Applications to Fourier series

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(February 19, 2005)

Applications to Fourier series

Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/˜garrett/

We showed that the Fourier series of periodic C¹ functions f on R¹ converge uniformly (that is, in the C^o metric), and also converge pointwise to the original f . Thus, in the one dimensional case, the Fourier series of periodic C¹ functions do converge to the functions.

From the general discussion of L² functions, we know that a sequence of L² functions has a subsequence which converges pointwise. (In fact, that is how the limit is constructed in proving completeness of L².) This applies to Fourier series, but pointedly does not say anything about the pointwise convergence of the whole sequence of partial sums.

Theorem:

There is a periodic continuous function f on [0, 1] (periodicity meaning that f (0) = f (1)) such that the Fourier series

X

n∈Z

f (n) eˆ ^2πinx

of f does not converge at 0, where as usual f (n) =ˆ

Z 1 0

e^−2πinxf (x) dx

The Banach-Steinhaus (Uniform Boundedness) theorem for Banach spaces V has as a special case that for a collection of continuous linear functionals {λα: α ∈ A} on V either there is a uniform bound M such that

|λα| ≤ M for all α ∈ A, or else there is v in the unit ball of V such that sup

α∈A

|λαv| = +∞

In fact, the collection of such v is dense in the unit ball, and is an intersection of a countable collection of dense open sets (called a G_δ).

Consider the functionals

λ_N(f ) = X

|n|≤N

f (n)ˆ

These are partial sums of the Fourier series of f evaluated at 0. Since

|λN(f )| ≤ Z 1

0

X

|n|≤N

e^−2πinx

|f (x)| dx

≤ |f |_∞· Z 1

0

X

|n|≤N

e^−2πinx

dx = |f |_∞· | X

|n|≤N

e^−2πinx|1

The point is to show, rather, that equality holds, namely

|λ_N| = | X

|n|≤N

e^−2πinx|₁

and that as n → ∞ the latter L¹-norms go to ∞.

First, summing the finite geometric series and doing a little adjustment gives X

|n|≤N

e^−2πinx=e^{−2πiN x}− e−2πi(−N −1)x

e^−2πix− 1 = e^{2πi(N +}¹²^)x− e^{−2πi(N +}¹²^)x

e^πix− e^−πix = sin 2π(N +¹₂)x sin πx

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Paul Garrett: Applications to Fourier series (February 19, 2005) Applying to this the elementary inequality

| sin t| ≤ |t|

gives

Z 1 0

sin 2π(N +¹₂)x sin πx

dx ≥ Z 1

0

| sin 2π(N + 1 2)x|1

xdx =

Z 2π(N +¹₂) 0

| sin x|1 xdx

≥

N

X

`=1

1

` Z 2π`

2π(`−1)

| sin x| dx ≥

N

X

`=1

1

` → ∞ as N → ∞. Thus, the L¹-norms do go to ∞.

Let g(x) be the sign of the kernel

X

|n|≤N

e^−2πinx=sin 2π(N +¹₂)x sin πx

from above. Let gj be a sequence of periodic continuous functions with |gj| ≤ 1 and going to g pointwise.

Then by the dominated convergence theorem

lim

j λ_N(g_j) = lim

j

Z 1 0

g_j(x) X

|n|≤N

e^−2πinxdx = Z 1

0

g(x) X

|n|≤N

e^−2πinxdx = Z 1

0

| X

|n|≤N

e^−2πinx| dx

That is, the L¹-norm of the kernel really is the norm of the functional.

The Banach-Steinhaus theorem implies that there is f ∈ C^o(R/Z) such that sup

N

|λ_N(f )| = +∞

That is, the Fourier series of f does not converge at 0. ///

The result can be strengthened by using Baire’s theorem again.

Theorem:

There is f ∈ C^o(R/Z) such that for a dense Gδ of x in [0, 1]

sup

N

X

|n|≤N

f (n) eˆ ^2πinx

= ∞

In fact, the set of such f is a dense G_δ in C^o(R/Z).

