PIPPERT, R. E. Math. Zeitschr. 85, 401--406 (1964)
On absolutely convergent trigonometric series By
RAYMOND E. PIPPERT
1. Introduction
In this note, we extend Theorem A of FATOU [1] and Theorem B of SALEM [3]. Using these results, we also obtain a theorem related to Theorem C of TsucmKURA and YANO [4].
Theorem A. Let the sequence {an} satisfy ]a,+ 1 [ =< t anl for all n. If either
(i) ~ ]an cos nXo ] < oo for some x o, or
(ii) ~ [a n sin nx o [ < oo for some x o 4=0 (rood n), then ~ [a,[ < oo.
Theorem B. Let ~ An(x)-ao/2+~(a, cos nx+b, sin nx), and write p , = (aZ,+b2) -~. If p,+l<pn for all n, and if ~ A , ( x ) converges absolutely at two (distinct) points x o and x 1 with [ Xo - xl [ < n, then ~ P, < oo.
Theorem C. Suppose that a trigonometric series ~ p, cos(nx+On) (pn>O) and its conjugate series ~ p. sin(nx+0n) are absolutely convergent at x = x o and x = x l , respectively. If x l - xo=(p /q )n (p/q irreducible) where p is an integer, positive, negative, or zero, and q is an odd integer, then ~ Pn< oo.
Remarks. ZYGMUND [5; p. 232--233] gives a proof of Theorem A due to Saks, and also gives a proof of Theorem B, similar to that of Salem, utilizing Theorem A. SALEM [3] actually proves a slightly more general result, replacing the condition that pn+l<=pn by the condition that Pn+v/P, be bounded inde- pendently of n a n d p > 0 . TsucmKURA and YANO [4] show that Theorem C is no longer true if p/q is replaced by p'/q' with an even q' and p' ~0, or by an irrational number. Hence, to improve the result of Theorem C, we require an additional hypothesis (in our case a restriction on {p.}).
2. Statement of results
Theorem 1. Let the sequence {an} satisfy ]an+l [ < Kla,] for all n, where K is a constant independent of n. I f either
(2.1) ~ [ancosnxol<Oo for some Xo, or
(2.2) ~ Jan sin nxo l < oo for some Xo +-0 (mod n),
then ~ [ a, [ < oo. Furthermore, if the constant K is replaced by any function q~ (n) such that q)(n)--*oo as n~oo, then the above results no longer hold.
Theorem 2. Let A.(x) --ao/2 + ~ ( a . cos nx+bn sin nx),
Bn(x) - Z (a. sin nx-bn cos nx),
402 RAYMOND E. PIPPERT;
and write p, = (a~ + b~) ~. Suppose the sequence {p.} satisfies p,+ , <= Kp , for all n, where K is a constant independent of n. I f one of the series ~,A.(x) or ~ B,(x) is absolutely convergent at two (distinct) points Xo and xl with ] x o - x l l +0 (mod zc), then ~ p, < oo and hence ~ ( l a . l + I b . l ) < ~ .
Theorem 3. Suppose {p,} is a sequence of non-negative numbers satisfying P.+ I <=Kp.for all n, where K is a constant independent of n. I f a trigonometrie series ~ p, cos (n x + 0,) and its conjugate series ~,p, sin (n x + 2,) are absolutely convergent at x = Xo and x = x~ respectively, then ~,p, < ~ .
Remark. The sequences considered in the above theorems obviously need satisfy the prescribed condit ions only after some finite n u m b e r of terms, i.e. only for n > no (constant).
3. Proofs of theorems
T o prove Theorem 1, we use
Lemma 1. Let a and b be any two f ixed numbers such that 0 < a < b; and denote by I[x] a closed interval of length a, centered at x. Then there exists an integer M such that whenever KaeI[kb] for some integers K and k, then (K+L) aeI[(k + 1) b]for some integer L< M. (Here M is independent of K and hence of k.)
Proof of Lemma 1. Since the length of 1Ix] is a, and l[x] is closed, I[x] contains at least one integer mult iple of a for every x. Thus I [kb] and I [(k + 1) b] each contain at least one integer mult iple of a, hence we need only choose M so as to assure tha t M steps of length a will take us f rom I[kb] to I [ ( k + 1) b]. Choosing M such tha t Ma>=b, i.e. M>=b/a, we have the desired result.
