Existence of sequence for an integral function in $\mathbf{R}$. Is it true that for a function $f\in L^1({\mathbf R})$, there exists a sequence $\,x_n\rightarrow\infty$ such that  $x_nf(x_n)\rightarrow 0$?
 A: Assume that no sequence as requested exists.
Then for each sequence $x_n\to \infty$ we have $S((x_n)_n):=\limsup |x_n f(x_n)|>0$.
Let $s$ be the infimum of all $S((x_n)_n)$ when $(x_n)_n$ runs over all such sequences.
If $s=0$, then for each $m\in \mathbb N$ there is a sequence $x_n^{(m)}\to \infty$ with $|x_n^{(m)}f(x_n^{(m)})|<\frac1m$ infinitely often.
Define recursively $y_0=0$ and $y_m=x_n^{(m)}$ with $n$ chosen big enough to gurarantee both $|x_n^{(m)}f(x_n^{(m)})|<\frac1m$ and $x_n^{(m)}>y_{m-1}+1$.
Then $y_n\to\infty$ and $|y_nf(y_n)|<\frac 1n$, i.e. $y_nf(y_n)\to0$, contrary to assumption.
Hence $s>0$. Then $|f(x)|>\frac s{2x}$ for all sufficiently big $x$ (for all $x>a$, say).
Indeed, if for every $n\in \mathbb N$ there were an $x_n>n$ with $|f(x_n)|\le\frac s{2x}$, then for this sequence $(x_n)_n$ we have $S((x_n)_n)\le \frac s2 < s$, contradiction.
Therefore $$\int_{\mathbb R}|f(x)|dx\ge \int_{a}^\infty \frac s{2x}dx=\infty$$
and $f$ is not $L^1$.

If I'm not mistaken, the proof above can easily be generalized to:
Assume $f\in L^1(\mathbb R)$, $a,b\in \mathbb R$, $b>0$, $g(a,\infty)\to(0,b)$ and $\int_a^\infty g(x)dx=\infty$. Then there exists a sequence $x_n\to \infty$ with $\frac{f(x_n)}{g(x_n)}\to 0$.
