uniqueness of rectangularly convergent trigonometric series

Annals of Mathematics

Uniqueness of Rectangularly Convergent Trigonometric SeriesAuthor(s): J. Marshall Ash, Chris Freiling and Dan RinneSource: Annals of Mathematics, Second Series, Vol. 137, No. 1 (Jan., 1993), pp. 145-166Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/2946621 .

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Annals of Mathematics, 137 (1993), 145-166

Uniqueness of rectangularly convergent trigonometric series

By J. MARSHALL ASH, CHRIS FREILING and DAN RINNE

Introduction

Georg Cantor proved in 1870 that if a 1-dimensional trigonometric series converges to 0 everywhere, then all of its coefficients must be 0. When one tries to extend Cantor's result to more dimensions, the first problem one en- counters is deciding what "converges" means. Take, for example, the double trigonometric series EM= n=-oo amneimx~inY There are several natural orderings for the summation over the lattice plane, each determining a different notion of convergence, each with its own uniqueness question. Some convergence methods that have been considered are:

(1) Iterated: E' c, E' _o means E' -oo (E' -oo).

(2) Circular: E e n=-oo ans limr-o Em2+n2?r.

(3) Rectangular: Em=-o E' means limmin(a,b)oo EZml<a EZnl<b

(4) Square: Em=-. En=-oomeans lima-o EZml<a EZnl<a

(For others see [AW2].) What we call rectangular convergence throughout this paper is often called unrestricted rectangular convergence. The problem of noncanonical ordering is not without analogy in 1 dimension, where we could take E' to mean ?_ - + E' o, even though it usually means lima---,=o Ea The latter is usually preferred, one reason being that more functions can be represented by the use of symmetrical partial sums. In 2 dimensions, none of the described methods of convergence is able to represent more functions than all of the others, although square convergence can cer- tainly represent more than rectangular convergence (see [AW2] for a discussion of these interrelationships). Along these lines, one might argue that square convergence has a certain natural strength because of the following result (proved independently by C. Fefferman [Fel], P. Sjolin [Sj] and N.R. Tevzadze [T]), which extends a famous result of Carleson to higher dimensions:

THEOREM (Fefferman, Sjolin, Tevzadze). Let f be a periodic function of n variables which is in LP for some p > 1. Then the Fourier series for f is square convergent almost everywhere to f.



146 J.M. ASH, C. FREILING AND D. RINNE

The corresponding theorem for rectangular convergence is false even when f is continuous (see [Fe2]). For circular convergence, the corresponding theorem is false for 1 < p < 2, but remains open for p > 2.

Uniqueness is an immediate consequence of Cantor's theorem in the case of iterated convergence. For nontrivial methods of convergence in 2 dimensions, the following uniqueness results have been proved (cf. [S], [Col], [AWl], [AW2], respectively):

THEOREM (Shapiro). If a multiple trigonometric series is spherically Abel summable to 0 everywhere and has coefficients {aM}MEZn satisfying

(S) IaMI = o(R) as R -+ oc, (R-1)2<M2+..+2 <R2

then all of its coefficients are 0.

THEOREM (Cooke). If a double trigonometric series converges circularly to 0 everywhere, then condition (S) must hold. Also circular convergence to a function implies circular Abel summability to the same function. Therefore, by the theorem of Shapiro, if a double trigonometric series converges circularly to 0 everywhere, then all of its coefficients are 0.

THEOREM (Ash-Welland). If a double trigonometric series converges rectangularly everywhere, then condition (S) must hold. Also rectangular convergence to a function implies circular Abel summability to the same function (although not necessarily circular convergence). Therefore, by the theorem of Shapiro, if a double trigonometric series converges rectangularly to 0 everywhere, then all of its coefficients are 0.

Neither of the theorems of Cooke and Ash-Welland readily extends to higher dimensions without the assumption of condition (S). Back in 2 dimensions, the corresponding uniqueness question for square convergence is still unknown, as is the question for rectangular convergence with bounds on the eccentricity of the rectangles. In dimensions larger than 2, no method of convergence has been previously shown to satisfy the analog of Cantor's theorem, except in the trivial case of iterated convergence. Our purpose in this paper is to prove the following theorem:

THEOREM. If a multiple trigonometric series is rectangularly convergent to 0 at every point, then all of its coefficients are 0.

Our proof will consist of finding higher-dimensional analogs of the major steps in Cantor's original uniqueness proof. His proof can be broken down into three parts. The first part uses a clever idea, due to Riemann ([Ri],



RECTANGULARLY CONVERGENT TRIGONOMETRIC SERIES 147

p. 245). Riemann associated with each trigonometric series, m ametmX, a second formal integral, aox2/2 + Em#O ameimx/(im)2, which, when the coefficients {am} are bounded, converges absolutely to a continuous function F(x). He then applied the following generalized second derivative (now called the Schwarz derivative) to F:

lim F(x - h) - 2F(x) + F(x + h) = lim + ameimx (sin(mh/2) 2

h--*Oh --O M-O mh/2J

He proved that if the original series has bounded coefficients and converges to a finite sum s at x, then the Schwarz derivative of F, evaluated at x, is also s (in such a case, the trigonometric series is called Riemann summable at x to the sum s).

The second part of Cantor's proof, due to Schwarz [Ca2], states that if a continuous function has a Schwarz derivative that is identically 0, then the function is linear (which is the same as saying that all Schwarz difference quotients (F(x - h) - 2F(x) + F(x + h))/h2 are identically 0). In the case of the second formal integral then, this says that

+ Ao?Ya imx (sin(mh/2) 2

is identically 0 for all h. Cantor put these two parts together with his early version of the Cantor-

Lebesgue theorem. (If a sequence amezmx + a-me-imx converges everywhere to 0, then limm >, am = limm >, a-m = 0.) Note, then, that if the trigonometric series Em=,o ame converges to 0 everywhere, the combined results of Riemann, Schwarz and Cantor immediately imply that

ado +Vameimx (sin(mh/2) 2

is identically 0 for all h. Setting h = 27r yields ao = 0. We reformulate the remainder of the proof as follows: Divide the original

series by eiix. This shifts the original series, but the Cantor-Lebesgue theorem provides sufficient desymmetrization of the original (n = 0 centered) convergence. In other words, the new series also converges to 0 everywhere. But now the constant term is a1. Hence aj = 0 for all j. (See [Al], [Ca2] or [Z], Ch. 9, for more detailed proofs of Cantor's theorem.)