Proof of Theorem 1. We m a y clearly assume 0=<Xo<~. We suppose first tha t (2.1) is satisfied, and we consider two cases.
Case 1. xo = 0 or Xo = ~. In this case the result is an immedia te consequence of the hypothesis.
Case 2. 0 < X o < n . F r o m the sequence {a,) we choose tha t subsequence (ank) of terms whose indices satisfy
nkxoeI[lk~] ( k = 1, 2, 3, ...)
where l k (k = 1, 2, 3 . . . . ) are some posit ive integers. By L e m m a 1 with a = xo and b = ~, we see tha t (n k+ 1 - n~) -- Mk < M (constant) for all k. F r o m the choice of {a,~} we see tha t
I cos n~ Xo I --> I cos (I~ ~ + Xo/2) 1 = cos Xo/2 for all k. N o w
Hence
lanCOS n)c o I < oO ~ y ' [ankCOS nkX o I< 00. n k
I a,~i cos Xo/2 < ~ I a,k cos nk Xo [ < o0, k k
so that
(3.1) Y~ i a . k [ < ~ .
On absolutely convergent trigonometric series 403
Now if nk< i<nk+ 1,
then ]ai]~K(i-nk) ]ank] < K M [an J ,
since we may clearly assume K>_ 1. Thus
~ [ a n l < ~ M k ( K ~ ] a n ~ l ) < ~ M K ~ l a , ~ [ < o o by (3.1), n k k
which proves the theorem if (2.1) is satisfied.
We now suppose that (2.2) is satisfied, and we have only the case 0 < Xo < z~. From the sequence {a,} we choose that subsequence {an~} of terms whose indices satisfy
t '- / ~ x 7
)
where l k (k = 1, 2, 3 . . . . ) are some positive odd integers. By Lemma 1 with
3~ a = xo and b = 2 ,
we see that (n k + 1 - n~) - Mk < 2 M (constant) for all k. From the choice of {a,,k} we see that
[s inn~xo] > \ \ 2 J Jl 2
The remainder of this part of the proof is similar to the first part and is omitted.
To prove the last statement of Theorem l, we suppose that the constant K is replaced by any function (p(x) such that 9 ( x ) ~ as x ~ , and we show that there exists a sequence (an} such that
]a~+l]<_9(n)]a~] for all n>=no and
but
(A similar result holds for the sine series.) To obtain this result we shall use
Lemma 2. I f 9 ( x ) is any funct ion such that ~o(x)--.oo as x-~oo, then there exists a funct ion ~t' (x) having the properties
(i) ~ (x) =< ~ (x) f o r all x ~ x o , (ii) ~g(x) is an increasing funct ion o f x ,
(iii) ~(x)-~oo as x-~oo, and
(iv) V ( x + 1 ) - ~g(x) is a non-increasing funct ion of x.
4 0 4 RAYMOND E. ~ P E R T :
~ cp(") n
Proof of Lemma 2. Choose a sequence of positive integers {nk} such that nk>=2nk_ l, and ~o(x)>_k+ 1 for x>nk ( k = l , 2, 3, ...). Define
T(x)=k-+ x -nk for n k ~ X < n k + 1 Ylk+ I - - Hk
i.e. T(x) is continuous, assumes the value k at X=nk, and is linear in each interval [nk, nk+l]. Let x be any number ~ n t . Then for some k, we have nk <=X<nk+1, and
~ ( X ) - = k q - X - - n k <k+l_-<q~(x). nk+ t -- nk
Thus (i) T(x)<=~o(x) for all x>nl,
(ii) T(x) is clearly an increasing function of x,
(iii) T (x) -~ ~ as x-~ o% and
(iv) the slope of T(x) is 1
tP'(x) = for nk<x<ne+l, n k + 1 - n k
which is non-increasing by our choice of {nk}, SO Lemma 2 is established.
Now to prove our assertion that the result of Theorem 1 is the best possible as explained above, we see from Lemma 2 that we may suppose q~(x) is a monotone increasing function of x, tending to infinity, whose slope is non- increasing. We write
bt =~p(1),
bk=q~(k)-q~(k-I ) (k=2, 3, ...).
