# Existence of sequence $a_n$ such that $a_n \to 0$ and $a_nf(a_n) \to \infty$ as $n \to \infty$

Suppose that for $$f: \mathbb{R} \to \mathbb{R}$$ we have that $$\int_{\Bbb R} |f| < \infty$$. Show that there exists a $$a_n \in \Bbb R$$ and $$n \in \Bbb N_1$$ for which $$a_n \to 0 \text{ and } a_nf(a_n) \to \infty, \text{ when n \to \infty}$$

Working towards contradiction we can suppose that $$a_n \to \infty$$ and $$a_nf(a_n) \to 0$$ as $$n \to \infty$$.

Now the first bit means that $$\exists n_0$$ such that $$|a_n| < \varepsilon$$, whenever $$n \ge n_0$$.

The second bit means that $$\exists n_1$$ such that for any $$M> 0$$ we have that $$a_nf(a_n) >M$$ for $$n \ge n_1$$.

I'm trying to combine these facts to get a contradiction for $$\int_{\Bbb R} |f|< \infty$$, but I don't know how?

• You are using the contradiction method wrong. You want to prove that if no $(a_n)_n$ exists with the desired properties, then $f$ is not absolutely integrable. Furthermore, the contrary of $a_n\rightarrow 0$ is not $a_n\rightarrow \infty$. Dec 16, 2021 at 15:43
• What you can do if you want to use contradiction, is to take a concrete sequence converging to 0 such as $a_n = 1/n$ and use that $a_n f(a_n)$ cannot diverge. Dec 16, 2021 at 15:50
• What if $f$ is the zero function? Dec 16, 2021 at 16:25
• Not only that, if $f$ vanishes on a neighborhood of zero, but yet integrable on the real line, for example, $f(x)=0$ on $|x|<1$ and $f(x)=1/x^{2}$ for $|x|\geq 1$, the statement does not hold. Dec 16, 2021 at 16:31

EDIT: Personally I think the question has some flaw, as David Mitra has pointed out that, a counterexample is simply the zero function, the following addresses on the result that $$a_{n}f(a_{n})\rightarrow 0$$ for some $$a_{n}\rightarrow 0$$.
Note that \begin{align*} \int_{-\infty}^{\infty}|f(x)|dx=\int_{0}^{\infty}|f(e^{-y})|e^{-y}dy. \end{align*} Since \begin{align*} \int_{0}^{\infty}|f(e^{-y})|e^{-y}dy<\infty \end{align*} we must have \begin{align*} \liminf_{y\rightarrow\infty}|f(e^{-y})|e^{-y}=0. \end{align*} Then there exists a sequence $$(b_{n})$$ such that $$b_{n}\rightarrow\infty$$ and \begin{align*} \lim_{n\rightarrow\infty}|f(e^{-b_{n}})|e^{-b_{n}}=0. \end{align*} Now we simply take $$a_{n}=e^{-b_{n}}$$.