In our proof we will try to copy, in some sense, Cantor's proof. A major difference is that we will need a new proof of Schwarz's theorem that avoids the maximum principle. In particular we will define a 2-dimensional, rectangular,




Schwarz derivative (or S-derivative) as follows:

f(x - h,y + k) - 2f(x,y + k) + f(x + h,y + k) -2f (x-h, y) + 4f (x, y) -2f (x + h, y)

lim + f (x - h, y- k) - 2f (x, y- k) + f (x + h, y- k) h,k--+O h2k2

If f E C4, then the S-derivative of f is equal to alf/(&2x92y). We could now try to prove a 2-dimensional version of Schwarz's theorem. For example, Ash and Welland conjectured that if a continuous function of two variables has an S-derivative identically 0, then it can be expressed in the form f (x, y) = a(y)x + b(y) + c(x)y + d(x), which is equivalent to saying that all of the 2- dimensional Schwarz differences are identically 0. It turns out, however, that we will need more. Consider the following example, which answers negatively the conjecture of Ash and Welland in [AW2]. (See also [ACFR].)

Example. Let F(x, y) = (x + y)Ix + yj. Since FXXYY = 0 at all points off the line x+y = 0, the function F has an S-derivative of 0 at these points. If the pair (x, y) is on this line, then F has an odd symmetry through (x, y), forcing all S-difference quotients centered at (x, y) to be 0. The S-derivative of F is therefore identically 0. Note, however, that if (x, y) = (0, 1) and h = k = 1, then the S-difference quotient is -2.

Such an example at first seems to throw cold water on our attempt to generalize Schwarz's theorem to higher dimensions, since the function F described is not only continuous, but is also continuously differentiable, whereas the second formal integral is only known to be continuous. In fact exten- sions of the example to higher dimensions get worse (in n dimensions, n > 2, F can be made to have continuous 2n -3 derivatives). We will get around this by means of a more general form of the S-derivative, which we call the "S- connector." It is the key idea that allows us to link one dimension to the next. Simply put, an S-connector acts like a Schwarz derivative in some directions and like a second forward difference in others.

In 1 dimension, for example, we have the usual Schwarz derivative

Sif(x) lim f(x - h) - 2f(x) + f(x + h)

and the connector

Sof(x) lim f(x) - 2f(x + h) + f(x + 2h).

Note that if f is continuous, then the connector is 0. We call this second forward continuity. (Actually second forward continuity at all points forces




continuity for measurable functions; see [ACFG1RdWa]. We will not need this fact here.)

If n = 2, then there are several possibilities. We have the Schwarz derivative

f(x - hy + k) - 2f(xy + k) + f(x + hyj y k) -2f (x-h, y) + 4f(x, y) -2f (x + h, y)

S f (X y) =lim +f(x -hy -k) -2f(xy-k)+ f(x+h,y-k) (1,1)~ h,k--+O h2k2I

as well as other connectors as follows:

f (x - h, y + 2k) - 2f (x, y + 2k) + f (x + h, y + 2k) -2f(x-h,y+k) +4f(x,y'k) -2f(x+h,y+k)

S(lo)f( y) hk urn + f(x-h,y) - 2f (x, y) + f (x + h, y)

f(x, y + k) - 2f(x + h, y + k) + f(x + 2h, y + k) -2f(x,y) + 4f(x + h,y) -2f(x + 2h,y)

S(o~l)f(xy) := rnim + f (XI y-k)-2f (x + h, y-k) + f (x + 2h, y-k)

f(x, y + 2k) - 2f(x + h, y + 2k) + f(x + 2h, y + 2k) -2f(x,y+k)+4f(x+hy+k) -2f(x+2h,y+k)

S(oo) f(x, y) h= urnk + f (X y) - 2f(x + h, y) + f(x + 2h, y)

Comparing these, we see that each numerator is created by applying a second difference to each coordinate. The square of the step size appears in the denominator only if the difference is symmetric in that direction. To motivate this somewhat, note that the 1-dimensional function A(x) = ceimx + de-imx has a second symmetric difference -4A(x)(sin(mh/2))2, whereas the second forward difference is -4A(x + h)(sin(mh/2))2. In other words, the second symmetric difference acts on a trigonometric series as a multiplier, while the second forward difference does this and, in addition, shifts each term by an amount dependent on the frequency. Since the symmetric differences are so nice, they can overcome the damage done to the quotient by the step-size squared term in the denominator; the forward differences are not as nice, but they do not have corresponding denominator terms fighting against their movement toward 0.

There will be one other important difference between our derivative and the usual S-derivative. As Riemann pointed out ([Ri], pp. 246-248), his "Schwarz" derivative is actually a special case of a 4-point derivative, which




may be defined for any fixed nonnegative number 6 < 1 as

lir f(x-h)-f(x-k)-f(x + k) + f(x + h) hOIk/hl< h2 - V

In order to force through our proof we will need to use this 4-point version. In fact we will choose 6 = 1/3.) We therefore define some notation as follows:

[a, b] This denotes the closed real interval bounded by a and b and is defined for b < a (in which case [a,b] := {x: b < x < a}) as well as for a < b.

2-interval This is an ordered pair of concentric closed intervals I = ([a, b], [c, d]) such that [c, d] c [a, b]. We call a and b the outer endpoints, c and d the inner endpoints, la - bl the length, and la - bl/2 the radius. If c = d, then I is called a Schwarz interval (or S-interval).

H, X, M, Z These are abbreviations for the n-tuples (hi ... ., hn)7 (xi,... , xn), (ml,... ,mnn) and (z1,... , Zn), respectively. The symbols H and X will denote elements of Rn, while M and Z denote elements of /n.

