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

Question

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?

Answer 1

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 $a_{n}=e^{-b_{n}}$.