Limit of bounded function

Question

Let $f:[0,\infty)\to\mathbb{R}$ be bounded on every interval $[0,b]$ for $b\gt0$. Assume that $$\lim_{x\rightarrow\infty}\left(f(x+1)-f(x)\right)=0.$$ How do I prove that $$\lim_{x\rightarrow\infty} \frac{f(x)}{x}=0?$$

Answer 1

The condition that $\lim_{x\to\infty} f(x+1)-f(x)=0$ can be written as: for every $\varepsilon>0$ there exists an $\alpha$ such that for all $x>\alpha$, we must have that $$f(x)-\varepsilon < f(x+1) < f(x)+\varepsilon .$$ By using induction, it can be proven that for all $n\in\mathbb{N}$, $$f(x)-n\varepsilon < f(x+n) < f(x)+n\varepsilon .$$

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Now let $\varepsilon$ be a positive real number and let $\alpha$ be any number such that above property holds.

Notice that any $x>\alpha$ can be written as $x=y+n$ with $y\in[\alpha,\alpha+1)$ and with $n\in\mathbb{N}$. It is obvious that $n\leq x$.

Also define $a$ and $b$ to be the infimum and supremum of $f(x)$ on $[\alpha,\alpha+1)$, respectively. These are finite due to $f$ being bounded on every interval.

Now using all these facts, we get that for any $\varepsilon>0$ there exists an $\alpha$ such that for any $x\to\infty$, we will have that $$\frac{f(x)}{x} = \frac{f(y+n)}{x} < \frac{f(y) + n \varepsilon }{x} \leq \frac{b}{x} + \frac{x\varepsilon}{x} = \frac{b}{x} + \varepsilon \to \varepsilon.$$ We can also see that for $x\to\infty$, $$\frac{f(x)}{x} = \frac{f(y+n)}{x} > \frac{f(y)-n\varepsilon}{x} \geq \frac{a}{x} - \frac{x\varepsilon}{x} = \frac{a}{x} - \varepsilon \to -\varepsilon .$$

Thus, for any $\varepsilon>0$, we must have that $-\varepsilon<\lim_{x\to\infty} f(x)/x<\varepsilon$, and therefore $\lim_{x\to\infty} f(x)/x=0.$

Answer 2

Note that

$$ \frac{f(x)}{x}=\frac1x\sum_{k=1}^{\lfloor x\rfloor}[f(x-\lfloor x\rfloor+k)-f(x-\lfloor x\rfloor+k-1)]+\frac{f(x-\lfloor x\rfloor)}x, $$

where $\lfloor x\rfloor$ is the integer part of $x$. Clearly, $$ \lim_{x\to+\infty}\frac{f(x-\lfloor x\rfloor)}x=0 $$ and given $\delta>0$, there exists $k_0\in\mathbb N$ such that $$ \big|f(x-\lfloor x\rfloor+k)-f(x-\lfloor x\rfloor+k-1)\big|<\delta $$ for $k\ge k_0$. So $$ \left|\frac1x\sum_{k=k_0}^{\lfloor x\rfloor}[f(x-\lfloor x\rfloor+k)-f(x-\lfloor x\rfloor+k-1)]\right|\le\frac\delta x. $$ Now you just need to put everything together.