Series and lim sup In the course of some calculations, I've come upon the following question: When is it true that
$$ \limsup_{t \to 0^+} \sum_m |f_m(t)| \leq \sum_m \sup_{0 < t < 2^{-m}} |f_m(t)|\; ? $$
It's clear that this does not hold in general; one could take for instance $f_m(t) = t/m$, which makes the LHS infinite, while the RHS converges.
I've considered using the reverse Fatou's lemma/dominated convergence theorem, but I don't have a dominating function available in my context.
I'd like to conclude that the inequality holds when the LHS is finite, but would welcome counterexamples/other conditions which make it correct. 
 A: Of course it's true if, say, $|f_m(t)|$ is a decreasing function of $t$. But that's sort of stupid. I doubt that there is any "interesting" condition that makes this true, since the LHS involves $f_m(t)$ for pairs $(m,t)$ that simply don't come up on the RHS.
No, assuming the LHS is finite is not enough. Say $$f_m(t)=\begin{cases}
1,&(t=2^{-m}),
\\0,&(t\ne 2^{-m}).\end{cases}$$
Then LSH $=1$, RHS $=0$. You could convert this to a continuous counterexample: Say $f_m$ is a continuous function supported on $[2^{-m},3\cdot 2^{-m}]$ with $0\le f_m\le 1$ and $f_m(2\cdot 2^{-m})=1$.
A: This does not hold in general even if both sides are finite.
To build a counterexample, it suffices to put a significant weight on the part that the RHS does not see. I will build an example with non continuous functions, but it is not difficult to smooth them.
I assume $m\geq 1$.
First, define $f_m(t)$ on the interval $[0,2^{-m})$ by $$f_m(t)=2^{-m}\qquad \text{if $0\leq t<2^{-m}$}$$
Clearly, your RHS is equal to $1$. Now choose the remaining parts of the $f_m$ large enough so as to make the LHS bigger than $1$, but small enough to make it finite. For instance:
$$f_m(t)=\cases{1&if $2^{-m}\leq t<2^{-m+1}$\\0&if $2^{-m+1}\leq t$}$$
will result in the LHS being equal to $2$.
