Let $I$ (integrable) be the set of continuous functions $f:\mathbb R_+\to\mathbb R_+$ that are integrable and nonincreasing. Let $D$ (divergent) be the set of continuous functions $g:\mathbb R_+\to\mathbb R_+$ with $\lim_{x\to\infty}g(x)=\infty$.
Claim. For any $f\in I$ there exists $g\in D$ such that $fg\in I$.
Proof. This is well-known, and the equivalent question for series appeared many times on this site: take $$ g(x) = \left[\int_x^\infty f(t)dt\right]^{-1/2}. $$ My question. Can we find a sequence $(f_n)$ of functions in $I$ for which there does not exist $g\in D$ such that $f_ng\in I$ for all $n$?
This question is closely related to the nonexistence of a boundary between divergent and convergent series. The sequence $f_n$ would somehow "converge" to that boundary, which does not exist, and intuitively this suggests that $(f_n)$ cannot exist.
A prototypical attempt of $f_n$ is as follows. Let $\log^{(1)}(x)=\max(1,\log x)$ and $\log^{(n)}(x)$ the $n$-fold composition of $\log^{(1)}$ (well-defined and $\ge1$ for all $x\ge0$). Set $$ f_n(x)=\frac1{x\log^{(1)}(x)\dots\log^{(n)}(x)} \frac1{\log^{(n)}(x)}. $$ Of course $f_n$ are very close to nonintegrable functions, which suggests that there is no corresponding $g$. However there is such a $g$: Let $a_n=\exp(\exp(\dots (\exp(n)))$ (exponentiation $n$ times) and define $g(a_n)=n$, and extend by linear interpolation. Clearly $g$ is monotone and divergent. Moreover $g(x)\le \log^{(n)}(x)$ for all $x\ge a_{n+1}$, which shows that $f_ng$ is integrable for all $n$.
I would happily accept an answer to the analogous question for series, as I believe that there shouldn't be a difference when dealing with functions instead of series.
Yes, an answer for series can easily be converted to an answer for(continuous) functions.
Notation for a sequence of sequences is awkward, so let's talk about functions $f_n,g:\Bbb N\to(0,\infty)$ instead.
Proof: Choose $N_1<N_2<\dots$ so that $$\sum_{j=N_k}^\infty f_n(j)<1/k^3,\quad(n=1,2,\dots, k).$$
Define $g$ by $$g(j)=k,\quad(N_j\le j<N_{k+1}).$$
Then for every $n$, if $k>n$ we have $$\sum_{j=N_k}^{N_{k+1}-1} f_n(j)g(j)\le1/k^2.$$