Can Stolz-Cesaro theorem be applied to this problem? If $\lim\limits_{x\to\infty}(f(x+1)-f(x))=l$, Prove that $\lim\limits_{x\to\infty}\frac{f(x)}x=l$

If $$f :(a , \infty ) \to \mathbb{R}$$ and $$f$$ is bounded on every $$(a,b)$$ such that $$a, prove that $$\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?

• 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. Feb 7 at 8:17
• 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. Feb 7 at 8:29

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 $$\displaystyle\lim_{x\to\infty} \frac{f(x)}x = l$$.

• 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?
– Jan
Feb 9 at 23:09
• I may be confused, but how do you argue “Since $s_{n+1} \geq f(y+1)$, we conclude that $s_{n+1}-s_n > l - 2\varepsilon$”?
– Jan
Feb 9 at 23:22
• 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). Feb 9 at 23:27
• @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?
– pie
Feb 10 at 14:47
• No, it's part of the definition of $s_n = \sup\{f(y)\colon y\in[n,n+1]\}$. Feb 10 at 18:50

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.