Let $ f:[1,\infty )\longrightarrow (0,\infty ) $ be a continuous function satisfying the following properties: $ \text{(i)}\exists\lim_{x\to\infty } \frac{f(x)}{x}\in\overline{\mathbb{R}} $ $ \text{(ii)}\exists\lim_{x\to\infty } \frac{1}{x}\int_1^x f(t)dt\in\mathbb{R}. $ a) Show that $ \lim_{x\to\infty } \frac{f(x)}{x}=0. $ b) Prove that $ \lim_{x\to\infty } \frac{1}{x^2}\int_1^x f^2(t)dt=0. $
Problem
Source: Romania National Olympiad 2014, Grade XII, Problem 3
Tags: function, real analysis
Aniv
03.03.2019 23:38
Considering $g(x)=\int_{1}^{x}f(t)dt$, then we have
$L=\lim_{x\to\infty } \frac{1}{x}\int_1^x f(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}{\int_1^x f(t)dt}}{\frac{d}{dx}x}=\lim_{x\to\infty}f(x)$
so $\lim_{x\to\infty}f(x)=L$, so $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^xf(x)}{e^x}=\lim_{x\to\infty}\frac{\frac{d}{dx}(e^xf(x))}{\frac{d}{dx}e^x}=\lim_{x\to\infty}(f(x)+f'(x))\Rightarrow \lim_{x\to\infty}f'(x)=L-L=0$
so we have shown that $\lim_{x\to\infty}f'(x)=0$
Now as we have $\lim_{x\to\infty}f'(x)=0$, so given any $p>0$, $\exists M>0$, such that $|f'(x)|<\frac{p}{2}$, $\forall x\ge M$
Now we have from mean value theorem
$|f(x)-f(M)|=|f'(c_x)||x-M|<\frac{p}{2}x$, $\forall x>M$, $\exists c_x\in (M,x)$ and $|f'(c_x)|<\frac{p}{2}$
Then we have $|\frac{f(x)}{x}-\frac{f(M)}{x}|<\frac{p}{2}$, $\forall x>M$
so $|\frac{f(x)}{x}|\le |\frac{f(x)}{x}-\frac{f(M)}{x}|+|\frac{f(M)}{x}|<2\frac{p}{2}=p$, $\forall x>\max \{M,\frac{2}{p}|f(M)|\}$
or $|\frac{f(x)}{x}|<p$, $\forall x>\max \{M,\frac{2}{p}|f(M)|\}$, Hence we have $\lim_{x\to\infty}\frac{f(x)}{x}=0$
Now considering $\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^{x}f^{2}(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}\int_{1}^{x}f^{2}(t)dt}{\frac{d}{dx}x^2}=\lim_{x\to\infty}\frac{f^{2}(x)}{2x}=\lim_{x\to\infty}\frac{\frac{d}{dx}f^{2}(x)}{\frac{d}{dx}(2x)}=\lim_{x\to\infty}f(x)f'(x)$
or $\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^{x}f^{2}(t)dt=\lim_{x\to\infty}f(x)f'(x)=0$, as $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=0$
Masac9
04.03.2019 06:35
a) suppose $\lim_{n\to \infty} f(x)/x \neq 0 $ then we have to prove $ \lim_{n\to \infty} f(x)/x =\lambda ,\lambda>0$ otherwise take $k=-\lambda$ but $|f(x)/x-\lambda |<|f(x)-x\lambda |(since x\in [0,\infty))<|f(x)|+|\lambda x| $ now we cannot find any big $M>0$ such that the RHS is $<\epsilon ,for all x\geq M$ contradiction .
