on the gibbs phenomenon for riesz means

ON THE GIBBS PHENOMENON FOR RIESZ MEANS. 153

ON THE GIBBS PHENOMENON FOR RIESZ MEANS

B. KUTTNER*.

1.1. Suppose that we consider a method of summability [such assummability (0, 1)] with the property that the means of the Fourierseries of any positive'] function are everywhere positive; this is equivalentto the statement that the kernel of the method is everywhere positive.Then the means of the kind considered of the Fourier series of anyfunction must fail to present the Gibbs phenomenon']. If, in consideringthe Gibbs phenomenon, we restrict ourselves to functions having a simplediscontinuity at the point under consideration, the converse of this resultis false. The kernel for summability (0, K) is everywhere positive if, andonly if, K ~ 1. On the other hand, a well-known result due to Cramer§ isthat there is a constant ro such that 0 < f o < 1, and such that, for afunction having a simple discontinuityl], the Gibbs phenomenon vanishesfor the means (0, K) if, and only if, K~fo' Cramer's constant f o hasbeen computed by Gronwallf] ; its value is 0·4395516....

I have recently pointed out** that although, for fixed K, the Rieszmethods of summability (R, n\ K) are, for different values of A, equivalentin the sense that a series summable by the one method is also summableby the other, the behaviour of their kernels is quite different. If A= 1,the kernel for summability (R, n\ K) is everywhere positive if, and only if,IC~ 1; if A~ 2, it takes negative values no matter how great IC may be.Moreover, if A> 2 (but not if A= 2), even the kernel for the Abel means(A., n h) takes negative values. In a later paper t ], I have investigated

* Received 14 July, 1944; read 16 November, 1944.t It is convenient to use the term" positive" in the wide sense, i.e., to mean" greater

than, or equal to, 0".: For the case of summability (0, 1) see, e.q., A. Zygmund, Trigonometrical series

(Warsaw, 1935), §§ 3.22 and 8.7, or G. H. Hardy and W. W. Rogosinski, Fourier series(Cambridge Tracts, No. 38), § 5.7. The general result, which is stated by Hardy andRogosinski, is easily proved.

§ H. Cramer, "Etudes sur 180 sommation des series de Fourier", Arkiv for Matematik,13 (1919), No. 20, 1-21. Or see Zygmund, lac. cit.

II It is not difficult to prove that the proposition that the means (0, Ie) of the Fourierseries of any function fail to exhibit the Gibbs phenomenon is true only when Ie;;;;' 1.

, T. H. Gronwall, "Zur Gibbs'schen Erscheinung ", Annals of Math., 3 (1931),233-240.

** B. Kuttner, "Note on the Riesz means of a-Fourier series ", Journal London Math.Soc., 18 (1943), 148-154. This paper will be referred to as R1.

tt B. Kuttner, "On the Riesz means of a Fourier series (II)", Journal London Math.Soc., 19 (1944), 77-84. This paper will be referred to as R2.

154 B. KUTTNER

more fully the case A< 2. These results evidently suggest that we shouldinvestigate for what values of Aand K the Gibbs phenomenon persists forthe means (R, n>", K), and that is the subject of the present paper.

I .2. The main results of this paper are that if A~ 2, then the Gibbsphenomenon persists for the means (R, n\ K) of the Fourier series of afunction having a simple discontinuity, and that if A> 2 the Gibbsphenomenon also persists for the means (A, n A) . In view of the remarksof § 1.1, it is clear that these results include the main results of Rl , Thecase A< 2 will also be considered, the results to be proved being givenby the following

THEOREM. If 0 < A< 2, there is a function r(A) such that the Gibbsphenomenon vanishes for the means (R, n\ K) of the Fourier series of afunction having a simple discontinuity if K ~ r(A), but not if K < r(A). Thefunction r(A) is continuous and (strictly) increasing, and is, for all A< 2,less* than the function k(A) defined in R2. It tends to 0 as A-?-O, equalsGramer's constant ro when A-:-l, and tends to infinity as A-?-2. If A=2,the Gibbs phenomenon persists for the means (R, n A, K) however large K maybe. If A> 2, it persists for the Abel means (A, n A) , and hence necessarilyfor the means (R, n/', K).