PHi Let H E Rn. Then JHI := max{Ihli,..., Ihnl} limHbo By this we mean HI -O 0 such that none of the coordinates

hi, ... ,hn is 0.

n2 This denotes the set of sequences of O's and l's of length n. (Note that n2 has 2n elements.)

HIl; a If H eR anda= (ai,...,aE) En2, then H T a = lnLih . We will be using a to keep track of the directions in which an S-connector is symmetric.

F' If F : Rn - R and I is an S-interval ([a, b], [c, c]), then F' denotes the function of n - 1 variables defined by F'(xi ... , Xn-) F(xi X* 7 Xn- X, a) - 2F(x, .... ., Xnl c) + F(xi X ... 7 Xn-1, b).

S-interval In higher dimensions we will try to avoid, for ease of notation, specific mention of an arbitrary 2-interval, even though the notion will be implicit in our proofs. We define an n-dimensional Schwarz interval (or just S-interval) as a sequence (I, . . . , In) of 1-dimensional S-intervals.

LSF Let I = (Ii,..., In) be an n-dimensional S-interval and F : Rn R. Then the Schwarz difference of F over I is ASF =

I(aXH) If X = (xi,...,xn) E Rn, H = (hl,...,hn) E Rn and a = (ai,... , an) E n2, then I(a, X, H) denotes the n-dimensional S-interval (I1,, . . ., In), where Ii := ([xi -aihi , xi +2hi - ihi] , [xi +hi -aihi , xi +hi - ihi]) . In other words, Ii is symmetric about xi when ai = 1 and has the endpoint xi when ai = 0.




SLTF(X) If X E Ri and a E '2, then the Schwarz connector SF(x) := limH~o Sx H)F/(H T a)2. This is a limit of a 3n-point S-difference quotient, which is symmetric only in dimensions where ai = 1. (Examples with a = 1, 07 (1, 1), (1,0 ), (0, 1) and (0,0 ) were given above.)

Note. In the case where n = 0, it will be convenient in Section 2 to consider a number k as a function of 0 variables and to let Ask = k and QtX= 1 sothat Sok=k.

Returning to the counterexample F(x, y) = (x + y) x + yl, one easily sees that all S-connectors of the form a = (1, 1) or (0,0 ) are 0, but the connectors for a = (1, 0) and (0, 1) are infinite at points on the line x + y = 0.

We will divide our proof, analogous to Cantor's original proof, into three parts. In Section 1, the Riemann part, we will show that if certain partial sums of a trigonometric series are bounded at X and the series converges rectangularly to 0 at X, then all S-connectors of its second formal integral are 0 at X. In Section 2 we will give a substantially different proof of Schwarz's theorem, which, by taking advantage of the S-connectors, we will generalize to higher dimensions. In particular we will show, using partition properties of 2-intervals, that if F is continuous and all S-connectors are 0, then all S- difference quotients are 0. In Section 3 we will extend a theorem of Ash and Welland (rectangularly convergent trigonometric series have bounded rectangular partial sums) to the more general partial sums of Section 1 (where one or more of the indices are fixed), using the Cantor-Lebesgue theorem. This will immediately give the first coefficient as 0. We will then shift the coefficients to finish the proof.

1. The Riemann part of the proof

To each multiple trigonometric series

T = E aml m eiM1X1+ +iMnXn

we associate a series E ... am,...,mn (eimlxl /(iml)2) ... (eimnxn/(imn)2), which we call the second formal integral of T in each dimension. Here, when mj = 0, expressions of the form eimixj /(imj)2 will be understood to represent x? /2. The following is immediate by the Weierstrass M-test:

PROPOSITION 1. 1. If the aM are bounded, then the second formal integral converges both absolutely and uniformly to a function F(X). It follows that F(X) is continuous and the terms in the series for F may be summed up in any order.




Definition. Am(h) := (sin(mh/2)/(mh/2))2 if m ? 0, and 1 if m = 0, and AAm(h) := Am(h) - Am+i(h).

Using elementary trigonometric identities (and the fact that addition and multiplication by a constant can be performed on any convergent series term by term), we easily establish the next result.

PROPOSITION 1.2. Let AM(X) := aml,,mneimlXl++imnXn. Let

T = ... AM(X)

be a trigonometric series with bounded coefficients. Then when a = (1, 1, .. ., 1), the S-difference quotient of the second formal integral of T is

ZAM(X)Am, (hi). Amn(hn). Note in particular that if h1 = -

hn = 27r, then this reduces to Ao,...,o(X), which is the same as ao... .

Definitions. Let pm(h) = Am(h) * h2eimh and, for a fixed a E n2, let

Am (h) .- pm(h) if aT(j) = 0, A'm~h) lAm(h) if a(j) = 1.

Then the analog of Proposition 1.2 for S-connectors is the following:

PROPOSITION 1.3. Under the conditions of Proposition 1.2, the S- connector SF(X) is equal to limHbO E Z.. E AM (X)Al,ml (hi) ... Animn (hn)

PROPOSITION 1.4. The Am(h) have the following property: (a) For some K < oo and for any h, E' z1AAm(h)I < K.

Also the pm(h) satisfy (b) liMhO Z p=00m (h)I = 0.

Proof. For part (a) we have

IL\Am(h)I = i|(m+l)h [2 (x/2)]ld

< A( [h (x/2)2] dx,

so that the sum in part (a) is bounded by K := fOfl (sin2 x/x2)'|dx < oo. For part (b) let e > 0 be given. Let N be so large that Z?m?N 4/N2 +

Z-N- 4/M2 < 6/2 and h be so small that EZN=$yLpm(h) < 6/2. Then Em=-00 IPm(h)l <6E/2 + 6/2 = 6. Li

Remark. If in the definition of S-connectors we made the lopsided directions look like ordinary differences f (x + h) -f (x) rather than second forward differences f (x) - 2f (x + h) + f (x + 2h), then pm(h) would be defined as




(eimh - 1)/rn2 when m + 0, and 0 when m = 0. Condition (b) of Proposi- tion 1.4 would still hold. However it will be sufficient (and in fact slightly more convenient) but not necessary to use the definition of S-connectors that we have given.

