Truncations of Fourier Series as Best Approximants1

M.THAMBAN NAIR

Department of Mathematics, IIT Madras

Chennai-600 036, INDIA,

E-Mail mtnair@iitm.ac.in

1. Trigonometric Polynomials and 2π-Periodic Functions

Expression of the form

pk(t) := c0 +k∑

n=1

(an cos nt + bn sin nt) , t ∈ R, k = 1, 2, . . . ,

for real numbers c0, an, bn are called trigonometric polynomials. We may observe that

pk(t) = pk(t + 2π) ∀ t ∈ R,

i.e., pk(t) are 2π-periodic functions on R. Recall that a function f : R → R is said to

be a 2π-periodic function on R if

f(t) = f(t + 2π) ∀ t ∈ R.

Writing cos nt = ent+e−nt

2, sin nt = ent−e−nt

2i, the trigonometric polynomial pk(t) can

be represented as

pk((t) =k∑

n=−k

cnent,

with cn = an−ibn

2, c−n = an+ibn

2for n ∈ {1, 2, . . . , k}. We may observe that

an =1

π

∫ π

−π

pk(t) cos nt dt, bn =1

π

∫ π

−π

pk(t) sin nt dt, n ∈ {1, . . . , k},

cn =1

2π

∫ π

−π

pk(t)e−intdt, n ∈ {0,±1,±2, . . . ,±k}.

1Lecture at a QIP-Short Term course at Department of Mathematics, IIT Madras, on July 7, 2004.1

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2. Fourier Series of 2π-Periodic Functions

Suppose f is a 2π-periodic function on R integrable on [−π, π]. Then the series

S(f, t) := c0 +∞∑

n=1

(an cos nt + bn sin nt) , t ∈ R,

with

c0 =1

2π

∫ π

−π

f(t)dt, an =1

π

∫ π

−π

f(t) cos ntdt, bn =1

π

∫ π

−π

f(t) sin ntdt, n ∈ N,

is called the Fourier series of f . The numbers c0, an, bn are called the Fourier coeffi-

cients of f .

If f is an even function, then we see that

c0 =1

π

∫ π

0

f(t)dt, an =2

π

∫ π

0

f(t) cos ntdt, bn = 0 ∀n ∈ N.

Similarly, if f is an odd function, then we see that for every n ∈ N, and in that case

c0 = 0, an = 0, bn =2

π

∫ π

0

f(t) sin ntdt ∀n ∈ Z.

In case it is known that f is defined only on [0, π], then it can be extended to [−π, π]

as an even function or an odd function, and obtain the Fourier coefficients accordingly.

The motivation for studying Fourier series initiated by Joseph Fourier (1768-1830)

was its use in the context of heat equation:

∂u

∂t=

∂2u

∂x2,

0 < x < π

0 < t ≤ τ

with the boundary condition

u(0, t) = 0 = u(π, t), 0 < t ≤ τ,

and with initial temperature u(x, 0) = f(x). Here, u(x, t) represents the heat pro-

file, the temperature, at time t at a point x on a thin metallic wire of length π,

represented by the closed interval [0, π]. It can be seen, by the method of separa-

tion of variables, that a functions of the form u(x, t) := bne−n2t sin nx, n ∈ N, are

solutions of the heat equation and the boundary conditions; but they will satisfy the

initial condition only when f is of the form f(x) = bn sin nx. Similarly functions

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of the form∑k

n=1 bne−n2t sin nx, are solutions of the heat equation and the bound-

ary conditions; but they will satisfy the initial condition only when f is of the form

f(x) =∑k

n=1 bn sin nx. This motivates us to consider the solutions of the form

u(x, t) =∞∑

n=1

bne−n2t sin(nx).

For this to satisfy the initial condition u(s, 0) = f(s), we must have

f(x) = u(x, 0) =∞∑

n=1

bn sin(nx),

the Fourier series of f . Thus, we must have

bn =2

π

∫ π

0

f(ξ) sin nξdξ.

We may also observe that for a fixed t, the Fourier coefficients of u(·, t) are bne−n2t wit

bn as above.

Now the important question is whether the series S(f, t) represents the function f ,

i,e., whether the sequence of partial sums of S(f, t),

Sk(f, t) := c0 +k∑

n=1

(an cos nt + bn sin nt) , k ∈ N,

converges to f in some sense. In this regard we have the following results.

THEOREM 2.1. (See Theorem 2.3.9, Bhatia [1]) Suppose f is integrable on [−π, π]

and Lipschitz continuous at t0 ∈ (−π, π). Then

limk→∞

Sk(f, t0) = f(t0).

