TITLE: $\lim\limits_{x\to \infty} xf(x)$ if $f(x)$ is a positive monotonically decreasing function.
QUESTION [2 upvotes]: There might be several simple solutions to the following problem (I would be interested in knowing them too). However the main intention of this post is verification. I wanted to know if the argument made here is flawless.

$\textbf{Question.}$   Let $f: \mathbb{R}^+ \to \mathbb{R}^+$ be a monotonically decreasing function. If $xf(x)$ does not converge to $0$,
then there exist $\epsilon > 0$ and $M>0$ such that for all $x\geq M$,
\begin{equation*}
  xf(x) \geq \epsilon. 
 \end{equation*}

$\textbf{Attempt.}$ Let us suppose the contrary. That is

$\textbf{Statement 1:}$  Given $\epsilon >0$, whatever the real number $M>0$ we choose, it is possible to find $x>M$ such that $xf(x) < \epsilon$.

From the given condition that $\lim\limits_{x\to\infty}xf(x) \neq 0$, we should have an $\epsilon_0 > 0$ for which

$\textbf{Statement 2.}$ Whatever $M>0$ is chosen, it is possible to find $x>M$, such that $xf(x) > \epsilon_0$.

Before moving further, we shall establish a fact for the given situation. Statement $2$ implies that it is possible to construct a increasing divergent sequence $\{x_k\}$ such that $x_k f(x_k) \geq \epsilon_0$. We shall show that for any such sequence $\{x_k\}$, the corresponding series $\sum f(x_k)$ diverges. Given any $x_k$ it is possible to choose a natural number $n_k$ such that $x_k < n_k$. We can thus assume to have constructed an increasing sequence $\{n_k\}$ such that $n_1 < n_2< \dots$ and $x_k < n_k$. Construct now a sequence $b_n$ such that $b_{n_k} = f(x_k)$ and $b_m = 0$ for the rest of all. Then
\begin{equation*}
 \sum b_n = \sum f(x_n). 
\end{equation*}
(The above means, if one diverges, other also diverges. If one converges, the other too converges to the same limit value). Clearly
\begin{equation*}
 n_kb_{n_k} \geq x_kf(x_k) \geq \epsilon_0. 
\end{equation*}
This implies $\sum b_n$ does not converge (because if $\sum a_n$ converges, with $a_n >0$, then $na_n \to 0$ as $n\to \infty$). Hence $\sum f(x_n)$ diverges.
We now see what happens if the Statement 1 is also true.  Statement 1 and Statement 2 allow us to construct two sequences $\{x_n\}$ and $\{y_n\}$ as follows:
\begin{equation*}
 x_1 < y_1 \quad \text{and} \quad y_{n-1} < x_n < y_n \quad \text{for all $n \geq 2$} , 
\end{equation*}
\begin{equation*}
 y_n > n^2
\end{equation*}
and
\begin{equation*}
 x_nf(x_n) \geq \epsilon_0 \quad \text{and} \quad y_nf(y_n) < \epsilon_0. 
\end{equation*}
Clearly
\begin{equation*}
 f(y_n) < \frac{\epsilon_0}{y_n}< \frac{\epsilon_0}{n^2}. 
\end{equation*}
This implies that $\sum f(y_n)$ converges. Now $y_{n-1} < x_n$ for all $n\geq 2$. Since $f$ is decreasing, this implies that $f(x_n) \leq f(y_{n-1})$. This implies that $\sum f(x_n)$ converges, which cannot happen in the given situation. Hence Statement 1 cannot hold good. $\blacksquare$

REPLY [1 votes]: The claim is false. Consider your favourite strictly increasing sequence $a_n\nearrow \infty$ such that $\lim_{n\to\infty}\frac{a_{n}}{a_{n+1}}=0$ (namely, $a_n=n^n$). Make it $a_0=0$ for convenience. Then call $n(x)=\min\{n\in \Bbb N\,:\, a_n\ge x\}$. Then call $$f(x)=\frac1{a_{n(x)}}$$
Notice that eventually $a_{n+1}> a_n+n^{-1}$ and $a_n-n^{-1}>a_{n-1}$, and therefore  $$\lim_{n\to\infty}(a_n+n^{-1})f(a_n+n^{-1})=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}+\frac1{na_{n+1}}=0\\ \lim_{n\to\infty}(a_n-n^{-1})f(a_n-n^{-1})=\lim_{n\to\infty}1-\frac{1}{na_n}=1$$
This function can be adapted into a $C^\infty$ function by smoothing out the jumps, without changing any of the relevant features. You can also make it strictly decreasing at any time, by adding $e^{-x^2}$.
{"set_name": "stack_exchange", "score": 2, "question_id": 4246561}