Let $f:[0,\infty]\rightarrow \mathbb{R}$ a bounded function in each bounded interval. If $\lim_{x\rightarrow \infty} [f(x+1)-f(x)]=L$ then prove that $\lim_{x\rightarrow \infty}\displaystyle\frac{f(x)}{x}=L$

I appreciate any hint to solve this problem.

Thanks a lot!


For any $a > 0$:

  Let $y$ be such that $f(x+1) - f(x) \in L + [-a,a]$ for any $x \ge y$

  Then $f(y+t+k) - f(y+t) \in k ( L + [-a,a] )$ for any $k \in \mathbb{N}$ and $t \in [0,1]$

  Also $f(y+t)$ is bounded over all $t \in [0,1]$

  Thus ${\large \frac{f(y+t+k)}{y+t+k} } \in {\large \frac{f(y+t)+k(L+[-a,a])}{y+t+k} } \to L+[-a,a]$ uniformly for all $t \in [0,1]$ as $k \to \infty$

  Thus ${ \large \frac{f(x)}{x} } \in L + [-2a,2a]$ as $x \to \infty$

Therefore ${ \large \frac{f(x)}{x} } \to L$ as $x \to \infty$

To explain what happens in the last two lines inside "For any $a>0$", we basically have $\large\frac{p(t)+k(L+q(t))}{y+t+k}$ for some $p$ that is bounded on $[0,1]$ and $q$ that is in $[-a,a]$. As $k$ increases, $p(t)$ becomes insignificant because it is bounded, and also $\frac{k}{y+t+k}$ approaches 1 uniformly over all $t \in [0,1]$, hence the expression approaches something in $L+[-a,a]$ uniformly over all $t \in [0,1]$. "Uniformly" means that for any desired margin of error there is a common cutoff point for $k$ beyond which the expression is within the error margin for all $t$. The next line follows from this because eventually the expression will be within $a$ of $L+[-a,a]$.

  • $\begingroup$ I get lose in your proof. I don't get it. $\endgroup$ – YTS Apr 24 '14 at 2:22
  • $\begingroup$ Which lines do you not get? $\endgroup$ – user21820 Apr 24 '14 at 2:32
  • $\begingroup$ In the second line: Why $f(y+t+k)-f(y+t)\in k(L+[-a,a]$? Why do you only "multiply" the interval? In the fourth and fifth: Why this is true? Why the fifth follow from the fourh? The others lines are "clear" :P $\endgroup$ – YTS Apr 24 '14 at 2:45
  • $\begingroup$ To get the second line, use induction on $k$. $f(x+1)-f(x)$, $f(x+2)-f(x+1)$, ... all differ by some amount in $L+[-a,a]$. $\endgroup$ – user21820 Apr 24 '14 at 2:57
  • 1
    $\begingroup$ I'll edit to expand what I wrote to clarify the other two lines. $\endgroup$ – user21820 Apr 24 '14 at 2:59

This is another proof:

Given $\epsilon>0$ there exist $y>0$ such that if $x>y$ then:


we can conclude the following inequalytis valid for every $n$





adding each of this inequalitys

$$nL-n\epsilon<f(x+n)-f(x)<nL+n\epsilon $$

taken $n=[x]$ and $h(x)=x-[x]$ then

$$[x]L-[x]\epsilon<f(x)-f(h(x))<[x]L+[x]\epsilon $$

since $0\leq h(x) <1$ and $[0,1)$ is bounded, then there exist $M>0$ such that $|f(h(x))|<M$. This implies that $\lim_{x\rightarrow\infty}\displaystyle\frac{f(h(x))}{x}=0$. Dividing by $x$:

$$\displaystyle\frac{[x]}{x}L-\displaystyle\frac{[x]}{x}\epsilon<\displaystyle\frac{f(x)-f(h(x))}{x}<\displaystyle\frac{[x]}{x}L+\displaystyle\frac{[x]}{x}\epsilon $$

and recalling that $\lim_{x\rightarrow\infty}\displaystyle\frac{[x]}{x}=1$. We take $x\rightarrow\infty$ then

$$L-\epsilon\leq\lim_{x\rightarrow\infty}\displaystyle\frac{f(x)}{x}\leq L+\epsilon$$

since $\epsilon>0$ is aritrary we get the result.

  • 1
    $\begingroup$ Your proof is wrong for two reasons. Firstly you cannot get down to $f(x) - f(x - \lfloor x \rfloor)$ since the inequality $f(x+1)-f(x)<L+ε$ only works for $x>y$. Secondly you cannot claim that the limit of $\frac{f(x)}{x}$ exists at your second last line. All you can say is that it lies within some larger interval such as $L+[-2ε,2ε]$ like I showed already. Always keep in mind the scope; your second error arises because you did not see that your $y$ depends on $ε$ and hence you cannot 'unbind' $ε$ to obtain any limits while still within the scope of "for any $ε$: for some $y$". $\endgroup$ – user21820 Apr 24 '14 at 17:22
  • $\begingroup$ Taken $k=[x-y]$ where $x=y+k+t$ and $m<|f(x-k)|<M$ and get the inequalities: $-\frac{[x-y]}{x}\epsilon+\frac{m}{x}<\frac{f(x)}{x}-\frac{[x-y]}{x}L<\frac{[x-y]}{x}\epsilon+\frac{M}{x}$ How do I conclude from here? $\endgroup$ – YTS Apr 24 '14 at 18:24
  • $\begingroup$ I can figure out what you are doing, but I advise you to try and make the scope clear for your own clearer understanding, rather than following the conventional "prose and centered formula" style, which is quite useless. In the inequality you now have, if you move out of the innermost scope binding $x$, you can see the following: "for some $c$: for any $x>c$: $\frac{k}{x} \in [1-ε,1]$ and $\frac{m}{x} \in [-ε,ε]$ and $\frac{M}{x} \in [-ε,ε]$", and hence: "for some $c$: for any $x>c$: $\frac{f(x)}{x} \in [1-ε,1] L + [1-ε,1] [-ε,ε] + [-ε,ε]$"... $\endgroup$ – user21820 Apr 25 '14 at 3:12
  • $\begingroup$ ... which now is free of $y$, so it can be pulled out of the "for some $y$" scope, which leaves it in the outer "for any $ε>0$" scope. We now move out of that scope and see that we can make the right-hand side of the inner expression as close to $L$ as we want by instantiating sufficiently small $ε>0$. To be specific, we create a new scope "for any $d>0$:" and inside it construct a suitable $ε$ such that "$[1-ε,1] L + [1-ε,1] [-ε,ε] + [-ε,ε] \subset L + [-d,d]$", and then use the above statement to get: "for any $d>0$: for some $c$: for any $x>c$: $\frac{f(x)}{x} \in L + [-d,d]$, as desired. $\endgroup$ – user21820 Apr 25 '14 at 3:36
  • $\begingroup$ On an unrelated note, use "\lfloor ... \rfloor" rather than "[...]" because some people use the latter for many other things including ceiling and it's quite confusing. $\endgroup$ – user21820 Apr 25 '14 at 3:39

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