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Consider a continuous function $f\colon [0,1]\to\mathbb{R}$, and the corresponding Riemann sum: $$R_n(f)=\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)$$ I am interested in the rate of convergence of the sequence $(\Delta_n(f))_{n\ge1}$ given by $$\Delta_n(f)\stackrel{\rm def}{=} R_n(f)-\int_0^1f(x)dx$$ What we know is that this sequence converges to zero, and it is $O(1/n)$ if $f$ is of bounded variation. What I am looking for are bad functions $f$, for example such that $ n^\epsilon |\Delta_n(f)|$ does not converge to $0$ for some $0<\epsilon<1$. Do you know of such creatures?

Clement C.
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Felix Klein
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  • With $\displaystyle\int_0^{1/2} \frac{dx}{x \log^2 x}$ you should get $|\Delta_n| \sim \frac{1}{\log n}$ – reuns Aug 15 '17 at 20:24
  • @reuns, do you think this is continuous at 0? – Felix Klein Aug 15 '17 at 20:36
  • A first naive idea would be to look at the usual suspects -- simple enough continuous functions which are not of bounded variation, such as $x\mapsto x\sin\frac{1}{x}$. What does it give for that one? – Clement C. Aug 15 '17 at 20:44
  • Relevant: https://math.stackexchange.com/questions/569750/speed-of-convergence-of-riemann-sums – Clement C. Aug 15 '17 at 21:09
  • @ClementC. what I want are exactly functions that do not satisfy the expansion in the question you mentioned, as for $x\mapsto x\sin(1/x)$ it seems that the calculations are not easy to be done, maybe I will try some numerical calculations. – Felix Klein Aug 15 '17 at 21:15
  • @FelixKlein Yes -- at least, the answer shows that such functions do exist. – Clement C. Aug 15 '17 at 21:19

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Suppose we consider a sum $f(x) = \sum_{j=1}^\infty f_j(x)$ to be defined inductively. I'll want to choose these so that for some increasing sequence $n_k$, $|\Delta_{n_k}(f)| > n_k^{-\epsilon}$. In fact, I'll take $$\left|\Delta_{n_k}\left(\sum_{j=1}^{k}f_j \right)\right| > 2 n_k^{-\epsilon}$$ and require $$ \sup_{x \in [0,1]}|f_j(x)| \le 2^{-j} n_k^{-\epsilon} \ \text{for}\ j > k $$ Given $f_1, \ldots, f_{k-1}$ and $n_1, \ldots, n_{k-1}$, let $B_k$ be the bound on $\sup_{x \in [0,1]} |f_k(x)|$ arising from these. Take $n_k$ large enough that $B_k > 2 n_k^{-\epsilon}$. Let $g$ be a continuous function whose graph consists of narrow triangles of height $B_k$ centred at the multiples of $1/n_k$ (so that $R_{n_k}(g) = B_k$), narrow enough that $\int_0^1 g \; dx < B_k - 2 n_k^{-\epsilon}$. Thus $\Delta_{n_k}(g) > 2 n_k^{-\epsilon}$. Take $f_k = \pm g$, the sign chosen to be the same as that of $\Delta_{n_k}\left(\sum_{j=1}^{k-1} f_j\right)$, so that $\Delta_{n_k}\left(\sum_{j=1}^k f_j \right) \ge 2 n_k^{-\epsilon}$.

Robert Israel
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  • So your function is $\sum_{k=0}^\infty B_k \text{ tri}(n_k x)$ with $\text{tri}(x) = |\lfloor x \rfloor - x -1/2|$ the periodic triangle ? And the series converges uniformly ? – reuns Aug 15 '17 at 21:56
  • No: narrow enough that ... So $g(x) = B_k \max{0, 1 - r_k |x - j/n_k| ; : j = 0 \ldots n_k }$ for sufficiently large $r_k$. And yes, it does converge uniformly. – Robert Israel Aug 16 '17 at 08:27
  • @RobertIsrael, This beautiful, thank you. – Felix Klein Aug 16 '17 at 15:07