Notation. Let T = E AM be an infinite series in n dimensions. If Z E Zn and a E n2, then we let sz,, denote the finite partial sum of T, where the mj vary from -Izi to zjI if a (j) = 1 and are fixed at mj = zj if a (j) = 0. Hence, if a = (1,1,... , 1), we have a rectangular partial sum of T, which we abbreviate as sz.

THEOREM 1.5. Let 7I = LAM be an n-dimensional series rectangularly convergent to 0. Suppose there is a B such that, for each Z in 7Z and each a E n2, Isz,XJI < B. Let

00 00

D(H): ...5? Ami . Mn Ai,m, (hi ) -An~mn (hn) ml=-oo mn=-0o

Then limHbo D(H) = 0.

Proof. By the symmetry of the coordinates we may assume, without loss of generality, that a consists of j l's followed by (n - j) O's. We have

00 00

{D(H) I = ...

mj+j=-oo mn=-00 N N

ilim . AMAm .Ami Pmj+ .Pmn.

ml=-N m3=-N

Fix mj+?, ..., and mn. Since Am(h) = A-m(h), we may write the expression in curly brackets as

N N / lim ... . . AK)Am *. Amj

mi=0 mj=O Iklil=rni kjl=mj

where kj+? = mj+? for v > 0. Summing by parts in all of the first j variables (see [AW2], p. 413, for the formula) yields

NV-1

lim S S SK,,A Am1 (hi) AaiAmj (hj), aEi2 {q: ?7q=1} nq=O

where r(v) = 1, v = 1, .. ., j and r(v) = 0 for v = j + 1, .. ., n; ki = N if 7i = 0, and ki = mi otherwise; and

i\ i Asmi(hi) Amij(hi) - Ami+i(hi) if C, = 1, =

.AN(hi) if ai=O0.




By hypothesis, all of the SKE are uniformly bounded and, whenever v= 1,

N-1 00

E Ai Ari (hi) I < JE|AIk(hi)I < K Mi=O k=O

by part (a) of Proposition 1.4. Since AN(hi) = O(N-2) = o(1) as N -x 00, if at least one vi = 0, then the limit is 0. Thus only the term corresponding to a = (1, ... , 1) contributes, and the expression in curly brackets in the formula for ID(H)l becomes

N-1 N-1

lim E SM,rAAmi (hi) .AAm (hj) ml=O mj =O

so that 00 00

ID(H) I= Z Z mj+3=-0o mn=-00

{ 00 00

S.. SMOAAmi (hl) * AAmj (hj) Pmj+* Pn ml=O mj =O

If j < n, use B as a bound for all of the sM,r and the K from Proposition 1.4(a) to get the estimate ID(H)l < BKj(EZ?O0O Ipi(hi)I)' i. Then from part (b) of Proposition 1.4 we see immediately that limjHjo D(H) = 0.

For the special case when j = n, we have 00 00

ID(H)I? < ... IsmAAM m AAmnIl m1=O m'=O

Let K be the bound from part (a) for each Zrnk=O I Amk(hk)I. Let e > 0 be given. Choose N large enough that ISMI < e/Kn whenever IMI > N. Let R be the subregion of Nn where each nk > N. Let Rk be the region where nk < N. Then

00 00 5 IsmAAmm AAmnI < E/K. 5 IAAmI I ... IAAmnI R ml=0 mn=O

< Burn Kn =,e.

Also N 00 00

5ISMAAmn AAmnI? < B 5 IAAmiI| 5 IAAM2I 5 IAAmnI, Ri mi=O m2=O mn=O




where the first sum on the right tends to 0 as h1 - 0 and the other sums on the right are bounded by K. Similarly each ERk ISMAAmi *

... AAmnI 0 as hk -a0. SinceN' = RUR1U .*URU , we have ZN n?< ER?+ER1 + +ERn when all of the terms of the sum are nonnegative. Therefore we can make D(H) small by choosing h1, . . ., hn all close to 0. El

The main result of this section is the following corollary:

COROLLARY 1.6. Let T be a trigonometric series in n variables converg- ing rectangularly to 0 at some point X. If for each a there is a bound on all the partial sums sz,(X), then all of the Schwarz connectors of its second formal integral in each dimension are 0.

Proof. This is immediate from Theorem 1.5 and Propositions 1.1, 1.3 and 1.4. 0

2. The Schwarz part of the proof*

Consider a function f(x, y) of two variables, a rectangle R = [a, b] x [c, d] and an associated difference operator ARf = f(a, c) - f(a, d) - f(b, c) + f(b, d). If R = U Ri is a finite partition of R into nonoverlapping rectangles with sides parallel to the coordinate axes, then it is easy to see that ARf =

Ei ARpf. This provides a kind of "partition" of the difference AR. Given a Schwarz interval I = ([a, b], [c, c]), where c = (a+b)/2, we wish to partition the 1-dimensional difference ASF = F(a) - 2F(c) + F(b) in a similar way. To do this we identify to the function F a function f(x, y) = F((x + y)/2). Thus we think of the domain of F as being a line of slope 1 and of the f created from F by sweeping the graph of F along a slope of -1. If R = [a, b] x [a, b], then ARf = ASF. This identification allows us to project a partition {Ri} of R into a partition {Ii } of I, as long as every rectangle Ri is a square. Note that oblong rectangles project onto 4 points rather than 3. This motivates our passing from Schwarz intervals to arbitrary 2-intervals ([a, b], [c, d]), where (a + b)/2 = (c + d)/2. Note that each 2-interval has several rectangles which project onto it. We will use this "partitioning" of 2-intervals to develop a new proof that if a continuous function f(x) has a Schwarz derivative 0, then it is linear. To form a 1-dimensional Schwarz difference quotient, we let I = ([x - h, x + h], [x, x]) and divide AS by h2. This denominator h2 is 1/4 the area of any square that projects onto I. The natural denominator for 2-intervals,

