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.


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*}


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.