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Let $f:[0,\infty )\rightarrow \mathbb{R}$ be an increasing function, such that $\lim_{x\rightarrow \infty }\frac{1}{x^{2}}\cdot \int_{0}^{x}f(t)dt=1$.

Prove that exists $\lim_{x\rightarrow \infty }\frac{f(x)}{x}$ and calculate this limit.

If $f$ would be continuous, we'd have $1=\lim_{x\rightarrow \infty }\frac{1}{x^{2}}\cdot \int_{0}^{x}f(t)dt=\lim_{x\rightarrow \infty }\frac{f(x)}{2x}$, therefore $\lim_{x\rightarrow \infty }\frac{f(x)}{x}=2.$
But we don't know if $f$ is continuous or not, and I wasn't able to find any other idea.

ztefelina
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  • Related to continuity of the funtion?https://math.stackexchange.com/questions/246973/continuity-of-a-monotonically-increasing-function – kimi Tanaka Sep 28 '17 at 15:11
  • It is easy to show that if $f(x) /x\to L$ then $L=2$. The main challenge is to show that $f(x) /x$ tends to a limit. – Paramanand Singh Sep 28 '17 at 18:42

1 Answers1

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Suppose $x=n$ is an integer. By the Stolz-Cesaro theorem, \begin{eqnarray} 1&=&\lim_{x\rightarrow \infty }\frac{\int_0^xf(t)dt}{x^2}=\lim_{n\rightarrow \infty }\frac{\int_0^nf(t)dt}{n^2}=\lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{(n+1)^2-n^2}\\ &=&\lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{2n+1}=\frac12\lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{n} \end{eqnarray} and hence $$ \lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{n}=2 $$ Since $f(t)$ is increasing, one has $$ f(n)\le\int_n^{n+1}f(t)dt\le f(n+1) $$ and hence $$ \frac{f(n)}{n}\le\frac{\int_n^{n+1}f(t)dt}{n}\le \frac{f(n+1)}{n+1}\frac{n+1}{n}\le \frac{f(n+1)}{n+1}. $$ Letting $n\to\infty$, one obtains $$ \lim_{n\rightarrow \infty }\frac{f(n)}{n}=\lim_{n\rightarrow \infty }\frac{\int_{n}^{n+1}f(t)dt}{n}=2.$$ For general $x>0$, let $\lfloor x\rfloor=n$ and then $$ n\le x<n+1.$$ Since $f(t)$ is increasing, one has $$ f(n)\le f(x)\le f(n+1) $$ and hence $$ \frac{f(n)}{n+1}\le \frac{f(x)}{x}\le \frac{f(n+1)}{x}\le\frac{f(n+1)}{n}. $$ By the Squeeze Theorem, one has $$ \lim_{x\to\infty}\frac{f(x)}{x}=\lim_{n\to\infty}\frac{f(n)}{n}=2.$$

xpaul
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    The Stolz-Cesaro test says the INVERSE! If $$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ exists, then so does $$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}$$ and they are equal. Not the other way around! – Yiorgos S. Smyrlis Sep 28 '17 at 18:38
  • @YiorgosS.Smyrlis, see https://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf – xpaul Sep 28 '17 at 19:11
  • The link you gave says exactly the same thing as the comment from @YiorgosS.Smyrlis. What is needed here is to show that the ratio $f(x) /x$ tends to a limit. – Paramanand Singh Sep 28 '17 at 19:36
  • @ParamanandSingh, see the remark in the link. – xpaul Sep 28 '17 at 20:28
  • All those remarks are equivalent formulations. There is nothing new in the remarks apart from the form of the equations involved. Perhaps you misinterpreted some of those remarks. – Paramanand Singh Sep 29 '17 at 02:44
  • @ParamanandSingh, I don't know what you are talking about. – xpaul Oct 01 '17 at 12:35
  • I think my comment is perfectly understandable and unambiguous. You have to understand that your usage of Cesaro-Stolz is invalid. The proper solution to the problem is given in the linked question. The pdf which you have mentioned in your comment discusses Cesaro-Stolz in detail with many variations but what you have used is not described in the pdf and is not true in general. – Paramanand Singh Oct 01 '17 at 14:00