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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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1 Answer 1

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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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  • $\begingroup$ Very nice answer. $\endgroup$
    – JohnD
    Dec 22, 2013 at 14:43
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    $\begingroup$ 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. $\endgroup$
    – D. Thomine
    Dec 22, 2013 at 18:14
  • $\begingroup$ @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. $\endgroup$
    – RRL
    May 23, 2016 at 22:40
  • $\begingroup$ @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 $k/n \pm \varepsilon$. $\endgroup$ May 24, 2016 at 8:41
  • $\begingroup$ Then $\int_0^1 f_{n,\varepsilon}(x)\,dx = -1 + 2n\varepsilon$, so $T_n f_{n,\varepsilon} = n + n(1-2n\varepsilon) = 2n - 2n^2\varepsilon$. $\endgroup$ May 24, 2016 at 8:41

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