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

I tried using the definition on the first limit; then, I attempted to use that inequality to apply the triangular inequality to arrive at the definition that gives me the second limit. It is supposed to be simple, but I'm not feeling safe as in how I should write it.

DrHAL
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  • define $g(x)=f(x)-Lx$. Then you need to show that $g(x+1)-g(x)\to 0$ implies $g(x)/x\to 0$. – A.S. Feb 06 '16 at 03:50
  • You just need to apply Stolz-Cesaro: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – A.S. Feb 06 '16 at 04:47
  • @A.S : Can you show how Stolz–Cesàro theorem can be used here? –  Feb 06 '16 at 08:11
  • @John What is there to show? Just apply it directly to all $f(\alpha+n)$ sequences (indexed by $\alpha\in[0,1)$). – A.S. Feb 06 '16 at 08:14
  • Um... Is $\lim_{n\to \infty} \frac{f(\alpha +n)}{\alpha + n} = L$ for all $\alpha \in [0,1)$ sufficient to imply that $\lim_{x\to \infty} \frac{f(x)}{x} = L$? @A.S. (In particular, how is the boundedness of $f$ comes into play?) –  Feb 06 '16 at 08:17
  • https://math.stackexchange.com/questions/192963/if-lim-x-to-inftyfx1-fx-ell-then-lim-limits-x-to-infty-frac – StubbornAtom Jun 10 '19 at 21:07

3 Answers3

10

Using the condition, for all $\epsilon >0$, there is $N=N_\epsilon$ so that

$$ |f(x+1) - f(x) -L|< \epsilon$$

for all $x\ge N$. Then for each $M = 1, 2, 3, \dots$, by induction on $M$ we have

$$ | f(x+ M) - f(x) -ML|< M\epsilon$$

if $x\ge N$. As $f$ is bounded on bounded intervals, there is $K >0$ so that $|f(x) | \le K$ when $x\in [N, N+1]$. So for all $x\in [N, N+1]$, $M\in \mathbb N$ we have

$$ |f(x+M)-ML| < M\epsilon +K.$$

Now for all $y\ge N$, there is $M$ so that $y=x+M$ for some $x\in [N, N+1]$. So for all $y\ge N$,

$$\begin{split} |f(y) - yL| &= |f( x+M) - ML + ML -yL| \\ &\le |f(x+M) - ML| + |(M-y)L| \\ &< M\epsilon +K +(N+1)|L| \\ &\le (y -N)\epsilon + K +(N+1)|L| \\ &= y\epsilon +C, \end{split}$$

where $C = -N\epsilon + K +(N+1)|L|$ is independent of $y$. So

$$ \left|\frac{f(y)}{y} -L\right| \le \epsilon + \frac{C}{y}.$$

Now choose $N_1 \ge N$ so that $\frac{C}{y} <\epsilon$ whenever $y\ge N_1$. So

$$\left|\frac{f(y)}{y} -L\right| \le 2\epsilon$$

whenever $y\ge N_1$. Thus

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

user26857
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2

For each $\epsilon > 0$, there exists $N > 0$, such that for $x > N$, we have $$ |f(x + 1) - f(x) - L | < \epsilon$$. Expand it, we have $$ L -\epsilon + f(x) < f(x + 1) < f(x) + L + \epsilon$$ Use above relation, for each $y > N+ 1$, we have for some $ N \le z \le N + 1 $, $$ f(z) - z (L - \epsilon) < f(y) < f(z) +(y - z) (L + \epsilon) $$ Rewrite the abouve as $$ f(z) + (- z) (L - \epsilon) < f(y) - yL < f(z) - z (L + \epsilon) $$ Furthermore $$ |f(y) - Ly| \le \max_{N\le z\le N+1}|f(z)| + (N+1)(L+\epsilon)$$ Now, let $M > N + 1 >0$ so that $\frac{max_{N<Z<N+1}|f(z)| + (N+1)(L+ \epsilon)}{M} \le \epsilon$

for $x > M$, $$|\frac{f(x)}{x} - L |= |\frac{f(x) - Lx}{x}|$$ $$ \le \frac {\max_{N\le z\le N+1}|f(z)| + (N+1)(L+\epsilon)}{M}$$ $$\le \epsilon$$

