrased,
‖f‖`∞(Z) ≤ ‖f‖L1[0,1),
which is an interesting relationship...Finally we turn to a claim mentioned at the start of this section. In fact, with
the theory we have developed so far, the proof is not completely straightforward.I give a proof below for the sake of completeness, but you may find it a littledifficult depending upon your background – do not worry if so.
Theorem 18. If f and g in L1[0, 1) are of period 1 and f(k) = g(k) for allk ∈ Z, then f(x) = g(x) almost everywhere.
.In our proof we will require
Lemma 19. For φ ∈ L1[0, 1],∫ 1
0
|φ| dx = supβ
∫ 1
0
βφ dx,
where the supremum is taken over all β ∈ C[0, 1] with |β(x)| ≤ 1 for all x.
11
This may be found in a number of analysis textbooks; I provide a sketch ofa proof here.
Proof. Since any integrable function may be approximated arbitrarily closelyby a continuous function, it is sufficient to prove this claim for φ ∈ C[0, 1].(You may want to fill in the details of the claim I’ve just made, if you’re notconvinced.) But, for φ continuous,∫ 1
0
|φ| dx =
∫ 1
0
sgn(φ) · φdx = limn→∞
∫ 1
0
βnφdx,
where sgn(φ(x)) is the sign function equal to 1 when φ(x) is positive, −1 whenφ(x) is negative, and 0 when φ(x) is 0, and where
βn(x) = sgn(φ(x)) ·min(1, n · dist(x,E)),
where dist(x,E) is the distance of x to the set E of zeros of φ.(The above equality follows from the fact that βn(x) → sgn(φ(x)). Why is
this fact true for continuous φ?)βn is continuous for all n, so∫ 1
0
|φ| dx ≤ supβ
∫ 1
0
βφ dx,
But clearly ∫ 1
0
|φ| dx ≥ supβ
∫ 1
0
βφ dx,
so our proof is complete.
Proof of Theorem 18. From the lemma above, by approximating each β uni-formly by trigonometric polynomials (which is possible by Theorem 13), onemay see that ∫ 1
0
|φ| dx = supP
∫ 1
0
P · φdx,
where the supremum is taken over all trigonometric polynomials P with |P (x)| ≤1 for all x.3
Hence, ∫ 1
0
|f(x)− g(x)| dx = supP
∫ 1
0
P · (f − g) dx.
But for any trigonometric polynomial P (x) =∑N−M cke
i2πkx,
∫ 1
0
P · (f − g) dx =
N∑k=−M
ck(f(−k)− g(−k)) = 0,
3See a discussion on the piazza discussion board for an elaboration on why, as well anargument due to Roland Prohaska that takes a slightly different path.
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as f(−k)− g(−k) = 0 for all k. This of course implies∫ 1
0
|f(x)− g(x)| dx = 0,
which implies f(x) = g(x) almost everywhere (by a well known result proved inAnalysis III).
Note: There is another proof of this theorem that depends upon a proof thatfor f ∈ L1, we have ‖f − σN (f)‖L1[0,1) → 0. We will not take this approach inthis class, but a proof may be found in Katznelson, section I.2 or Krantz section1.4.
[Lecture 3]
6 Summability kernels
In this section we generalize Fejer’s theorem to other methods of summability.
Definition 20. A summability kernel is a sequence {kn} of continuous func-tions, all of period 1, such that
(1)∫ 1
0kn(x) dx = 1
(2)∫ 1
0|kn(x)| dx is bounded as n→∞
(3) For δ > 0,limn→∞
kn(x) = 0
uniformly for δ ≤ |x| ≤ 1/2.
Remark: Sometimes in other texts the condition (3) is replaced with theweaker condition
(3’)∫δ≤|x|≤1/2 |kn(x)| dx→ 0,
depending on the application the author has in mind.Remark: The sequence {kn}, with n = 1, 2, ... may be replaced by a contin-
uous parameter family {kr} with r ↗ 1, for instance. We will see an exampleof this later with the Poisson kernel.
We also introduce the notation of convolutions:
Definition 21. For two functions of period 1, f ∈ L1[0, 1) and g ∈ L∞[0, 1),
f ∗ g(x) :=
∫ 1
0
g(x− t)f(t) dt
Proposition 22.
f ∗ g(x) =
∫ 1
0
g(τ)f(x− τ) dτ.
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This is a simple exercise in making the change of variable τ := x−t. Anotherway to write the above proposition is f ∗ g = g ∗ f .
Note thatSn(f, x) = Dn ∗ f(x),
σN (f, x) = KN ∗ f(x).
Remark: The condition that g ∈ L∞ can be weakened to g ∈ L1, and f ∗ gwill still be well defined almost everywhere, but we do not prove this fact inthese notes.
Summability kernels are of interest because of the following result.
Theorem 23. If {kn} is a summability kernel and f is continuous and of period1, then
kn ∗ f(x)→ f(x)
uniformly.
Proof. Exactly as for the Fejer kernel.
Aside from the Fejer kernel, one other example of a summability kernel thatis of particular importance is the Poisson kernel.
Definition 24. For |r| < 1, and f continuous and of period 1,
ρr(f, x) :=
∞∑k=−∞
r|k|f(k)ei2πkx.
Remark: The notation ρr to denote this sum is not universal.
