Periodic Functions

Chapter 1 The Classical Theory

1.3 Rate of Convergence

1.3.1. Periodic Functions

. Then, for any non-overlapping, finite, exact coverC of [0,1] and any choice mapξ∈Ξ(C), (1.3.1)

(R) Z

[0,1]

f(x)dx− R(f;C,ξ)

≤ kf⁰kukCk.

1The contents of this section are adapted from the article “Some Riemann sums are better than others,” that I wrote with V. Guillemin and which appeared in the bookRepresen- tations, Wavelets, and Frames, edited by P. Jorgensen, K. Merrill, and J. Packer and published in 2008 by Birkh¨auser.

Moreover, at least qualitatively, (1.3.1) is optimal. To see this, takef(x) =x, and observe that

1 n

n

X

m=1

fm n

=n(n+ 1) 2n² = 1

2 + 1 2n = (R)

Z

[0,1]

f(x)dx+ 1 2n. Hence, in this case,

(R) Z

[0,1]

f(x)dx− R(f;Cn,ξn)

≥ kf⁰kukCnk

2 ,

where Cn=_m−1

n ,^m_n

: 1≤m≤n andξn

_m−1

n ,^m_n

= ^m_n.

In spite of the preceding, in this subsection we will see that (1.3.1) is very far from optimal when f is asmooth(i.e.,C^∞), periodic function. That is, when f is the restriction to [0,1] of a smooth function onRthat has period 1. In fact, I will show that if

(1.3.2) Rn(f)≡ 1

n

n

X

m=1

f ^m_n andf on [0,1] is a smooth, periodic function, then

(1.3.3) lim

n→∞n^`

Rn(f)−(R) Z

[0,1]

f(x)dx

= 0 for all`∈Z⁺. Before proceeding, one should recognize how essential both periodicity and the selection of the choice map are. Indeed, the preceding example shows that periodicity cannot be dispensed with. To see the importance of the choice map, take f(x) = e^i2πx, where i=√

−1. This function is certainly smooth and periodic. Next, take α_n = 1−_n¹. Then _m−1

n ,^m_n

∈ Cn 7−→ ^mα_nⁿ ∈R is an allowable choice map ξ_n and

R(f;Cn,ξn) = e^i2παnn n

1−e^i2παn 1−e^i2παnn . Thus, since

(R) Z

[0,1]

f(x)dx= 0 and lim

n→∞e^i2παnn 1−e^i2παn 1−e^i2παnn =−1, we conclude that

n→∞lim n

(R) Z

[0,1]

f(x)dx− R(f;Cn,ξn)

= 1.

1 The Classical Theory

Turning to the proof of (1.3.3), let f be a smooth, periodic function, and note that

(R) Z

[0,1]

f(x)dx− Rn(f) =

n

X

m=1

(R) Z

Im,n

f(x)−f ^m_n dx, where Im,n =_m−1

n ,^m_n

. Integrating by parts, one finds that the mth sum- mand equals

−(R) Z

Im,n

x−^m−1_n

f⁰(x)dx, and so

(R) Z

[0,1]

f(x)dx− Rn(f) =−

n

X

m=1

(R) Z

I_m,n

x−^m−1_n

f⁰(x)dx.

So far, we have used only smoothness but not periodicity. But we now use periodicity to see that the integral off⁰ over [0,1] is 0 and therefore that the sum is unchanged when

x−^m−1_n

is replaced by

x−^m−1_n −c

for anyc∈R. In particular, by taking c = _2n¹ , which is the average value of x− ^m−1_n on Im,n, each of the summands can be replaced by

(R) Z

I_m,n

h

x−^m−1_n

−_2n¹ i

f⁰(x)dx, in which case f⁰(x) can be replaced by f⁰(x)−f^{0 m}_n

. After making these replacements, one arrives at

(R) Z

[0,1]

f(x)dx−Rn(f) =−

n

X

m=1

(R) Z

I_m,n

f⁰(x)−f^{0 m}_nh

x−^m−1_n − 1 2n

i dx.

