Positivity and boundedness of trigonometric sums

Analysis in Theory and ApplicationsVolume 23 Number 4, 2007, 380–388DOI10.1007/s10496-007-0380-6

POSITIVITY AND BOUNDEDNESSOF TRIGONOMETRIC SUMS∗

Gavin Brown(University of Sydney, Australia)

Received Oct. 28, 2006

Abstract. We give a systematic account of results which assure positivity and bounded-ness of partial sums of cosine or sine series. New proofs of recent results are sketched.

Key words: partial sum, cosine series, sine series, orthogonal polynomials

AMS (2000) subject classification: 42A16, 42A32

1 Consine Sums

The typical result is of the form:

ak ≥ 0, ak+1 ≤ Rkak =⇒n

∑k=0

ak coskx ≥ 0,

for 0 ≤ x ≤ π and n = 1,2, · · · . Here Rk are the ratios ck+1/ck of some standard sequence (ck) ofdecreasing positive numbers.

In fact we need only check the standard sequence. For, if ak = ρkck, then

n

∑k=0

ak coskx = (ρ0 −ρ1)c0 +(ρ1 −ρ2)(c0 + c1 cosx)+ (ρ2 −ρ3)(c0 + c1 cosx+ c2 cos2x)+ · · ·+ ρn(c0 + c1 cosx+ · · ·+ cn cos nx).

Of course (ρk) is decreasing as

ρk+1

ρk=

ak+1/ak

ck+1/ck≤ Rk

ck+1/ck= 1.

In general the larger Rk the sharper the result.Let us set out the basic results and their chronology:

1912: R0 = 1, Rk = 1− 1k+1 . (Young’s Inequality[24]

c0 = 1, ck =1k

, k = 1,2, · · · .

∗This is a plenary report on the International Symposium of Approximation Theory and Remote Sensing Applicationsheld in Kunming, China in April 2006.

Analysis in Theory and Applications, Vol. 23, No.4 (2007) 381

1969: R0 = 1, Rk = 1− 1k + α + 1

. (Szego-Rogosinski-Gasper[22],[17]

c0 = 1, ck =k−1+ α

k + α, k = 1,2, · · · .

This is true precisely for α ≤ 4.5678 · · · . (Szego-Rogosinski obtained the result for α = 1in 1928)

1988: R0 = 1, Rk =(

1− 1k + 1

)α. (Brown, Wang, Wilson[12]

c0 = 1, ck =1kα , k = 1,2, · · · .

The cosine sums are positive precisely when α ≥ α0 = .30844 · · · . Here α0 is the solution

of∫ 3π/2

0

cos ttα dt = 0. This is the Salem-Littlewood-Izumi constant, and gives also the

criterion for the sums to be bounded below. See [25].

Note further that a general result of Belov [4], shows that positive cosine sums must haveak = O(k−α0).

2006: Rk =(

1− αk + 1

). (Koumandos, Ruscheweyh[21]

ck =Γ(1+ k−α)Γ(1−α)k!

.

The result is positive precisely when α ≥ α0 = .30844 · · · .2007: R0 = 1, Rk = (1− α

k+1), k = 1,2, · · · . (Brown, Dai, Wang, [5], [8])

c0 = 1, ck = Γ(1+k−α)Γ(2−α)k! , k = 1,2, · · · .

Sums are non-negative precisely when α ≥ α∗ = .33542 · · · . Here α∗ is the unique α in(0,1) such that the seventh partial sum has a zero. The partial sums are bounded belowprecisely when α ≥ α0.

The last result is stronger than the Brown-Wilson-Wang result, where it is valid, because theratios are larger: (

1− 1k + 1

)α≤ 1− α

k + 1,

but this is not true for the Koumandos-Ruscheweyh result because of the first ratio. It yields

11−α

+n

∑k=1

coskαkα > 0,

as the authors note. In turn the B-W-W result gives the K-R and B-D-W results in the restrictedrange where

1− αk + 1

≤(

1− 1k + 1

)α0, k = 1,2, · · · .

It is easy to see this requires α ≥ .38498 · · · .The K-R result gives a lower bound

−α1−α

for the B-D-W partial sums in the range α0 ≤α ≤ α∗. Their result is natural for applications to starlike functions.

382 G. Brown : Positivity and Boundedness of Trigonometric Sums

In the final section I will sketch how to derive the B-D-W result more efficiently and how toprove the basic K-R result by using our methods.

Although I will confine detailed discussion to sums of cosines and sines, let me make anaside at this point. Similar techniques to those developed in the trigonometric case yield resultsfor a wider class of orthogonal polynomials. Let me quote one example from Brown, Kouman-dos, Wang, [11].

