# There exists a sequence of real numbers $(x_n)$ such that $\lim_{n \rightarrow \infty}x_n = \infty , \lim_{n \rightarrow \infty}x_nf(x_n)=\infty.$

Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $$\int_{-\infty}^{\infty}|f(t)|dt<\infty.$$ Prove that there exists a sequence of real numbers $(x_n)$ such that $$\lim_{n \rightarrow \infty}x_n = \infty , \lim_{n \rightarrow \infty}x_nf(x_n)=\infty.$$

This is one of my past exam questions. I have no idea how to start at all. Any hint would be appreciated.

• Why is $f(x)=0$ not a counterexample? – user156213 May 5 '17 at 1:07
• I strongly suspect the actual question has $\lim_{n\to \infty} x_n f(x_n) = 0$. The point being that if this did not hold, there would be some $\epsilon > 0$ such that $|f(x)| \ge \epsilon/ x$ for all sufficiently large $x$. And then use the comparison test... – Robert Israel May 5 '17 at 1:14
• @RobertIsrael Yes, an argument like $f \in L^1 \Rightarrow f \rightarrow 0$ faster than $\frac{1}{x}$ (loosely speaking) should work. Is the converse (which you asked about and deleted in comments) true? – Solomonoff's Secret May 5 '17 at 1:23
• No, e.g. $1/(x \log x)$ goes to $0$ faster than $1/x$ but is still not integrable at $\infty$. – Robert Israel May 5 '17 at 3:48