Proposition 25. For |r| < 1,
ρr(f, x) = Pr ∗ f(x),
where
Pr(t) =
∞∑k=−∞
r|k|ei2πkt
=1− r2
1− 2r cos 2πt+ r2.
Proof. We have
ρr(f, x) =∑k
r|k|ei2πkx∫f(t)e−i2πkt dt =
∫f(t)
∑k
r|k|ei2πk(x−t) dt
where swapping the order of integration and summation (in the second equality)follows from dominated convergence.
14
Moreover, letting ω = ei2πx,
∞∑k=−∞
r|k|ei2πkx = 1 +∑k≥1
rkωk +∑k≥1
rkω−k
= 1 +rω
1− rω+
rω−1
1− rω−1
=1− r2
1− r(ω + ω−1) + r2.
The Poisson kernel Pr is particularly useful in complex analysis, where itcorresponds to power series.
It is left to the reader to verify from the definition that
Theorem 26. Pr is a summability kernel.
This is done in much the same way as for the Fejer kernel.
[Section 7 below is somewhat more difficult and optional]
7 The Dirichlet kernel revisited: a theorem ofdu Bois-Reymond
Theorem 27 (du Bois-Reymond). There exists a continuous function f ofperiod 1, such that the partial sums Sn(f, x) diverge at some point x.
Clearly, given Theorem 23 and Lemma 9, this can only be the case if theDirichlet kernel Dn is not a summability kernel. This is in fact the case. Con-dition (2) is not met.
Lemma 28. ‖Dn‖L1[0,1] ≥ 4π2 log n.
Proof. Note that | sinx| ≤ |x| for all x, so that
‖Dn‖1 =
∫ 1/2
−1/2|Dn(x)| dx ≥ 2
∫ 1/2
0
| sin((n+ 1/2) · 2πx)
)|
πxdx
= 2π
∫ n+1/2
0
| sinπt|dtt≥ 2
π
n∑k=1
1
k
∫ k
k−1| sinπt| dt =
4
π2
n∑k=1
1
k
≥ 4
π2
n∑k=1
∫ k+1
k
dτ
τ=
4
π2log(n+ 1).
15
In fact, it is not hard to modify this proof to show that
‖Dn‖1 =4
π2log n+O(1),
though we do not need this fact in what follows.For us, the lemma in particular implies by Lemma 19 that for each n there
exists a continous function βn such that ‖βn‖∞ ≤ 1, yet
Dn ∗ βn(0) ≥ 2
π2log n (5)
Clearly each function βn has a Fourier series that is ‘badly behaved’ near theorigin, at least for the n-th partial sum. Our approach will be to add togethera large number of these functions to obtain a single function badly behaved foran infinite number of partial sums.
This approach is very powerful, and foreshadows what is called “the uniformboundedness principle” in functional analysis. In fact, it is by an appeal to theuniform boundedness principle that Theorem 27 is usually proved in textbooks.(See Rudin, Real and Complex Analysis, ch. 5, or Katznelson section II.2, fromwhich our proof is adapted.) The proof we give below, instead of appealingdirectly to this somewhat abstract theorem, borrows the technique of its proofto construct a concrete counterexample. The price we pay for being concreteis that our proof appears a little messier than what one would find in Rudin’stextbook for instance.
Proof of Theorem 27. We have constructed functions βn above in (5). We nowconstruct trigonometric polynomials αn(x) := σn2(βn, x) = Kn2 ∗ βn(x). Be-cause ‖Kn2‖1 = O(1), we have that
‖αn‖∞ = O(1),
and
|Sn(αn, x)− Sn(βn, x)| ≤∑|k|≤n
|αn(k)− βn(k)|
=∑|k|≤n
|k|n2 + 1
|βn(k)|
≤ 2.
In particular,
|Sn(αn, 0)| ≥ 2
π2log n− 2. (6)
We define λn := 23n
and
f(x) :=
∞∑n=1
1
n2αλn(λnx). (7)
16
We will show that f is continuous, yet lim supSn(f, 0) =∞.The continuity follows because the sum (7) is absolutely convergent for all
t.On the other hand, we have that
αλj (λjx) =∑|k|≤λ2
j
αλj (k)ei2πλjkx,
so that
|Sλ2n(f, 0)| =
∣∣∣∣Sλ2n
( n∑j=1
1
j2αλj (λjx), 0
)+
∞∑j=n+1
1
j2αλj (0)
∣∣∣∣=
∣∣∣∣ n−1∑j=1
1
j2αλj (0) +
1
n2Sλn(αλn , 0) +
∞∑j=n+1
1
j2αλj (0)
∣∣∣∣≥ 2
π2
log λnn2
−O(1)
→∞,
with the inequality in the third line following from (6).
[Resumption of lecture 3]
8 Fourier coefficients and smoothness
We briefly expand on an exercise in the homework, which showed that a twicecontinuously differentiable function has Fourier coefficients that decay quadrat-ically.
General principle: The smoother a function f is, the faster f(k) → 0, andvice-versa.
Likewise, in general, the smoother a function is, the more quickly the partialsums (or summability kernels) of its Fourier series will converge to the functionitself.
An instance of this principle, proved in the same manner as your homeworkis the following:
Proposition 29. For f ∈ C`(R) and of period 1,
f(k) = O( 1
1 + |k|`).
Because this is so similar to your homework we do not outline the proof indetail, but only say that it follows easily from the fact that
(f (`)) (k) = (i2πk)`f(k),
where f (`) is the `-th derivative of f .