1.3. Let 4>(x) denote the function of period 217 and equal to !(17-X)in the interval (0, 217). Then 4>(x) has a simple discontinuity at theorigin; its Fourier series is

00 sinnx~

n=l n(1)

It is familiar that, in order to investigate the Gibbs phenomenon for themeans (0, r) of any function having a simple discontinuity, it is sufficientto investigate the behaviour of the means (0, r) of (1); this follows fromthe facts that any function having a simple discontinuity at x = gis (exceptperhaps for its value at the point g itself) the sum of a function continuous at x = gand of a constant multiple of 4>(x-g), and that the means(0, r) cannot present the Gibbs phenomenon at a point of continuitywhen r > o. It is evident that a similar remark applies to the methodsof summability considered in the present papert. In what follows we

* It follows from the remarks of § 1.1 that r(A) '" k(i\), but the strict inequalityr(i\) < k(i\) is less trivial. We remark also that in consequence of Rl, Theorem 3, theGibbs phenomenon cannot occur for the means (A, n2 ) .

t See, e.g., Hardy and Rogosinski, Theorem 70 (ii), which shows that similar remarksapply to any method of summability satisfying the conditions of that theorem.

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shall therefore consider throughout the means of the series (1). Moreover,we shall, as we clearly may, restrict ourselves to the case when x is (smalland) positive.

2 . 1. The proof of the theorem is based on the following

LEMMA. Suppose that f(x) is of bounded variation in (0, (0), and thatf(x) Ix is absolutely integrable in (1, (0). Suppose also that there are constants A, 0, greater than 0, such that If(x)-j(y) I~ Olx-yl whenever x, yboth lie in (0, A). Let

(z) = J~ j(x) Si:zx dx,

and let 'Y(u, 8) = ~ j(!!:'-) sin n8.n=l U n

Then 'Y(u, 8)-(u8)-+0 as 8-+0 and u-+oo independently.

We remark that the hypotheses imply that

n~l~f(:)is absolutely convergent for all u ;» 0, so that the sum (3) exists.

In order to prove the lemma, we distinguish two cases; that in whichu ~ If82, and that in which u < IJ82• In the first case, writing

s(n, 8) = ~ sin v8,,,=n V

we nave

\f(1t, 8) = n~l f(:) [s(n, 8)-s(n+ 1, 0)]

=f(O)s(l, 8)+n~ls(n, 8) [f(:)-f(n u 1) ]= !(1T-8)f(O) +2:1 +2:2,

say, where 2:1 denotes the sum of those terms for which n < Au, and ~z

the sum of those terms for which n ~ Au. Since we clearly have

s{n, 8) = O[lj(n8)]

for all n, 8, we see that

~1 = 0 ( 2: -(1

) = 0 (log8U

) = 0 (u-! log u) -+ 0l~n<.Au. n u u

where

156 B. KUTTNER

as u -+ 00. Further,

L 2 =O{ L ~rn Idf(~)I·}=O(~ItrJ ~1iJ,f(~)lln*"AU nO In - 1 U 10 Au-l xu)'

On integrating by parts, we deduce without difficulty that

L2 = O(:O)~O

as u-e-o», whence it follows that

'F(u, 0)~ !7Tf(O)

as O~O, u~oo. On the other hand, it is easily proved that

(z) ~!7T f(O)

as Z~ 00, and, since in the case now under consideration UO~OO, the lemmais evident.

In the case in which u < 1/02, we have

I(uO)- 'F(u, O)! = I ~ In/u {f(x) sin uOx -f(!!.-) sin nO} ax In=l (n-l)/u X u n/u

II = U ~ In/u If(X)-f(~) I' sin nO I. dx;n=1 (n-l)!u U n

12= ~ rn

/u

If(x) IIsinuOx _ sin nO !dX.n=l J(n-l)/u x n/u

Now, since IsinnOI ~ nO, we have

II ~uO ~ In/u If(x)-f(!!.-) Idx ~OV~f(x)~On=l (n-l)/u u

as 0-+ O.In order to investigate 12 , we observe that, for x in ((n-l)/u, nju),

Isin uOx _ sin nO I~ 1..- max I~ -{Sin uOt} \.