THEOREM 2.2. (See Exercise 2.3.10, Bhatia [1]) Suppose f is piecewise continuously

differentiable on [−π, π], and continuous at t0 ∈ (−π, π). Then

limk→∞

Sk(f, t0) = f(t0).

If f is discontinuous at t0 ∈ (−π, π), then

limk→∞

Sk(f, t0) =f(t0+) + f(t0−)

2.

THEOREM 2.3. (Dirichlet’s Theorem, See Theorem 2.3.18, Bhatia [1]) Suppose

f is of bounded variation on [−π, π] and continuous at t0 ∈ (−π, π). Then

limk→∞

Sk(f, t0) = f(t0).

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If f is continuous, then the convergence of {Sk(f, t)} to f(t) is uniform.

THEOREM 2.4. (Carleson, 1966) Suppose f is square integrable, then {Sk(f, t)}converges to f(t) almost everywhere. In particular, if f is continuous, then {Sk(f, t)}converges to f(t) except on a set of measure zero.

Here, a set E ⊆ R is said to be of measure zero if it can be covered by a countable

number of open intervals whose total length is zero.

Along with the above positive results we also have a number of negative results.

THEOREM 2.5. (See Nair [2], Theorem 6.10) There exists a dense subset X0 of

X := {f ∈ C[−π, π] : f(−π) = f(π)} w.r.t. the supremum-norm ‖ · ‖∞ such that for

every f ∈ X0, {Sk(f, 0)} diverges.

THEOREM 2.6. (Kolmogorov, 1926) There exists an integrable function f on

[−π, π] such that {Sk(f, t)} diverges at every t ∈ [−π, π].

THEOREM 2.7. (Kahane and Katznelson) If E is subset of [−π, π] is of measure

zero, then there exists a continuous 2π-periodic function f such that its fourier series

diverges for all t ∈ E.

3. Best Approximation Property of Sk(f, ·)

Now we show that if f ∈ C[−π, π], then its Fourier series converges to f in the

least-square sense. Moreover, we show that

‖f − Sk(f, ·)‖2 ≤ ‖f − ϕ‖2

for every trigonometric polynomial of order k, i.e., for every ϕ of the form

ϕ(t) = γ0 +k∑

n=1

(αn cos nt + βn sin nt)

with real numbers γ0, αn, βn.

A few preliminaries from Functional Analysis would simplify the exposition to a

great deal:

Let V be an inner product space with inner product 〈·, ·〉 and corresponding norm

‖·‖. Recall that vectors u, v in V are said to be orthogonal if 〈u, v〉 = 0; and a set S ⊆ V

is said to be an orthogonal set if every pair of distinct vectors in S are orthogonal. An

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orthogonal set in which every vector is of unit norm is called an orthonormal set. It is

easily seen that every orthonormal set is linearly independent.

As an example consider V = C[−π, π] with inner product

〈u, v〉 =

∫ π

−π

u(t)v(t) dt, u, v ∈ C[−π, π].

Denote

un(t) =eint

√2π

, n ∈ Z,

ϕ0(t) =1√2π

, ϕn(t) =

{cos nt√

πif n even,

sin nt√π

if n odd.

Then it is seen that the sets

(3.1) {un : n ∈ Z}, {ϕn : n ∈ {0} ∪ N}

are orthonormal sets in C[−π, π].

Now, let {v1, . . . , vk} be an orthonormal set in V and V0 be its span. For u ∈ V , let

(3.2) Pu :=k∑

n=1

〈u, vn〉vn.

We observe that Pvj = vj for every j = 1, . . . , k so that Pv = v for every v ∈ V0. Also

we observe that

〈Pu, u〉 = 〈u, Pu〉 = 〈Pu, Pu〉 =k∑

n=1

|〈u, vn〉|2.

Hence, for every u ∈ V ,

(3.3) ‖u− Pu‖2 = 〈u, u〉 − 〈u, Pu〉 − 〈Pu, u〉+ 〈Pu, Pu〉 = ‖u‖2 − ‖Pu‖2.

Now let u ∈ V . Hence, for u ∈ V and v ∈ V0, we have

‖(u− v)− P (u− v)‖2 = ‖u− v‖2 − ‖P (u− v)‖2 ≤ ‖u− v‖2.

But, since v ∈ V0, ‖(u− v)− P (u− v)‖ = ‖u− v − Pu + Pv‖ = ‖u− Pu‖. Thus we

have proved the following.

THEOREM 3.1. For every u ∈ V ,

‖u− Pu‖ ≤ ‖u− v‖ ∀u ∈ V0.