17
9 An application: uniform distribution
As an application of the theory we have developed, we consider the distributionof the fractional parts of the sequence ϑ, 2ϑ, 3ϑ, ..., for numbers ϑ.
Definition 30. bxc = max{k ∈ Z : k ≤ x} and {x} = x− bxc.
bxc and {x} are called the floor and fractional part of x respectively.If ϑ = p/q, then {nϑ} must be one of 0, 1/q, 2/q, ..., (q − 1)/q so we see that
Proposition 31. If ϑ is rational, then the sequence{{nϑ}
}is discrete.
On the other hand, for ϑ irrational, the behavior of this sequence is moreinteresting.
Proposition 32. If ϑ is irrational, then the sequence{{nϑ}
}is dense in [0, 1].
This theorem will be familiar to anyone who as a child ever tried to avoidthe periodic cracks between sidewalk tiles while walking. By walking in theordinary fashion (with each step the same length, ϑ), unless you carefully stepin ratio to the length of tiles, you’re certain sooner or later to step on one ofthe sidewalk’s cracks – no matter in what location you begin.
We will prove more general theorem than Proposition 32 below, but for themoment let us at least outline a self-contained proof of this proposition thatcould be filled in by interested readers. Place the points {nϑ} where n ∈ Naround the unit circle by marking points at 2πnϑ radians. If you supposethat this collection of points in not dense, then there must be some non-emptyopen interval on the unit circle in which none of these points occur. By acompactness argument, there exists some such interval I which is of the longestlength. Consider I rotated by 2πϑ radians clockwise. By assumption, thisinterval too must contain none of the points that have been marked on the circle.Continue rotating I in this way, an infinite number of times. None of theserotated intervals can contain any points that have been marked on the circle.Because ϑ is not rational, none of these rotated intervals can exactly coincide.But on the other hand, none of the rotated intervals can overlap without exactlycoinciding, since this would produce a larger interval that contains no points.We have thus produced an infinite collection of disjoint intervals, all of the samesize and contained on the unit circle – plainly a contradiction.
Whether you have chosen to fill in the details of the above outline or not, isnatural having thought about Proposition 32 to ask whether anything else canbe said about the distribution of the points
{{nϑ}
}that that they are dense.
In fact more can be said:
Theorem 33. For (a, b) ⊂ [0, 1),
limN→∞
1
N#{{nϑ} ∈ (a, b) : 1 ≤ n ≤ N
}= b− a.
Note that this implies Proposition 32, as (a, b) may be centered around anynumber and of arbitrarily small length. We will prove Theorem 33 from
18
Theorem 34. For f continous and of period 1,
limN→∞
1
N
∑n≤N
f({nϑ}) =
∫ 1
0
f(x) dx.
Proof of Theorem 34. Step 1: It is trivial to verify that the theorem is truewhen f is constant.
Step 2: We see that the theorem is true when f(x) = ei2πkx, for any fixedk 6= 0. For, ∫ 1
0
ei2πkx dx = 0.
Yet, for α = ei2πkϑ,
1
N
N∑n=1
ei2πk(nϑ) =1
N(α+ α2 + · · ·+ αN )
=1
N
α− αN+1
1− α
= O( 1
N
)→ 0.
Here we have made critical use of the fact that ϑ is irrational, so α 6= 1 for anyk.
Step 3: For general f , for all ε > 0, there exists a trigonometric polynomialsuch that ∣∣∣∣f(x)−
M2∑k=−M1
ckei2πkx
∣∣∣∣ < ε/2, ∀x.
From this we have,∣∣∣∣ ∫ f(x) dx− c0∣∣∣∣ =
∣∣∣∣ ∫ f(x) dx−∫P (x) dx
∣∣∣∣ =
∣∣∣∣ ∫ f(x)− P (x) dx
∣∣∣∣≤∫|f(x)− P (x)| dx
≤ ε/2.
Moreover, ∣∣∣∣ 1
N
N∑n=1
f({nϑ})− 1
N
N∑n=1
P ({nϑ})∣∣∣∣ ≤ ε/2.
Yet
limN→∞
1
N
N∑n=1
P ({nϑ}) = c0.
19
So
lim supN→∞
∣∣∣∣ 1
N
N∑n=1
f({nϑ})− c0∣∣∣∣ ≤ ε/2,
implying
lim supN→∞
∣∣∣∣ 1
N
N∑n=1
f({nϑ})−∫f(x) dx
∣∣∣∣ ≤ ε,As ε was arbitrary, this proves the theorem.
Proof of Theorem 33. It is easy to see that for all ε > 0, there exist continuousfunctions of period 1, f− and f+ such that
f−(x) ≤ 1(a,b)(x) ≤ f+(x),
for all x, yet
0 ≤∫ 1
0
f+(x)− 1(a,b)(x) dx ≤ ε
0 ≤∫ 1
0
1(a,b)(x)− f−(x) dx ≤ ε.
Note that ∫ 1
0
1(a,b)(x) dx = b− a.
Hence we have that,
1
N
∑n≤N
f−({nϑ}) ≤ 1
N
∑n≤N
1(a,b)({nϑ}) ≤1
N
∑n≤N
f+({nϑ}),
with1
N
∑n≤N
f−({nϑ})→∫ 1
0
f−(x) dx ≥ (b− a)− ε,
and1
N
∑n≤N
f+({nϑ})→∫ 1
0
f+(x) dx ≤ (b− a) + ε.