QED

runaround
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0

Intuitive approach when $l \in \mathbb{R}$:

let $a:= x- \lfloor x\rfloor$ , $n :=\lfloor x\rfloor $

$$f(x)=f(a+n) = f(a)+ \sum\limits_{k=0}^{n-1} f(a+k+1)- f(a+k)$$

$$\frac{f(x)}{x}=\frac{f(a)+ \sum\limits_{k=0}^{n-1} f(a+k+1)- f(a+k)}{x}$$

Since $ \lim\limits_{n \to \infty} f(x+1)- f(x) =l $ choose $N \in \mathbb {N}$ st for all $x>N$, $|f(x+1)- f(x)-l |< \epsilon$

$$= \frac{f(a)+ \sum\limits_{k=0}^{N-1} f(a+k+1)- f(a+k) }{x}+\frac{ \sum\limits_{k=N}^{n-1} f(a+k+1)- f(a+k) }{x} $$ Note that the first fraction goes to $0$ as $x$ goes to $\infty$ , the second fraction $\frac{n-1-N}{x}(l -\varepsilon )< \frac{ \sum\limits_{k=N}^{n-1} f(a+k+1)- f(a+k) }{x} < \frac{n-1-N}{x}(l +\varepsilon )$ as $x \to \infty$ the this goes to $l$

Rigorous proof:

Given $0<\varepsilon<1$ since $ \lim\limits_{n \to \infty} f(x+1)- f(x) =l $ then $\exists N \in \mathbb {N}$ s.t for all $x>N$, $|f(x+1)- f(x)-l |< \frac{\varepsilon}{3+|l|}$.

Since $f$ is bounded on $[0, N+1]$ then let $k = N(1+3 \sup {|f(x)|})$ where $ 0\le x\le N+1 $. Now choose $M$ such $\frac{k}{M} \le \frac{\varepsilon}{3+|l|}$

let $a:= x- \lfloor x\rfloor$ , $n :=\lfloor x\rfloor $

$$f(x)=f(a+n) = f(a)+ \sum\limits_{k=0}^{n-1} f(a+k+1)- f(a+k)$$

for all $x >M$

$$\frac{f(x)}{x}= \frac{f(a)+ \sum\limits_{k=0}^{n-1} f(a+k+1)- f(a+k) }{x}+\frac{ \sum\limits_{k=N}^{n-1} f(a+k+1)- f(a+k) }{x} $$

$$-\frac{\varepsilon}{3+|l|}+ \frac{n-1-N}{x}(l -\frac{\varepsilon}{3+|l|} )< \frac{f(x)}{x} <\frac{\varepsilon}{3+|l|} +\frac{n-1-N}{x}(l +\frac{\varepsilon}{3+|l|}3)$$

$$-\frac{\varepsilon}{3+|l|}+ \left (1- \frac{a+N}{x}\right )(l -\frac{\varepsilon}{3+|l|} )< \frac{f(x)}{x} <\frac{\varepsilon}{3+|l|} +\left (1- \frac{a+N}{x}\right )(l +\frac{\varepsilon}{3+|l|} )$$

$$l -3\frac{\varepsilon}{3+|l|} -|l| \frac{\varepsilon}{3+|l|} < \frac{f(x)}{x} < l +3\frac{\varepsilon}{3+|l|} +|l| \frac{\varepsilon}{3+|l|} $$

$$-(3 +|l|)\frac{\varepsilon}{3+|l|} < \frac{f(x)}{x}-l < (3 +|l|)\frac{\varepsilon}{3+|l|}$$

$$-\varepsilon<\frac{f(x)}{x}-l <\varepsilon$$

$$\left|\frac{f(x)}{x}-l \right|<\varepsilon$$

pie
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