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CONVERGENCE OF FOURIER SERIES

SOPHIA XUE

Abstract. The subject of Fourier analysis starts as physicist and mathemati-cian Joseph Fourier’s conviction that an ”arbitrary” function f could be givenas a series. In this expository paper, we build up from the basic definitions ofFourier analysis to answer the question in what sense does the Fourier seriesof a function converge to the function itself. Di↵erent criteria for convergencewill be introduced along the way. The proof of Mean Square Convergence willconclude this paper.

Contents

1. Introduction to Fourier Series 12. Uniqueness of Fourier series 33. Convolutions 64. Mean-square convergence of Fourier series 11Acknowledgments 15References 15

1. Introduction to Fourier Series

Throughout this paper, an integrable function should be interpreted as integrablein the Riemann sense. In this section, we will go through the basic definitions ofFourier Analysis that will appear throughout this paper. Examples and variationsare supplied to aid understanding.

Definition 1.1. If f is an integrable function given on an interval [a, b] of lengthL (that is, b� a = L), the nth Fourier coe�cient of f is defined by

f(n) =1

L

Zb

a

f(x)e�2⇡inx/Ldx, n 2 Z.

Note that more often than not, we will be working with functions that are integrableon a circle, which means that the function is integrable on every interval of length2⇡.

Definition 1.2. The Fourier series of f is given normally by

1X

n=�1f(n)e2⇡inx/L.

The notation

f(x) ⇠1X

n=�1f(n)e2⇡inx/L

1

2 SOPHIA XUE

means that the series on the right-hand side is the Fourier series of f .

Definition 1.3. The N th partial sum of the Fourier Series of f , for N a positiveinteger, is given by

SN

(f)(x) =NX

n=�N

f(n)e2⇡inx/L.

The notion of N th partial sum of the Fourier Series of f is very important inthe study of Fourier Analysis. Using the partial sums of the Fourier series, we canview the convergence of Fourier series as the ”limit” of these symmetric sums as Ntends to infinity . Indeed, the basic question can be reformulated as follows:

Question 1.4. In what sense does SN

(f) converge to f as N ! 1?

Following are some simple examples to familiarize ourselves with above defini-tions.

Example 1.5. Let f(✓) = ✓ for �⇡ ✓ ⇡. To calculate the Fourier coe�cientsof f , we should split it into two cases.

First, when n 6= 0,

f(n) =1

2⇡

Z⇡

�⇡

✓e�in✓d✓

=1

2⇡[✓e�in✓

�in

��⇡

�⇡

�Z

⇡

�⇡

e�in✓

�ind✓]

= �e�in⇡ + ein⇡

2in+

e�in⇡ � ein⇡

(2⇡in)(�in)

=�cos(n⇡)

in+

sin(n⇡)

i⇡n2

=(�1)n+1

in.

When n=0,

f(n) =1

2⇡

Z⇡

�⇡

✓d✓ = 0.

Hence, the Fourier series of f is

f(✓) ⇠X

n 6=0

(�1)n+1

inein✓ = 2

1X

n=1

(�1)n+1 sin(n✓)

n.

Furthermore, by the alternating series test, we can easily see that the Fourier seriesof f is convergent.

Example 1.6. Let f(✓) =| ✓ | for �⇡ ✓ ⇡. First, when n 6= 0,

f(n) =1

2⇡

Z⇡

0✓e�in✓d✓ � 1

2⇡

Z 0

�⇡

✓e�in✓d✓.

CONVERGENCE OF FOURIER SERIES 3

Using the result from last example, we get that

f(n) =1

2⇡[✓e�in✓

�in

��⇡

0

�Z

⇡

0

e�in✓

�ind✓ � ✓

e�in✓

�in

��0

�⇡

+

Z 0

�⇡

e�in✓

�ind✓]

=1

2⇡[⇡

e�in⇡

�in� 0 +

e�in⇡ � 1

(in)(�in)� 0 + (�⇡)

ein⇡

�in� 1� e�in⇡

(in)(�in)]

=1

2⇡[⇡

cos(�n⇡) + isin(�n⇡)� cos(n⇡)� isin(n⇡)

�in+

e�in⇡ � 1� 1 + e�in⇡

(in)(�in)]

= 0 +e�in⇡ � 1

⇡n2

=cos(n⇡)� isinn⇡ � 1

⇡n2

=(�1)n � 1

⇡n2.

