(a) if Aa + Bb + Cc^O,fe tx(x) for almost every xeE. (b) //^a+£6-l-Cc=0,/G. Ky{x), i.e., ... Part (a) was first proved by J. P. Kahane. It can be rephrased: if fe Gy(x, {A, ...
GENERALIZATIONS OF THE RIEMANN DERIVATIVE^) BY
J. MARSHALL ASH
Introduction. In §1 of this paper a derivative generalizing the Riemann derivative is considered. The existence of this derivative on a set is shown to imply the existence of the Peano derivative almost everywhere on the set. In §2 the W norm (1 ^p 0.
We say / is Peano bounded of order k at x, i.e., fe TAx), if there are constants
foix),. . .,fk-Ax) such that /(*+')
-/o(*)+/i(*)*+
Let A={a0, ax,...,
• • • +|^1§
ak+l; A0,...,
f-' + Oit«)
ast^O.
Ak+¡} be a set of real numbers with a^üj
if ijtj satisfying
2^
= 0,
; = 0, 1,..., k-1,
= k\,
j = k.
We say that / has a kth generalized derivative with respect to A at the point x, i.e.,fegAx, A), if there is a constant/(W(x)=/(fc)(x, A) such that »-ri
2
2 AJix + att) = fk)ix)tk + oitk)
as t -+ 0.
i=0
Received by the editors May 23, 1966. (1) The author wishes to express his gratitude to Professor Antoni Zygmund who suggested the problem. Professor Mary Weiss helped complete some of the results. The research was supported by the National Science Foundation.
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182
J. M. ASH
A function / is generalized-bounded
[February
of order k with respect to A at x, i.e.,
fe Gk{x,A), if k+l
I
2 A,f{x+att) = 0(tk)
as i->0.
¡= 0
To demonstrate the reason for the conditions on ¿l,i¿¿ ^fil, let/e fc+ i
r k
1= 0
Lj = 0
tk(x). Then
fAx) 2 Ajix+aj)=2A< ZJJW-M+o(ñ i=0
J-
+ o(tk)
j = o J-
Li= o
= fk(x)tk + o{tk)
asi-^0.
In other words, the conditions assure that if the Peano derivative exists, the generalized derivative will exist and be equal to it. If /=0 and if the a,'s are given, since the k + 1 ,4¡'s must satisfy the k +1 conditions, and since the matrix ((af)) is a Van der Monde matrix and hence invertible, it follows that the A¡'s can be expressed in terms of the a¡'s. To be precise (see Denjoy [1])
At= [ni*-«*)]-1-*!
i = 0,l,...,k.
If, on the other hand, />0, the a¡s and the k+l conditions do not uniquely determine the Ay's. I will be called the excess. Probably the most important example of the generalized derivative is the Riemann derivative. The kth Riemann derivative is obtained by setting
ay = -2+/,
i = 0, 1,..., k.
Since /=0, we find that
Mn[(-H-(-N])~'-C) 0 for all xeE.
i=0
Then there is an integer s = 0 such that D sf is {k+s) Riemann-bounded at almost every xeE.
Proof. First let us suppose that all the a,'s are integers and that /=0. We may suppose that a02 and ^]{¿oAi¡{ai—2)=0, then after sliding to the left by 2, we have a (k + l)st derivative based on the original k+l+1 af's so that the excess is immediately /—1.) Set r=k+l+l. Recalling that we may suppose ay-y = i, lHi¿¡r, we may now write our assumption
(1.1)
2 A*- ¿i*+") = W
as í -> 0,
i=l
for all xeE. (2) If 2kí¿ (Ata{)is equal to 0 when j < k, is equal to &! when j=k, then 2?=)* íÍJ>OI*ígjriKrtf*) •
Hence from the hypothesis of Theorem 2 we may deduce that
(2.1)
If
J AJix+aiOdt^ Oihk) at/i^O
" Jo (=o
for every x e E. If some a,=0, say a0=0, then i
k+l
^— l i2= 0 Ai[f{x+2ait)-f{x+ait)] z still is a kth generalized derivative, since
-±-kfAia{{2>-l) = 0, L
jdt¿2 f
«}}.
\Kt)\'dt,
since
Aix,h)S Ü Aix,h) ¡=i where At{x, h)={se[0,h-x]\x+aiseF}={te[x,h]\xil-ai)+aiteF} equality coming from the substitution s = x+t). Suppose that we can prove:
(the last
(2.7) There is an x e [-A/2, 0] n E such that
f
Ja¡(x,h)
\¡it)\'dtÚ± F \lit)\»dt, ¿n jo
i=l,2,...,n.
