# $\lim_{x\rightarrow\infty}f(x)=0$ given that f is continuously differentiable, $f'(x)$ is bounded and $\int_{a}^{\infty}\left|f(x)\right|dx$ exists

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!

• I don't follow your reasoning that shows that $\lim_M f(M)^2$ exists. Feb 4, 2020 at 20:08
• $\lim_{x\rightarrow\infty}f(x)^2 = 2\int_{a}^{\infty}f(x){f'(x)}dx +f(a)^2$. $\int_{a}^{\infty}|f(x)f'(x)|dx \leq \int_{a}^{\infty}|f(x)||f'(x)|dx \leq \int_{a}^{\infty}|f(x)|Kdx = K\int_{a}^{\infty}|f(x)|dx$ thus the right side of the first equation converges (it is given that the last integral exists). Feb 4, 2020 at 20:35
• 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. Feb 4, 2020 at 21:16

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