aa r X i v : . [ m a t h . C A ] S e p Journal ofC lassical& Analysis
A PROBLEM CONCERNING RIEMANN SUMS I OSIF P INELIS
Submitted to J. Classical. Anal.Abstract.
An open problem concerning Riemann sums, posed by O. Furdui, is considered.
Let f : [ , ] → R be a continuous function. For natural n , let x n : = n − ∑ k = f (cid:16) kn (cid:17) and y n : = x n + − x n . (1)One may note here that x n / n is a Riemann sum approximating the integral R f ( x ) dx .Part (a) of Problem 1.32 in the book [3] is to find lim n → ∞ y n if the function f iscontinuously differentiable. It is not hard to do a bit more:P ROPOSITION Whenever f is absolutely continuous, one has lim n → ∞ y n = Z f ( x ) dx . (2)On the other hand, it is even easier to show this:P ROPOSITION Whenever lim n → ∞ y n exists, equality (2) holds. Propositions 1 and 2 will be proved at the end of this note.Part (b) of Problem 1.32 in [3] is the following question, which has so far remainedapparently unanswered:
What is the limit [in (2)] when f is only continuous?
By Proposition 2, this limit, if it exists, may only be R f ( x ) dx . However, we haveT HEOREM There are continuous functions f : [ , ] → R for which lim n → ∞ y n does not exist. This is a pure existence theorem, and its proof, given below, is non-constructive.So, the problem of explicitly constructing a continuous function for which lim n → ∞ y n does not exist remains open. Mathematics subject classification (2010): 26A06, 26A27, 26A42, 26A46, 60Exx, 60G15.
Keywords and phrases : Riemann sums; probabilistic method; Wiener process; Brownian motion; in-tegrals; sequences; limits; continuity; absolute continuity; fractional part integrals. roof of Theorem . It may come as a surprise that this proof uses a probabilisticmethod. Let f be the random function W that is a standard Wiener process (Brownianmotion) over the interval [ , ] . Then the probability that f is continuous everywhereon [ , ] is 1 ; see e.g. [4]. In fact, without loss of generality we may assume that allrealizations of the random function f = W are everywhere continuous.Informally, the idea of this proof of Theorem 3 is that, while all realizations of W are everywhere continuous, they are rather non-smooth, not only in the sense of beingnowhere differentiable, but also not being H¨older continuous with any exponent ≥ / EMMA For f = W , the distribution of the random variable y s − y s con-verges to the centered normal distribution with variance / . (cid:0) The convergence here and in the rest of the proof of Theorem 3 is as N ∋ s → ∞ . (cid:1) In-deed, if Theorem 3 were false, we would have y s − y s → Proof of Lemma . Since y s − y s is a centered normal random variable, it sufficesto show that E ( y s − y s ) −→ / . (3)The proof of (3) consists in direct calculations, which are somewhat involved, though,as we have to deal carefully enough with the discreteness in the definition of x n . Incarrying out this task, the choice of indices, 4 s and 2 s , in the statement of Lemma 4turns out to be sufficiently convenient.The just mentioned calculations are based on the formula E W ( u ) W ( v ) = u ∧ v for all u , v in [ , ] . By (1), for f = W , E x n = n − ∑ j , k = E W (cid:16) jn (cid:17) W (cid:16) kn (cid:17) = n − ∑ j , k = (cid:16) jn ∧ kn (cid:17) = n − n + . (4)Somewhat similarly, E x n x n + = n − ∑ j = n ∑ k = (cid:16) jn ∧ kn + (cid:17) = n − ∑ j = n ∑ k = j + jn + n − ∑ j = j ∑ k = kn + = n − n − . (5)It follows from (1), (4), and (5) that E y n = E x n + + E x n − E x n x n + = / . (6)2ow take any natural s . Similarly to (5), we have E x s x s = s − ∑ j = s − ∑ k = (cid:16) j s ∧ k s (cid:17) = s − ∑ k = k ∑ j = j s + s − ∑ k = s − ∑ j = k + k s = s − s + , E x s + x s + = s ∑ j = s ∑ k = (cid:16) j s + ∧ k s + (cid:17) = s ∑ k = k − ∑ j = j s + + s ∑ k = s − ∑ j = k k s + = s + s s + , E x s + x s = s ∑ j = s − ∑ k = (cid:16) j s + ∧ k s (cid:17) = s − ∑ k = k ∑ j = j s + + s − ∑ k = s ∑ j = k + k s = s − s − s − s + , E x s x s + = s − ∑ j = s ∑ k = (cid:16) j s ∧ k s + (cid:17) = s ∑ k = k − ∑ j = j s + s ∑ k = s − ∑ j = k k s + + s ∑ k = s + k − ∑ j = j s + s ∑ k = s + s − ∑ j = k − k s + = s + s − s − s + . So, E y s y s = E x s x s + E x s + x s + − E x s + x s − E x s x s + = s + s + s + s + −→ . Thus, in view of (6), E ( y s − y s ) = E y s + E y s − E y s y s −→ + − × = , so that (3) is verified, which completes the proof of Lemma 4.The proof of Theorem 3 is now complete as well.To conclude this note, it remains to prove Propositions 1 and 2. Proof of Proposition . Since f is absolutely continuous, there is a function g ∈ L [ , ] such that f ( x ) = f ( ) + Z x g ( u ) du = f ( ) + Z g ( u ) I { u < x } du (7)3or all x ∈ [ , ] , where I {·} denotes the indicator. So, by (1), x n = ( n − ) f ( ) + Z g ( u ) n − ∑ k = I n u < kn o du and hence y n = f ( ) + I n ( g ) − J n ( g ) , (8)where I n ( g ) : = Z g ( u ) I n u < nn + o du , J n ( g ) : = Z g ( u ) h n ( u ) du , and h n ( u ) : = n − ∑ k = I n kn + ≤ u < kn o . (9)Clearly, I n ( g ) = f (cid:16) nn + (cid:17) − f ( ) −→ f ( ) − f ( ) . (10)Here and in the rest of this proof, the convergence is as n → ∞ .To deal with J n ( g ) , take any real ε > g ∈ L [ , ] , by [1, Corollary 4.2.2], R | g ( u ) − ˜ g ( u ) | du ≤ ε for some continuous function ˜ g : [ , ] → R . Note also that0 ≤ h n ≤ [ kn + , kn ) ⊂ [ k − n , kn ) for k = , . . . , n − | J n ( g ) − J n ( ˜ g ) | ≤ Z | g ( u ) − ˜ g ( u ) | du ≤ ε . (11)Introduce now the function ˜ g n by the formula˜ g n ( u ) : = n − ∑ k = ˜ g (cid:16) kn (cid:17) I n kn + ≤ u < kn o for u ∈ [ , ] . Since the function ˜ g is continuous, it is uniformly continuous on [ , ] ,so that, in view of (9), k ˜ gh n − ˜ g n h n k ∞ = k ˜ gh n − ˜ g n k ∞ → | J n ( ˜ g ) − J n ( ˜ g n ) | −→ . (12)On the other hand, using the continuity of ˜ g and integration by parts, we have J n ( ˜ g n ) = n − ∑ k = ˜ g (cid:16) kn (cid:17) kn n nn + −→ Z ˜ g ( u ) u du = ˜ f ( ) − Z ˜ f ( u ) du , (13)where ˜ f ( x ) : = f ( ) + R x ˜ g ( u ) du for x ∈ [ , ] . By (7) and the second inequality in (11),we have | ˜ f − f | ≤ ε and hence | R ˜ f ( u ) du − R f ( u ) du | ≤ ε . Collecting now (8), (10),(11), (12), and (13), we see thatlim sup n → ∞ (cid:12)(cid:12)(cid:12) y n − Z f ( u ) du (cid:12)(cid:12)(cid:12) ≤ ε , for any real ε > roof of Proposition . The Stolz–Ces`aro theorem ([5, pages 173–175] and[2, page 54]) states the following: if ( a n ) and ( b n ) are sequences of real numberssuch that b n is strictly increasing to ∞ and a n + − a n b n + − b n −→ ℓ ∈ R , then a n b n −→ ℓ . NowProposition 2 follows immediately by applying the Stolz–Ces`aro theorem with a n = x n and b n = n , since x n n −→ R f ( x ) dx . R E F E R E N C E S[1] V. I. Bogachev.
Measure theory. Vol. I, II . Springer-Verlag, Berlin, 2007.[2] E. Ces`aro. Sur la convergence des s´eries.
Nouvelles annales de math´ematiques , 7:49–59, 1888.[3] O. Furdui.
Limits, series, and fractional part integrals . Problem Books in Mathematics. Springer, NewYork, 2013. Problems in mathematical analysis.[4] P. M¨orters and Y. Peres.
Brownian motion , volume 30 of
Cambridge Series in Statistical and Probabilis-tic Mathematics . Cambridge University Press, Cambridge, 2010. With an appendix by Oded Schrammand Wendelin Werner.[5] O. Stolz.
Vorlesungen ¨uber allgemeine Arithmetik: nach den Neueren Ansichten . Teubners, 1885.
Iosif Pinelis, Department of Mathematical SciencesMichigan Technological UniversityHoughton, Michigan 49931, USAe-mail: [email protected]