# Enforcing an inequality on series

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?

• do you meant that condition $\sum p_n \le 1$ rather than equality , because If so the sequence $p_n = \epsilon,0,0 ...$ is the wrong order of magnitude for a function like $x^2$
– user83457
Feb 2 '17 at 15:54
• As stated, I think the answer is no for 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 $f(b_0p_0)<a_0p_0$. Feb 2 '17 at 15:54
• You are of course both right, I edited accordingly. Feb 2 '17 at 16:24
• benblumsmith's objection remains valid even with the weaker condition $\sum_{n=0}p_n=1$. Indeed we should have in particular $f(tb_0+(1-t)b_n)\ge ta_0+(1-t)a_n$ for all $0\le t \le 1$, but letting $n\to+\infty$ also $f(tb_0)\ge ta_0$ so that $f'_+(0)\ge a_0/b_0>0$ (in fact, the condition with $=$ is only weakly weaker). Feb 2 '17 at 21:32

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 $n$.
• Sorry, there is something I don't get: denoting by $Q$ the quadratic form, the positivity of the discriminant shows that the equation $Q(p_1,p_n)<0$ has solutions, but why should any of these solutions satisfy $p_1=1-p_n\in [0,1]$ (recall that the $p_k$ are non-negative)? Feb 2 '17 at 20:33
• Our inequality is homogeneous (this is why I multiplied it may $p_1+p_n$), thus we may forget about $p_1+p_n=1$ condition. And they are of the same sign when $Q(p_1,p_n)<0$, since $b_1^2\geqslant a_1$, $b_n^2\geqslant a_n$ (take $p_1=1$ in the initial inequality), $2b_1b_n<a_1+a_n$ for large $n$. Feb 2 '17 at 22:29