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Suppose that the function $f$ is continuous in $\mathbb{R}$ and $$\lim_{x\to \infty}\left(f(x+1)-f(x)\right)=0$$ then does this mean that $\lim_{x\to \infty}\left(\frac{f(x)}{x}\right)=0$ ?

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marked as duplicate by Paramanand Singh calculus Dec 25 '17 at 16:57

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Yes, that is true. By hypothesis, for any $\epsilon>0$, there is some $t>0$ such that for all $x>t$, we have $|f(x+1)-f(x)|<\epsilon$. Let $M=\max\limits_{x\in[t,t+1]}|f(x)|$, which exists since $f$ is continuous.

For $x=t+h$ with $n+1>h\geq n$, we have that $$|f(x)-f(t+h-n)|\leq |f(t+h)-f(t+h-1)|+\cdots +|f(t+h-n+1)-f(t+h-n)|< n\epsilon$$ and $$|f(t+h-n)|\leq M$$ hence $$|f(x)|\leq|f(x)-f(t+h-n)|+|f(t+h-n)|< n\epsilon+M\leq (x-t)\epsilon+M$$ so that for any $x=t+h$ with $h>0$, $$\left|\frac{f(x)}{x}\right|=\left|\frac{f(t+h)}{x}\right|<\left|\frac{(x-t)\epsilon+M}{x}\right|=\left|\epsilon+\frac{M-t\epsilon}{x}\right|\leq \epsilon+\left|\frac{M-t\epsilon}{x}\right|$$ Thus, for $x>\max\{t,|\frac{M-t\epsilon}{\epsilon}|\}$, we have that $\left|\frac{f(x)}{x}\right|\leq 2\epsilon$. This suffices to show that $\lim_{x\to\infty}\left|\frac{f(x)}{x}\right|=0$.

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  • $\begingroup$ Ah... In the third line from the bottom, middle equality: that cleared the whole darn thing up! Also, the choice of epsilon...which isn't fixed until the last step. Very nice indeed. +1 $\endgroup$ – DonAntonio Dec 25 '17 at 10:31
  • $\begingroup$ How we can prove it using sequences ? $\endgroup$ – S.H.W Dec 25 '17 at 11:52
  • $\begingroup$ A very nice proof Zev! As far as I understand, continuity is too much here, we can only demand the function to be bound! $\endgroup$ – dmtri Dec 25 '17 at 16:07
  • $\begingroup$ ....to be bounded in any interval [t,t+1]. $\endgroup$ – dmtri Dec 25 '17 at 16:30
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Fix $\varepsilon > 0$. There exists $K\in\mathbb{Z}$ such that $$ |f(x+1)-f(x)| < \varepsilon \qquad \forall x \geq K. $$ Let $M := \max_{x\in [K, K+1]} |f(x)|$. Let $x > K$ and $n\in\mathbb{Z}$ such that $n \leq x < n+1$. Then $$ |f(x)| \leq |f(x-(n-K)| + \sum_{j=1}^{n-K} |f(x+1-j) - f(x-j)| \leq M + (n-K)\varepsilon \leq M + (x-K)\varepsilon, $$ so that $$ \limsup_{x\to +\infty} \frac{|f(x)|}{x} \leq \varepsilon. $$

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  • $\begingroup$ (1) In the 4th line, how is that first inequality obtained? (2) Assuming this same line is correct, you got $\;|f(x)|\le M+(x-K)\epsilon\;$ How from here you get the last inequality with the lim sup? $\endgroup$ – DonAntonio Dec 25 '17 at 10:20
  • $\begingroup$ (1) $|f(x)| \leq |f(x-(n-K))| + |f(x) - f(x-(n-k)| \leq \ldots$. (2) Divide the inequality by $x$ (that you can assume positive) and take the limsup. $\endgroup$ – Rigel Dec 25 '17 at 10:23
  • $\begingroup$ I tried to: getting $$\frac{|f(x)|}x\le \frac Mx+\frac{x-K}x\epsilon ....\text{still unclear}$$ Only clear that the second summand on the right tends to $\;\epsilon\;$ when $\;x\to\infty\;$ . $\endgroup$ – DonAntonio Dec 25 '17 at 10:28
  • $\begingroup$ See my comment under the answer by Zev... $\endgroup$ – DonAntonio Dec 25 '17 at 10:32
  • $\begingroup$ I don't understand your point. If $h(x) \leq g(x)$, then $\limsup h(x) \leq \limsup g(x)$. In this case, if $g(x)$ is the r.h.s. of the inequality, then $\limsup g(x) = \lim g(x) = \varepsilon$. $\endgroup$ – Rigel Dec 25 '17 at 10:40
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Subtle question. By Cesàro-Stolz we have $$ \lim_{n\to +\infty}\frac{f(n)}{n} = \lim_{n\to +\infty}\frac{f(n+1)-f(n)}{(n+1)-n} = 0 $$ and the same holds if we consider $\lim_{n\to +\infty}\frac{f(n+\theta)}{n+\theta}$ with $\theta\in(0,1)$. On the other hand continuity plus $\lim_{x\to +\infty}f(x+1)-f(x)$ ensure that $f$ is uniformly continuous over $\mathbb{R}^+$, since they ensure that $f$ is continuous and approximately $1$-periodic. In particular, for any $\theta\in(0,1)$,

$$ \left|\frac{f(n+\theta)}{n+\theta}-\frac{f(n)}{n}\right|\leq\frac{n|f(n+\theta)-f(n)|+\theta|f(n)|}{n^2}=O\left(\frac{1}{n}\right)$$ and $$ \lim_{x\to +\infty}\frac{f(x)}{x}=0$$ holds.

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  • $\begingroup$ (1) How "the same holds...", as CS theorem is for sequences? I suppose $\;\theta\;$ is fixed here, so I understand how the extension of CS holds (please do correct me if I'm wrong). (2) Also, why can you deduce unif. continuity? As far as I understand for UC we should evaluate the differences $\;|f(x)-f(y)|\;$ , and here the values of the varialbes are $\;l1\;$ unit away: x,\,x+1\;$ ...I guess that "approx. 1-periodic" thingy is unclear to me. (3) The last equality with the big "O" is incomprehensible to me, too. $\endgroup$ – DonAntonio Dec 25 '17 at 10:26
  • $\begingroup$ @DonAntonio: is it clear that a continuous and $1$-periodic function is uniformly continuous? Well, here it is almost the same. We have $\lim_{n\to +\infty}\frac{f(n+\theta)}{n+\theta}=0$ for any $\theta$, we just have to "stick these limits together" by exploiting UC. $\endgroup$ – Jack D'Aurizio Dec 25 '17 at 10:39

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