$\int_0^\infty f(x) dx \geq \sum_{n=1}^\infty (x_{n+1}-x_n)f(x_{n+1})$ then $\lim_{x\rightarrow \infty} \frac{f(x)}{x}=0$

Question

Suppose $f:[0,\infty)\rightarrow [0,\infty)$ is continuous and decreasing. I have shown that for any increasing sequence $\{x_n\}$ in $[0, \infty)$ that $$\int_0^\infty f(x) dx \geq \sum_{n=1}^\infty (x_{n+1}-x_n)f(x_{n+1}).$$ Now I want to use this to prove that if $f$ is integrable, then $$\lim_{x\rightarrow \infty} \frac{f(x)}{x}=0.$$

I understand intuitively that since $f$ is decreasing and its values lie on $[0, \infty)$, then $f$ is getting closer to zero. If we divide $f(x)$ by $x$ then surely $f(x)/x\rightarrow 0$. I just don't know how to use the inequality to show this is true formally.

Answer 1

Suppose $\lim\limits_{x\to \infty} \dfrac{f(x)}{x} \neq 0$. There exists a positive number $C$ such that to every positive number $M$, there corresponds an $x \ge M$ for which $f(x) > Cx$. Choose $x_1 > 1$ such that $f(x_1) > Cx_1$. Choose $x_2 > \max\{2, 2x_1\}$ such that $f(x_2) > Cx_2$. Having chosen $x_1,\ldots x_{k-1}$, choose $x_{k} > \max\{k,2x_{k-1}\}$ such that $f(x_k) > Cx_k$. This inductively defines an increasing sequence $(x_n)$ in $[0,\infty)$. Since $x_{n+1} - x_n > \dfrac{x_{n+1}}{2}$ and $f(x_{n+1}) > \epsilon x_{n+1}$ for all $n$, then by your claim, $$\int_0^\infty f(x)\, dx \ge \sum_{n = 1}^\infty (x_{n+1} - x_n)\,f(x_{n+1}) \ge \frac{C}{2}\sum_{n = 1}^\infty x_{n+1}^2 \ge \frac{C}{2}\sum_{n = 1}^\infty (n+1)^2 = \infty$$