2

Let $f$ be a non negative continuous function on $[0,\infty)$ such that $$\lim_{x\to\infty}\int_x^{x+1}f(y)dy=0.$$ How do we prove that $$\lim_{x\to\infty}\frac{\int_0^{x}f(y)dy}{x}=0.$$ If we see this question using the primitive function, do we have the following result for a continuous function $F$: $$\lim_{x\to\infty}F(x+1)-F(x)=0$$ implies that $$\lim_{x\to\infty}\frac{F(x)}{x}=0.$$

Martin Sleziak
  • 56,060
user165633
  • 3,001

2 Answers2

2

Hint:

  1. Since $\lim\limits_{x\to\infty}\int_x^{x+1}f(t)dt$ then, by the definition of the limit and since $f$ is non-negative, for any $\epsilon>0$ there exist an $x_*$ s.t. if $x>x_*$ then $ 0 \leq \int_x^{x+1}f(t)dt < \frac{\epsilon}{2}$.

  2. We have $\frac{\int_0^xf(t)dt}{x} = \frac{\int_0^{x_*}f(t)dt}{x} + \frac{\int_{x_*}^xf(t)dt}{x} $

  3. The last term in 2) is by 1) bounded by $\frac{(x-x_*)\epsilon}{2x} < \frac{\epsilon}{2}$ since $\int_{x_*}^xf(t)dt = \int_{x_*}^{x_*+1}f(t)dt +\int_{x_*+1}^{x_*+2}f(t)dt + \ldots + \int_{x_*+\lfloor x-x_*\rfloor}^{x}f(t)dt$.

  4. The first term in 2) can be made as small as possible ($<\frac{\epsilon}{2}$) by taking $x$ large enough.

Winther
  • 25,313
  • Is it possible to formalize this proof rigorously like: for $\epsilon>0$, find $x_\epsilon$ such that for all $x>x_\epsilon$, $\frac{\int_0^x f(y)dy}{x}\leq \epsilon$. I just can't find that $x_\epsilon$ – user165633 Feb 14 '15 at 18:43
  • @user170243 Yes it is. Try to take $x_\epsilon = \frac{2\int_0^{x_}f(t)dt}{\epsilon}$ then if $x>x_\epsilon$ then the first term in 2. is smaller than $\epsilon/2$. Now if $x>x_$ and $x>x_\epsilon$ then you get $|\int_0^xf(t)dt / x| < \epsilon/2 + (1-\frac{x_*}{x})\epsilon/2 < \epsilon$ – Winther Feb 14 '15 at 18:49
  • How much is k? can k tend to $\infty$ and leave you with sum of too many terms? – Arashium Feb 14 '15 at 20:26
  • @Arashium This is just bad notation on my behalf. Given $x,x,*$ then there is a unique integer $k$ s.t. $x = x* + k + \delta$ where $\delta < 1$. So $k = \lfloor x-x_*\rfloor$. This is the $k$ I'm talking about. – Winther Feb 14 '15 at 20:34
  • @Winther I think there is no problem with your notation. $x_*$ is constant for each $\epsilon$ but as $x$ increases, $k$ increase too (boundlessly). – Arashium Feb 14 '15 at 20:38
  • 1
    @Arashium There are $\lfloor x- x_\rfloor$ terms in the sum and each term is less than $\epsilon/2$ so $\int_{x_}^xf(t)dt < \lfloor x- x_\rfloor\cdot \epsilon/2$ which gives $\frac{\int_{x_}^xf(t)dt}{x} < \frac{\lfloor x- x_*\rfloor}{x} \cdot \frac{\epsilon}{2}$. The $1/x$ kills this growth. – Winther Feb 14 '15 at 20:40
0

Continuing from your last step in question. see that $F(x)$ is a differentiable function.

Then $F(x+1)-F(x)=F'(\eta_x)(x+1-x)=F'(\eta_x)$ where $x\leq \eta_x \leq x+1$.(By mean value theorem)

since $\lim_{x\rightarrow \infty}(F(x+1)-F(x))=0\implies \lim_{x\rightarrow \infty}F'(\eta_x)=0$

Now $$ \lim_{x\rightarrow \infty}(\frac{F(x)}{x})=\lim_{x\rightarrow \infty}F'(x)=\lim_{\rightarrow \infty}F'(\eta_x)=0 $$

Use L'Hospital's rule in first inequality

Harish
  • 858
  • 6
  • 17
  • $\lim_{x\to \infty }F'(x)=\lim_{x\to \infty }F'(\eta_x)$ is not true a priori ! Moreover the limit $\lim_{x\to\infty }\frac{F(x)}{x}$ Is not necessarily indeterminate, and thus l'Hopital rule is not useable a priori ! – Surb Feb 14 '15 at 17:52
  • Thanks, For correcting me. – Harish Feb 14 '15 at 17:57