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(February 19, 2005 )

Applications to Fourier seriesPaul Garrett  garrett@math.umn.edu http://www.math.umn.edu/˜garrett/

We showed that the Fourier series of periodic C 1 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 C 1 functions do converge to the functions.From the general discussion of  L2 functions, we know that a sequence of  L2 functions has a subsequencewhich 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 thewhole sequence of partial sums.

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

n∈Z

f̂ (n) e2πinx

of  f  does not converge at 0, where as usual

f̂ (n) = 10

e−2πinx f (x) dx

The Banach-Steinhaus (Uniform Boundedness) theorem for Banach spaces V  has as a special case that fora 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

supα∈A

|λα v| = +∞

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 Gδ).

Consider the functionals

λN (f ) =|n|≤N 

f̂ (n)

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

|λN (f )| ≤

 10

|n|≤N 

e−2πinx

|f (x)| dx

≤ |f |∞ ·

 10

|n|≤N 

e−2πinx

dx = |f |∞ · ||n|≤N 

e−2πinx|1

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

|λN | = ||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

|n|≤N 

e−2πinx =e−2πiNx − e−2πi(−N −1)x

e−2πix − 1=

e2πi(N +1

2)x − e−2πi(N +

1

2)x

eπix − e−πix=

sin2π(N  + 12

)x

sin πx

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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 Gδ is uncountable.Proof: Suppose that

E  =n

U n = {x1, x2 . . .}

is a dense Gδ, where the sets U n are (necessarily dense) open. Let

V n = U n − {x1, . . . , xn}

Then V n is still dense (by the assumption that there are no isolated points) and open, butn V n = φ,

contradicting the Baire theorem. ///

Theorem: (Riemann-Lebesgue Lemma) Let f  ∈ L1[0, 1]. Then

f̂ (n) → 0

Proof: Finite linear combinations of exponentials are dense in C o(R/Z) in sup norm, and C o[0, 1] is densein L1[0, 1]. That is, given f  ∈ L1 there is g ∈ C o(R/Z) such that |f − g|1 < ε and a finite linear combinationh 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 distinctexponentials. Thus,

|f̂ (n)| =

 10

(f (x) − h(x)) e−2πinx dx

≤ |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 thecollection of Fourier coefficients of an L1 function.

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

limn→∞

an = 0

with norm|{an}| = sup

n|an|

This norm makes c0 a Banach space. Let T  : L1 → c0 by

T f  = {f̂ (n) : n ∈ Z} ∈ c0

Since

|f̂ (n)| =

 10

f (x) e−2πinx dx

≤

 10

|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  f̂ (n) = 0 for every n for some L1 function f , then for any trigonometricpolynomial h we have  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 continuoush. From Lusin’s theorem and (again) dominated convergence, the same applies with h being a characteristicfunction 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 > 0such that for every L1 function f 

|f̂ |∞

≥ δ |f |1

Letf N (x) =

|n|≤N 

e−2πinx

On one hand, the sup norm of  f̂ N  is certainly 1. On the other hand, the computation above shows thatthe 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

C o(X ) = C oo (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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