1
$\begingroup$

Suppose $(a_k)_{k\geq0}$ is a sequence of real numbers. Consider the operator $\mathcal{L}a_k=a_k^2-a_{k-1}a_{k+1}$.

We say $(a_k)_k$ is log-concave (resp. log-convex) provided $\mathcal{L}a_k\geq0$ (resp. $\mathcal{L}a_k\leq0$) for all $k$.

We say $(a_k)_k$ is $n$-fold log-concave (resp. $n$-fold log-convex) as long as the iterates $\mathcal{L}^ja_k\geq0$ (resp. $\mathcal{L}^ja_k\leq0$) for each $1\leq j\leq n$ and for all $k$. These concepts play a valuable role in Combinatorics, Number Theory, etc.

I am exclusively interested in sequences of positive integers $(a_k)_k$.

It is clear that $(a_k)_k$ is log-concave iff $(\frac1{a_k})_k$ is log-convex since $\mathcal{L}\frac1{a_k}=\frac1{a_k^2}-\frac1{a_{k-1}a_{k+1}}=-\frac{\mathcal{L}a_k}{a_{k-1}a_k^2a_{k+1}}$.

I would like to ask:

QUESTION. Is it true that $(a_k)_k$ is $2$-fold log-concave iff $(\frac1{a_k})_k$ is $2$-fold log-convex?

Improve this question
$\endgroup$
4
  • $\begingroup$ Take for ease of notation $a,b,c,d,e$ as consecutive terms. Suppose $b^2-ac\ge0$, $c^2-bd\ge0$, $d^2-ce\ge0$. $(a,b,c,d,e)$ is 2-fold log-concave iff $$(c^2-bd)^2\ge(b^2-ac)(d^2-ce),$$ the reciprocals are 2-fold log-convex iff $$(c^2-bd)^2\le\frac{c^2}{ae}(b^2-ac)(d^2-ce).$$ Now it seems to me that for $(1,4,8,16,31)$, the reciprocals are 2-fold log-convex but the numbers themselves not 2-fold log-concave. Have I missed something? $\endgroup$
    – Wolfgang
    Commented Aug 20, 2022 at 9:08
  • $\begingroup$ @Wolfgang: you brought up an excellent point and a counter-example. Now, does it help to insist that $a_k, \mathcal{L}a_k, \mathcal{L}^2a_k$ do not have "internal zeroes"? $\endgroup$
    – T. Amdeberhan
    Commented Aug 20, 2022 at 15:21
  • $\begingroup$ if you mean that in the sense that all inequalities should be strict, just replace 16 by 15.9 and/or 4 by 3.9. These are all continuous functions... $\endgroup$
    – Wolfgang
    Commented Aug 20, 2022 at 15:47
  • $\begingroup$ Thank you, really. I drop the claim. $\endgroup$
    – T. Amdeberhan
    Commented Aug 20, 2022 at 15:51

0

Reset to default

You must log in to answer this question.