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If $f :(a , \infty ) \to \mathbb{R}$ and $f $ is bounded on every $(a,b)$ such that $a<b <\infty$, prove that $\lim\limits_{x \to \infty }(f(x+1) - f(x))=l$ implies $\lim\limits_{x \to \infty }\frac{f(x)}{x}=l$.

The first thing that came to my mind was

Stolz-Cesàro theorem case $\frac{*}{\infty}$:- If $b_n $ is a monotone increasing sequence and $\lim \limits_{n \to \infty} b_n = \infty $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

Does this really solve this problem? There are uncountably many sub-sequences with the limit $l$, i.e., there is a subsequence for all $x\in (a, a+1]$ with limit $l$, but this doesn't imply that the sequence has the limit $l$.

One famous example where there are infinitely many sub-sequences with the same limit but the limit of the sequence doesn't exist is:

$$a_n = \begin{cases} \frac{1}{r}, & \text{if $n$ is a power of prime $n = p^r \ : r\ge1$} \\[2ex] 0, & \text{if $n$ is not a power of prime } \end{cases}$$ One can make infinitely many sub-sequences that converge to $0$ although $a_n $ doesn't converge to $0$.

After a lot of time thinking, I couldn't prove this problem, so I searched on MSE for a solution and found this, which gives a general proof for this problem. But my question is: Can we use the Stolz-Cesàro theorem to solve this problem?

If we can use the Stolz-Cesàro theorem, how do we complete the proof?

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    $\begingroup$ Somebody voted to close this question as a duplicate of math.stackexchange.com/q/1642662/42969. However, the question here is whether the Stolz-Cesàro theorem can be applied, and I do not see that answered in the suggested duplicate target. $\endgroup$
    – Martin R
    Feb 7 at 8:17
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    $\begingroup$ One might be able to convert the function limits into appropriate SFR relations and use Stolz-Cesaro for sequences. For what I mean by an SFR relation, see here. Of course, the appropriate relation would require monotonicity of convergent sequences so that Stolz-Cesaro may be applied. $\endgroup$ Feb 7 at 8:29

2 Answers 2

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TL;DR: apply Stolz–Cesàro separately to $\sup_{[n,n+1]}f(x)$ and $\inf_{[n,n+1]}f(x)$.

Let $f$ be a function that is bounded on every $(a,b)$ and satisfies $\lim_{x\to\infty} (f(x+1)-f(x)) = l$. Define $$ s_n = \sup\{ f(x)\colon n\le x\le n+1\} \quad\text{and}\quad i_n = \inf\{ f(x)\colon n\le x\le n+1\}, $$ which are well defined by the boundedness assumption on $f$.

Claim: $\lim_{x\to\infty} (s_{n+1}-s_n) = l$ and $\lim_{x\to\infty} (i_{n+1}-i_n) = l$.

Proof for $\{s_n\}$: Let $\varepsilon>0$ be arbitrary. Choose $x_0$ such that $|f(x+1)-f(x)-l| < \varepsilon$ for $x>x_0$.

For any $n>x_0+2$, consider $s_{n+1}-s_n$. There exists $y\in[n,n+1]$ such that $f(y) > s_n - \varepsilon$; then $y+1\in[n+1,n+2]$ and $f(y+1)-f(y)>l-\varepsilon$, so that $f(y+1) > l+s_n-2\varepsilon$. Since $s_{n+1} \ge f(y+1)$, we conclude that $s_{n+1}-s_n > l-2\varepsilon$. On the other hand, $f(z) \le s_n$ for all $z\in[n,n+1]$. Since $f(z+1)-f(z) < l+\varepsilon$ for all such $z$, we conclude that $f(z+1) < s_n+l+\varepsilon$ for all $z+1\in[n+1,n+2]$, and hence $s_{n+1} \le s_n+l+\varepsilon$.

We have thus shown that $|s_{n+1}-s_n-l| < 2\varepsilon$ for all $n>x_0+2$. Since $\varepsilon$ was arbitrary, this implies that $\lim_{n\to\infty} (s_{n+1}-s_n) = l$.

(A similar proof works for $i_n$, or just take the negative of everything in sight.)

Now applying Stolz–Cesàro with $a_n=s_n$ and with $a_n=i_n$ (and with $b_n=n$ in both cases) allows us to conclude that $$ \lim_{n\to\infty} \frac{s_n}n = l = \lim_{n\to\infty} \frac{i_n}n. $$ From here another (more straightforward) $\varepsilon$-$x_0$ argument can be used to show that $\displaystyle\lim_{x\to\infty} \frac{f(x)}x = l$.

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  • $\begingroup$ I’m assuming there’s a typo: $|f(x+1)-f(x)-l|<\varepsilon$ instead of $|f(x+1)-f(x)|<\varepsilon$ in the first sentence of the proof? $\endgroup$
    – Jan
    Feb 9 at 23:09
  • $\begingroup$ I may be confused, but how do you argue “Since $s_{n+1} \geq f(y+1)$, we conclude that $s_{n+1}-s_n > l - 2\varepsilon$”? $\endgroup$
    – Jan
    Feb 9 at 23:22
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    $\begingroup$ Thanks for spotting the typo. The second inequality in your last comment follows from the first inequality combined with $f(y+1) > l+s_n-2\varepsilon$ (which also had a typo that I now fixed). $\endgroup$ Feb 9 at 23:27
  • $\begingroup$ @GregMartin "There exists $y\in[n,n+1]$ such that $f(y) > s_n - \varepsilon$" I don't understand why this is true, Does this assume continuity? $\endgroup$
    – pie
    Feb 10 at 14:47
  • $\begingroup$ No, it's part of the definition of $s_n = \sup\{f(y)\colon y\in[n,n+1]\}$. $\endgroup$ Feb 10 at 18:50
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By Stolz-Cesaro we have $\frac{f(n+1)-f(n)}{(n+1) - n} = f(n+1)-f(n) \to \ell$ implies $\frac{f(n)}{n} \to \ell$. Let $[\cdot]$ be the integer part of a number and note that $$ \frac{f(x)}{x} = \frac{f([x])}{x} + \frac{f(x)-f([x])}{x} = \frac{f([x])}{[x]} \cdot \frac{[x]}{x} + \frac{f(x)-f([x])}{x}. $$ By simple inequalities $x-1 \le [x] \le x$ we see that $\frac{[x]}{x} \to 1$ and hence by the above application of Stolz-Cesaro, $\frac{f([x])}{[x]}\cdot \frac{[x]}{x}\to \ell$.

It remains to deal with the last part. Fix $\varepsilon>0$ and let $m \in \mathbb N$ be such that $|f(y+1)-f(y)-\ell|<\varepsilon$ for $y\ge m$. Given $x\in [m+n,m+n+1)$ we then have $$|f(x)-f([x])| \le |f(x)-f(x-1)-f([x])+f([x]-1)| + |f(x-1)-f([x]-1)|$$ $$ \le 2\varepsilon + |f(x-1)-f([x]-1)| \le ... \le 2n\varepsilon +2\sup\{ |f(y)| : y\in [m,m+1)\}.$$ In particular $$ \frac{|f(x)-f([x])|}{x} \le 2\varepsilon + \frac{2\sup\{|f(y)| : y\in [m,m+1]\}}{x} \to 2\varepsilon,$$ and since $\varepsilon>0$ was arbitrary we can conclude.

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