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This question is inspired by a previous question. It was shown that, for all function $f \in \mathcal{C} ([0, 1])$,

$$ \lim_{n \to + \infty} \sum_{k=0}^{n} f \left( \frac{k}{n+1} \right) - \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = \int_0^1 f (x) \ dx.$$

A stronger statement would be that there exists some constant $a(f)$ such that:

$$\sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = n \int_0^1 f (x) \ dx + a(f) + o(1),$$

or, in other words, that there is an asymptotic development at order $1$ of the Riemann sums:

$$\frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = \int_0^1 f (x) \ dx + \frac{a(f)}{n} + o(n^{-1}).$$

Given $f$, can we always find such a constant $a(f)$? If this is false, can we find a counter-example? If this is true, can $a(f)$ be written explicitely?

I have had a quick look at the litterature, but most asymptotics for the Riemann sums involve different meshes, which depend on the function $f$.

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D. Thomine
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  • Try it for $f(x) = x^2$. The error term is then easy to get. (In general, always try $f(x) = x^2$:) – marty cohen Nov 16 '13 at 23:52
  • If the function is Lipschitz continuous, this is what you need: https://math.stackexchange.com/questions/1924478/rate-of-convergence-of-riemann-sum-under-lipschitz – Carlos Pinzón Jan 22 '20 at 22:17

1 Answers1

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On the subspace $\mathcal{C}^1([0,1])$ of continuously differentiable functions, we have

$$\lim_{n\to\infty} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - n\int_0^1 f(x)\,dx = \frac{f(0) - f(1)}{2}.$$

We can see that by computing

$$\begin{align} &\Biggl\lvert\frac12\left(f\left(\frac{k}{n}\right) + f\left(\frac{k+1}{n}\right)\right) - n\int_{k/n}^{(k+1)/n} f(x)\,dx\Biggr\rvert\\ &\qquad = \frac{n}{2}\left\lvert \int_{k/n}^{(k+1)/n} \left(f\left(\frac{k+1}{n}\right)-f(x)\right) - \left(f(x) - f\left(\frac{k}{n}\right)\right)\,dx\right\rvert\\ &\qquad = \frac{n}{2}\left\lvert\int_{k/n}^{(k+1)/n}\int_x^{(k+1)/n} f'(t)\,dt - \int_{k/n}^x f'(t)\,dt\,dx\right\rvert\\ &\qquad = \frac{n}{2}\left\lvert\int_{k/n}^{(k+1)/n}\left(t-\frac{k}{n}\right)f'(t) - \left(\frac{k+1}{n}-t\right)f'(t)\,dt\right\rvert\\ &\qquad = n\left\lvert\int_{k/n}^{(k+1)/n} \left(t-\frac{k+\frac12}{n}\right)f'(t)\,dt\right\rvert\\ &\qquad = n\left\lvert\int_{k/n}^{(k+1)/n} \left(t-\frac{k+\frac12}{n}\right)\left(f'(t)- f'\left(\frac{k+\frac12}{n}\right)\right)\,dt\right\rvert\\ &\qquad \leqslant n \cdot\omega_{f'}\left(\frac{1}{2n}\right) \int_{k/n}^{(k+1)/n} \left\lvert t-\frac{k+\frac12}{n}\right\rvert\,dt\\ &\qquad = \frac{1}{4n}\cdot\omega_{f'}\left(\frac{1}{2n}\right), \end{align}$$

where

$$\omega_{f'}(\delta) = \sup \left\lbrace \lvert f'(s) - f'(t)\rvert : s,t\in [0,1], \lvert s-t\rvert \leqslant \delta\right\rbrace$$

is a modulus of continuity of $f'$. Summing up we obtain

$$\left\lvert \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - \frac{f(0)-f(1)}{2} - n\int_0^1 f(x)\,dx\right\rvert \leqslant \frac14 \cdot\omega_{f'}\left(\frac{1}{2n}\right),$$

and the continuity of $f'$ means $\lim\limits_{\delta\searrow 0} \omega_{f'}(\delta) = 0$.

