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Suppose that $g$ is a continuous, nonincreasing and nonnegative function on $(0,1)$. The question is whether one can characterize the integrability of such functions at zero by their decay rates at zero? Will the iterated logarithms suffice?

More precisely, will $\int_{0+}g(s)d s<\infty$ imply that for some $\alpha>1$, $m>1$, $C>0$, and $\epsilon\in(0,1)$,

$$ g(s)\le \frac{C}{s \left[\log^m(1/s)\right]^\alpha \prod_{i=1}^{m-1}\log^i(1/s)}, \qquad\text{for all $s\in (0,\epsilon)$,} $$

where $\log^n(x):=\log\dots\log(x)$ (the n-th iterated logarithm) and we use the convention that $\prod_{i=1}^0\equiv 1$? The reverse is clearly true.

Anand
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  • This is hopeless. If $g\in L^1$, then so is $fg$ for a suitable $f$ with $f\to\infty$. Also, the question would have been better suited for MSE. –  Jun 19 '16 at 17:09
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    Thanks Professor Christian Remling for your comments, but I am sorry that I cannot get the point. I have more conditions on $g$, such as continuity and monotonicity. Do you think these extra conditions cannot help? I appreciate for any more hints. – Anand Jun 19 '16 at 17:18
  • Thanks Professor Gerald Edgar for this link! – Anand Jun 19 '16 at 17:35
  • You can also look at the analogous problems for series as a warm up: If $a_n>0$, $\sum a_n<\infty$, then there are $b_n\to \infty$ such that $\sum b_na_n$ still converges. –  Jun 19 '16 at 17:44
  • Thanks Professor Christian Remling for this explanation. I didn't see any contradictions with my proposed statement. – Anand Jun 19 '16 at 18:29
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    This is a slight variant of http://math.stackexchange.com/questions/452053/is-there-a-slowest-rate-of-divergence-of-a-series – GEdgar Jun 20 '16 at 00:25

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