If $\lim_{x\rightarrow\infty}(f(x+1)-f(x))=L$ prove that $f=O(x)$.

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\lim_{x\rightarrow\infty}(f(x+1)-f(x))=L$$ Prove that $$\lim_{x\rightarrow \infty}\dfrac{f(x)}{x}=L$$

This was an exam question that I was given and got nowhere on it. Going back now, I don't think I'm any closer.

This is my idea so far. We know that $$\lim_{x\rightarrow\infty}\dfrac{f(x+1)-f(x)}{x}=0$$

I think I'm supposed to add the apprapraite $0$ to $$\left\vert \dfrac{f(x+1)-f(x)}{x}\right\vert$$ but I just keep getting a lower bound. A hint would be much appreciated. Thanks

Answer 1

The result does not hold. Try $f(x)=0$ if $x$ is an integer and $f(x)=1/(x-\lfloor x\rfloor)$ otherwise, where $\lfloor\ \rfloor$ denotes the integer part. Then $f(x+1)=f(x)$ for every $x$ but $f(x)/x$ has no limit at $+\infty$.

Edit: To answer a comment, if the function $f$ is assumed continuous, a very weak argument works: keep only from the continuity assumption the fact that the function $f$ is bounded on $[0,1]$, say $|f(x)|\leqslant C$ for every $x$ in $[0,1]$. Then, use, for every $x\geqslant0$, the decomposition $$ f(x)=f(x-\lfloor x\rfloor)+\sum_{k=1}^{\lfloor x\rfloor}f(x-k+1)-f(x-k). $$ The first term on the RHS is uniformly bounded and the average of the terms in the sum converges to $L$ by Cesàro. Since dividing by $x$ or by $\lfloor x\rfloor$ makes no difference in the limit $x\to\infty$, the result follows.

Answer 2

You need to impose that $f$ maps bounded intervals onto bounded intervals:

By the way, the proof by contraposition seems to be the more appropriate. But first, we recall the basic identity

$$ (f(x+1)-f(x)-L)^2=\left([f(x+1)-L(x+1)]+[Lx-f(x)]\right)^2= \\ \ \\ =(f(x+1)-L(x+1))^2+(f(x)-Lx)^2- \\ \ \\ -2(f(x)-Lx)(f(x+1)-L(x+1)). $$

So, if $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} \neq L$, one can find $\varepsilon_0,\varepsilon_1>0$ s.t. for every $\delta>0$, one can find $x>\delta$ satisfying the set of inequalities

$$ |f(x)-Lx|\geq \varepsilon_0 |x| ~~~\&~~~ |f(x+1)-L(x+1)|\geq \varepsilon_1 |x+1| .$$

Now pick $\varepsilon=\min \{ \varepsilon_0,\varepsilon_1\}$. Then one gets $$(f(x)-Lx)^2\geq \varepsilon^2 x^2~~~\& ~~~~ (f(x)-Lx)^2\geq \varepsilon^2 (x+1)^2 $$ so that $$ (f(x+1)-f(x)-L)^2\geq \varepsilon^2(x+1)^2+\varepsilon^2 x^2-2|x||x+1|\varepsilon^2. $$

Remark: Although it is not said, the inequality $-2(f(x)-Lx)(f(x+1)-L(x+1))\geq -2\varepsilon^2 |x|~|x+1|$ only fulfils only in case that $f$ maps a bounded interval onto a bounded interval. That is the case when one impose the condition $x>\delta$ [the set $f(]\delta,\infty[)$ is always bounded]. That means that the set of inequalities $f(x)-Lx\leq -\varepsilon |x|$ and $f(x+1)-L(x+1)\leq -\varepsilon |x+1|$ are always satisfied.

Finally, a short computation based on the binomial identity $(a-b)^2=a^2+b^2-2ab$ [$a=\varepsilon |x|$ & $b=\varepsilon |x+1|$] allows us to conclude that $$ (f(x+1)-f(x)-L)^2\geq \varepsilon^2. $$

From the arbitrary of $\delta>0$, that is equivalent to say that

$\displaystyle \lim_{x \rightarrow \infty} (f(x+1)-f(x)) \neq L$.

In conclusion, we have proved that $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} \neq L$ implies that $\displaystyle \lim_{x \rightarrow \infty} (f(x+1)-f(x)) \neq L$.