x n/u u z~t~n/" dt t

Further, for all relevant values of the parameters,

and also

~ fsin uOt} = 0 (uO) = 0 (uO)dt 1 t t x '

(4)

(5)

since t::( 2x, except in the case n = 1, when we replace the last term in(5) by 0(U2 03 ) . Applying (5) for x::( 1/(uO), and (4) for x> 1/(uO), wededuce that

f I1/U } (Jl/(UII) )12 = O lu03

0If(x)/dx +01U203 l/u If(x)[xdx/

+o.foroo If(x)1 dx). (6)( 1/(ulI) x r

Since f(x) is bounded, it is evident that the first two terms in (6) tend tozero as 0-+ o. The third term in (6) is

oJO roo If(x) IdX}+O (0 [1 dX}l 1 x 1 Jl!(U8) x

(where the second term in (7) is omitted when uO < 1), and this is

O(O)+O(OloguO) = 0(0)+0 (0 log (1/0»)-+0

as 0 -+ 0; and the proof of the lemma is now completed.

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2 . 2. Suppose that we consider any method of summability in whichthe sum of the series ~an is defined by

lim ~ f (!!:-) an11-+00 0=0 'U

(if it exists), where f(x) satisfies the conditions of the lemma. It followsat once from the lemma that a necessary and sufficient condition thatthe Gibbs phenomenon should vanish for the means of the series (1) ofthe kind considered is that (z) ::( t7T for all z > 0, where (z) is definedby equation (2). It is clear that summability (A, n") is of this type, withf(x) = exp(-xA

) ; so is Riesz summability (R, n\ x), with f(x) = (I-xA)/Cfor x ::( I and f(x) = 0 for x ~ 1. It may be remarked that for ordinaryconvergence we must put f(x) = 1 for x ~ 1 and f(x) = 0 for x> 1; inthis case the result of the lemma is well known*.

3. We shall now prove that part of the theorem which deals with theAbel means. We consider the function (z) defined by equation (2) withf(z) = exp( -XX). Clearly

}oo sinzx

17T-(z) = [l-exp(-xA)] -- dx .. 0 x

--- --"-~----""-----_.. ---------_..._~

... See, e.q., Zygmund, loco cit., § 8.5.

158 B. KUTTNER

Supposing that A> 1, so that [l-exp(-x>')]lx~O as X~O, we mayintegrate this equation by parts, when we obtain

Joo ax

z[t7T-cI>(z)] = cos ZX{(AxA+1) exp(-xX)-I} 2.o x

If we now suppose A?:::: 2, then sufficient conditions for Fourier's repeatedintegral theorem are satisfied. It follows that, for x?:::: 0,

J~Z(i 7T - cI> (z») cos zxdz = 2:2 {(Ax>'+I) exp(-x>')-l}, (8)

where, when x = 0, the expression on the right of (8) is to be replacedby its limit. If A> 2 (but not if A= 2), this limit is zero. Hence, puttingx = 0 in (8), we obtain

J~ Z (l7T-cI> (z») dz = o. (9)

Since the integrand in (9) is clearly not equivalent to zero, it follows thatit must take both signs, and the result is evident.

4.1. We now consider the Riesz means. Writing

Jl sinxz

<1>(z) = <1> (A, K; z) = 0 (l-x>')1C -x- dx,

a necessary and sufficient condition that the Gibbs phenomenon shouldvanish for the means (R, n\ K) of (1) is that cI>(z) ~ t7T for all z > o. Itis clear that, if the Gibbs phenomenon vanishes for the means (R, n\ K)of (1), then it must also vanish for the means (R, n\ K'), where K' >K.It follows that there must exist a (possibly infinite) function r('\) suchthat the Gibbs phenomenon for the means (R, n\ K) vanishes if K>r(A),but not if K < r(A). Since (,{>(A, K; z) is, for fixed A, z, a continuous functionof K, it follows that the Gibbs phenomenon must also vanish for K = r(A),if r(A) is finite. Further, since r(A) <k(A), and since, by R2, k(A) is finitefor A< 2, it follows that r(A) must also be finite for A< 2.

