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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?

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  • $\begingroup$ 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 $a_n\rightarrow \infty$. $\endgroup$ Dec 16, 2021 at 15:43
  • $\begingroup$ 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. $\endgroup$ Dec 16, 2021 at 15:50
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    $\begingroup$ What if $f$ is the zero function? $\endgroup$ Dec 16, 2021 at 16:25
  • $\begingroup$ 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. $\endgroup$
    – user284331
    Dec 16, 2021 at 16:31

1 Answer 1

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

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