On the other hand, when n = 0,

f(n) =1

2⇡

Z⇡

�⇡

| ✓ | d✓ =1

2⇡⇤ 2⇡2

2=

⇡

2.

Hence,

f(n) ⇠ 1

2⇡ein✓ +

X

n 6=0

(�1)n � 1

⇡n2ein✓

=1

2⇡ein✓ +

X

n=2k+1,k2N

�4

⇡n2ein✓

=1

2⇡(cos(n✓)� isin(n✓))� 4

⇡

X

n=2k+1,k2N

cos(n✓)� isin(n✓)

n2.

2. Uniqueness of Fourier series

If we were to believe that the Fourier series of functions f somehow converge tof , then we could infer that the Fourier coe�cient of a function uniquely determinesthe function. In other words, for our assumption to be true, if f and g have thesame Fourier coe�cients, then f and g are necessarily equal. The statement canbe reformulated as following by taking the di↵erence f � g:

Proposition 2.1. If f(n) = 0 for all n 2 Z, then f = 0.

It is obvious that this proposition cannot be true without reservation. Since cal-culating Fourier coe�cients requires integration and any two functions that are dif-ferent at finitely many points can have the same integration, one particular Fourierseries can be shared by two functions that are di↵erent at finitely many points.However, we do have the following positive result regarding continuous points.

Theorem 2.2. 1 Suppose that f is an integrable function on the circle with f(n) = 0for all n 2 Z. Then f(✓0) = 0 whenever f is continuous at the point ✓0.

This result is really nice. It shows that f vanishes for ”most” values of ✓.Following is a staightforward corollary.

1Rigorous proof of this theorem can be found in Elias M. Stein and Rami Shakarchi’s FourierAnalysis– an Introduction (2003), p39-41 following Theorem 2.1.

4 SOPHIA XUE

Corollary 2.3. If f is continuous on the circle and f(n) = 0 for all n 2 Z, thenf ⌘ 0.

Moreover, this corollary is useful in answering Question 2.4 under the conditionthat the Fourier series converges absolutely.

Corollary 2.4. Suppose that f is a continuous function on the circle and that

the Fourier series of f is absolutely convergent,P1

n=�1 | f(n) | < 1. Then, theFourier series converges uniformly to f , that is,

limN!1

SN

(f)(x) = f(x)

uniformly in x.

Proof. Let g(x) =P1

n=�1 f(n)einx. SinceP1

n=�1 | f(n) | < 1, g(x) is definedeverywhere.By triangle-inequality, we get that

| SN

(f)(x)�g(x) |=|X

|n|N

f(n)einx�1X

n=�1f(n)einx |=|

X

|n|>N

f(n)einx |X

|n|>N

| f(n) | .

Since the Fourier series of f is bounded, for any ✏ > 0, we can find large N suchthat

P|n|>N

| f(n) |< ✏. Hence, SN

(f)(x) uniformly converges to g(x) as N ! 1.Note that for large N , the Fourier coe�cients of g(x) are

g(n) =1

2⇡

Z 2⇡

0g(x)e�inxdx

=1

2⇡

Z 2⇡

0(X

nN

f(n)einx)e�inxdx

= f(n).

In other words, f and g have identical Fourier coe�cients. Now let’s define a newfunction h = f � g. By distributivity of the integral, it is easy to see that h(n) = 0.Apply Corollary 3.3 to h, we get that h = f � g = 0 or f = g. Combining thisresult with the uniform convergence of S

N

(f)(x) to g(x), we get the desired result,i.e. S

N

(f)(x) uniformly converges to f(x) as N ! 1. ⇤

Now that we have proven when the Fourier series of a continuous function isabsolutely convergent, its Fourier series converge uniformly to the said function,what conditions on f would guarantee the absolute convergence of its Fourier series?

Corollary 2.5. Suppose that f is a twice continuously di↵erentiable function onthe circle, then

f(n) = O(1/(| n |)2as | n |! 1,

so that the Fourier series of f converges absolutely and uniformly to f. The notation

f(n) = O(1/(| n |)2as | n |! 1 means that the left-hand side is bounded by a

constant multiple of the right-hand side, i.e. there exists C > 0 with | f(n) | C/(|n |)2 for all large | n | .