Finally, ∑n≤N
1(a,b)({nϑ}) = #{{nϑ} ∈ (a, b) : 1 ≤ n ≤ N
}.
Again because ε is arbitrary, we have shown that the limit in the theorem is asclaimed.
20
10 Another application: the Weierstrass approx-imation theorem
We use Fejer’s theorem to give a quick proof of another approximation theoremdue originally, with a different proof, to Weierstrass.
Theorem 35 (Weierstrass). For f ∈ C[0, 1/2], for all ε > 0 there is a polyno-mial Q(x) = a0 + a1x+ · · ·+ ahx
h such that
|f(x)−Q(x)| ≤ ε, ∀x ∈ [0, 1/2].
Here we have used the interval [0, 1/2] out of convenience, but it is not hardto see that this same theorem may be extended with any arbitrary interval [a, b]in place of [0, 1/2] (since re-scalings and translations of polynomials remainpolynomials).
Proof. Define g(x) := f(|x|) for x ∈ [−1/2, 1/2] and define g elsewhere by peri-odicity so that it is of period 1. Then g ∈ C(R), and can be approximated uni-formly by trigonometric polynomials. But each term ei2πkx that would appearas a summand in a trigonometric polynomial can be uniformly approximatedon [0, 1/2] by a sum
M∑`=0
1
`!(i2πkx)`,
for some M , by a Taylor approximation. Since a finite linear combination ofsuch sums remains a polynomial, we can approximate trigonometric polynomialsarbitrarily closely by ordinary polynomials, and thus we may approximate thefunction we began with as well.
[Lectures 4 and 5]
11 Inner product spaces
We have seen that the partial sums Sn(f, x) may not converge pointwise tof(x), even for a continuous function f . However, in the next two sections wewill show that for f ∈ L2[0, 1), we do have∫ 1
0
|f(x)− Sn(f, x)|2 dx→ 0.
In this section and the next we review some facts about inner product spacesand introduce/review the notion of Hilbert spaces, the latter of which cameabout historically in large part from the study of Fourier series. We will usethese notions to prove the convergence result above. (Throughout this sectionH will denote an arbitrary inner product space.)
Definition 36. A complex vector space H is an inner product space if thereexists an inner product 〈·, ·〉 : H2 → C such that for all x, y ∈ H, and α ∈ C,
21
(i) 〈x, y〉 = 〈y, x〉
(ii) 〈x+ y, z〉 = 〈x, z〉+ 〈y, z〉
(iii) 〈αx, y〉 = α〈x, y〉
(iv) 〈x, x〉 ≥ 0
(v) 〈x, x〉 = 0 if and only if x = 0.
From these definitions it is trivial to see that
Proposition 37. We have
i. 〈0, y〉 = 0
ii. 〈x, αy〉 = α〈x, y〉
iii. 〈x, y + z〉 = 〈x, y〉+ 〈x, z〉.
One can define a real inner product space in the same way, with scalarsalways real instead of complex.
The reader should check that that following are examples of inner productspaces:
(a) Cn with
〈x, y〉 =
n∑i=1
xiyi.
(Similarly Rn is a real inner product space.)
(b) L2[0, 1) with the inner product
〈f, g〉 =
∫ 1
0
fg dx,
as long as we make the identification that f1 = f2 whenever f1(x) = f2(x)almost everywhere.
(c) (somewhat less canonically) the subspace of L2[0, 1) consisting of polynomialfunctions
∑anx
n.
Definition 38. ‖x‖ :=√〈x, x〉.
‖x‖ is always non-negative and is called the norm of x, and we will demon-strate shortly that it possesses the properties of a norm. We first prove apreliminary but very important result:
Theorem 39 (Cauchy-Schwarz). For x, y ∈ H
|〈x, y〉| ≤ ‖x‖ · ‖y‖.
Before proving this, we first prove
22
Lemma 40. For u, v ∈ H
|〈u, v〉| ≤ 12‖u‖
2 + 12‖v‖
2.
One may note the similarity of this inequality to the inequality 2ab ≤ a2 +b2
for a, b positive reals. This follows from the fact that (a−b)2 ≥ 0. This strategyof proof we imitate in the proof below.
Proof of Lemma. Let |α| = 1 be such that α〈u, v〉 = |〈u, v〉|. Note that it followsfrom the axioms of an inner product space that |〈u, v〉| = 〈αu, v〉 = 〈v, αu〉 aswell. Then we have
0 ≤ 〈αu− v, αu− v〉 = ‖u‖2 − 2|〈u, v〉|+ ‖v‖2.
Proof of Cauchy-Schwarz. Clearly the claim is true if x = 0 or y = 0. If x, y 6= 0,let u = x/‖x‖ and v = y/‖y‖. Then ‖u‖ = ‖v‖ = 1, and the lemma tells usthat
〈u, v〉 ≤ 1.
But,
〈u, v〉 =〈x, y〉‖x‖ · ‖y‖
,
giving us the inequality we wanted.
One consequence is that inner products are continuous in their arguments.
Corollary 41. For any z ∈ H, x 7→ 〈x, z〉 is continuous.
Proof. |〈x1, z〉 − 〈x2, z〉| = |〈x1 − x2, z〉| ≤ ‖x1 − x2‖ · ‖z‖.
