If $f\in C^1[0,1]$, then $\left|\int_0^1f(x)dx-\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)\right|\leq\frac{\int_0^1|f'(x)|dx}{n}$

Question

I want to show that if $f\colon[0,1]\to\mathbb{R}$ is continuously differentiable, then $$\left|\int_0^1f(x)dx-\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)\right|\leq\frac{\int_0^1|f'(x)|dx}{n}$$

I'm not quite sure where to start - some things I thought about are that the left hand side approaches 0 because Riemann sums approach the integral as $\lambda(\Pi)\to0$, and that it's almost tempting to use the Newton-Leibniz theorem on the right hand side but I can't because of the absolute value.

I would love to get a hint.

Answer 1

There's even no need to use the mean value theorem. For any $t\in\left(0, \tfrac{1}{n}\right)$ we have: \begin{align} \left\vert f\left(\tfrac{k}{n} + t\right) - f\left(\tfrac{k}{n}\right) \right\vert = \left\vert \int_{0}^{t} f'\left(\tfrac{k}{n} + s\right)\ ds \right\vert \leq \int_{0}^{t} \left\vert f'\left(\tfrac{k}{n} + s\right) \right\vert\ ds \leq \int_{0}^{1/n} \left\vert f'\left(\tfrac{k}{n} + s\right) \right\vert\ ds \end{align} Note that the expression on the right does not depend on $t$ anymore. Thus: \begin{align} \int_{0}^{1/n} \left\vert f\left(\tfrac{k}{n} + t\right) - f\left(\tfrac{k}{n}\right) \right\vert\ dt \leq \frac{1}{n} \int_{0}^{1/n} \left\vert f'\left(\tfrac{k}{n} + s\right) \right\vert\ ds \end{align} Now, focusing on the main problem, try to apply the above considerations: \begin{align} \left\vert \int_0^1 f(x)\ dx - \frac{1}{n} \sum_{k=0}^{n-1} f\left(\tfrac{k}{n}\right) \right\vert &= \left\vert \sum_{k=0}^{n-1} \int_{0}^{1/n} f\left(\tfrac{k}{n} + t\right)\ dt - \sum_{k=0}^{n-1} \int_{0}^{1/n} f\left(\tfrac{k}{n}\right)\ dt \right\vert \\&= \left\vert \sum_{k=0}^{n-1} \int_{0}^{1/n} \left( f\left(\tfrac{k}{n} + t\right) - f\left(\tfrac{k}{n}\right)\right)\ dt \right\vert \\&\leq \sum_{k=0}^{n-1} \int_{0}^{1/n} \left\vert f\left(\tfrac{k}{n} + t\right) - f\left(\tfrac{k}{n}\right) \right\vert\ dt \\&\leq \frac{1}{n} \sum_{k=0}^{n-1} \int_{0}^{1/n} \left\vert f'\left(\tfrac{k}{n} + s\right) \right\vert\ ds \\&= \frac{1}{n} \int_{0}^{1} \left\vert f'\left(x\right) \right\vert\ dx \end{align}

Answer 2

I managed to finally solve it thanks to the wonderful hints I was given, so I suppose I'll post a full solution.

We will first note that $$\int_0^1 f(x)dx=\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx$$ $f$ is continuous, and for every interval $\left[\frac{k}{n},\frac{k+1}{n}\right]$ we will use the integral mean value theorem to conclude that there exists a $c_k\in\left(\frac{k}{n},\frac{k+1}{n}\right)$ such that $\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx=\frac{1}{n}f(c_k)$. Therefore, it suffices to show that $$\left|\sum_{k=0}^{n-1}f\left(c_{k}\right)-f\left(\frac{k}{n}\right)\right|\leq\int_{0}^{1}\left|f'\left(x\right)\right|dx$$ Indeed, $$\int_{0}^{1}\left|f'\left(x\right)\right|dx\geq\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{c_{k}}\left|f'\left(x\right)\right|dx\geq\sum_{k=0}^{n-1}\left|\int_{\frac{k}{n}}^{c_{k}}f'\left(x\right)\right|$$ applying the Newton-Leibniz theorem in each interval $\left[\frac{k}{n},c_k\right]$, we have $$\sum_{k=0}^{n-1}\left|\int_{\frac{k}{n}}^{c_{k}}f'\left(x\right)\right|=\sum_{k=0}^{n-1}\left|f\left(c_{k}\right)-f\left(\frac{k}{n}\right)\right|\geq\left|\sum_{k=0}^{n-1}f\left(c_{k}\right)-f\left(\frac{k}{n}\right)\right|$$ $\blacksquare$