Proof. Through integrating by parts twice , we have

f(n) =1

2⇡

Z 2⇡

0f(✓)e�in✓d✓

=1

2⇡f(✓)

�e�in✓

in

��2⇡

0

+1

2⇡in

Z

02⇡f 0(✓)e�in✓d✓

=1

2⇡in

Z

02⇡f 0(✓)e�in✓d✓

=1

2⇡inf 0(✓)

�e�in✓

in

��2⇡

0

+1

2⇡(in)2

Z

02⇡f 00(✓)e�in✓d✓

=�1

n2

Z 2⇡

0f 00(✓)e�in✓d✓.

Since f and f 0 are both periodic, f(✓)�e

�in✓

in

��2⇡

0

= 0 and f 0(✓)�e

�in✓

in

��2⇡

0

= 0 Let B

be a bound for f 00. Thus,

2⇡ | n |2| f(n) ||Z 2⇡

0f 00(✓)e�in✓d✓ |

Z 2⇡

0| f 00(✓) | d✓ 2⇡B.

Since B is independent of n, we have that

f(n) = O(1/(| n |)2as | n |! 1

⇤

There are also stronger versions of Corollary 2.5 such as the following.

Corollary 2.6. If f is 2⇡-periodic and has an integrable derivative, then its Fourierseries converges absolutely and uniformly to f .

More generally,

Theorem 2.7. The Fourier series of f converges absolutely (and hence uniformlyto f) if f satisfies a Holder condition of order ↵, with ↵ > 1/2, that is

sup✓

| f(✓ + t)� f(✓) | A | t |↵ for all t.

Proof. First, we will prove that for any 2⇡-periodic and integrable function f ,

f(n) = � 1

2⇡

Z⇡

�⇡

f(x+ ⇡/n)e�inxdx,

and thus

f(n) =1

4⇡

Z⇡

�⇡

[f(x)� f(x+ ⇡/n)]e�inxdx.

By definition, we have

f(n) =1

2⇡

Z⇡

�⇡

f(y)e�inydx.

6 SOPHIA XUE

Since f is periodic and we are integrating over the period of f , we can substitutex+ ⇡n for y without changing the bounds of integration. Hence,

f(n) =1

2⇡

Z⇡

�⇡

f(x+ ⇡/n)e�in(x+⇡/n)dx

=1

2⇡

Z⇡

�⇡

f(x+ ⇡/n)e�inxe�⇡idx

= � 1

2⇡

Z⇡

�⇡

f(x+ ⇡/n)e�inxdx.

The last equality follows from applying Euler’s Identity. Since

2f(n) =1

2⇡(

Z⇡

�⇡

f(x)e�inxdx�Z

⇡

�⇡

f(x+ ⇡/n)e�inxdx),

a little algebraic manipulation gives the desired result, that

f(n) =1

4⇡

Z⇡

�⇡

[f(x)� f(x+ ⇡/n)]e�inxdx.

It follows

| f(n) | = 1

4⇡|Z

⇡

�⇡

[f(x)� f(x+ ⇡/n)]e�inxdx |

1

4⇡

Z⇡

�⇡

| f(x)� f(x+ ⇡/n) || e�inx | dx.

Apply Holder and we get

| f(n) | 1

4⇡|Z

⇡

�⇡

C | ⇡/n |↵| e�inx | dx.

SinceR⇡

�⇡

| e�inx | dx = 2⇡,

| f(n) | C⇡↵

2 | n |↵ ,

which proves the theorem. ⇤

3. Convolutions

Definition 3.1. Given two 2⇡-periodic integrable functions f and g on R, theirconvolution f ⇤ g on [�⇡,⇡] is given by

(f ⇤ g)(x) = 1

2⇡

Z⇡

�⇡

f(y)g(x� y)dy.

Note that since the product of two integrable functions is again integrable, theabove integral makes sense for all x. Furthermore, since the functions are periodic,we can change variables to see that

(3.2) (f ⇤ g)(x) = 1

2⇡

Z⇡

�⇡

f(x� y)g(y)dy.

The notion of convolution plays a fundamental role in Fourier analysis. It re-duces the problem of understanding S

N

(f) to understanding the convolution of twofunctions as we will see in the following example.


Example 3.3. DN

is the N th Dirichlet kernel given by

DN

(x) =NX

n=�N

einx.

