Let $f$ be a non-negative differentiable function such that $f'$ is continuous and $\int_{0}^{\infty}f(x)\,dx$ and $\int_{0}^{\infty}f'(x)\,dx$ exist.

Question

Let $f$ be a non-negative differentiable function such that $f'$ is continuous and $\displaystyle\int_{0}^{\infty}f(x)\,dx$ and $\displaystyle\int_{0}^{\infty}f'(x)\,dx$ exist.

Prove or give a counter example: $f'(x)\overset{x\rightarrow \infty}{\rightarrow} 0$

Note: I think it is not true but I couldn't find a counter example.

Answer 1

Let $\varphi(x)=\exp\left(\dfrac{1}{3}-\dfrac{1}{4-x^{2}}\right)$ for $|x|\leq 2$, $\varphi(x)=0$ for $|x|>2$.

Let $f(x)=\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}\varphi\left(2^{n}(x-n)\right)$, one may check that $f\in C^{\infty}(0,\infty)$ and that $f,f'\in L^{1}(0,\infty)$.

For all $x$ with $1<2^{n}(x-n)\leq 2$, that is, $n+\dfrac{1}{2^{n}}<x\leq n+\dfrac{2}{2^{n}}$, we have \begin{align*} f'(x)&=\dfrac{1}{2^{n}}\exp\left(\dfrac{1}{3}-\dfrac{1}{4-(2^{n}(x-n))^{2}}\right)\cdot-\dfrac{2(2^{n}(x-n))}{(4-(2^{n}(x-n))^{2})^{2}}\cdot 2^{n}\\ &=-\dfrac{2(2^{n}(x-n))}{(4-(2^{n}(x-n))^{2})^{2}}\exp\left(\dfrac{1}{3}-\dfrac{1}{4-(2^{n}(x-n))^{2}}\right), \end{align*} localizing to $x=n+1/2^{n}$ we have $f'(n+1/2^{n})=-2/9$.

Answer 2

Let $g_n(x) = \sin (n^2 x) 1_{[0,{2 \pi \over n^2}]}$ and note that $g_n$ is continuous, $\int |g_n| = {2 \over n^2}$, $\int g_n = 0$ and if $f_n(x)=\int_0^x g_n$, then $f_n(x) \ge 0$ for all $x$. Furthermore, $\int |f_n| \le {4 \pi \over n^3}$.

Define $f(x) = \sum_{n \ge 1} f_n(x-n)$, note that $f$ is integrable and non negative. Furthermore, $f'(x) = \sum_{n \ge 1} g_n(x-n)$ and it is straightforward to check that $f'$ is integrable as well.