Then {bk} is a positive, non-increasing sequence such that ~ b, = ~ , and hence
b, = ~ [ 2 ; p . 14], but
We now take
Then
and
B u t
an= i bn-1
t q,(n-1)
~ 02@n) < oo [2;p. 14].
n even
n odd, n > l .
a . > 0 for all n > 2 ,
a.+l <=q~(n)a.,
n even n even ~P kr~J
bn Za.>. =
On absolutely convergent trigonometric series 405
since the series consists of every other term of a monotone divergent series of positive terms, and our assertion is established.
Proof of Theorem 2. We suppose first that ~ I A. (x) I < oo for x = Xo and X=Xl, and write ~ A . ( x ) - ~ p . sin (n x + 0.). Now let h=xo-xl . Then
nh=(nxo + O,)-(nxl + O,) and
sin n h = sin (n Xo + 0,) cos (n xi + O . ) - cos (n Xo + 0,) sin (n xl + 0,). Thus
I sin n h [ < I sin (n Xo + 0,) I + [ sin (n xi + 0,) [ and
~. [p, sin nhl <~ IP. sin(nxo +O,)[ + ~ ]p, sin(nxl +O,)[ <
by our assumption. But h + 0 (mod n), so by Theorem 1, ~ p , < oo.
Now suppose ~ IB,(x) l < oo for X=Xo and x=xl. We need only note that we can write
Z B.(x)=Z p. sin(nx+ 0.)
and the proof follows as above.
Proof of Theorem 3. Write xl - Xo = h. Then
~,p, lcos(n(xt-h)+O.)l= ~ p, lcos(nxo +O,)l < ~ . (3.2) Now
sin (n xl + 0,) = cos n h sin (n (xt - h) + 0,) + sin n h cos (n (xl - h) + 0,).
Thus
(3.3) ~ p. I cos nh sin(n (x i - h ) + 0 , ) / < oo.
But from (3.2), we have
(3.4) ~ p, I cos n h cos (n (xi - h) + 0,)] < m.
Thus
Z P, Ic~ lcos nh] [sinZ(n(xi-h)+O,)+cos2(n(x~-h)+O.)]
< Z P , [cos nh[ [Isin(n(x~-h)+O,)l+lcos(n(x~-h)+O,)l] < co by (3.3) and (3.4).
So by Theorem 1,
2 Pn < ~ "
4. Additional remarks
Using the information obtained in the proof of Theorem 1, we can easily obtain from Lemma 1 the following two results:
A. (i) cos nxo+-~O as n ~ c o , and
(ii) sin nxo+-~O as n ~ if Xo+0 (mod n).
B. Let the sequence {a,} satisfy [ a, + 11 < K I a, I for all n, where K is a con- stant independent of n. If either
Mathematlsche Zeitschrfft, Bd. 85 28
406 RAYMOND E. PIPPERT: On absolutely convergent trigonometric series
(i) an cos nxo--*O as n ~ o o f o r s o m e Xo, o r
(ii) a n sin nxo~O as n ~ o o fo r s o m e x 0 4=0 ( rood ~), t hen an-~0 as n - ~ m .
T h e resul t (A) can be o b t a i n e d in a d i f fe ren t m a n n e r [5; p. 34, Ex. 2].
Professor S. M. SI~AH has made valuable suggestions concerning the subject matter of this note, and has, in addition, given guidance in its preparation.
Added in Proof. For a different proof of the first part of Theorem 1, see
(i) SZASZ, O.: On the absolute convergence of trigonometric series. Ann. of Math. 47, 213--220 (1946).
(ii) Stmoucm, G., and S. YANO: Notes on Fourier Analysis. XXX. On the absolute convergence of certain series of functions. Proc. Amer. Math. Soc. 2,380-- 389(1951).
References [1] FATOU, P.: Sur la convergence absolue des s6ries trigonom6triques. Bull. Soc. Math.
France 41, 47-- 53 (1913). [21 HoBsoN, E.: The Theory of Functions of a Real Variable, Vol. 2. New York: Dover
Publications 1957. [3] SALEM, R. : The absolute convergence of trigonometrical series. Duke Math. J. 8, 317--
334 (1941). [4] TsucmKtraA, T., and S. YANO: On the absolute convergence of trigonometrical series.
Proc. Amer. Math. Soc. 1, 517--521 (1950). [5] ZYGMUND, A.: Trigonometric Series, Vol. 1. New York: Cambridge Univ. Press 1959.
Department of Mathematics, University of Kansas, Lawrence, Kansas, U.S.A.
(Received May 13, 1964)