Another consequence of Cauchy-Schwarz:
Theorem 42 (Triangle Inequality). For x, y ∈ H,
‖x+ y‖ ≤ ‖x‖+ ‖y‖.
This is just the statement that H is a normed linear space.
Proof.
‖x+ y‖2 = 〈x+ y, x+ y〉 = ‖x‖2 + 〈x, y〉+ 〈y, x〉+ ‖y‖2
≤ ‖x‖2 + 2‖x‖ · ‖y‖+ ‖y‖2 = (‖x‖+ ‖y‖)2.
Corollary 43. For all x, y, z ∈ H
‖x− z‖ ≤ ‖x− y‖+ ‖y − z‖.
23
This corollary of the triangle inequality is just a reminder that d(x, y) =‖x− y‖ defines a metric.
Inner product spaces have slightly more structure that most normed linearspaces however. In particular, we have a notion of orthogonality:
Definition 44. For x, y ∈ H, we say x and y are orthogonal and write x ⊥ yif 〈x, y〉 = 0.
Definition 45. For x ∈ H and S ⊂ H, we say x is orthogonal to S and writex ⊥ S if 〈x, y〉 = 0 for all y ∈ S.
Definition 46. A set E is an orthogonal set if 〈x, y〉 = 0 for all x, y ∈ Esuch that x 6= y.
Definition 47. A set E ⊂ H is an orthonormal system if it is an orthogonalset and ‖x‖ = 1 for all x ∈ E.
The reader should check that the following are examples of orthonormalsystems:
(a) the set of all basis vectors {e1, e2, ..., en} ⊂ Cn,
(b) the set of functions {ei2πnx}n∈Z ⊂ L2[0, 1)
(c) the set {e1, e2, e3} ⊂ C4
(d) the set of functions {ei2πj2x}j∈N ⊂ L2[0, 1).
The following result is very important:
Theorem 48 (Pythagorean Theorem). If x, y ∈ H are orthogonal,
‖x+ y‖2 = ‖x‖2 + ‖y‖2.
Proof. 〈x+ y, x+ y〉 = ‖x‖2 + 〈x, y〉+ 〈y, x〉+ ‖y‖2 = ‖x‖2 + ‖y‖2.
By induction, this clearly implies
Corollary 49. If {x1, ..., xn} ⊂ H is an orthogonal set (xi ⊥ xj for all i 6= j),then
‖x1 + x2 + ·+ xn‖2 = ‖x1‖2 + ‖x2‖2 + · · ·+ ‖xn‖2
Corollary 50. Let {φ1, φ2, ..., φn} be a finite orthonormal system in H, andα1, α2, ..., αn ∈ C. Then ∥∥∥∥ n∑
j=1
αjφj
∥∥∥∥2 =
n∑j=1
|αj |2.
A deeper consequence is as follows:
24
Corollary 51 (Best approximation). Let {φ1, ..., φn} be an orthonormal system,and u ∈ H. Define
aj := 〈u, φj〉.
Then ∥∥∥∥u− n∑j=1
ajφj
∥∥∥∥ ≤ ∥∥∥∥u− n∑j=1
bjφj
∥∥∥∥,for all b1, b2, ..., bn ∈ C, with equality if and only if aj = bj for all j.
For reasons that will become clear in the next section, the coefficients 〈u, φn〉are called “Fourier coefficients” of the vector u, and the best approximationcorollary above says that the closest one can approximate u by linear combina-tions of the vectors {φj} is a linear combination whose coefficients are exactlythe Fourier coefficients.
Proof. Note that ⟨u−
n∑j=1
ajφj , φk
⟩= ak − ak = 0,
for all k. That is to say, u−∑ajφj is orthogonal to the set {φ1, φ2, ..., φn}. Let
δj = aj − bj , so that
u−∑
bjφj = u−∑
ajφj +∑
δjφj .
Then the Pythagorean Theorem and its corollaries imply∥∥∥∥u− n∑j=1
bjφj
∥∥∥∥2 =
∥∥∥∥u− n∑j=1
ajφj
∥∥∥∥2 +
n∑j=1
|δj |2
≥∥∥∥∥u− n∑
j=1
ajφj
∥∥∥∥2
Another consequence of the observation that u −∑ajφj is orthogonal to
the set {φ1, φ2, ..., φn} is Bessel’s inequality:
Theorem 52 (Bessel’s inequality). For an := 〈u, φn〉,n∑j=1
|aj |2 ≤ ‖u‖2.
Proof. Applying the Pythagorean Theorem and its corollaries as before,
‖u‖2 =
∥∥∥∥u− n∑j=1
ajφj
∥∥∥∥2 +
n∑j=1
|aj |2 ≥n∑j=1
|aj |2.
25
Remark: In fact we know that Bessel’s inequality does not express the ‘wholetruth’, in the sense that we know in Cn or Rn this inequality is in fact an equality:
‖α1e1 + · · ·αnen‖Cn =√|α1|2 + · · ·+ |αn|2.
On the other hand, if we began with the orthonormal system {e1, e2, e3} inC4, we would indeed have a strict inequality; To take for example the vectoru = e1 + e2 + e3 + e4 = (1, 1, 1, 1), we have
‖u‖2 = 4,
while|〈u, e1〉|2 + |〈u, e2〉|2 + |〈u, e3〉|2 = 3.