For any 2⇡-periodic integrable function f , its convolution with the N th Dirichletkernel is

(f ⇤DN

)(x) =1

2⇡

Z⇡

�⇡

f(y)(NX

n=�N

ein(x�y))dy

=NX

n=�N

(1

2⇡

Z⇡

�⇡

f(y)e�inydy)einx

=NX

n=�N

f(n)einx

= SN

(f)(x).

As we have shown above, Question 2.4 can be again reformulated as following:

Question 3.4. In what sense does (f ⇤DN

)(x) converge to f as N ! 1?

But before we dive in, let’s look at some nice properties of convolution.

Proposition 3.5. Suppose that f , g, and h are 2⇡-periodic integrable functions.Then:

(i) f ⇤ (g + h) = (f ⇤ g) + (f ⇤ h)

(ii) (cf) ⇤ g = c(f ⇤ g) = f ⇤ (cg) for any c 2 C

(iii) f ⇤ g = g ⇤ f

(iv) (f ⇤ g) ⇤ h = f ⇤ (g ⇤ h)

(v) f ⇤ g is continuous

(vi) [f ⇤ g(n) = f(n)g(n).

Part (i) and (ii) give linearity. Part (iii) gives commutativity while part (iv)demonstrates associativity. Part (v) shows that convolution is a smoothing opera-tion in the sense that even though f and g are merely integrable, f ⇤g is continuous.Part (vi) converts convolution to multiplication, which is key in the study of Fourierseries.

Now we move on to the proof of Proposition 3.5. Part (i) and (ii) follow directlyfrom the linearity of integration. Part (iii) follows from Equation (3.2). Part(iv) can be easily proven through interchanging two integral signs and changingvariables. Though Part (v) and (vi) are easily deduced if f and g are continuous,to prove these properties under the condition that f and g are merely integrable,we are going to need the following approximation lemma.

8 SOPHIA XUE

Lemma 3.6. Suppose f is integrable on the circle and bounded by B. Then thereexists a sequence {f

k

}1k=1 of continuous functions on the circle so that

supx2[�pi,⇡]

| fk

(x) | B

for all k = 1, 2, ..., andZ

⇡

�⇡

| f(x)� fk

(x) | dx ! 0 as k ! 1.

Proof of Lemma 3.6. When f is real, given ✏ > 0, 9 a partition P of the interval[�⇡,⇡] such that U(f, P )�L(f, P ) < ✏ since f is integrable. Now we define a stepfunction

g(x) = supxj�1yxj

f(y)

if x 2 [xj�1, xj

) for j 2 [1, N ]. We can see that g is bounded by B andZ

⇡

�⇡

| g(x)� f(x) | dx =

Z⇡

�⇡

(g(x)� f(x))dx < ✏.

To modify the step function g and make it continuous, we will take small � > 0and define g*(x) = g(x) when the distance between x and any partition point in Pis at least �. When the distance between x and any partition point in P is more than�, define g*(x) as the linear function that connects g(x� �) and g(x+ �). It is easyto see that g* is continuous. Now let g*(x) = 0 when x 2 [�⇡ � �,�⇡ + �]andx 2[⇡� �,⇡+ �], which extends g* to a 2⇡-periodic function. Since g* only di↵ers fromg in N intervals of length s� and is also bounded by B, we have

Z⇡

�⇡

| g(x)� g*(x) | dx 4BN�.

With � su�ciently small, we haveZ

⇡

�⇡

| g(x)� g*(x) | dx < ✏.

By triangle inequality,Z

⇡

�⇡

| f(x)� g*(x) | dx Z

⇡

�⇡

| g(x)� f(x) | dx+

Z⇡

�⇡

| g(x)� g*(x) | dx < 2✏.

Let 2✏ = 1k

and denote g* by fk

, the sequence {fk

} is what we are looking for.When f is complex, we can apply the above proof to the real part and the

imaginary part separately to get the same result. ⇤

Applying Lemma 3.6, we can complete the proof of part (v) and (vi), the ideahere is to substitute f

k

and fk

for f and f respectively and use the continuity offk

to complete the proof.

Proof of Part (v) and (vi). Let’s apply Lemma 3.6 to 2⇡-periodic integrable func-tions f and g. We get sequences {f

k

} and {gk

} of approximating continuous func-tions. Note that with a bit of algebraic manipulation, we have

f ⇤ g � fk

⇤ gk

= (f � fk

) ⇤ g + fk

(g � gk

).