Proof:

Take a dense countable set of points xj in the interval. Let λj,N be the continuous linear functionals on C^o(R/Z) defined by

λ_j,N(f ) = X

|n|≤N

f (n) eˆ ^2πinx^j

As in the previous proof, the set Ej of functions f where sup

N

|λj,Nf | = +∞

is a dense Gδ, so the intersection

E = ∩jEj

is a dense Gδ. Also

s(f, x) = sup

N

| X

|n|≤N

f (n) eˆ ^2πinx^j|

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Paul Garrett: Applications to Fourier series (February 19, 2005) is the sup of continuous functions, so is lower semicontinuous, and so for any f ∈ C^o(R/Z)

{x ∈ [0, 1] : s(f, x) = +∞}

is a Gδ in [0, 1]. Then for f ∈ E, the corresponding Gδ contains a dense subset (the xi). ///

Lemma:

In a complete metric space X, d (with no isolated points), a dense G_δ is uncountable.

Proof:

Suppose that

E =\

n

Un = {x1, x2 . . .}

is a dense Gδ, where the sets Un are (necessarily dense) open. Let V_n= U_n− {x₁, . . . , x_n}

Then V_n is still dense (by the assumption that there are no isolated points) and open, but T

nV_n = φ,

contradicting the Baire theorem. ///

Theorem:

(Riemann-Lebesgue Lemma) Let f ∈ L¹[0, 1]. Then f (n) → 0ˆ

Proof:

Finite linear combinations of exponentials are dense in Cô(R/Z) in sup norm, and Cô[0, 1] is dense in L¹[0, 1]. That is, given f ∈ L¹there is g ∈ Cô(R/Z) such that |f − g|1< ε and a finite linear combination h of exponentials such that |g − h|_∞< ε. Since the sup-norm dominates the L¹-norm, |f − h|1< 2ε.

Given such h, for large-enough n the Fourier coefficients are simply 0, by orthogonality of distinct exponentials. Thus,

| ˆf (n)| =

Z 1 0

(f (x) − h(x)) e^−2πinxdx

≤ |f − h|1< 2ε

This proves the Riemann-Lebesgue Lemma. ///

Theorem:

(corollary of Baire and Open Mapping) It is not true that every sequence an→ 0 occurs as the collection of Fourier coefficients of an L¹function.

Proof:

Let c0 be the collection of sequences {an: n ∈ Z} of complex numbers such that

n→∞lim an= 0 with norm

|{an}| = sup

n

|an| This norm makes c₀a Banach space. Let T : L¹→ c₀ by

T f = { ˆf (n) : n ∈ Z} ∈ c₀ Since

| ˆf (n)| =

Z 1 0

f (x) e^−2πinxdx

≤ Z 1

0

|f (x)| dx = |f |1

it is clear that |T | ≤ 1. In fact, taking f (x) = 1 shows that |T | = 1.

We check that T is injective: if ˆf (n) = 0 for every n for some L¹ function f , then for any trigonometric polynomial h we have

Z 1 0

f (x) h(x) dx = 0

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Paul Garrett: Applications to Fourier series (February 19, 2005)

and by dominated convergence (and density of trigonometric polynomials) the same holds for all continuous h. From Lusin’s theorem and (again) dominated convergence, the same applies with h being a characteristic function of a measurable set. Thus f = 0, proving injectivity of T .

If T were surjective from L¹ to c₀, then the Open Mapping Theorem would assure the existence of δ > 0 such that for every L¹function f

| ˆf |_∞≥ δ |f |₁ Let

fN(x) = X

|n|≤N

e^−2πinx

On one hand, the sup norm of ˆf_N is certainly 1. On the other hand, the computation above shows that the L¹ norm of f_N goes to ∞ as N →= ∞. Thus, there is no such δ > 0. Thus, T cannot be surjective.

///

Remark:

For any locally compact Hausdorff space X we can define

Co(X) = C_o^o(X) = {f ∈ C^o(X) : for each ε > 0 there is compact K such that |f (x)| < ε off K}

and show that with sup norm this space is complete.

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