This paper is closely related to a recent study of intersections of typical con- tinuous functions with certain classes of functions. The term "typical continuous.
proceedings of the american mathematical Volume
103, Number
2, June
society 1988
CONTINUOUS FUNCTIONS WITH EVERYWHERE INFINITE VARIATION WITH RESPECT TO SEQUENCES Z. BUCZOLICH (Communicated
ABSTRACT.
by R. Daniel Mauldin)
We prove that if {onl^Lj
is such that an \
lim (a„+i/on)
n —oo
then for the typical
continuous
function
0 and
= 1,
/ we have
oo
Sn0 :=
Yl \f(xn + l) - f{xn)\ = +00 n = nn
whenever x S [0,1 — ano] and x„ € [x + a„+i, x + an}. Based on our result in a previous paper, we know that
fails to hold if an+i/an
the above theorem
= A < 1.
We also prove that if {an}^°=1 is such that an \ 0, then for the typical continuous function / we have Sno = +oo if xn = x + an and x € [0,1 — ano].
This paper is closely related to a recent study of intersections of typical continuous functions with certain classes of functions. The term "typical continuous function" means that the set of all functions which have the property under consideration is a residual subset of the complete metric space G[0,1]. In [HL], P. Humke and M. Laczkovich proved that every continuous function agrees with an absolutely continuous function on a set which is not bilaterally strongly x1+6 porous (Theorem 3). In fact, the authors proved that for any / G G[0,1] and ¿ > 0 there is a point x G [0,1] and a sequence xn —* x such that x.n G [x + ((n + I)!)"1"*,
x + (n!)"1-*]
and \f(xn)
- /(a;)|
< n"1"*/2
Let g G G[0,1] be linear on the intervals [0,i], [i„+i,in]
for n > n0.
(n = 1,2,...
), [xi, 1),
and agree with / at the points x„. Since J2™=no \f(xn+i) - f(xn)\ < +oo, g is absolutely continuous and / = g on a set which is not bilaterally strongly xl+6
porous. In [BU] we proved the following related result. For any Darboux function /: [0,1] —♦R and any 0 < ¿ < 1, there is a point x G [0,1 — ¿] and a sequence xn such that xn G [x + ¿"+1,2: + ¿"] (n = 1,2,...) and £~=1 |/(a:n+1)-/(in)| < +oo. This theorem with a function g defined similarly to the foregoing definition yields
the following corollary: For {x: f(x) The we can
every f G G[0,1] there is an absolutely continuous function g such that = g(x)} is not bilaterally strongly porous. last exponential type restriction on xn seems to be the strongest for which obtain a theorem of the above type. In this paper we prove the following
theorem. Received by the editors March 10, 1987. 1980 Mathematics Subject Classification (1985 Revision). Primary
26A15.
©1988
American
0002-9939/88
497
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Z. BUCZOLICH
498
THEOREM 1. Let {an}^L,
be a sequence such that an \ lim (an+i/an)
n—+oo
0 and
= 1;
i/ien i/iere exists f G G[0,1] suc/i that, for all s G N, 2: G [0,1 —as] and for all sequences
{xn}, xn G [x + a„+1,2: + an], we have 00
X! l/fan +l) -/(^n)l = +00. n=s
Furthermore, the set of the functions f having the above property is dense G¿; that is, the typical continuous function has this property. If we choose to let xn = x + an, then the restriction
that lim„_oc(o„+i/a„)
= 1
may be deleted as follows: THEOREM 2. Let {an}'%L1 be a sequence with an \
0; then for any s G N,
x G [0,1 - as], we have J2^LS \f(x + o.n+i) ~ f(x + an)\ = +00 for the typical continuous
function
f G G[0,1].
LEMMA l. Suppose that we have a sequence (an}^=1 such that an \ 0 and limn_00(an-)-i/on) = 1. Furthermore, let K € N and Mx G N. Then there exists
g G G(R) and M2 G N such that
(1) supieR \g{x)\ = 1/K, (2) for all x E R and {zn}£L,,
xn G [x + an+i,x
+ a„], we have
M2-i
X] Ifffan+i)-0(zn)| > 1, n=Mi
(3) ¡7 ¿s periodic and piecewise linear.
PROOF. First we choose an no > Mi such that
(4)
1 —üj+i/oi < 1/(16 ■Ä") for i > no,
and let (5)
d:=ano/8K.
We define the function g G G(R) by
'0 g(x) := I 1/K
ifxG[4k-d,(4k + l)-d],kGZ, if x G [(4k + 2) ■d, (4k + 3) ■d], k G Z and g is linear on the intervals contiguous to the above intervals.
Obviously, (1) and (3) are fulfilled. Let x G R be given and x G [T d, (I + 1) • d\. We claim that for any k G {(I + 1),..., (/ + 8 • K - 2)} there is an ik such that (6)
[2:+ aik + i,x + alk] C [k ■d, (k + l)d\.
Indeed, if x + a3 G [(I + l)d, (I + 8K —l)d], then aj < ano and hence aj-i Consequently, by (4) and (5),
x + aj-i - (x + aj) = a3-i - a3 < Jg-^-!
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< jf¡^
= ^
< a„0.