In order to establish the remaining properties of r(A), we require thedominant terms in the asymptotic expansion of <1>(z). Integrating(l-s),,)"eisZjs·round a rectangle with vertices at 0, 1, I+Ri, Hi, (with anindentation at the origin), making R tend to infinity, and then taking theimaginary part, we see that

<1>(z) =~7T+JOCJ [~(I_e!".iXyA)K] e-lIZ

dy2 0 Y

-~{ieiZ J~ [1-(l+iyY]K le~~ dy}, (10)

from which it is easy to deduce that, for large z,

(z) = t7T-K sin p.\ r(.\) Z-)'_.\K F'(x-} I) cos (Z-tK7T) Z-K-l

+°(Z-2A) +°(Z-K-2). (11)

Further, this result holds uniformly in .\ for .\ in any finite closed intervalcontained in .\ > o. We next observe that when .\ = 2 the first integralin (10) vanishes. It follows that, in this case, the term O(Z-2A) in (II)may be omitted, and hence that, however large K may be, t7T-(Z) takesboth signs for sufficiently large values of z. Thus '1'(2) = 00.

In R2, the proof that, if .\ < 2, then F(.\, k(A); z) = 0 for some z was(except in the case .\ = 1) based on the results that k(.\) >.\, but that,for any K > A, F(A, K; z) ~ 0 for sufficiently large z, so that we can finda K < k(A) such that F(.\, K; z) ~ 0 for sufficiently large z. Nowwhen1 <.\ < 2, it follows from (11) that t7T-(.\, A-I; z) takes both signs forlarge z, so that '1' (A) > A-I, while when 0 < A< 1 it is evident that '1'(A) > O.On the other hand, it follows from (11) that, if 1 < A< 2, K > A-I, or if0< A~ I, K > 0, then <1>(.\, K; z) < t7T for sufficiently large z. It is thusclear that we can show that, if A< 2, then (A, '1'(.\); z) = t7T for some z,by arguments similar to those of E2 (except that the case A= 1 does notnow require to be dealt with separately). Now, if F(z) = F(.\, K; z) isdefined as in R2, it is easily verified that

(z) = t7T- J~ F(u)du.

Since it is clearly false that F (z) is equivalent to zero for sufficiently largez, it follows that F(.\, '1'(.\}; z) takes both signs. Hence k(.\) > '1'(.\).

It may easily be verified that

where

(I2)

It follows* from this that, if (A, K; z) < t7T for all z, and if I-'- < A, then<1>(1-'-, K; z) < t7T for all z. Combining this with the result that

 (,\, '1'(.\); z) = t7T

for some z, we deduce that '1'(A) is (strictly) increasing.

* Since the expression on the right of (12) is, apart from a trivial factor, of the sameform as the expression (15) in R2.

160 B. KUTTNER

Using the results already obtained, the continuity of r.(A) may now beestablished by arguments similar to those of R2.

4.2. We next show that r(A)~O as A~O. Let K be any (fixed)assigned number greater than O. We have to show that, for sufficientlysmall A, (A, K; z) ~ trr for all z. Since

which tends to zero as A~ 0 uniformly in z for z in any finite interval, wemay clearly, without loss of generality, restrict ourselves to values of zgreater than (say) 1.

Throughout what follows, z and A are both allowed to vary, and theconstants implied by the "0" symbols are uniform in z > 1 and A< 1.We consider the expression for (A, K; z) given by equation (10); wedenote the first integral in (10) by 11' and the second by 12, If we writeIl-ellTiAyAI = a, and arg {1_ellTiAyA} = -0 (the value of the argumentbetween 0 and -71' being taken), then, supposing that K < 1, we have

~(I_el,..iAyA)1e= -ale SinKO < -Kale sin 0 = _Kyl.ale-1sin !1TA.

Hence

II ~ -K sin!1TA J~ yA-lalC-1e-YZdy ~ -K sin!71'A J:yA-l aK-1e- lIzdy

= -AZ-A( (1+0(A»),

where A is a constant greater than O. On the other hand, since

it follows that

Hence

and the result is evident.Finally, the result that r(l) is equal to Cramer's constant is trivial.

For since the general term of the Fourier series of any functionj(x) is 0(1)

on the gibbs phenomenon for riesz means

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