Let Cλn (x) be the ultraspherical polynomial of degree n and order λ . The partial sums

n

∑k=0

Cλn (x)

Cλn (1)

, −1 < x < 1,

are positive for λ ≥ λ0 = β0 +12

and are unbounded for any smaller λ , where β0 is the solution

of ∫ z2

0t−αJα(t)dt = 0,

where Jα is the Bessel function of the first kind of order α and z2 is its second positive root.(λ0 = .23061297 · · · .)

The lower bound had been a recurring source of speculation since Szego’s publication,posthumously on behalf of Feldheim, of the latter’s contribution to the problem.

Let me return to cosine sums, although we shall see that this is a precursor to discussion ofsine sums.

1958: R2k = 1, R2k−1 =12k

. (Vietoris[23]

c2k = c2k+1 = 2−2k(2k

k

).

2005: R2k = 1, R2k−1 = 1− 12k + α

. (Brown-Dai-Wang[6],[9]

c2k = c2k+1 =Γ(1+ α/2)Γ(k + 1/2+ α/2)Γ(1/2+ α/2)Γ(k + 1+ α/2)

.

The partial sums are non-negative if and only if α ≥ α0 = 2.3308 · · · , where α0 is uniquelydefined by the fact that the sixth partial sum has a zero.(Note: Vietoris corresponds to α = 0, Brown-Hewitt (1984), [10], to α = 1, and Brown-Yin (2001), [14], to α = 2)

2006: R2k = 1, R2k−1 =βk

. (Koumandos[20], Brown-Dai-Wang[7]

c2k = c2k+1 =Γ(k + 1−β )

Γ(1−β )Γ(k + 1).

The partial sums are positive when β ≥ α0 = .30844 · · · , and fail to be bounded below ifβ < α0.

2 Sine Sums

Now we have sequences ak, ck, Rk, k = 1,2, · · · and seek results of the form

ak ≥ 0, ak+1 ≤ Rkak =⇒n

∑k=1

ak sinkx ≥ 0, for 0 ≤ x ≤ π , n = 1,2, · · · .


Again, Rk = ck+1/ck.There is a much briefer chronology:

1911: Rk = 1− 1k + 1

; ck =1k

, conjectured by Fejer, proved by Gronwall and Jackson, [18],

[19]. (The proofs were given in 1910 but I have used the publication dates for consistencythroughout)

1958: R2k = 1, R2k−1 = 1− 12k

; ck = c2k+1 = 2−2k(2k

k

). (Vietoris[23])

An improved proof was given by Askey-Steinig 1974, [3]. Again a general result of Belov,[4], shows that positive partial sums require ak = O(ck).

If one maximises Rk at each step to obtain positive partial sums then one finds the Vietorissequence.

Unlike in the cosine case, the Victoris result is canonical and best possible. It is sad that itwas ignored by the mathematical community for almost 20 years and we owe Dick Askey a debtof gratitude for intuitively recognising its importance and bringing it to wider attention. Indeed,as Askey tells the story in [2], Mathematical Reviews sent the paper out three times only tohave it returned on each occasion without review. Richard Askey then prepared and submitteda review (incorporating some applications) but the editor turned it down on the grounds that thepaper had appeared long before. The resourceful Askey then wrote the paper with John Steinig,[3], which improved Victoris proof but, more importantly, publicised the result.

One could make a case for the Victoris paper being jinxed. The first sequel article with whichI was associated was the last paper I wrote with Edwin Hewitt before he suffered a disablingstroke. More recently, the papers with Dai and Wang, which had been circulated as preprints, satfor some years on my desk unsubmitted for publication because of a misunderstanding whichwas entirely my fault! Following Askey and Steinig, Hewitt and I gave some applications ofVictoris-type results to the location of zeros of trigonometric polynomials. In a postscript toour paper Askey, [1], gave direct applications of our results to sums of Jacobi polynomials. Onreflection, I believe that the Victoris ideas stand by themselves and that the most significantapplication has been the stimulus to tackle positivity results for a wide range of partial sums ofmore general series. I find it ironical that, as a university president, I regularly have to arguefor basic research on its own terms, whereas the Victoris work received so little attention evenwithin the pure mathematical community. That is partly why I have confined this article to basicresults for sine and cosine series. I want the results and proofs to speak for themselves. Althoughthe result of Victoris is the last word on positivity of partial sums of sine series in the pure sensethat I have specified, it is possible to establish a genre of less sharp results. In particular, Fejer

(1925), he[15], observed thatn−1

∑k=1

sinkxk + 1

+12

sinnx > 0. This allows a weaker type of theorem

where the hypothesis is that a2k = R2ka2k+1, a2k−1 ≤ R2k−1a2k, and the conclusion is

(i)2n+1

∑k=1

ak sin kx > 0;

(ii)2n

∑k=1

ak sinkx > 0, for 0 < x < π − 12nπ;

(iii)2n−1

∑k=1

ak sinkx+Ca2n sin2nx > 0, for some explicit constant C.