In C4, what distinguishes the orthonormal system {e1, e2, e3} – where Bessel’sinequality is a strict inequality – from the orthonormal system {e1, e2, e3, e4} –where it would be an equality? The answer, of course is that the linear span of{e1, e2, e3, e4} is C4, while the linear span of {e1, e2, e3} is not.
Can we reproduce this sort of reasoning in more complicated inner productspaces like L2? We can, but we must work with a certain refined type of innerproduct space known as a Hilbert space.
12 Hilbert space
12.1 General theory
Definition 53. If an inner product space H is complete (with respect to themetric induced by ‖ · ‖), then H is called a Hilbert space.
The reader should check that the following are examples of Hilbert spaces
(a) Cn with
〈x, y〉 =
n∑i=1
xiyi.
(Similarly Rn is a real Hilbert space.)
(b) L2[0, 1) with the inner product
〈f, g〉 =
∫ 1
0
fg dx,
as long as we make the identification that f1 = f2 whenever f1(x) = f2(x)almost everywhere.
(c) `2(N) := {((a1, a2, ...)) :∑|ai|2 < +∞} with
〈x, y〉 =
∞∑i=1
aibi.
26
In verifying that (b) and (c) are Hilbert spaces, one must use the non-trivial result, proved in Analysis III and also in most real analysis textbooks(for instance Rudin’s Real and Complex Analysis, pp. 67-68):
Theorem 54. L2[0, 1) and `2(N) are both complete.
(Recall that a complete metric space is one which contains the limit pointof every Cauchy sequence.)
The definitions of orthogonal vectors, sets, and orthonormal systems remainsthe same in Hilbert spaces as in inner product spaces.
We can quickly prove the following result:
Theorem 55. Let {φ1, φ2, ...} be an orthonormal system in a Hilbert space H,and ((α1, α2, ...)) ∈ `2(N). Then
∞∑j=1
αjφj (8)
converges in H. Moreover, ⟨ ∞∑j=1
αjφj , φn
⟩= αn. (9)
Proof. H is complete, so to see that (8) converges, we need only verify thatSN :=
∑j≤N αjφj is a Cauchy sequence:
‖SN − SM‖2 =
∥∥∥∥ N∑j=M+1
αnφn
∥∥∥∥2 =
N∑n=M+1
|αn|2,
from Corollary 50. But this is no more than
∞∑n=M+1
|αn|2 → 0
as M →∞, because α ∈ `2(N).To verify (9), note that because the inner product is continuous (Corollary
41), ⟨ ∞∑j=1
αjφj , φn
⟩= limN→∞
⟨ N∑j=1
αjφj , φn
⟩= αn.
The following sorts of orthonormal systems are of particular importance:
Definition 56. A complete orthonormal system in a Hilbert space H isan orthonormal system {φj} such that
span{φj} = H.
27
The reader should verify that the following are examples of complete or-thonormal systems:
(a) {e1, e2, ..., en} in Cn
(b) {((1, 0, 0, ...)), ((0, 1, 0, ..)), ((0, 0, 1, ...)), ...} in `2(N).
The following is an abstract version of the theorem about Fourier series weset out to prove.
Theorem 57. Let H be a Hilbert space, and {φj} be a complete orthonormalsystem. For any u ∈ H, define
Sn(u) =
n∑j=1
ajφj ,
where aj = 〈u, φj〉 as before. Then
‖u− Sn(u)‖ → 0.
The idea of our proof is this: we can approximate u arbitrarily closely bylinear combinations of elements φj , because these elements compose a completeorthonormal system. But by the best approximation corollary, for any linearcombination we use to approximate u, there will be a better approximation ofthe sort Sn(u).
Proof. Choose ε > 0. Because span{φj} = H, there exists a finite set of scalarsα1, α2, ..., αN such that ∥∥∥∥u− N∑
j=1
αjφj
∥∥∥∥ < ε.
But by Corollary 51 (best approximation), for n ≥ N ,∥∥∥∥u− n∑j=1
ajφj
∥∥∥∥ ≤ ∥∥∥∥u− N∑j=1
αjφj
∥∥∥∥.As ε is arbitrary this is just the statement that ‖u− Sn(u)‖ → 0.
We can also refine Bessel’s inequality,
Theorem 58 (Parseval). For all u ∈ H, with H a Hilbert space, and for aj asin Theorem 57, with {φj} a complete orthonormal system,
‖u‖2 =
∞∑j=1
|aj |2. (10)
Moreover, if v ∈ H and bj := 〈v, φj〉,
〈u, v〉 =
∞∑j=1
ajbj . (11)
28
Proof. We begin by proving (10) as in the proof of Bessel’s inequality. We have
‖u‖2 =
∥∥∥∥u− n∑j=1
ajφj
∥∥∥∥2 +
n∑j=1
|aj |2 =
∞∑j=1
|aj |2,
by letting n→∞, since ‖u− Sn(u)‖ → 0.(13) is similar but requires a little more bookkeeping. It is easy to verify
that
〈Sn(u), Sn(v)〉 =
n∑j=1
ajbj (12)
and as u = (u− Sn(u)) + Sn(u) and v = (v − Sn(v)) + Sn(v), we have
〈u, v〉 =〈u− Sn(u), v − Sn(v)〉+ 〈u, v − Sn(v)〉+ 〈u− Sn(u), v〉+ 〈Sn(u), Sn(v)〉
=
n∑j=1
ajbj + o(1). (13)
This last line follows from (12) and the relations n→∞,
|〈u− Sn(u), v − Sn(v)〉| ≤ ‖u− Sn(u)‖ · ‖v − Sn(v)‖ = o(1),
|〈u, v − Sn(v)〉| ≤ ‖u‖ · ‖v − Sn(v)‖ = o(1),
|〈u− Sn(u), v〉| ≤ ‖u− Sn(u)‖ · ‖v‖ = o(1).
