Let $f:[0,1]\to\mathbb{R}\_+$ be a convex, strictly increasing function such that $f(0)=0$ (typically, $f$ is very flat at $0$, i.e. increases very slowly). I would like to prove or disprove the following statement:



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> For all decreasing sequence $(a\_n)\_{n\ge 0}$ of positive reals such that $\lim a\_n = 0$, there exist a sequence $(b\_n)\_{n\ge 0}$ of positive reals such that (1) $\lim b\_n =0$ and (2) for all sequence $(p\_n)\_{n\ge 0}$ of non-negative reals such that $\sum p\_n = 1$:
> $$f\big(\sum b\_n p\_n \big) \ge \sum a\_n p\_n.$$
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(**Edit**: *replaced $\sum p\_n\le 1$ by $\sum p\_n =1$, since the former condition makes the answer obviously negative, as pointed out in comment. This modification should have no bearing on my intended application*)


When $f$ is very flat, we thus want to take $b\_n$ decreasing very slowly, to ensure $f\big(\sum b\_n p\_n \big)$ is large.


The motivation lies in the thermodynamical formalism, and would be quite long to explain, but a positive answer would have very fun consequences. There might be a simple obstruction though, but I could not find one yet.


In fact, I don't really need to prove this for all $(a\_n)$, one in particular could suffice.



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> **Q'**: is the above claim true if we replace "for all $(a\_n)\_{n\ge0}$" with "there exist $(a\_n)\_{n\ge0}$" (still decreasing, tending to $0$).
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> **Q''**: is the above claim true if we take $a\_n=1/n$, or $a\_n=1/n^2$, say?
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Enforcing an inequality on series
No even for $f(x)=x^2$. We should have $(b\_1p\_1+b\_np\_n)^2\geqslant (a\_1p\_1+a\_np\_n)(p\_1+p\_n)$ (take all $p\_i$'s for $i\ne 1,n$ equal to $0$). But the discriminant of the quadratic form $(b\_1p\_1+b\_np\_n)^2- (a\_1p\_1+a\_np\_n)(p\_1+p\_n)$ becomes positive for large $n$.