Masac9
04.03.2019 06:40
Aniv wrote:
Considering $g(x)=\int_{1}^{x}f(t)dt$, then we have
$L=\lim_{x\to\infty } \frac{1}{x}\int_1^x f(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}{\int_1^x f(t)dt}}{\frac{d}{dx}x}=\lim_{x\to\infty}f(x)$
so $\lim_{x\to\infty}f(x)=L$, so $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^xf(x)}{e^x}=\lim_{x\to\infty}\frac{\frac{d}{dx}(e^xf(x))}{\frac{d}{dx}e^x}=\lim_{x\to\infty}(f(x)+f'(x))\Rightarrow \lim_{x\to\infty}f'(x)=L-L=0$
so we have shown that $\lim_{x\to\infty}f'(x)=0$
Now as we have $\lim_{x\to\infty}f'(x)=0$, so given any $p>0$, $\exists M>0$, such that $|f'(x)|<\frac{p}{2}$, $\forall x\ge M$
Now we have from mean value theorem
$|f(x)-f(M)|=|f'(c_x)||x-M|<\frac{p}{2}x$, $\forall x>M$, $\exists c_x\in (M,x)$ and $|f'(c_x)|<\frac{p}{2}$
Then we have $|\frac{f(x)}{x}-\frac{f(M)}{x}|<\frac{p}{2}$, $\forall x>M$
so $|\frac{f(x)}{x}|\le |\frac{f(x)}{x}-\frac{f(M)}{x}|+|\frac{f(M)}{x}|<2\frac{p}{2}=p$, $\forall x>\max \{M,\frac{2}{p}|f(M)|\}$
or $|\frac{f(x)}{x}|<p$, $\forall x>\max \{M,\frac{2}{p}|f(M)|\}$, Hence we have $\lim_{x\to\infty}\frac{f(x)}{x}=0$
Now considering $\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^{x}f^{2}(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}\int_{1}^{x}f^{2}(t)dt}{\frac{d}{dx}x^2}=\lim_{x\to\infty}\frac{f^{2}(x)}{2x}=\lim_{x\to\infty}\frac{\frac{d}{dx}f^{2}(x)}{\frac{d}{dx}(2x)}=\lim_{x\to\infty}f(x)f'(x)$
or $\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^{x}f^{2}(t)dt=\lim_{x\to\infty}f(x)f'(x)=0$, as $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=0$
I think you can't use LMVT as here don't say that $f$ is differentiable on $(1,\infty)$
Aniv
04.03.2019 09:05
Masac9 wrote: Aniv wrote:
Considering $g(x)=\int_{1}^{x}f(t)dt$, then we have
$L=\lim_{x\to\infty } \frac{1}{x}\int_1^x f(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}{\int_1^x f(t)dt}}{\frac{d}{dx}x}=\lim_{x\to\infty}f(x)$
so $\lim_{x\to\infty}f(x)=L$, so $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^xf(x)}{e^x}=\lim_{x\to\infty}\frac{\frac{d}{dx}(e^xf(x))}{\frac{d}{dx}e^x}=\lim_{x\to\infty}(f(x)+f'(x))\Rightarrow \lim_{x\to\infty}f'(x)=L-L=0$
so we have shown that $\lim_{x\to\infty}f'(x)=0$
Now as we have $\lim_{x\to\infty}f'(x)=0$, so given any $p>0$, $\exists M>0$, such that $|f'(x)|<\frac{p}{2}$, $\forall x\ge M$
Now we have from mean value theorem
$|f(x)-f(M)|=|f'(c_x)||x-M|<\frac{p}{2}x$, $\forall x>M$, $\exists c_x\in (M,x)$ and $|f'(c_x)|<\frac{p}{2}$
Then we have $|\frac{f(x)}{x}-\frac{f(M)}{x}|<\frac{p}{2}$, $\forall x>M$
so $|\frac{f(x)}{x}|\le |\frac{f(x)}{x}-\frac{f(M)}{x}|+|\frac{f(M)}{x}|<2\frac{p}{2}=p$, $\forall x>\max \{M,\frac{2}{p}|f(M)|\}$
or $|\frac{f(x)}{x}|<p$, $\forall x>\max \{M,\frac{2}{p}|f(M)|\}$, Hence we have $\lim_{x\to\infty}\frac{f(x)}{x}=0$
Now considering $\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^{x}f^{2}(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}\int_{1}^{x}f^{2}(t)dt}{\frac{d}{dx}x^2}=\lim_{x\to\infty}\frac{f^{2}(x)}{2x}=\lim_{x\to\infty}\frac{\frac{d}{dx}f^{2}(x)}{\frac{d}{dx}(2x)}=\lim_{x\to\infty}f(x)f'(x)$
or $\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^{x}f^{2}(t)dt=\lim_{x\to\infty}f(x)f'(x)=0$, as $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=0$