* For an introduction to the the methods of this section, see the expository treatment of the 1-dimensional case in [A3J. For an application to the theory of integration and the coefficient problem, see [FRThJ.




then, is 1/4 the area of any rectangle that projects onto it. Thus we will consider difference quotients of the form

f(x-h)-f(x-k)-f(x + k) + f(x+ h) h2 - k2

We can still get the result on Schwarz intervals, however, since the Schwarz difference on any 2-interval is merely the difference of Schwarz differences on two concentric Schwarz intervals. Applying a similar subtraction to difference quotients will show, as long as the ratio k/h is bounded away from 1, that if the difference quotients on the two concentric Schwarz intervals is small, so is the difference quotient on the 2-interval. The restriction that k/h be bounded below 1 turns out to be equivalent to the requirement that the associated rectangles have bounded eccentricity. Thus we will be able to show that every Schwarz difference quotient is small by partitioning the desired S-interval into small (not too eccentric) 2-intervals, which are close enough to their center to have a small difference quotient. This motivates the following definitions:

Definitions. (1) If K is the 2-interval ([x - h, x + h], [x - k, x + k]), then JIKiI := h2 - k2 and IKI := Jhi. We say that K is 3-fine for some real function 6 if JKJ < 6(x). We call K r-regular if Jkl < rjhl.

(2) Let ]P be a finite multiset of 2-intervals (i.e., some 2-intervals may be counted more than once). Let k(x) denote the number of elements of P containing x as an outer endpoint minus the number of elements of P containing x as an inner endpoint. Then IP is a partition of ([a, b], [c, d]) if k(a) = k(b) = 1 and k(c) = k(d) = -1 and k(x) = 0 for all other x.

(3) A rectangle is a subset of R2 of the form [a, b] x [c, d]. The eccentricity of a rectangle is the maximum ratio of its 2 dimensions.

(4) Let 1? = {R1, .. ., Rn be a finite set of rectangles, whose interiors are disjoint. Then IP partitions a rectangle R if U Ri = R.

(5) For R a rectangle, ir(R) is the 2-interval formed by the projection of R on the line y = x; that is, if R = [a,b] x [c,d], then 7r(R) = ([(a + c)/2, (b+d)/2], [(a+d)/2, (b+c)/2]). Note that if R has eccentricity < e, then ir(R) is (e - 1)/(e + 1)-regular. In particular, if R has eccentricity < 2, then 7r(R) is 1/3-regular. Also notice that each partition Pi of R induces a partition of 7r(R) by projecting the rectangles in P onto the line y = x.

(6) If 6 is a positive function on IR, then the function -y is called an r-regular error function for 6 if

(i) ty is defined for all 6-fine, r-regular 2-intervals, and 'y(I) = 0 for such intervals;

(ii) -y is defined and nonnegative for all Schwarz intervals and, for any sequence {Jn} of Schwarz intervals shrinking to a common outer endpoint




(this means Jn = ([x, b], [cqcn]), where cn = (x+bn)/2 and bn -* x), we have limnoo y(Jn) = 0.

LEMMA 2.1. Let L be a closed set of lines with slope -1 in the x-y- coordinate plane and let D be the complement of L. Then for any a < b and any y, there is a c such that (b - a)/2 < (c - y) < 2(b - a) and the rectangle R = [a, b] x [y, c] can be partitioned into subrectangles such that each rectangle in the partition is one of the following types:

(a) center in L and eccentricity < 2, (,3) interior in D.

Proof. Assume without loss of generality that a = 0, y = 0 and b = 1. Let E be the intersection of L with the y-axis. If some line e in L intersects the y-axis in [3/4,3/2], then e also intersects the vertical line x = 1/2 at some point (1/2, p), where 1/4 < p < 1. If we let c = 2p, then the rectangle R has center (1/2, p) and is of type (a), and we are done. Therefore assume that Ef n[3/4,3/2] = X

From now on let c = 1/2 so that R = [0, 1] x [0, 1/2]. If Efn [O 3/4] is also empty, then R is of type (f3), and we are done. Therefore let e be the line in L with the largest y-intercept < 3/4. If E n [1/2,3/4] $ b, then e intersects the horizontal line y = 1/4 at some point (p, 1/4), where 1/4 < p < 1/2. Then R can be partitioned into two rectangles [0, 2p] x [0, 1/2], [2p, 1] x [0, 1/2], where the first is of type (a) with center (p, 1/4) and the second is of type (0f) by the maximality of e.

Therefore we may assume that e intersects the y-axis at a point (0,p), where p < 1/2. But then the square S = [O,p] x tO,p] is of type (a) and int(R - S) C D and can be partitioned into two rectangles of type (3). The square and two rectangles partition R as required. LI

LEMMA 2.2. Let L and D be as in the previous lemma and let S be the square [a, b] x [a, b]. Then there is a subrectangle R = [a, d] x [a, b] of S, where d - a > 1/12(b - a), and R can be partitioned into rectangles, each of type (a) or (p3).

Proof. Assume without loss of generality that a = 0 and b = 1. As before, we let E be the set of y-intercepts of lines in L. If there is a line ? with a y-intercept in [3/4,1], then f intersects the horizontal line y = 1/2 at some point (p, 1/2), where 1/4 < p < 1/2. But if we let d = 2p, then R is of type (a) with center (p, 1/2), and we are done. Therefore assume that En [3/4,1] = A.

From now on let d = 1/12 so that R = [0,1/12] x [0,1]. By the previous lemma we may partition in the desired way a rectangle R1 = [0,1/12] x [0, cl], where 1/24 < cl < 1/6. Applying the lemma again, we partition




R2 = [0,1/12] x [c1, C2], where 1/24 < c2 - c1 < 1/6. Repeated applica- tions of the lemma yield on top of each Ri = [0, 1/12] x [ci1, cj] a partition of Rj+1 = [0,1/12] x [cici+i], where 1/24 < ci+l - ci < 1/6. Let n be such that cn e [9/12, 11/12]. This will occur for some n, since ci - ci- is always < 2/12. We have therefore partitioned in the desired way a rectangle R1 U R2 U ... U Rn = [0, 1/12] x [0, cn]. We will be done if we can partition R1 = [0,1/12] x [Cn, 1].