CONTINUOUS
FUNCTIONS
499
and hence plainly, for any k G {(l + l),..., (1+ 8K —2)}, there is an ik fulfilling (6). It follows that for Xik G [x + aik + i, x + a¿J C [k-d,(k+ l)d] we have g(xik) = 0 if 4\k and g(xlk) — 1/K if 2|fc and 4 f k. We choose M2 such that üm2 < d and we have AÍ2-1
53 \o(xn+i)- g(xn)\ n=M\
[(/ + 8/C-2)/2] ï
J2
£
g(xn+i)- g(xn)
fc'=[(J+l)/2]+l [(i+8K-2)/2]
53
lí(* 21- JV|_! - (Vî-i + 3),
and applying Lemma 1 with K := Ki and Mx := Ni-i we get gi and Ni := M2. Let /, := /,_! + (Vi-i + 3) ■gi\[o,\\- By (7) and (1) we have (8)
sup |/f(x)-/t-i(x)|
k for any x € [0,1 - as] n=s
and {xn}™=i,
xnG[x
+ an+i,x
+ an\\
are dense open sets in G[0,1]. For any / G C[0,1], the function fm(x)
:=min
ln = s
J
is continuous on [0,1 — a3]. Plainly fm(x) < fm+i(x) for rn > s, and for f G GI, x G [0,1 — as] we have limm_oo fm(x) > k and by the compactness of [0,1 — as] there exists an ?r¿o and Ç > 0 such that fmo(x) > k + ç > k for any x G [0,1 — as]. Thus, taking any g such that \\g — /||max < f/2mo, we have mo
mo
mo
53 \f(xn+l)- /(*«)!- 53 \s(Xn +l) - g(Xn)\< 53 2' 0^~ ^ *i n=5
n=3
n=s
that is, gmo(x) > /mJi) - ç > k and hence jgG|. Let h G G[0,1] and e > 0 be given. By the density of G^O, 1] in G[0,1] there is g G G1 [0,1] such that
Hff- /»Umax< e/2 (where CM0.1] := {/ G G[0,1]: /' G G[0,1]}). We have, by the first part of this proof, an / G G[0,1] fulfilling the property of Theorem 1. Obviously, for any a G R, a ■f also fulfills the property of Theorem 1, and taking a0 = £/2||/||max
we have
For any {xn} such that OO
||a0 ■/||max
< e/2.
xn G [x + an+i,x °°
rXn
53 \°(Xn+l)-g(Xn)\ = 53 / n=l
+ an], we have r\
q'(x)dx
Qj+i
Ni+i + (i + l)/k -
1
> .
bi ' Nt+i + 1 -
and Cj+l _ Q¿+ i
Ci
Ci
1
Ni+i+i/k
O-i+l
Ci
1
Nl+i
~
that is, [o¿+i,c,+ i] C [o¿,c¿]. For every d G (o¿+i,c,+i) we obtain also [a¿+i/d] = Nt+i and i/k < {a¿+i/c/} < (i + l)/k. In this way we get the intervals • ■• D \bk,Ck] and, plainly, every d G (bk,Ck) satisfies (10) and (11). LEMMA 1'.
Suppose
that we have a sequence {an} such that an \
[öi,ci]
2
0, K G N
and Mi G N. Then there exists g G G(R) and M2 G N such that (13) suplen \g{x)\ = 1/K, (14) Y.u1m\ \g(x + an+i) - g(x + an)\ > 1, and (15) g is periodic and piecewise linear.
PROOF. We put k := K + 5. Let {an,}k=1 be a subset of {a„}£L,
such that
rtk > Mi, an¡ < < ank, and o„, > 3 • k ■a„,_, for i = 2,..., k. From Lemma 2 we get d G [ani/2,an¡] which fulfills (10) and (11). We define the continuous
function g by
0 g(x) := { 1/K
if x G [21■d, (21+ l)d - d/k], l G Z, if x G [(21+ l)d, (21+ 2) ■d - d/k], l G Z and linear on the intervals contiguous
to the above intervals.
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Z. BUCZOLICH
502
By definition, g fulfills (13) and (15). We put M2 := ni. Let x G R be given. By (11) in Lemma 2 there are at most two subsequent
(mod k) indices j,j
+ 1 such
that {x + anJd} > (k - l)/k (for t = j or j + 1); that is, ¡7(2;+ ani) — 0 or 1/K for z € {1,2,...,
k}\{j,j
+ 1}. Furthermore,
x + a„H
also by (11) we have
x + an,
Ld for i G {1,2,...,
k}\{j - l,j,j
+ 1}. That is, by (10) and the definition of g, we
have \g(x + a„j+1 ) - g(x + a„,)| = 1/Ä" for the above ¿'s. Thus fc-1
A/2-1
(?(x+ a„+i) - 0(2:+ on)| > 53
£
¿=1
n = M\
n, —1
53 9(a:+ a„+i) - 3(2:+ a„ n=n¿+i
fc-4 = 53i?(a: + an-) -?(a;+a™.+i)i 2. -^- > 1 ¿=1
and this proves Lemma 1'. PROOF OF THEOREM 2. Using Lemma l' instead of Lemma 1 we can repeat
the proof of Theorem 1.
References [BU]
Z. Buczolich,
For every continuous
f there is an absolutely continuous
g such that [/ = 3] is not
bilaterally strongly porous, Proc. Amer. Math. Soc. 100 (1987), 485-488. [HL] P. Humke
and M. Laczkovich,
Typical continuous
functions
are virtually nonmonotone,
Proc.
Amer. Math. Soc. 94 (1985), 244-248.
Department of Analysis, H-1088, Hungary
Eötvös
University,
Budapest,
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Muzeum krt.
6-8,