Moreover, (f � fk

) ⇤ g uniformly converges to 0 as k ! 1.

| (f � fk

) ⇤ g(x) 1

2⇡

Z⇡

�⇡

| f(x� y)� fk

(x� y) || g(y) | dy

1

2⇡supy

| g(y) |Z

⇡

�⇡

| f(y)� fk

(y) | dy

! 0 as k ! 1.

Similarly, fk

⇤ (g� gk

) ! 0 uniformly in x. Therefore, as k ! 1, fk

⇤ gk

uniformlyconverges to f ⇤ g. By continuity of each f

k

⇤ gk

, f ⇤ g is also continuous, whichproves Part (v).

For each integer n, since fk

⇤ gk

uniformly converges to f ⇤ g as k ! 1, we have

\fk

⇤ gk

(n) ! [f ⇤ g(n) as k ! 1.

By continuity of fk

and gk

, we have

\fk

⇤ gk

(n) =1

2⇡

Z⇡

�⇡

(fk

⇤ gk

)(x)einxdx

=1

2⇡

Z⇡

�⇡

1

2⇡(

Z⇡

�⇡

fk

(y)gk

(x� y)dy)einxdx

=1

2⇡

Z⇡

�⇡

fk

(y)e�iny(1

2⇡

Z⇡

�⇡

gk

(x� y)e�in(x�y)dx)dy

=1

2⇡

Z⇡

�⇡

fk

(y)e�iny(1

2⇡

Z⇡

�⇡

gk

(x)e�inxdx)dy

= f(n)g(n).

Hence,

| f(n)� fk

(n) | = 1

2⇡|Z

⇡

�⇡

(f(x)� fk

(x))e�inxdx |

1

2⇡

Z⇡

�⇡

| f(x)� fk

(x) | dx

! 0 as k ! 1,

which means fk

(n) ! f(n) as k ! 1. Again, similarly, gk

(n) ! g(n) as k ! 1.By substituting f

k

(n) and gk

(n) for f(n) and g(n), we can easily get Part (vi). ⇤

Now that we have familiarized ourselves with the basic properties of convolution,we will introduce the concept of good kernels.

Definition 3.7. A family of good kernels is a sequence of functions {Kn

(x)}1n=1

that satisfies the following properties:

(a) For all n � 1,1

2⇡

Z⇡

�⇡

Kn

(x)dx = 1.

(b)There exists M > 0 such that for all n � 1,Z

⇡

�⇡

| Kn

(x) | dx M.

10 SOPHIA XUE

(c) For every � > 0,Z

�|x|⇡

| Kn

(x) | dx ! 0, as n ! 1.

The concept of good kernels is very important in the study of Fourier seriesbecause of the nice property good kernels have, which is stated in the followingtheorem.

Theorem 3.8 (Approximation to the Identity). Let {Kn

(x)}1n=1 be a family of

good kernels, and f an integrable function on the circle. Then

limn!1

(f ⇤Kn

)(x) = f(x)

whenever f is continuous at x. If f is continuous everywhere, then the above limitis uniform.

Proof. If f is continuous at x, given ✏ > 0, 9� such that if | y |< � then | f(x� y)�f(x) |< ✏. By property (a) of good kernel, we have

(f ⇤Kn

(x)� f(x)) =1

2⇡

Z⇡

�⇡

Kn

(y)f(x� y)dy � f(x)

=1

2⇡

Z⇡

�⇡

Kn

(y)[f(x� y)� f(x)]dy.

Thus, let B be a bound for f , by triangle inequality, we get the following:

| (f ⇤Kn

)(x)� f(x) | =| 1

2⇡

Z⇡

�⇡

Kn

(y)[f(x� y)� f(x)]dy |

1

2⇡

Z

|y|<�

| Kn

(y) || f(x� y)� f(x) | dy

+1

2⇡

Z

�|y|⇡

| Kn

(y) || f(x� y)� f(y) | dy

✏

2⇡

Z

�⇡

⇡ | Kn

(y) | dy + 2B

2⇡

Z

�|y|⇡

| Kn

(y) | dy.