But there is no map $\alpha \colon \mathcal{C}([0,1]) \to \mathbb{C}$ such that for every $f \in \mathcal{C}([0,1])$ we have

$$\lim_{n\to\infty} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - n\int_0^1 f(x)\,dx = \alpha(f),$$

or equivalently

$$\frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) = \int_0^1 f(x)\,dx + \frac{\alpha(f)}{n} + o\left(\frac1n\right).$$

For if there were, since $\mathcal{C}([0,1])$ is a Banach space under the supremum norm, the Banach-Steinhaus theorem (uniform boundedness principle) would assert that the family

$$T_n \colon f \mapsto \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - n\int_0^1 f(x)\,dx$$

is equicontinuous, or norm-bounded.

However, it is easy to see that

$$\lVert T_n\rVert = 2n.$$

Thus the set of $f \in \mathcal{C}([0,1])$ such that

$$\lim_{n\to\infty} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - n\int_0^1 f(x)\,dx$$

exists is meagre (it is, however, strictly larger than $\mathcal{C}^1([0,1])$).

Note that the previous question only considered continuously differentiable functions, i.e. $\mathcal{C}^1([0,1])$. The Banach-Steinhaus theorem shows that

$$\lim_{n\to\infty} \sum_{k=0}^n f\left(\frac{k}{n+1}\right) - \underbrace{\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right)}_{S_n(f)}$$

does not exist for all $f\in\mathcal{C}([0,1])$, since $\lVert S_{n+1} - S_n\rVert = 2n-1$ is not bounded.

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Daniel Fischer
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  • Very nice answer. – JohnD Dec 22 '13 at 14:43
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    Thank you very much, it completely answer my question. I didn't think at all about using functionnal analysis tools à la Banach-Steinhaus; that's very nice. – D. Thomine Dec 22 '13 at 18:14
  • @DanielFischer: I see $||T_n|| \leqslant 2n$. How do you show the norm equals $2n$, i.e for what $f \in C([0,1])$ with $\sup_{[0,1]} |f(x)| = 1$, do we have $|T_n f| = 2n$? Thanks. – RRL May 23 '16 at 22:40
  • @RRL The norm is not attained, we have $\lvert T_n f\rvert < 2n \lVert f\rVert$ for all $f \neq 0$ (and all $n \in \mathbb{N}\setminus {0}$). But we can come arbitrarily close. Fix $n$, pick $0 < \varepsilon < \frac{1}{2n}$ and consider the function $f_{n,\varepsilon}$ with $f_{n,\varepsilon}(k/n) = 1$ for $k \in {0,1,\dotsc,n}$, $f_{n,\varepsilon}(x) = -1$ for $\frac{k}{n} + \varepsilon \leqslant x \leqslant \frac{k+1}{n}-\varepsilon$, $0 \leqslant k < n$, and that linearly interpolates between $k/n$ and $k/n \pm \varepsilon$. – Daniel Fischer May 24 '16 at 08:41
  • Then $\int_0^1 f_{n,\varepsilon}(x),dx = -1 + 2n\varepsilon$, so $T_n f_{n,\varepsilon} = n + n(1-2n\varepsilon) = 2n - 2n^2\varepsilon$. – Daniel Fischer May 24 '16 at 08:41
  • @DanielFischer: Thanks for the clarification. – RRL May 24 '16 at 15:48
  • I know this is an old question, but how do we conclude that the set referenced in the answer is meagre? – Fimpellizzeri Sep 12 '18 at 21:50
  • @Fimpellizieri (Just came here from another question and happened to see your comment by chance, sorry for the late reply.) If the set were non-meagre (of the second category), then the sequence would be pointwise bounded on the whole space, and therefore uniformly bounded (and thus the limit would exist everywhere). The hypothesis in the Banach–Steinhaus theorem need not be that the sequence of operators (it needn't even be a sequence, can be an arbitrary family) is pointwise bounded on the whole space, the conclusion follows if it is pointwise bounded on a set of the second category. – Daniel Fischer Nov 18 '19 at 19:56
  • That is for example the version Rudin has (as Theorem 2.5.) in his Functional Analysis . This version is of course only useful if the domain of the operators is of the second category. Another version (that requires pointwise boundedness everywhere for meagre domains) demands the domain be barrelled (and the codomain locally convex) to deduce the equicontinuity of the family. – Daniel Fischer Nov 18 '19 at 19:56