Brown-Wilson[13], gave these for

(a) Rk = 1− 1k + 1

, ck =1

k + 1. (Best constant C = log2)

(b) Rk = 1, R2k−1 = 1− 12k + 1

, c2k = c2k+1 =1

k + 1(22k/

(2k+1k

)). (Best constant C =

34

)

The point is that the condition a2k = R2ka2k+1 eliminates the even partial sums in the typicalsummation by parts.

3 Proof Sketches

In this section we will show how to obtain the main results of [8], [21] using a slight

improvement of the methods of [8]. Accordingly, let us write Sn(x,α) =n

∑k=0

ck(α)cos kx and

ck(α) =Γ(k + 1−α)Γ(2−α)k!

for k = 1,2, and c0(α) = 1.

Note thatck(β1)ck(β2)

≥ ck+1(β1)ck+1(β2)

, for k = 0,1,2, · · · , β2 ≤ β1. Hence

n

∑k=0

ck(β1)coskx =n−1

∑k=0

(ck(β1)ck(β2)

− ck+1(β1)ck+1(β2)

)Sk(x,β2)+

cn(β1)cn(β2)

Sn(x,β2).

In particular, if all partial sums for β2 are non-negative then the same is true for β1.Recall that α∗ is defined by minx S7(x,α∗) = 0, so as soon as we have proved that Sn(x,α∗) >

0, n �= 7, we will know thatSn(x,α) ≥ 0, α ≥ α∗,

with equality only when α = α∗ and n = 7.Set α1 = .33542. This is smaller than α∗, but we shall proveLemma . minx Sk(x,α1) > 0 if k �= 7.Assume the truth of the lemma for now.Summation by parts gives positivity of Sk(x,α∗) for k = 0,1, · · · ,6. If it were the case that

S7(x,α1) ≥ 0 then summation by parts would force S7(x,α∗) > 0. So S7(x,α1) < 0, and now

n

∑k=8

ck(α∗)cos kx

= −c8(α∗)c8(α1)

S7(x,α1)+n−1

∑k=8

(ck(α∗)ck(α1)

− ck+1(α∗)ck+1(α1)

)Sk(x,α1)+

cn(α∗)cn(α1)

Sn(x,α1)

> 0.

Now we have Sn(x,α∗) > 0, n �= 7, subject to the lemma.Remark. Another simple argument using summation by parts allows us to deduce that

S7(x0,α) < 0 for α < α∗, where x0 is the point where S7(x0,α∗) = 0.We can no longer avoid calculations. I will simply ask you to believe a lemma from [8] that

allows us to switch to a simpler sequence plus a perturbation factor.Lemma. For n ≥ 100 and α ∈ (0,1),

Γ(n+ 1−α)Γ(n+ 1)

=1

(n+ 14)α

+A(α)

(n+ 14)1+α

+12A2(α)+B(α)

(n+ 14)2+α

+R(n,α)

(n+ 14 )3+α

,


where

A(α) =12

α2 − 14

α ,

B(α) = − 196

α − 18

α2 +16

α3,

|R(n,α)− e(α)| ≤ .0301,

e(α) =1

64α +

95128

α2 − 23384

α3 − 1192

α4 +5

96α5 +

148

α6.

Now let us write∞

∑k=n+1

ck(α)cos kx =1

Γ(2−α)

∞

∑k=n+1

cos kx

(k + 14)α

+1

Γ(2−α)

∞

∑k=n+1

dk cos kx

(k + 14 )α

,

where dk = A(α)+12A2(α)+B(α)

(k + 14)

+R(k,α)(k + 1

4)2. The last term in the equality, call it un(α ,x),

has modulus less than

1Γ(2−α)

∞

∑k=n+1

|dk|(k + 1

4)α≤ ρn(α)

Γ(2−α)α(n+ 14)α

,

where

ρn(α) = |A(α)|+ |12A2(α)+B(α)|

(n+ 54)

+|e(α)|+ 0.0301

(n+ 54)2

.