If int(R') c D, then [0, 1/12] x [cn, 1] is of type (3), and we are done. Otherwise there is a line in L that intersects the horizontal line y = 1 in a point (p, 1), where 0 < p < 1/12. Since L is closed, we may let ? be the line in L with the smallest such p. Since E n [3/4, 1] = A, it follows that p > 0. Then the square S = [p, 1/12] x [11/12 +p, 1] has its center on e and is of type (a). Again taking into account E n [3/4, 1] = A, we see that the minimality of e leads to int(R' \ S) c D. Thus R' \ S can be partitioned into two rectangles of type (,3). These two rectangles together with S form the desired partition of R', and we are done. R

Remark. By using symmetry or by repeating the proof of Lemma 2.1, one can prove an "upside down" and two "sideways" versions of that lemma. The upside-down version starts with a horizontal line segment and creates a partitionable rectangle of eccentricity < 2, whose top edge is that segment. The second (resp. third) version starts with a vertical line segment and creates a partitionable rectangle of eccentricity < 2, whose right (resp. left) edge is that segment. One can then prove three other versions of Lemma 2.2, wherein at least the right (resp. top, bottom) 1/12 of a given square may be partitioned. The point of this remark is to justify the various appeals to symmetry occurring in the proof of the next theorem.

THEOREM 2.3. Let R be a rectangle, L a closed set of lines of slope -1 in R, and D the complementary open set of lines. Then R can be partitioned into rectangles each of type (a) or (3).

Proof. If L is all of R, we are done, since every 2-regular subrectangle of R is then type (a). If D $& X, partition R by means of a vertical line and a horizontal line so that the common corner c of the four resulting subrectangles is in D. It will suffice to show that one of these four can be partitioned as required, the argument for the others being symmetric. Let T be the subrectangle with the lower right corner c E D. Let S, be the largest square in T containing the corner of T that is opposite c. If Si contains the left edge of T, we partition a subrectangle T1 c Si that contains at least the left 1/12 of S1, using Lemma 2.2. If S, contains the top edge of T, we can similarly partition a subrectangle T1 containing at least the top 1/12 of S1. After T1 is




chosen, we let S2 be the largest square contained in the rectangle cl(T \ Ti) that contains the corner cl(T \ Ti) opposite c, and we partition a subrectangle T2 that includes at least the left 1/12 or the top 1/12 of S2. Continuing in this manner, we get a sequence of partitioned rectangles such that, for some n, T' = cl(T \ (Ti U ... U Tn)) is a rectangle containing c that is completely in D, so that T' is of type (i). Thus {Ti, T2, .. ., T?, T'} is the required partition of T. ?

THEOREM 2.4. Let 6 be a positive upper-semicontinuous function (i.e., for all x, 6(x) > lim supy, 6(y)) and let -y be an associated 1/3-regular error function. Then for any E > 0 and 2-interval I there is a partition PE of I so that EKEP 7y(K) <E.

Proof. Let P be the set of x such that 6 is bounded above 0 on some neighborhood of x. Here P is open by definition and P is dense. (For if we let I be an open interval, then upper-semicontinuity of 6(x) implies for each n that En := {x E I: 6(x) > 1/n} is closed in I. Since I = Un=i En, the Baire Category Theorem says that there is at least one Em with a nonempty interior. Any interior point of Em is in P.) Say that a 2-interval ([x - h, x + h], [x - k,x + k]) is contained in a set S if [x - h,x + h] c S. Say that a set N has property (*) if for any E > 0 and 2-interval I C N, there is a partition P of I so that EKEP a(K) < E. Define D by x E D if some neighborhood of x has property (*). We will now show that P c D. If 6(x) > 6 > 0 on some interval containing a 2-interval I, and R is a rectangle such that 7r(R) = I, then let P be any partition of R into rectangles, whose eccentricity is < 2 and which are small enough that their projections have radii less than 6. Then 7r(P) partitions I so that EKE,(P) a(K) = 0.

Observe also that any component of D has property (*): Let I be any 2-interval completely contained in D. By compactness, a rectangle whose projection is I may be partitioned into n rectangles, each of which is so small that its projection lies in a region with property (*). Then one can apply (*) to each of these using E/n.

Let L = R \ D. We show that L = b. Suppose not. By the Baire Category Theorem, 6 | L is bounded away from 0 on some nondegenerate interval that intersects L. Thus we may pick an interval J so that J n L $& 0 and 6 is greater than the length of J on J n L. Let I be any subinterval of J. Let B be any rectangle with 7r(B) = I. Let e > 0 and partition B into {A1, ... , An} U {B,, .. ., Bm}, where each rectangle Ai is of type (ca) and each Bi is of type (i3), using Theorem 2.3, with 7r'l[L] for the closed set of lines. We further partition each Bi = [a, b] x [c, d] into two squares Si1 = [a,a r+] x [c,c r+] and Si2 = [b-ri,b] x [d-r1,d], where q is chosen so that y(7r(Sil)) + y(7r(Si2)) < E/2m, and into three rectangles Ti1, Ti2 and




Ti3, which project entirely into D. Now let 1P = {ir(Ai)}, 'P2 = {1r(Sij)} and P3 = {ir(Tij)}. Then I is partitioned by Pi U P2 U I3 and, since each interval in P3 is in D, there is a refinement IP of 13 such that EKeP' y(K) < e/2. Also EKEP1 ay(K) = 0, since each such interval is 1/3-regular and within 6 of its center, and ZKEP2 -y(K) < e/2 by the choices of {Sil} and {Si2}. Then P = 1P1 U P2 U IP partitions I and EKEP y(K) < E. Since I was an arbitrary 2-interval in J, it follows that J c D, which is a contradiction. O

Definitions. Let us define two more notational uses:

X Ax Let X = (Xi , Xn) E in. If x E Rt, then X ax := (xl,.... ., Xn, X).

