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Prove that $\lim_{x \rightarrow \infty} \frac{f(x)}{x}= \alpha$ when $f$ is continuous and $\lim_{x \rightarrow \infty} (f(x+1) - f(x)) = \alpha$

The given hint suggests to consider the case of $\alpha=0$.

Given $\varepsilon \gt 0$, there exists $M$ such that $x \ge M $ implies that $|f(x+1) - f(x)| \lt \varepsilon$. Then for any $y$ which is larger than $M$, we can write it as $y=x+N$ where some $x \in [M, M+1]$.

And $f$ is continuous on $[M, M+1]$, it has a maximum $M_0$ on this interval, i.e., $|f(x)| \le M_0$ on this interval. Then $|f(x+N)| \le M_0 + N\varepsilon$. So $|\frac{f(x+N)}{x+N}| \le \frac{M_0 + N\varepsilon}{x+N}$. Then as $N \rightarrow \infty$, $\lim_{y \rightarrow \infty} \frac{f(y)}{y} \le \varepsilon$, which means that $\lim_{y \rightarrow \infty} \frac{f(y)}{y} =0$.

Is this argument valid? And I’d like to know whether there is a simple argument.

Arctic Char
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glimpser
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  • You are taking $N \in \mathbb{N}$ ? – user-492177 Jul 19 '20 at 04:01
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    Don't mind me a lot, I am tired and I might be about to write something false but I don't seem to get why are you using $M_0$. I don't quite see why your estimate $|f(x+N)|\leq M_0+N\varepsilon$ holds. Even if it holds, wouldn't this be saying that $f(y)/y$ is bounded by $M_0 + \varepsilon?$. What I think is true is that $|f(x+N)| \leq N\varepsilon$ whenever $x\geq M$. Then, $|f(x+N)/(x+N)|\leq N\varepsilon/(x+N)$ for $x\geq M$ and then the result follows. – Alonso Delfín Jul 19 '20 at 04:15
  • @user710290 Yes, I am. – glimpser Jul 19 '20 at 04:19
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    @AlonsoDelfín Ah, I see. That’s much simpler. – glimpser Jul 19 '20 at 04:21
  • @AlonsoDelfín Then what about the case of $\alpha \ne 0$? – glimpser Jul 19 '20 at 04:26
  • I am not sure whether what I thought is right, but here it goes: taking $ \alpha = 0$, one can substitute $\frac{1}{0}$ for $\infty $. . $f(\infty)$ may become anything, but that divided by $ \infty $ is equal to multiplying it with $0$, so the overall answer may be $0$. – Spectre Jul 19 '20 at 04:30
  • Maybe use a similar argument but now looking at $f(x)-\alpha x$. Then use that $|frac{f(x)}{x}-\alpha|=|\frac{f(x)-\alpha x}{x}|$. – Alonso Delfín Jul 19 '20 at 05:20

2 Answers2

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I am thinking in terms of your argument. I think what might be confusing in your answer is the simultaneous variation of $x \in [M,M+1]$ and $N \in \mathbb{N}$ as $y \to \infty $

Let $y \gt M+1$ . Then $\exists x_y\in [M,M+1], N_y\in \mathbb{N}$ such that

$y=x_y+N_y$ Then

$|f(y)-f(x_y)|=|f(x_y+N_y)-f(x_y)|$

$=|f(x_y+N_y)-f(x_y+N_y-1)+f(x_y+N_y-1)-f(x_y+N_y-2)...+f(x_y+1)-f(x_y)|\le N_y \epsilon$ (By Triangle Inequality)

So $ |f(y)|-|f(x_y)| \le |f(y)-f(x_y)|\le N_y \epsilon$

(since $|a-b| \ge ||a|-|b|| \ge |a|-|b|$)

which gives

$|f(y)| \le M_0+N_y \epsilon$

Now, $\big| \frac {f(y)}{y}\big| \le \frac {M_0/N_y+\epsilon}{x_y/N_y+1}$

Since $x_y$ is bounded , we must have as $N_y \to \infty $ , $\displaystyle\lim_{y\to \infty} \big| \frac {f(y)}{y}\big| \le \epsilon$

user-492177
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  • This is what I thought. But my writing was a bit more ambiguous. Thanks for clarifying it. – glimpser Jul 19 '20 at 06:56
  • Yes the dependence of $x,N$ on $y$ was missing in your answer. Also for $\alpha \neq 0$ , consider $g(x)=f(x)-\alpha x$ as already suggested to you. – user-492177 Jul 19 '20 at 07:21
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Apologies if this is an overly tortured derivation. Let me know if there are any hand-wavey parts.

We know that for any $\epsilon > 0$ there exists $x_{\epsilon}$ such that $\left| f(x+1) - f(x)-\alpha\right| < \epsilon\;\forall x>x_{\epsilon}$.

For brevity, take $x'=\max\{x_{\epsilon},1\}$

This implies $ -\epsilon <f(x'+\delta+1) - f(x'+\delta)-\alpha < \epsilon\;\;\forall \delta > 0$

Dividing by $x'+\delta$ we get:

$$\frac{-\epsilon}{x'+\delta} < \frac{f(x'+\delta+1) - f(x'+\delta)-\alpha}{x'+\delta} < \frac{\epsilon}{x'+\delta}$$

Taking the limit as $\delta \to \infty$ we get:

$$0 < \frac{f(x'+\delta+1) - f(x'+\delta)-\alpha}{x'+\delta} < 0$$

Therefore,

$$\lim_{x\to \infty} \frac{f(x+1) - f(x)-\alpha}{x} = 0$$

This implies that $f(x) = c x + o(x)$ for some $c$

If we assume $c\neq 0$, then $c>\alpha \implies \lim_{x\to \infty} \frac{f(x+1) - f(x)-\alpha}{x} > 0$ and analogously for $c<\alpha$, therefore, $c=\alpha$.

If we assume $c=0$ then $f(x) = o(x)$ and $\lim_{x\to\infty} f(x+1) - f(x) = 0$. If $\alpha =0$ this works out fine, but if not then we have a contradiction with a prior assumption.

Therefore, $c=\alpha$ and $f(x) = \alpha x + o(x) \implies \lim_{x\to\infty}\frac{f(x)}{x} = \alpha$