Inequalities involving integrals

Question

I'm trying to tackle the following question:

Let $\displaystyle f:[0,1]\to \Bbb{R} \ \text{s.t} \ f(0)=0,f(1)=1$ and $f$ is increasing.

1. Show that $\displaystyle \int_{0}^1 f(x)\text{d}x\le \frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right)\le \int_{0}^1f(x)\text{d}x+\frac{1}{n}$.
2. Show that $\displaystyle \int_0^1 f(x)\text{d}x\le \frac{2}{n^2}\sum_{j=1}^n jf\left(\frac{j^2}{n^2}\right)$.

My try:

First of all, $f$ is integrable as increasing function in close interval. Now, $\displaystyle \frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right)$ is the upper Darboux sum of $f$ where the partition $P$ is $\left(0,\frac{1}{n},\frac{2}{n},...,1\right)$ and in each segment $\displaystyle \sup_{x\in[\frac{j-1}{n},\frac{j}{n}]}f(x)=f\left(\frac{j}{n}\right)$ because $f$ is increasing.

Thus, $\displaystyle \int_0^1 f(x)\text{d}x\le \frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right)$.

Second, we know that if $f$ is integrable, then the upper Darboux sum converges to the integral of $f$, hence $\displaystyle \forall \varepsilon>0, \ \exists N\in\Bbb{N}, \ \text{s.t} \ \forall n>N: \ \left|\bar{\Sigma}(f,P)-I \right|<\varepsilon$, or in other words $\displaystyle \bar{\Sigma}(f,P)<I+\varepsilon$. Let $\displaystyle \varepsilon=\frac{1}{n}$ and we get $\displaystyle \frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right)\le \int_0^1 f(x)\text{d}x+\frac{1}{n}$ as needed.

Now, for part $(2)$ I thought to write $\displaystyle \frac{1}{n}\sum_{j=1}^n \frac{2j}{n}f\left(\frac{j^2}{n^2}\right)$. I know that as $n\to \infty$ this sum becomes the Riemann integral $\displaystyle \int_0^1 \left[2xf(x^2)\right]\text{d}x$. I don't know how to continue or even this helps me somehow...

Is my reasoning for part $(1)$ is good? How to continue with part $(2)$?

Please help, thank you!

Answer 1

Your first part of the reasoning for (1) is right, for the second part: This does not work, note that $N$ (and hence the $n$) for which your conclusion holds depends on $\epsilon$. This forbids to use $\epsilon = \frac 1n$ (first the $\epsilon$, then the $N$). But: As $f$ is increasing, we have $$ f(x) \ge f\left(\frac{j-1}n\right), \quad x \in \left[\frac{j-1}n, \frac jn\right] $$ So (lower sum) $$ \frac 1n\sum_{j=1}^n f\left(\frac{j-1}n\right) \le \int_0^1 f(x)\, dx\tag{$*$} $$ Now \begin{align*} \frac 1n \sum_{j=1}^n f\left(\frac{j-1}n\right) &= \frac 1n \sum_{j=0}^{n-1} f\left(\frac{j}n\right)\\ &= \frac 1n \sum_{j=1}^{n-1} f\left(\frac{j}n\right) & \text{$f(0)= 0$}\\ &= \frac 1n \left(\sum_{j=1}^n f\left(\frac{j}n\right)\right) - \frac 1n& \text{$f(1) = 1$} \end{align*} Hence, adding $\frac 1n$ to both sides of $(*)$ gives $$ \frac 1n\sum_{j=1}^n f\left(\frac{j}n\right) \le \int_0^1 f(x)\, dx + \frac 1n $$

(2) For the second part, use your reasoning and recall that $2xf(x^2)$ is also increasing, giving $$ \int_0^1 2xf(x^2) \, dx \le \frac 2{n^2}\sum_{j=1}^n jf\left(\frac{j^2}{n^2}\right) $$ and now invoke integration by substitution to get $\int_0^1 2xf(x^2)\, dx = \int_0^1 f(t)\, dt$ (if $t = x^2$, $dt = 2x\, dx$).