Now

1Γ(2−α)

∞

∑k=n+1

cos kx

(k + 14)α

=1

Γ(2−α)2sin x2

∞

∑k=n+1

∫ (k+ 12 )x

(k− 12 )x

cos t

(k + 14)α

dt

=xα

Γ(2−α)2sin x2

∫ ∞

(n+ 12 )x

cos ttα dt + vn(α ,x),

where

vn(α ,x) =xα

Γ(2−α)2sin x2

∞

∑k=n+1

∫ (k+ 12 )x

(k− 12 )x

( 1

((k + 14)x)α

− 1tα

)cos t dt

and a standard estimate gives

|vn(α ,x)| ≤ x

Γ(2−α)4sin x2(n− 1

2 )α.

Thus∞

∑k=n+1

ak(α)cos kx =xα

Γ(2−α)2sin x2

∫ ∞

(n+ 12 )x

cos ttα dt +un(α ,x)+ vn(α ,x).

Note further that∞

∑k=0

ak(α)cos kx =sin(πα

2 +(1−α) x2)

(1−α)(2sin x2)1−α − α

1−α.


Summarizing the calculations, we have

Sn(α ,x) =−α

1−α+

sin(πα2 + (1−α)x

2 )(1−α)(2sin x

2)1−α

− xα

2sin x2

1Γ(2−α)

∫ ∞

(n+ 12 )x

cos ttα dt + pn(x,α),

where

|pn(x,α)| ≤ ρn(α)Γ(2−α)α(n+ 1

4)α+

x

Γ(2−α)4sin x2(n− 1

2)α.

The last adjustment is to recall

sinπα2

=1

Γ(1−α)

∫ ∞

0

cos ttα dt.

So we have

Sn(α ,x) =−α

1−α+

[sin(πα

2 + (1−α)x2 )

(1−α)(2sin x2)1−α − xα

2(1−α)sin x2

sinαπ2

]

+xα

2Γ(2−α)sin x2

∫ (n+ 12 )x

0

cos ttα dt + pn(x,α).

Let me remark further that the term in square brackets, let us call it M(x,α), is positive andbounded on 0 < x ≤ π for any α in (0,0.4). Moreover,

M(x,α) ≥ (0.7)xα

(1−α)

(x/2

sin(x/2)

), 0 < x ≤ π .

Let us prove that Sn(α ,x) is not uniformly bounded below for α < α0. To this end note

the boundedness of pn(α) and M(x,α), and choose x = xn =3π

2n+ 1. The dominant term in

Sn(α ,xn), as n → ∞, isxn xα−1

n

2Γ(2−α)sin xn2

∫ 3π2

0

cos ttα dt.

Because α < α0, the integral is negative. Moreover, xα−1n → ∞. The required result follows.

Now let us show that

Sn(x)+α

1−α≥ 0, for α ≥ α0.

Using summation by parts, it is enough to check the case α = α0. The integral is non-negative,so it suffices to show that

M(x,α0) ≥ |pn(α0)|.In other words, for α = α0, we need

(0.7)xα

(1−α)

(x/2

sin(x/2)

)ρn(x,α)

Γ(2−α)α(n+ 14)α

+(

x/2sin(x/2)

)1

2Γ(2−α)(n− 12 )α

.


Assume x > π/2n, so we need

πα(0.7)(

12n

)αΓ(2−α)α ≥ ρn(α)

(n+ 14)α

(sin(x/2)

x/2

)+

α2(n− 1

2)α.

It suffices to showπα(0.7)Γ(2−α)α

2α ≥ ρn(α)+α

2(1− 1200)α

.

In fact ρn(α0) < 0.71 for n > 100, and the inequality checks out.The assumption that x > π/2n is, of course, harmless since all the cosines are otherwise

positive. The assumption that n > 100 is inelegant but valid provided we check the first 100cases numerically.

It remains to prove the lemma concerning the minimum of Sn(x,α1) for n �= 7. Once morewe take n > 100, and for α = α1 = .35542 show that

−α1−α

+M(x,α)+ pn(x,α)+xα−1

Γ(2−α)

(x/2

sin(x/2)

)∫ (n+ 12 )x

0

cos ttα dt

has a positive minimum.Calculation gives ∣∣∣∣ −α

1−α+ pn(x,α)

∣∣∣∣ ≤ 0.592, for n > 100,

1Γ(2−α)

∫ (n+ 12 )x

0

cos ttα dt ≥ 1

Γ(2−α)

∫ 3π/2

0

cos ttα dt > 0.7,

provided x >π2n

. So we need only check

(0.7)(1−α)

xα +(0.7)xα−1 > 0.592, 0 < x ≤ π,

and this turns out to be true.

References

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The University of SydneyNSW 2006Main Quadrangle(A14)Australia

E-mail: [email protected]

Positivity and boundedness of trigonometric sums

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