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How do I prove that $$\lim_{n\to\infty}n\left[\int_{0}^{1}f(x){d}x - \frac{1}{n}\sum_ {i=0}^{n}f\left(\frac{i}{n}\right)\right] = -\frac{1}{2}$$ for any $ f(x)$ in $C^0([0,1])$ such that $f(0)=0$ and $f(1)=1$ ?

Suggested steps:

  1. Set $f(x)=x+g(x)$;
  2. Prove that for any $g\in C^0([0,1])$ such that $g(0)=g(1)=0$, the above limit is zero$\ldots$
  3. $\ldots$ by approximating $g$, uniformly over $[0,1]$, with a trigonometric polynomial of the form $$ \tilde{g}(x)=\sum_{m=1}^{M} c_m \sin(\pi m x) $$
  4. For any function like $\sin(\pi m x)$, both the integral and the sum are simple to compute.
Jack D'Aurizio
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Lily
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    Interesting question, but many users here tend to downvote so-called PSQs (Problem Statement Questions) like yours, and answers, too (some god will forgive them, but the poor human Jack won't). So it is in your (and community's) best interest to improve your actual question by adding some context (your attempts, why this question is relevant to you, something along these lines). Cheers. – Jack D'Aurizio Oct 27 '16 at 03:49
  • Here's a duplicate (for the $C^1$ case). – metamorphy Aug 25 '20 at 06:16

1 Answers1

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For the more restricted case, $f \in C^1([0,1])$ this can be shown as follows.

Consider

$$x_n = n\left(\int_0^1f(x) \, dx - \frac{1}{n}\sum_{i=1}^nf(i/n)\right).$$

Applying the mean value theorem,

$$\begin{align}x_n &= n\sum_{i=1}^n\int_{(i-1)/n}^{i/n}[f(x) - f(i/n)] \, dx \\ &= n\sum_{i=1}^n\int_{(i-1)/n}^{i/n}f'(\xi_{i,n})(x - i/n) \, dx \end{align}.$$

Define

$$M_{i,n} = \sup \{f'(x): (i-1)/n \leqslant x \leqslant i/n \}, \\ m_{i,n} = \sup \{f'(x): (i-1)/n \leqslant x \leqslant i/n \}. $$

Hence,

$$n \sum_{i=1}^nm_{i,n}\int_{(i-1)/n}^{i/n}(x - i/n) \, dx \leqslant x_n \leqslant n \sum_{i=1}^nM_{i,n}\int_{(i-1)/n}^{i/n}(x - i/n) \, dx. $$

Note that

$$\int_{(i-1)/n}^{i/n}(x - i/n) \, dx = -\frac{1}{2}\left(\frac{i}{n} - \frac{i-1}{n} \right)^2 = - \frac{1}{2n^2},$$

and

$$-\frac{1}{2n}\sum_{i=1}^nm_{i,n} \leqslant x_n \leqslant -\frac{1}{2n}\sum_{i=1}^nM_{i,n}. $$

Now take the limit as $n \to \infty$, noticing the upper and lower Riemann sums above, to obtain

$$\lim_{n \to \infty} x_n = - \frac{1}{2} \int_0^1f'(x) \, dx = \frac{f(0)-f(1)}{2}= -\frac{1}{2}.$$

This link suggests it is not generally true for $f \in C^0([0,1]):$

Speed of convergence of Riemann sums

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