Then from (2.2), (2.5), (2.6), and (2.7) we will have
I f \l{t)\pdt^ Mih-x)a = M{2Kf= 2aMha which when combined with (2.4), yields
f \lit)\pdtS 2 Jof \K0\pdt2a+2Mha
J -ft
which in turn implies (2.3), as required. Only (2.7) remains to be proved. Set M=maxieN {l/2 + (3/2)|a¡|}. Define e = eQi) by J çMh
E= r
« J -Mb.
Xf(0 ^
where xf is the characteristic function of 7*.Note that 0 is a point of rarefaction of F, so e can be made arbitrarily small by choosing h sufficiently small. Set
Kx) = •Mi(x.ft) f |/(0|*dt = Jxf |/(0|pxrix[l-aj +aj) dt.
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194
J. M. ASH
[February
Then
f
J-h/2
It(x)dxzi¡0 if {¡(Ol'xpixV-ayi + ayOdAdx J-ft/2 LJ-h J i
rMh
rn
= TT3T\\
|1—c7¡| J -Mh J-h
\l{u)\pXF{v)dudv,
setting u = t, v = ayt+ (l -at)x and noting \(d(t, x)¡d(u, v))\ =(1/| 1 -a¡|).
Hence
L«*>*sir^LC^H[J>'H using Fubini's theorem and its converse freely since all functions are positive and
integrable. By (2.4) and the definition of e,
f IMdxi-J-rehlÇ \I{t)\>dt. l1—flf| JO
J-h/2
By Tchebycheff's inequality, if
By= [xe [-J o] n E\It(x)> ¿ £ |/(0|"*}, then
^•¿r^^^iï^ir'w^ i.e.,
4neh
l*.|= \T=a7\ Picking h so small that min |1—a¡| e
0, leLP for all xeE /•ft I "
P
2 AJix+aiO dt = OQi") as h -+ +0
Jo 1,^0
where a> 1, /»^ 1, then
\l'x+01" dt = OQf)
as h-^ +0for almost every xeE.
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Pick
196
J. M. ASH
[February
Proof. Because of Lemma 6, it suffices to show
|/(x+0|p dt = 0(ha)
ash->+0
for almost every x e E. Assume that a0 > 0. As in the proof of Lemma 5, we may then assume A0 = a0= 1. (If a0 < 0, the hypothesis can be reduced to
f° I\l(x+t)+y «ÔAyl(x+ att)
J-h\
¡tl
dt = 0(ha)
as h -+ +0 for all xeE.
The proof then proceeds to the conclusion that i
J -h
|/(x+0|p
dt = 0(h")
as h^- +0 for almost every x e E.
Now apply the second part of Lemma 6 to produce the desired conclusion.) As in Lemma 5, without loss of generality we may assume
¡¡(x+^+XAdix
Jo |
(tl
+ ait) dt ú Mha,
0 < h < 8 for all x e E.
By discarding a subset of measure zero we may further assume that every point of £ is a point of differentiability for /£ |/(0|p dt, so that for all xeE,
Z>(Jj/(0|pI(0,h)¡2. In this case the proof is very much like that of Lemma 5. Pick x e [0, A/2] n E. As in Lemma 5,
r(H=Lmpdt- ¡lmvdt (2.8)
g f X\l(x+t)+yAtl(x + ayt)Pdt+y f JO
y= y
(tl Ja,(x,M
|/(0r"*
where Ay(x, h)={te [x, h] \ [1 -ay]x+ayt $ £}. As in Lemma 5, if h is sufficiently small, an x e [0, h¡2] n £ may be found such that for all / = 1, 2,..., n
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1967]
GENERALIZATIONS OF THE RIEMANN DERIVATIVE
197
For this x, (2.8) implies
7(0,h) < 2l(j, h\ í 4 f * \lix+t)+ 2 AJix+aA)]"dt g 4Af(A-x)aá 4Mb." if h is sufficientlysmall. Case II.
but
iMh-W^
»9)
i = 0,1,..., fc—1,
,(^4) >>,(„,*).
In this case we have
7(0, A) = /(fj, 2¿t) +7^,
A)
for all /,
so
'KM«-^H'(
