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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?

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  • $\begingroup$ 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 $x^2$ $\endgroup$ – user83457 Feb 2 '17 at 15:54
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    $\begingroup$ 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 $f(b_0p_0)<a_0p_0$. $\endgroup$ – benblumsmith Feb 2 '17 at 15:54
  • $\begingroup$ You are of course both right, I edited accordingly. $\endgroup$ – Benoît Kloeckner Feb 2 '17 at 16:24
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    $\begingroup$ 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). $\endgroup$ – Pietro Majer Feb 2 '17 at 21:32
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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 $n$.

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  • $\begingroup$ 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)? $\endgroup$ – Benoît Kloeckner Feb 2 '17 at 20:33
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    $\begingroup$ 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 $n$. $\endgroup$ – Fedor Petrov Feb 2 '17 at 22:29

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