Fx If F : Rn- R, then for each fixed x, the Fx denote the function of n-1 variables Fx(xl,. . ., xn-1) := F(xl... . Xnli x)

Note. It follows from the definition of partition that if P partitions a 2-interval I and f : R- R, then EKEP ASKf = ASf. In the case where f(x) = x2/2, this shows that EKEP IIKiI = IIIII.

THEOREM 2.5. Let E, E' be greater than 0 and let 6 be a positive upper- semicontinuous function with -y an associated r-regular error function. Let P be a partition of a 2-interval I such that EKEP y(K) < E'. Suppose that f is a function of one real varmable such that

(a) if K is a 6-fine S-interval, then IAS f/IIKIII < E and (b) if K E P is not 6-fine, then IA f I < a(K). Then

(i) IAS f/IIKIII < E(l+r2)/(1-r2) for any r-regular, 6-fine 2-interval K; and

(ii) JASf / lIIIII < E(l + r2)/(l -r2) + E'/JJJJJ.

Proof. Let K = ([x - h, x + h], [x - k, x + k]) be an r-regular, 6-fine 2-interval. Then

IAkf/IIKIJI = l(f(x - h) + f(x + h) - f(x - k) - f(x +k))/(h2-k2) = j(h2[(f (x - h) - 2f(x) + f(x + h))/h2]

- k2[(f (x - k) - 2f(x) + f(x + k))/k2])/(h2 - k2)I < (h2E + k2E)/(h2 - k2)

< E(1 + r2)/(1-r2).

Since P partitions I, it follows that 1111 = ZKEPIIKiI and zIf =

ZKEIPSKf. Let K1,... ,Kn denote those intervals in P that are 6-fine and Kn+l X... , Km denote those that are not. Then




however the first quotient on the right is the weighted average of terms

_Kf fKn 0 0

11K, 11 ' 'jKnjj ' JjKn+1 11 X jIKmjjl'

each no bigger than E(1 + r2)/(1- r2) in absolute value, while the second quotient has an absolute value no bigger than (ZKEP-y(K))/1gIII. It follows that LASf/IIIIII < E(1 + r2)/(1 _r2) + E'/lIIII L

THEOREM 2.6. Let F: Rn -E R be continuous in the nth coordinate, with X E Rn-1 and a E n-12. Suppose that, for every real number x,

(1) ScAlF(X A X) = 0; and (2) SaAoF(X AX) = O.

Then, for any S-interval I, SFI(X) = 0.

Proof. Fix X E Rn-1, an S-interval I and e > 0. We will prove the theorem by showing that whenever H E Rn-1 and JHJ is small enough, then AS(XH)FI/(H T a)21 is less than e.

For each real x let 6(x) be the largest number 6 < 1 such that, whenever H E Rn-1 h E R and IH A hi < 6, then

AS ~F I(aA1,XAXHAh) (2/5)E

((H A h) T (U A 1))2 |- 111 Note that 6 is upper-semicontinuous by the continuity of F and is positive by supposition (1). We now define a function -y. If K is 1/3-regular and 6-fine, then y(K) = 0. If K is an S-interval ([x, x + 2h], [x + h, x + h]), which is not 6-fine, then

-y(K) = sup A1(oAOXAx'HAh)FI |H| < |K| L((H A h) T(a AO))2 ''

It follows by property (2) that y is an associated 1/3-regular error function for 6.

Let P be a partition of I (by Theorem 2.4) such that ZKEP y(K) < E/2. Let H be any element of Rn-1 such that IHJ < minKEP 1K!. For x E R let G(x) = (As(X, H)Fx)/(H T a)2. (Recall that if n = 1, this means that

G(x) = F(x).) If K is the S-interval ([x - h, x + h], [x, x]), we then have

A\s G Ks(AI(Ox)H)Fx IIKII (H T a)211KII

(zAI(aA1,XAx,HAh) )F _ (AI(7A1,XAx,HAh) )F (H T o7)211KII ((H A h) t (a A 1))2




so that if K is 6-fine, then

[ As G [ < (2/5)E IIKII - IIIII

On the other hand, if K E P is the S-interval ([x, x + 2h], [x + h, x + h]), then

4(Lhs ,HFx) AS xLHhF As G = K( I(oXH) I) I(oAO,XAxHAh)

K ~(H Iao)2 (H- a) 2

However H T a = (H A h) T (a A 0) SO that JAS GI < -y(K). It follows from Theorem 2.5 that

JASGI 1 + (1/3)2 (2/5)E E/2 e < .

l'HII - 1 - (1/3)2 IIIII IIIII 'IIII

and so zASGI <,e. However

,ASG = (I(oe,X,H)Fs (H t o)2

_ I(a,X,H) Iu X) I(aXH)

(H T a)2 (H T a)2 which finishes the proof. E

COROLLARY 2.7. Let F be a function of n variables, continuous in the nth coordinate, and I be any 2-interval. If all Schwarz connectors of F are identically 0, then all Schwarz connectors of F' are identically 0.

Proof. This is immediate from Theorem 2.6. 0

This brings us to the main result of this section:

COROLLARY 2.8. Let F be a continuous function of n variables. If all Schwarz connectors of F are identically 0, then all Schwarz differences of F are identically 0.

Proof. Apply Corollary 2.7 repeatedly. E

3. The Cantor part of the proof

Recall the original Cantor-Lebesgue theorem as well as the Ash-Welland

theorem ([AW2], pp. 410-412).

THEOREM (Cantor-Lebesgue). Let Am(x) = ameimx + a-me-imx. (1) If for each x in some set of positive measure there is some B(x) such

that, for each m E 7, IAm(x)l < B(x), then am and a-m are bounded.




(2) If for each x in some set of positive measure, limm-,o(Am(x) = 0, then limmO am = limmOO arm = O.

