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

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