By property (b) of good kernels, ✏

2⇡

R�⇡

⇡ | Kn

(y) | dy ✏M

2⇡ . By property (c) of

good kernels, for large n, B

⇡

R�|y|⇡

| Kn

(y) | dy < ✏. Let C > 0 be some constant.For large n, we have

| (f ⇤Kn

)(x)� f(x) | C✏.

If f is continuous everywhere, by compactness of [�⇡,⇡], f is uniformly contin-uous, which completes the proof. ⇤

Let’s recall what we have shown in Example 3.3, that

(f ⇤DN

)(x) = SN

(f)(x).

Having proven the above theorem, it is natural to wonder whether or not theDirichlet kernels are a family of good kernel. If the Dirichlet kernels are a familyof good kernels, Theorem 3.8 would imply that the Fourier series of f is point-wiseconvergent to f at points of continuity.


Example 3.9 (the Dirichlet kernels is not a family of good kernel). Let’s firstremind ourselves that

DN

(✓) =X

k=�N

Neik✓ =sin((N + 1/2)✓)

sin(✓/2).

The last part of the equation is the closed form formula for the N th Dirichlet kernel.Since x

sin(x) � 1 for x 2 [�⇡/2,⇡/2],

| DN

(✓) |� 2| sin(N + 1/2)✓ |

✓for ✓ 2 [�⇡, pi].

It follows then,Z

⇡

�⇡

| DN

(✓) | d✓ � 4

Z⇡

0

| sin(N + 1/2)✓ |✓

d✓

= 4

Z (N+1/2)⇡

0

| sin(✓) |✓

d✓

� 4

ZN⇡

0

| sin(✓) |✓

= 4N�1X

k=0

Z (k+1)⇡

k⇡

| sin(✓) |✓

d✓

� 4N�1X

k=0

1

(k + 1)⇡

Z (k+1)⇡

k⇡

| sin(✓) | d✓

=8

⇡

N�1X

k=0

1

k + 1

=8

⇡log(N + 1)

� 8

⇡logN,

which meansR⇡

�⇡

| DN

(✓) | d✓ is not bounded and thus the Dirichlet kernel does notsatisfy property (b) of good kernels. This result is disheartening as it suggests thatthe point-wise convergence of Fourier series may even fail at points of continuity.

4. Mean-square convergence of Fourier series

Although we hit a dead end at the end of last section when we saw that Dirichletkernels are not good kernels, in this section, we will show that regarding the overallbehavior of a function f over the entire interval [0, 2⇡], there are still nice results.

Orthogonality is at the heart of the theorem we are about to prove. We first layout the definitions and notations that will be used throughout this section.

Consider a space R of integrable functions on the circle. The inner product oftwo integrable functions f and g is defined as

(f, g) =1

2⇡

Z 2⇡

0f(✓)g(✓)d✓.

12 SOPHIA XUE

The norm ||f || is defined as

||f ||2 = (f, f) =1

2⇡

Z 2⇡

0| f(✓) |2 d✓.

Notation-wise, for convenience, we use an

to denote the Fourier coe�cients off , where f is an integrable function on the circle. As ein✓ would appear over andover in this section, for each integer n, let e

n

= ein✓.Now that we have everything defined, let’s get down to some easy examples to

familiarize ourselves with the definitions and notations.

Example 4.1. Consider the family {en

}n2Z.

When n = m,

(en

, em

) =1

2⇡

Z 2⇡

0ein✓eim✓d✓

=1

2⇡

Z 2⇡

0ei(n�m)✓d✓

=1

2⇡

Z 2⇡

01d✓

= 1.

When n 6= m,

(en

, em

) =1

2⇡

Z 2⇡

0ein✓eim✓

= frac12⇡

Z 2⇡

0ei(n�m)✓d✓

=1

2⇡

ei(n�m)✓

i(n�m)

��2⇡

0

=1

2⇡

ei(n�m)2⇡ � e0

i(n�m)

=1

2⇡

1 + 0� 1

i(n�m)

= 0.

Therefore, we have shown that the family {en

}n2Z is orthonormal.

Besides the orthonormal property of the family {en

}n2Z, another important ob-

servation is that the Fourier coe�cients of the function f can be represented as theinner products of f with elements of the above family:

(f, en

) =1

2⇡

Z 2⇡

0f(✓)e�in✓d✓ = a

n

.

Moreover, it is also important to note that

SN

(f) =X

|n|N

an

en

.

With these in mind, we are ready to prove a simple but useful lemma.