Taking the limit n→∞ in (13) completes the proof.
12.2 Application to Fourier series
We begin this section by adding one more set to our list of complete orthonormalsystems.
Lemma 59. {ei2πjx}j∈Z is a complete orthonormal system in L2[0, 1).
Proof. We need only show that for all f ∈ L2[0, 1) and all ε > 0, there exists atrigonometric polynomial
P (x) =
N∑k=−M
ckei2πkx
such that‖f − P‖2 < ε.
We define the function TrimA(f) : x 7→ sgn(f(x)) ·min(A, |f(x)|). Note that bydominated convergence
limA→∞
∫ 1
0
(f − TrimA(f))2 dx = 0,
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so that there exists g ∈ L∞[0, 1) such that
‖f − g‖2 ≤ ε/3.
(Just take g = TrimA(f) for sufficiently large A, for instance.) Let B := ‖g‖∞.We can find φ ∈ C[0, 1) such that
‖φ− g‖1 ≤(ε/3)2
B,
and ‖φ‖∞ ≤ B. Hence,
‖φ− g‖22 =
∫ 1
0
|φ− g|︸ ︷︷ ︸≤B
·|φ− g| dx ≤ B · (ε/3)2
B
implying ‖φ− g‖2 ≤ ε/3. (To this point, what we have shown, in effect, is thatcontinuous functions are dense in L2[0, 1).)
This established, we can find a trigonometric polynomial P such that
|φ(x)− P (x)| ≤ ε/3, ∀x,
implying that‖φ− P‖2 ≤ ‖φ− P‖∞ ≤ ε/3.
Hence,‖f − P‖2 ≤ ‖f − g‖2 + ‖g − φ‖2 + ‖φ− P‖2 ≤ ε,
as we wanted.
With this lemma proved, Theorem 57 immediately implies
Theorem 60. For f ∈ L2[0, 1),
‖f − Sn(f)‖2 → 0.
Said another way,
f(x) = limn→∞
n∑k=−n
f(k)ei2πkx in the L2[0, 1) norm.
Because this is such an important result, we review the proof of Theorem57, now put in the concrete language of Theorem 60. The proof went as follows:choose any f ∈ L2[0, 1) and any ε > 0. From Lemma 59, we have that there is
a trigonometric polynomial P (x) =∑N−M cke
i2πkx such that
‖f − P‖2 ≤ ε.
But by best approximation (Corollary 51) for n ≥ N,M ,
‖f − Sn(f)‖2 ≤ ‖f − P‖2 ≤ ε.
This is the same claim that the statement in the theorem makes.Other results we have proved in the abstract setting of Hilbert spaces now
translate directly to results about Fourier series as well. Theorem 55 implies
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Theorem 61. For any {cj} ∈ `2(Z), there exists a unique f ∈ L2[0, 1) such
that ck = f(k) for all k ∈ Z.
Likewise Theorem 58 implies
Theorem 62 (Parseval). For f ∈ L2[0, 1), we have ‖f‖L2 = ‖f‖`2 . That is,∫ 1
0
|f(x)|2 dx =
∞∑k=−∞
|f(k)|2.
Moreover for f, g ∈ L2[0, 1),∫ 1
0
f(x)g(x) dx =
∞∑k=−∞
f(k)g(k).
[Lecture 6 - optional]
13 Riemann’s localization principle
In this section we prove a few last convergence results for Fourier series.
Theorem 63. For f of period 1 and in L1[0, 1), suppose f is differentiable atx0. Then
limn→∞
Sn(f, x0) = f(x0).
Proof. We define for t ∈ [−1/2, 1/2),
F (t) :=
{f(x0−t)−f(x0)
t if t 6= 0, t ∈ [−1/2, 1/2)
−f ′(x0) if t = 0.
As f is differentiable at x0, F (t) is bounded in a neighborhood of 0 andtherefore we see that F is integrable on [−1/2, 1/2). Note that
Sn(f, x0)− f(x0) =
∫ 1/2
−1/2f(x0 − t)Dn(t) dt− f(x0)
=
∫ 1/2
−1/2(f(x0 − t)− f(x0))Dn(t) dt (as
∫Dn(t) dt = 1)
=
∫ 1/2
−1/2F (t)
(tDn(t)
)dt.
Yet
tDn(t) =t
sinπtsin(2π(n+1/2)t
)= ei2πnt
eiπt
2i
( t
sinπt
)−e−i2πnt e
−iπt
2i
( t
sinπt
),
and both eiπt
2i ( tsinπt ) and e−iπt
2i ( tsinπt ) remain bounded in t, so that the theorem
is implied by the Riemann-Lebesgue lemma (Theorem 16).
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Note that the differentiability of a function at a point is an entirely localcondition, so that the convergence of Sn(f, x) depends in the above theoremupon how f behaves only at x. This leads to the following localization principleof Riemann, which is far from obvious from the definition of Sn(f, x0), whichdepends on the behavior of f everywhere, not just at x0.