THEOREM (Ash-Welland). If a trigonometric series converges rectangularly at each X, then for each X there is a bound on all rectangular partial sums calculated at X.

Let us also recall some notation from Section 1:

Notation. Let T = E AM be an infinite series in n dimensions. If Z E En and a E n2, then we let sz,a denote the finite partial sum of T, where the mj vary from -IZjI to IZjI if a(j) = 1, and are fixed at mj = zj if a(j) = 0. Hence, if a = (1, 1, ... ,1), we have a rectangular partial sum of T, which we abbreviate as sz.

From the Cantor-Lebesgue theorem we derive the following desymmetrization theorem:

THEOREM 3.1. If for some trigonometric series, for all X there exist B(X) such that, for all Z, Isz(X)I < B, then for all X there exist B'(X) such that, for all Z and all a, I sz,a(X) < B'(X).

Proof. Since there are only finitely many a's in n2, it suffices to establish a different bound for each a and X. By symmetry assume that a is of the form j O's followed by (n - j) 1's. We use induction on j. If j = 0, then sz,,(X) =

sz(X), and we are done. To complete the induction let a* have (j - 1) O's followed by (n - j + 1) 1's, and assume that for all X there exist B'(X) such that, for all Z, 1sz,a*(X)j < B'(X). Fix X = (xi,... ,xn) and assume toward a contradiction that there is a sequence {Zk} = {(Zkl, ... ,zkn)} such that lim supkoo ISZk,a(X)I = 00. By passing to a subsequence if necessary, we may assume that each Zkj > 0. (To see this note that if Zkj = 0, then Szk,a,(X) = SZk,7* (X) < B'(X) so that the lim sup is still oc after discarding all Zk with Zkj = 0. Then at least one of lim SUpkooZkj>0 and lim supkNOZkj<o is still oc. By symmetry we may as well assume it is the former.) We may also assume that the sequence {zkj }j' 1 is either constant or strictly increasing. By adding more elements to the sequence if necessary, we may assume that either

(1) there exists z > 0 such that, for all k, Zkj = Z, or (2) for all k, Zkj = k. Let Zkj be the same as Zk, except that the jth coordinate (Zkj) is replaced

by -Zkj. Let Z,71 be the same as Zk, except that the jth coordinate is replaced by Zkj - 1. Let r(x) be the same as X, except that the jth coordinate (x;) is replaced by x. Put

Ck := - Zz and Ck =- eiz




Note that Ck and c-k do not actually vary with x and that {I Ck I} is unbounded. Let

Ak(X) := Ckeikj + C-ke kj = SZk,0c*(T(X)) - Sz-1 *(T(X))

so that, by the induction hypothesis, IAk(x)I < 2B'(r(x)). If there exists z > 0 such that, for all k, Zkj = Z, then Ck = Ak(O)/2 + Ak(7r/2z)/(2i), contradicting the unboundedness of {Ck}. Similarly, if for all k, Zkj = k, then we immediately violate part (1) of the Cantor-Lebesgue theorem. E

The main purpose of Theorem 3.1 is to combine it with Corollary 1.6 and the theorem of Ash and Welland to get the following result:

COROLLARY 3.2. Let T be a trigonometric series, which converges rectangularly to 0 everywhere. Let F be its second formal integral in each dimension. Then all Schwarz connectors of F are 0.

Definition. For Z E Zn we have IZI := minj=,...,n Izjl. Since we are interested in Z -x oc, min is used here, whereas earlier we were interested in H -O 0 so that max was used in defining I I on Rn.

The next theorem is an analog of Theorem 3.1, where "bounded" is replaced by "approaching 0". We need only prove it for a = (0,1,1, . .. , 1).

THEOREM 3.3. Let a = (0,1,1,... , 1). If a trigonometric series converges for all X, then limpz1+) sz,a,(X) = 0 at each X.

Proof. Fix X and assume that there is a sequence {Zk} such that lim supko ISZk,a (X) I > 0, while limkoo IZk I = oc. Assume without loss of generality that Zkl = k. Define Zk, ZY1, T(x), Ck and C-k as before, with j replaced by 1. Then we have limsup{ICkI} > 0 and Ak(x) = ckeiZkjX cCke-iZkjx =

SZk(T(x)) - SZ-1(T(X)), which is the difference of two partial sums and hence tends to 0. This contradicts part (2) of the Cantor-Lebesgue theorem. E

The purpose of Theorem 3.3 is to prove the following theorem:

THEOREM 3.4. Let T be any translation of the space Zd (i.e., f(Z1, v Zn) = (Zi + t1. 'Zn + tn)). If the trigonometric series E aMeiMX converges rectangularly to 0 for all X, then so does the series E aT(M)e .

Proof. By successive application we may assume without loss of generality that T translates in only one coordinate. By symmetry of the coordinates we may assume without loss of generality that T translates in only the first coordinate (O = t2 = ... = tn). Again, by symmetry, we may assume that t1 > 0 and, by successive application, that tj = 1.




Let sz be a partial sum of the original series, where each zi > 0. Let sz be the corresponding partial sum of the translated series so that s eix-sz =

ST(Z),a(X) - SZ-,,(X), where Z- = (-z1, Z2,... , Zn) and a = (0, 1,1... ., 1). If IZI -- 00, then sz -O 0 by assumption; and both ST(Z)-,(X) 0 0 and sz-,(X) )-- 0, by Theorem 3.3, yielding s' -- 0. E

We are now ready to complete the proof of our uniqueness theorem. Let E aMeimx converge rectangularly to 0 for all X. We will show that

each aM is 0. By Theorem 3.4 it suffices to show that a(oo...,O) = 0. However this fact follows immediately from Corollaries 3.2 and 2.8 and Proposition 1.2 (when we set h = 27r). 0

DEPAUL UNIVERSITY, CHICAGO, ILLINOIS CALIFORNIA STATE UNIVERSITY, SAN BERNARDINO, CALIFORNIA

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(Received May 15, 1991)



uniqueness of rectangularly convergent trigonometric series

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