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Volume 79, Number 1, May 1980



All~eT. A construction is given of a rightcontinuous function which satisfies an arithmetic continuity condition, but which is not continuous. The problem is motivated by a familiar criterion for the continuity of sample paths of a stochastic process.

In order to prove the continuity of sample functions of a stochastic process, such as Brownian motion, one sometimes applies a result like the following proposition from the forthcoming book of K. L. Chung.
Let f be a realvalued function defined on (oo, oo) and continuous from the right everywhere. Suppose thatf also has left limits and satisfies
lim max k + 1 ) _ ~A)11 = 0;
n~oo 1 oo thenf continuous in (oo, oo).
Professor Chung recently brought to the author's attention the following question: Can one give an example which shows that the assumption of the existence of left limits cannot be dropped from the preceding proposition? There does not seem to be an example in the literature, and the arithmetical character of (1) rules out the trivial candidates. The main objective of this note is to provide a construction which settles the question and which is sufficiently systematic to prove useful in related problems.
If CO(R) denotes the class of continuous functions with compact support, one can define a new norm on 60(R) by
* = K (p +
1M1 plq q q
The required example will be constructed as an application of a lemma which
establishes a relation between ilf 11* and Jif 1100.

LEmmA. Given any integer M and 0 < a < b < 1, there is a real continuous 4, with support contained in (a, b) for which (1) lixpliw > M and (2) 11011* Q 5.

Supposing this lemma for the moment, we take disjoint intervals I. = (a., b.) with a. .;,v 1 and b. x 1. By normalizing the functions provided by the lemma, we obtain functions,0, such that supp oj C 1j, 114Pjli,, = 1, and 1JS5j11, < I/j2 forj  1, 2 .....

Received by the editors March 6, 1979. AMS (MOS) subject c4wvificatio (1970). P~ 60005, 6OG17; Secondary 20A15, 10B45. Key words andphraws. Continuous path, continuity criterion, quadratic irrational.

0 1980 American mathematical Society

108 J. M. STEELE

On setting f(x) 1.0,(x), one immediately sees f(x) is rightcontinuous but not
continuous; to check (1) we note

(k + 1) k ) 1 < y _ 0
(k + 1) k )l +

n n jJ

The finite sum gives a uniformly continuous function, and the infinite sum can be made as small as we like. This completes the verification of (1) and construction of the desired example. The heart of the matter thus rests in proving the lemma we have just used.
By FN we denote the fractions in [0, 1] which have denominators no larger than N. We then define a step as a transition from one pI q e Q to another element of the form plq llaq c, where a is an integer and c is any rational for which lel < IIN. Next ip is defined at each element of FN n (a, b) as the least number of steps needed to leave the interval (a, b), and %p is defined to be zero at each element of F n (a, by. The definition of %p is then extended to R by linear interpolation.
To see that Ilipil, < 5, we note 145((p + 1)1q)  ip(plq)l < 1 if plq EE FN since
at most one more step is needed to leave (p + 1)1q than was required to leave from plq. Next consider plq (Z FIv and choose rationals rj, r2 EE F,, such that
Irl  pIqI < 1IN and 1r2  (p + I)Iqi < IIN. Since q > N, one also has jr, 
r21 < 31N, and consequently lip(rl)  4,(r2)l < 3. The fact that lip(rl)  ip(plq)i <
1 and lip(r2)  0((p + 1)1q)l < 1 then completes the proof that litpil, < 5.
Now we attend to the more serious matter of showing that N can be chosen so that ilipil,, > M. To this end consider the interval 10 consisting of the middle third of (a, b). Let 1 denote the length of 10, and choose a quadratic irrational a EE I0. The only fact needed from diophantine analysis is that for any quadratic irrational there is a constant c > 0 such that Iplq  al > clq 2 for allplq e Q [1, p. 451.
First take anypOIN such that lpOIN  al < IIN, and let dk denote the largest distance one can go from p.IN in k steps. We then have

dk, l < dk + +

where Q is the smallest integer so that for somep one has 1PIQ  pOIN1 < d.. We have IplQ  a] < dk + 1IN by the choice of po, so by bound on Iplq  al mentioned above we have

1 1 )112C_ 1/2
~i < (dk +

This proves the basic recursion inequality,

1 1/2
dk+l < dk + W + dk + c 1/2.
N )

By choosing N prime we can guarantee that pol N is irreducible and thus d, 4
21N. The basic recursion now immediately shows that N can be taken sufficiently




large to guarantee that d.,, M, so 11011,, > M as claimed.


1. A. Ya. Khintchin, Contimedfractions, Univ. of Chicago PrM, Chicago, Ifi., 1964.

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