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let $a$ be a real number, $f: [a,\infty)\rightarrow \mathbb{R}$ be continuously differentiable, $f'(x)$ is bounded and $\int_{a}^{\infty}\left|f(x)\right|dx$ exists. Prove that $\lim_{x\rightarrow\infty}f(x)=0$. Hint: try using $\int_{a}^{\infty}f(x){f'(x)}dx$.

In order to solve this we can look at $\int_{a}^{M}f(x){f'(x)}dx$, and by using Integration by parts and the Newton-Leibniz formula we get that $\int_{a}^{M}f(x){f'(x)}dx = \frac{{f(M)}^2-{f(a)}^2}{2}$, thus $\lim_{M\rightarrow\infty}f(M)^2$ exists since $\int_{a}^{M}f(x){f'(x)}dx$ is bounded by $K \int_{a}^{M}\left|f(x)\right|dx$ which is given exists when $M\rightarrow \infty$ ($K$ is the bound of $|f'(x)|$).

Due to the continuity of $\sqrt{x}$ and limit rules, if we let $L^2$ be the limit of $f(x)^2$ then the limit of $|f(x)|$ is $|L|$.

We are left to prove that $L$ is $0$, what i tried was using the fact that $f(x)$ is Lipschitz continuous since $f'(x)$ is bounded, thus $f(x)$ is uniformly continous. We can then define $g(x)=|f(x)|$ and prove that $\lim_{x\rightarrow\infty}g(x)=0$.

In class we had proven that if $f(x)$ is non negative and uniformly continuous on $[1,\infty)$ and $\int_{1}^{\infty}f(x)dx<\infty$ then $\lim_{x\rightarrow\infty}f(x)=0$, and this fits the description of $g(x)$. (a proof can be also found here).

Is this proof correct and is there an easier way to prove this without having to use what i said we proved in class?

Thanks for the help!

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  • $\begingroup$ I don't follow your reasoning that shows that $\lim_M f(M)^2$ exists. $\endgroup$
    – copper.hat
    Feb 4, 2020 at 20:08
  • $\begingroup$ $\lim_{x\rightarrow\infty}f(x)^2 = 2\int_{a}^{\infty}f(x){f'(x)}dx +f(a)^2$. $\int_{a}^{\infty}|f(x)f'(x)|dx \leq \int_{a}^{\infty}|f(x)||f'(x)|dx \leq \int_{a}^{\infty}|f(x)|Kdx = K\int_{a}^{\infty}|f(x)|dx$ thus the right side of the first equation converges (it is given that the last integral exists). $\endgroup$
    – Titan3
    Feb 4, 2020 at 20:35
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    $\begingroup$ I see what you are trying to do (your limits are off), but it seems a roundabout way of doing things. You know that $|f|$ is Lipschitz and $\int |f| < \infty$.so you can just apply your in class result directly. $\endgroup$
    – copper.hat
    Feb 4, 2020 at 21:16

2 Answers 2

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Note that if $f'$ is bounded then $f$ is uniformly Lipschitz, with some constant $K$.

Note that if $|f(x_0)| >0$ then $\int_{|x-x_0| \le {1 \over K} |f(x_0)|} |f(x)| dx \ge {1 \over K} |f(x_0)|^2$.

Let $\epsilon>0$. If there is a sequence $x_n \to \infty$ such that $|f(x_n)| \ge \epsilon$ infinitely many times then $\int_a^\infty |f(x)|dx = \infty$, a contradiction.

Hence for any sequence $x_n \to \infty$ there are at most a finite number of $x_n$ such that $|f(x_n)| \ge \epsilon$.

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Why do you need all the stuff about the limit existing and converging to $L$? Why not just use that $f'$ bounded implies Lipschitz implies $f$ is uniformly continuous implies $|f|$ is uniformly continuous, since $|f(x)|=g(f(x)),$ with $g(x)=|x|$ being uniformly continuous (see here). Then just apply the theorem in class.

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  • $\begingroup$ because there was a hint saying " try looking at $\int_{a}^{M}f(x){f'(x)}dx$ ". $\endgroup$
    – Titan3
    Feb 4, 2020 at 20:12

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