I think you can't use LMVT as here don't say that $f$ is differentiable on $(1,\infty)$ In that case, Lets try without considering $f(x)$ to be Differentiable !! consider $g(x)=\int_1^x f(t)dt$, and we have $g'(x)=f(x)$ as $g(x)$ is Differentiable we have $a=\lim_{x\to\infty}\frac{g'(x)}{x}=\lim_{x\to\infty } \frac{f(x)}{x} $ and $b=\lim_{x\to\infty } \frac{1}{x}\int_{1}^xf(t)dt=\lim_{x\to\infty } \frac{g(x)}{x}\ $ ; $a,b\in \mathbb{R}$ as we have both $\lim_{x\to\infty}\frac{f(x)}{x}$ and $\lim_{x\to\infty } \frac{1}{x}\int_{1}^xf(t)dt$ exists! now $g(x)$ is differentiable and $b=\lim_{x\to\infty } \frac{g(x)}{x} =\lim_{x\to\infty } \frac{xg(x)}{x^2}=\lim_{x\to\infty } \frac{g(x)+xg'(x)}{2x}=\frac{b}{2}+\frac{1}{2}\lim_{x\to\infty }g'(x)\ $, by using L'Hospital's Rule so $\lim_{x\to\infty }g'(x)=b=\lim_{x\to\infty } \frac{g(x)}{x}$, hence $\lim_{x\to\infty}\frac{g'(x)}{x}=\lim_{x\to\infty } \frac{b}{x} =0$ or $L_1=\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}\frac{g'(x)}{x}=0$........((a) Proved) now again by using L'Hospital's Rule $L_2=\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^xf^2(t)dt=\lim_{x\to\infty}\frac{\frac{d}{dx}\int_{1}^xf^2(t)dt}{\frac{d}{dx}x^2}=\lim_{x\to\infty}\frac{f^2(x)}{2x}=\frac{1}{2}\lim_{x\to\infty}f(x)\left(\frac{f(x)}{x}\right)=\lim_{x\to\infty}g'(x)\left(\frac{f(x)}{x}\right)$ or $L_2=\lim_{x\to\infty}\frac{1}{x^2}\int_{1}^xf^2(t)dt=\lim_{x\to\infty}g'(x)\left(\frac{f(x)}{x}\right)=b\lim_{x\to\infty}\left(\frac{f(x)}{x}\right)=0$...........((b) Proved) So $L_1=L_2=0\ $ $\square$
Tsikaloudakis
11.03.2019 12:25
$\begin{array}{l} \left. \begin{array}{l} \forall \mathop {}\nolimits_{}^{} x \ge 1\mathop {}\limits_{}^{} :{}_{}^{} f(x) > 0 \\ {}_{} \\ F(x) = \int_1^x {f(t)dt} \\ \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} \forall \mathop {}\nolimits_{}^{} x > 1\mathop {}\limits_{}^{} :\mathop {}\limits_{}^{} F(x) > 0 \\ \mathop {}\nolimits_{}^{} \\ \forall _{}^{} x \ge 1\mathop {}\nolimits_{}^{} :f(x) > 0\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} \mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = l_1 \ge 0 \\ \mathop {}\nolimits_{}^{} \\ If\mathop {}\nolimits_{}^{} \mathop {}\limits_{}^{} l_1 > 0\mathop {}\nolimits_{}^{} then\mathop {}\nolimits_{} \mathop {\lim }\limits_{x \to + \infty } f(x) = + \infty \\ \mathop {}\limits_{}^{} \\ Then\mathop {}\limits_{}^{} \left\{ \begin{array}{l} l_2 = \mathop {\lim }\limits_{x \to + \infty } \frac{{F(x)}}{x} = \mathop {\lim }\limits_{x \to + \infty } f(x) = + \infty \\ l_2 \in ( - \infty , + \infty )\mathop {}\limits_{}^{} \mathop {}\nolimits_{}^{} {\rm{ }}\mathop {}\nolimits_{}^{} {\rm{contradiction}} \\ \end{array} \right.\mathop {}\nolimits_{}^{} and\mathop {}\nolimits_{}^{} then\mathop {}\nolimits_{}^{} \mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = l_1 = 0 \\ \end{array}$
Tsikaloudakis
11.03.2019 14:00
But I think there is no such continuous function: $f:[1, + \infty ) \to (0, + \infty )$ would be more correct: $f:(0, + \infty ) \to (0, + \infty )$
Aniv
11.03.2019 14:25
Tsikaloudakis wrote: But I think there is no such continuous function: $f:[1, + \infty ) \to (0, + \infty )$ would be more correct: $f:(0, + \infty ) \to (0, + \infty )$ Actually i think it would be more accurate if $f:(1,\infty)\to(0,\infty)$