Lemma 4.2. For any complex number bn

,

(f �X

|n|N

an

en

)?X

|n|N

bn

en

.

Proof. We will first show that (f �P

|n|N

an

en

)?en

.

(f �X

|n|N

an

en

, en

) = (f, en

)� (X

|n|N

an

en

, en

)

= an

� an

= 0

For any complex number bn

,

(f �X

|n|N

an

en

,X

|n|N

bn

en

) = (f �X

|n|N

an

en

, b�N

e�N

) + ...+ (f �X

|n|N

an

en

, bN

eN

)

= b�N

(f �X

|n|N

an

en

, e�N

) + ...+ bN

(f �X

|n|N

an

en

, eN

)

= 0 + ...+ 0

= 0.

⇤

While this lemma may not seem too interesting by itself, it has two very usefulresults.

For the first result, let’s recall the Pythagorean theorem, which states that if Xand Y are orthogonal, then

||X + Y ||2 = ||X||2 + ||Y ||2.

Let bn

= an

, we get

||f �X

|n|N

an

en

+X

|n|N

an

en

||2 = ||f �X

|n|N

an

en

||2 + ||X

|n|N

an

en

||2,

which can be rewritten as following:

||f ||2 = ||f � SN

(f)||2 +X

|n|N

| an

|2 .

The second result is best known as the best approximation lemma, which statesthat

Lemma 4.3 (Best approximation lemma). If f is integrable on the circle withFourier coe�cients a

n

, then for any complex numbers cn

||f � SN

(f)|| ||f �X

|n|N

cn

en

||.

Moreover, equality holds precisely when cn

= an

for all | n | N .

Proof. Let bn

= an

� cn

8n,

f �X

|n|N

cn

en

= f �X

|n|N

an

en

+X

|n|N

bn

en

.

14 SOPHIA XUE

By Lemma 4.2, we know that (f �P

|n|N

an

en

)?(f �P

|n|N

bn

en

). Hence, wecan apply the Pythagorean theorem and get the following result:

||f � (f �X

|n|N

cn

en

)||2 = ||f � SN

(f)||2 + ||(f �X

|n|N

bn

en

)||2.

When cn

= an

, bn

= 0

||f � (f �X

|n|N

cn

en

)|| = ||f � SN

(f)||.

When cn

6= an

, bn

6= 0

||X

|n|N

bn

en

||2 =1

2⇡

Z 2⇡

0| b

n

en

|2 d✓ > 0.

Hence,

||f � SN

(f)|| < ||f �X

|n|N

cn

en

||.

⇤Theorem 4.4 (Mean Square Convergence). Suppose f is integrable on the circle.Then

1

2⇡

Z 2⇡

0| f(✓)� S

N

(f)(✓) |2 d✓ ! 0 as N ! 1.

Proof. Apply Lemma 3.6 and choose a continuous function g on the circle such that

sup✓2[0,2⇡]

| g(✓) | sup✓2[0,2⇡]

| f(✓) |= B

and Z 2⇡

0| f(✓)� g(✓) | d✓ < ✏2.

Since supx2[�⇡,⇡] | f(x) |= B,

||f � g||2 =1

2⇡

Z 2⇡

0| f(✓)� g(✓) |2 d✓

=1

2⇡

Z 2⇡

0| f(✓)� g(✓) || f(✓)� g(✓) | d✓

B

⇡

Z 2⇡

0| f(✓)� g(✓) | d✓

C✏2

where C = B

⇡

is some constant.Moreover, we can approximate g by a trigonometric polynomial P of degree M

such that| g(✓)� P (✓) |< ✏, for all ✓.

After taking squares and integrating, we get

||g � P || < ✏.

It follows directly that ||f � P || < C1/2✏. We can then get the desired result byapplying the best approximation lemma.

⇤


Acknowledgments. It is a pleasure to thank my mentor, Mariya Sardarli andDolores Walton, for all their help. Without their patient support and guidance,this paper would not be possible. I would also like to thank Professor Peter Mayfor organizing this amazing program and Professor Lszl Babai for his interestinglectures that made my summer meaningful.

References

[1] Elias M. Stein and Rami Shakarchi. Fourier analysis-an introduction, Princeton UniversityPress. 2003.

[2] Maria Christina Pereyra and Lesley A. Ward. Harmonic Analysis: from Fourier to Haar

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