To see that we have already made progress toward (1.3.3), note that the absolute value of each summand in the preceding expression is dominated by

kf⁰⁰ku

4n³ and therefore that we have shown that

(R) Z

[0,1]

f(x)dx− Rn(f)

≤kf⁰⁰ku

4n² .

Before attempting to go further, it is best to introduce the following nota- tion. For ` ≥ 1, let C<sub>1</sub><sup>`</sup> [0,1];C

be the space of f ∈ C^` [0,1];C

with the property that f^(k)≡∂^kf takes the same value at 0 and 1 for each 0≤k < `, and set C₁^∞ [0,1];C

=T∞

`=0C<sub>1</sub><sup>`</sup> [0,1];C

. Next, givenk∈N, set (1.3.4) ∆^(k)_n (f) = 1

k!

n

X

m=1

(R) Z

I_m,n

h

x−^m−1_n i^k

f(x)−f ^m_n dx.

Then

∆⁽⁰⁾_n (f) = (R) Z

[0,1]

f(x)dx− Rn(f), and the preceding calculation shows that

∆⁽⁰⁾_n (f) = 1

2n∆⁽⁰⁾(f⁰)−∆⁽¹⁾(f⁰).

More generally, integration by parts shows that themth term in the expression for ∆^(k)(f) equals

− 1 (k+ 1)!(R)

Z

I_m,n

h

x−^m−1_n i^k+1

f⁰(x)dx.

Next, assuming thatf ∈C₁¹ [0,1];C

and using periodicity in the same way as we did above, one sees that the sum of these is the same as the sum of

− 1 (k+ 1)!(R)

Z

Im,n

h

x−^m−1_n ^k+1

−_(k+2)n¹ k+1

i

f⁰(x)−f^{0 m}_n dx

= 1

(k+ 2)!n^k+1(R) Z

I_m,n

f⁰(x)−f^{0 m}_n dx

− 1 (k+ 1)!(R)

Z

Im,n

h

x−^m−1_n ik+1

f⁰(x)−f^{0 m}_n dx.

Hence, we have now shown that (1.3.5) ∆^(k)_n (f) = 1

(k+ 2)!n^k+1∆⁽⁰⁾_n (f⁰)−∆^(k+1)_n (f⁰) for any f ∈C₁¹ [0,1];C

.

Working by induction on `∈N, one can use (1.3.5) to check that, for any f ∈C<sub>1</sub><sup>`</sup> [0,1];C

,

(1.3.6) ∆⁽⁰⁾_n (f) = 1 n^`+1

`

X

k=0

(−1)^kb<sub>`−k</sub>n<sup>k+1</sup>∆<sup>(k)</sup><sub>n</sub> (f<sup>(`)</sup>), where {b_k: k≥0} is determined inductively by

(1.3.7) b₀= 1 and b_`+1=

`

X

k=0

(−1)^k (k+ 2)!b_`−k. Noting that

(1.3.8) n^k+1

∆^(k)_n (f)

≤ kf⁰k_u (k+ 2)!,

1 The Classical Theory it follows from (1.3.6) that, for any f ∈C₁^`+1 [0,1];C

,

(1.3.9)

(R) Z

[0,1]

f(x)dx− Rn(f)

≤K`+1

n^{`+1</sup>kf<sup>(`+1)}ku, where

(1.3.10) K_`+1=

`

X

k=0

|b_`−k| (k+ 2)!.

Obviously, (1.3.9) does more than just prove (1.3.3); it even gives a rate.

To get a more explicit result, one has to get a handle on the numbers bk and the associated quantities K`+1. As we will see in§7.2.2, the numbersbk are intimately related to a famous sequence known as the Bernoulli numbers (so named for their discoverer, Jacob Bernoulli). However, here we will settle for less refined information and be content with knowing that

(1.3.11) lim

`→∞K<sub>`</sub>^1` = 1 2π.

