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Let $f$ be continuous on $\left[0,1\right]$ and positive and assume also that for every $\varepsilon>0$ there exists some $\overline{n}>0$ such that for all $n\geq\overline{n}$ we have $$\left|\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right)-\int_{0}^{1}f\left(u\right)du\right|\leq\frac{\varepsilon}{n}.\tag{1}$$ For a problem I need to prove that $(1)$ implies that $f$ is constant. How is it possible to prove it? I tried to search bounds for the Riemann sums error but they require regularity hypotheses, like $f$ is twice differentiable.

Thank you.

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The statement is false.

Take $f(x) = x$ then we have $$\int_0^1 f(u) du = \frac{1}{2}$$ and \begin{align*}\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right) &= \frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}+\frac{1}{2n}\right) \\\ &= \frac{1}{n}\left(\sum_{k=0}^{n-1}\frac{k}{n}+\sum_{k=0}^{n-1}\frac{1}{2n}\right) \\\\ &= \frac{1}{n}\left(\frac{n-1}{2} + \frac{1}{2}\right) \\\\ &= \frac{1}{n}\cdot\frac{n}{2} = \frac{1}{2}\end{align*}

hence:

$$\left|\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right)-\int_{0}^{1}f\left(u\right)du\right| = \left|\frac{1}{2} - \frac{1}{2}\right| = 0\leq\frac{\varepsilon}{n}.\tag{1}$$ for all $\varepsilon > 0, n\in\mathbb{N}$

Although $f(x) = x$ is not positive on $[0,1]$ consider that a shift upwards doesn't change anything (it's added as a constant to the LHS and to the integral as well so it will cancel out).

So each function $f(x) = x + c$ with $c>0$ contradicts your claim.

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