Theorem 64 (Riemann). For f, g ∈ L1[0, 1) and of period 1, if f(x) = g(x)for all x in some neighborhood U of x0, then
Sn(f, x0)− Sn(g, x0)→ 0.
Proof. Note that
Sn(f, x0)− Sn(g, x0) = Sn(f − g, x0),
but (f − g)(x0) = 0 and moreover (f − g)(x) = 0 for x ∈ U , so that certainlyf − g is differentiable at x0. Theorem 63 then applies.
These theorems only scratch the surface of what is known about the conver-gence of Fourier series for various sorts of functions. We have discussed somemore advanced results in class, but any proofs of them exceed the scope ofthis class. The books by Katznelson or Krantz are an excellent place to startlearning about them in more depth.
14 Flat polynomials
We conclude these notes with a discussion of a problem due to Littlewood that,despite its elementary character, remains unsolved. The problem is as follows:
Do there exist positive constants c1 and c2 such that for all n ≥ 1 there issome choice of signs ε1 = ±1, ε2 = ±1, ..., εn = ±1 such that
c1√n ≤
∣∣∣∣ n∑k=1
εkei2πkx
∣∣∣∣ ≤ c2√n, for all x? (14)
(Here {ε1, ..., εn} = {ε(n)1 , ..., ε(n)n } may vary with n.)
Polynomials of this sort, such that∑εkz
k are roughly of order√n for all
|z| = 1 are known as flat polynomials. Why is√n the right constant here? The
reason is that for any choice {εk},∫ 1
0
∣∣∣∣ n∑k=1
εkei2πkx
∣∣∣∣2 dx =
n∑k=1
|εk|2 = n.
In particular, for at least some x, the sum∑εke
i2πkx must be at least√n, and
for some x the sum must be no more than√n. (If you don’t see why, assume
for instance the sum is always less than√n and derive a contradiction.)
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This reasoning shows that if ε can be chosen so that the sum is nearlyconstant for all x, the value it must always be close to is
√n.
I should make at this point two confessions about the above flat polynomialproblem. The first is that there is no consensus among experts that the prob-lem has an affirmative answer. Indeed, there does not seem to be any specialheuristic evidence either for it or against it. However, if we allow the coefficientsεk to be any complex scalar of absolute value 1, rather than just ±1, then it isknown that for all n there do exist flat polynomials of the sort (14). Indeed,something even stronger is true, that for any δ > 0 one can choose c1 = 1−δ andc2 = 1 + δ. That is, for all sufficiently large n there exist complex coefficients ε′
depending on n such that |ε′k| = 1 for all k, and
(1− δ)√n ≤
∣∣∣∣ n∑k=1
ε′kei2πkx
∣∣∣∣ ≤ (1 + δ)√n, for all x.
Results of this sort are due to Byrnes, Korner, Kahane, and Bombieri &Bourgain, among others. The restriction that εk takes only the values −1 or 1is more severe, but one may hope it is not so severe that a result of this sort islost entirely.
I leave the second confession to Tom Korner (who as listed above has workedon this problem and from whom I take much of the exposition below):
“Admittedly the problem has no deep mathematical significance, but I willhappily send the cost of a bottle of good champagne to the first person to answerit.”
I mention the problem at the end of our course mainly to convey that evenat a very elementary level, nice problems can be asked about trigonometricpolynomials and Fourier series, to which a solution is still not known. Perhapsyou might like to think up some of your own!
What we will demonstrate in what follows answers one half of the questionabove.
Theorem 65 (Golay-Shapiro-Rudin). For each n there exists ε(n)1 , ε
(n)2 , ...ε
(n)n ∈
{−1, 1} such that ∣∣∣∣ n∑k=1
ε(n)k ei2πkx
∣∣∣∣ ≤ A√n,where A is the absolute constant 1/(
√2− 1).
Proof. We define the polynomials Pm and Qm of degree 2m inductively by
P0(z) = z, Q0(z) = z,
andPm+1(z) = Pm(z) + z2
m
Qm(z),
Qm+1(z) = Pm(z)− z2m
Qm(z).
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It is easy to verify inductively that the coefficients of Pm and Qm remaineither −1 or 1, and for |z| = 1,
|Pm+1(z)|2 + |Qm+1(z)|2 = 2(|Pm(z)|2 + |Qm(z)|2
).
But this implies inductively that for |z| = 1,
|Pm(z)|2 + |Qm(z)|2 = 2m(|P0(z)|2 + |Q0(z)|2
),
and so|Pm(z)| ≤ 2(m+1)/2,
for all |z| = 1. Pm(z) is thus a polynomial of degree 2m satisfying the hypothesisof the theorem for n = 2m and A =
√2.
Yet for any integer n we may write
n =∑k=0
2m(k),
by expanding n into its binary representation. It is not difficult to verify that
P (z) :=zm(1)+m(2)+···+m(`−1)Pm(`)(z)
+ zm(1)+m(2)+···+m(`−2)Pm(`−1)(z)
+ · · ·+ zm(1)Pm(2)(z)
+ Pm(1)(z),
is equal ton∑k=0
ε(n)k zk,
for some coefficients ε(n)k that take the values −1 or 1 for all k. Moreover, for
|z| = 1,
|P (z)| ≤∑k=0
2(m(k)+1)/2
≤ 2(m(`)+1)/2(
1 +1
21/2+
1
22/2+ · · ·
)≤√n√
2− 1,
as we wanted.
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