To prove (1.3.11), first observe if f(x) =e^i2πx, then ∆⁽⁰⁾₁ (f) = −1,kf^(`)ku

= (2π)<sup>`</sup>, and so (1.3.9) shows that K<sub>`+1</sub> ≥ (2π)^{−`−1</sup> and therefore that lim<sub>`→∞</sub>K<sub>`</sub><sup>1`} ≥ (2π)⁻¹. To prove the corresponding upper bound, begin by using (1.3.7) to inductively check that |b_k| ≤β^−k, whereβ is the element of (0,∞) that satisfiese^β = 1 + 2β. As a consequence, we know that the gen- erating function B(λ)≡P∞

k=1b_kλ^k−1 is well defined forλ∈Cwith|λ|< β.

Moreover, from (1.3.7), for|λ|< β, B(λ) =

∞

X

`=0

b<sub>`+1</sub>λ<sup>`</sup>=

∞

X

k=0

(−λ)^k (k+ 2)!

∞

X

`=k

b<sub>`−k</sub>λ<sup>`−k</sup>= 1 +λB(λ)e^−λ−1 +λ λ² , and therefore

(1.3.12) B(λ) =1−e^λ+λe^λ

λ(e^λ−1) ,

where it is to be understood that the expression on the right is defined at 0 by analytic continuation. In other words, it equals ¹₂ atλ= 0. Sincee^λ 6= 1 for 0<|λ|<2πande^i2π = 1, it follows that 2πis the radius of convergence forB(λ) and therefore that

(1.3.13) lim

k→∞|b_k|^1k ≤ 1 2π.

Finally, plugging this into (1.3.10), we see that, for each θ ∈(0,1), 1> c >

(2π)⁻¹, and sufficiently large`, K<sub>`+1</sub>≤c^(1−θ)` X

0≤k≤θ`

1

(k+ 2)!+M^` X

`≥k>θ`

1

(k+ 2)! ≤ec^{(1−θ)`</sup>+ eM<sup>`} (bθ`c+ 1)!, whereM ≡sup_k≥0|bk|^k1 ∈[1,∞),btcdenotes theinteger partoft∈R(i.e., the largest integer dominated byt), and Taylor’s remainder formula has been applied to see that

X

k>n

1 k! =e−

n

X

k=0

1 k! = 1

n!(R) Z

[0,1]

(1−t)ⁿe^tdt≤ e (n+ 1)!. Using the trivial estimaten!≥ ⁿ₂ⁿ₂

, one can pass from the preceding to

K`+1≤e 1 + 2

θ`

^θ₂ M c^1−θ

^{`</sup>! c<sup>(1−θ)`},

from which it is clear that lim_`→∞K

1

`

` ≤c^1−θ for every θ ∈ (0,1) and 1>

c > (2π)⁻¹. Thus, lim`→∞K

1

`

` ≤ (2π)⁻¹ follows after one lets θ & 0 and c&(2π)⁻¹.

These findings are summarized in the following.

Theorem 1.3.14. If`∈N and f ∈C<sub>1</sub><sup>`+1</sup> [0,1];C

, then (1.3.9)holds. In particular, iff ∈C₁^∞ [0,1];C

andn∈Z⁺, then

`→∞lim kf<sup>(`)</sup>ku^1` <2πn =⇒ (R) Z

[0,1]

f(x)dx=Rn(f).

Proof: The first assertion needs no further comment. As for the second, it is an immediate consequence of the first combined with (1.3.11).

At first sight, the concluding part of Theorem 1.3.14 looks quite striking.

However, as will be shown in§7.2.2, it really only reflects the fact that there are relatively few smooth, periodic functions whose successive derivatives grow at most geometrically fast.

§1.3.2. The Non-Periodic Case: It is interesting to see what can be said