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:

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 $\sum p_n = 1$: $$f\big(\sum b_n p_n \big) \ge \sum a_n p_n.$$

(**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.

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 $0$).

Q'': is the above claim true if we take $a_n=1/n$, or $a_n=1/n^2$, say?

nofor every sequence $a_n$. Take $p_n=0$ for $n>0$. Then the LHS is $f(b_0p_0)$ and the RHS is $a_0p_0$. If $f$ is once-differentiable at zero, then $f(b_0p_0) = f'(0)p_0 + O(p_0)^2$ regardless of how you choose $b_0$, so if $f'(0)=0$, then you can make $p_0$ small enough that $f(b_0p_0)<a_0p_0$. $\endgroup$ – benblumsmith Feb 2 '17 at 15:54