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From (3.3), we have

k∑n=1

|〈u, vn〉|2 = ‖Pu‖2 ≤ ‖u‖2, (Bessel′s inequality)

Pu = u ⇐⇒k∑

n=1

|〈u, vn〉|2 = ‖Pu‖2 = ‖u‖2. (Parseval′s identity)

It can also be seen that P is a linear transformation from V into itself, and satisfies

P 2 = P, 〈u, v〉 = 0 ∀u ∈ R(P ), v ∈ N(P ).

Such a map is called an orthogonal projection onto V0.

Recall that {un : n ∈ Z} and {ϕn : n ∈ {0} ∪ N} are orthonormal sets in C[−π, π].

Let

Vk := span{un : n = 0,±1,±2, . . . ,±k} = span{ϕ0, ϕ1, . . . , ϕ2k}.

We may observe that for f ∈ C[−π, π],

〈f, un〉 =1√2π

∫ π

−π

f(t)e−intdt, n ∈ Z,

〈f, ϕ2n〉 =1√π

∫ π

−π

f(t) cos nt dt, 〈f, ϕ2n−1〉 =1√π

∫ π

−π

f(t) sin nt dt, n ∈ N.

Thus,

〈f, un〉un(t) = cneint, 〈f, ϕ2n〉ϕ2n(t) = an cos nt, 〈f, ϕ2n−1〉ϕ2n−1(t) = bn sin nt.

Therefore, if we denote by Pk the orthogonal projection onto Vk, defined as in (3.2),

then we see that Pkf = Sk(f, ·). As a consequence of Theorem 3.1, we have the

following.

THEOREM 3.2. For every f ∈ C[−π, π],

(3.4) ‖f − Sk(f, ·)‖2 ≤ ‖f − ϕ‖2

for every trigonometric polynomial of order k, i.e., for every ϕ ∈ Vk.

Remarks. (a) Theorem 3.1 is not only true for projections of the form (3.2), but also

for any orthogonal projection P : V → V , i.e., for any linear transformation P : V → V

which satisfies

P 2 = P, R(P ) ⊥ N(P ).

Here, R(P ) is the range of P , N(P ) is the null space of P and by R(P ) ⊥ N(P ) we

mean that 〈u, v〉 = 0 for any u ∈ R(P ), v ∈ N(P ). (cf. Nair [2].)

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(a) Theorem 3.2 is not only true for functions f in C[−π, π], but also for any func-

tions f in L2[−π, π]. Here, L2[−π, π] denotes the space of all (Lebesgue measurable)

square-integrable functions on [−π, π]. (cf. Nair [2].)

4. Convergence of Sk(f, ·)

Next we may recall that, by Weierstrass’ approximation theorem, the set of all

trigonometric polynomials is dense in C[−π, π] w.r.t. the surpemum norm. Hence,

for every ε > 0, there exists a trigonometric polynomial ϕ, say of order k, such that

‖f − ϕ‖∞ < ε. Hence

‖f − ϕ‖22 =

∫ π

−π

|f(t)− ϕ(t)|2dt ≤ 2π‖f − ϕ‖2∞ <

√2π ε.

Therefore, by the relation (3.4) above,

‖f − Sk(f, ·)‖2 <√

2π ε.

We may note that Sk+1(f, ·)− Sk(f, ·) ∈ Vk+1 so that

‖f − Sk+1(f, ·)‖2 = ‖ (f − Sk(f, ·))− (Sk+1(f, ·)− Sk(f, ·)) ‖2

= ‖ (f − Sk(f, ·))− Pk+1 (Sk+1(f, ·)− Sk(f, ·)) ‖2

≤ ‖f − Sk(f, ·)‖2.

Thus, ‖f − S`(f, ·)‖2 ≤ ‖f − Sk(f, ·)‖2 ∀ ` ≥ k. Hence, for every ε > 0, there exists a

k such that

‖f − S`(f, ·)‖2 <√

2π ε ∀ ` ≥ k.

Thus

THEOREM 4.1. For every f ∈ C[−π, π],

‖f − Sk(f, ·)‖2 → 0 as k →∞.

Remarks. Theorem 4.1 is not only true for functions f in C[−π, π], but also for any

functions f in L2[−π, π]. Here, L2[−π, π] denotes the space of all (Lebesgue measur-

able) square-integrable functions on [−π, π]. (cf. Nair [2].)

References

[1] R. Bhatia, Fourier Series, Hindustan Book Agnecy, New Delhi, 1993.[2] M.T. Nair, Functional Analysis: A First Course, Prentice-Hall of India, New delhi, 2002.