Let $0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx).$ For each $ k\in \mathbb{N},$ set $$f_k(a;x):=\frac{x^k}{(1-a)(1-a^2)\dots (1-a^k)}\,\psi_a(x).$$

Question. Is it true that, for mutually prime $j$ and $m$, where $1<j<m,$ the following inequality holds: $$f_{mk}(a^j;x)\leq f_{jk}(a^m;x) \;\;\mathrm{for\;\;all}\;\;k\in \mathbb{N}\;\;\mathrm{and\;\;all}\;\;x\in [0,1]?$$

Remark. When $j=1, m=2,$ the answer in provided in

Inequality for functions on [0,1]

and when $j=1, m>2$ in arXiv:1708.07669.

  • $\begingroup$ True and trivial in the regime $e^{-x}+e^{-T}=1$ (when the corresponding integral inequality is an identity). Still need to think what happens when we have an overlap. On one hand, it strongly pushes in favor of the inequality. On the other hand it screws up the current argument a bit. So give me some more time :-) (I'm posting that just to let you know that I still remember about this problem). $\endgroup$ – fedja Dec 1 '17 at 3:17
  • $\begingroup$ True and almost trivial in all cases. Now it remains to find some time for posting... $\endgroup$ – fedja Dec 1 '17 at 6:44

OK, here goes.

We start with changing the notation ($z\to 20z^2$, $-z-3\to r$, $20rz\to y$ means that what was denoted by $z$ will be denoted by $20z^2$ from now on, $r$ is $-z-3$ with new $z$, so it is $-\sqrt{z/20}-3$ in terms of old $z$, and $y$ denotes $20rz$ with just (re)defined $r,z$).

So, execute the following sequence (the order matters!) $$ x\to e^{-x},\ a\to e^{-a},\ mkja\to T,\ ja\to a $$ Denote $\varphi(t)=-\log(1-e^{-t})$. For a function $f$ denote by $LRS(f,u,v; h)$ and $RRS(f,u,v;h)$ the left and the right Riemann sums of $f$ on the interval $[u,v]$ with step $h$ respectively (we will always assume that $(v-u)/h\in \mathbb N\cup\{\infty\}$).

Then the above substitutions induce the substitution $$ -\log f_{mk}(a^jx)\to \frac{LRS(\varphi,x,\infty;a)+xT-RRS(\varphi,0,T;a)}a $$
The claim to prove reduces to the statement that this function decreases in $a$ as long as we run $a$ along admissible steps for $[0,T]$ with fixed $T$.

We have already seen that replacing Riemann sums by integrals results in the area $A\ge 0$ of the "excess triangle" formed by the vertical line through $x$, the horizontal line through $T$, and the graph of $\varphi$ in the coordinate system where $x$ runs over the horizontal axis and $T$ over the vertical one in the numerator. Thus, the function to investigate is $$ \frac Aa+\frac {\Phi(a)}a+\frac{\Psi(a)}a $$ where $\Phi(a)=LRS(\varphi,x,\infty;a)-\int_x^\infty \varphi(t)\,dt$ and $\Psi(a)=\int_0^T\varphi(t)\,dt-RRS(\varphi,0,T,a)$

Note that we have a nice series expansion $$ \varphi(t)=\sum_{k\ge 1}\frac 1ke^{-kt} $$ The integration and the Riemann sum computation are linear operations with respect to the function, so we have $$ \Phi(a)=\sum_{k\ge 1}\frac 1k\Phi_k(a),\qquad \Psi(a)=\sum_{k\ge 1}\frac 1k\Psi_k(a) $$ where $\Phi_k$ and $\Psi_k$ are defined in the same way but using $e^{-kt}$ instead of $\psi(t)$.

Now, we (or, if you prefer, at least I) cannot integrate $\varphi$ or to sum it along an arithmetic progression. However everybody can do it for an exponential function. The direct computation yields $$ \Phi_k(a)=e^{-kx}\left[\frac a{1-e^{-ka}}-\frac 1k\right],\qquad \Psi_k(a)=(1-e^{-kT})\left[\frac 1k-\frac a{e^{ka}-1}\right] $$ Note that in the computation of $\Psi_k(a)$ we have used the fact that $T/a$ is an integer. However the resulting answer is formally defined for all $a>0$. So we will not use the "fitting condition" anywhere below.

Notice also that $$ \left[\frac a{1-e^{-ka}}-\frac 1k\right]+\left[\frac 1k-\frac a{e^{ka}-1}\right]=a $$ so $$ \frac{\Phi_k(a)+\Psi_k(a)}a=\frac{1-e^{-kx}-e^{-kT}}k\left[\frac 1a-\frac k{e^{ka}-1}\right]+ \operatorname{const} $$ Thus we need to show that $$ \frac Aa+\sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kT}}{k^2}\left[\frac 1a-\frac k{e^{ka}-1}\right] $$ is decreasing or, equivalently, that $$ \frac A{a^2}+\sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kT}}{k^2}\left[\frac 1{a^2}-\frac {k^2}{e^{ka}+e^{-ka}-2}\right]\ge 0\,. $$ When the excess triangle is absent or is above the graph, we have $e^{-x}+e^{-T}\le 1$ and the factor $k\ge 1$ can only improve this inequality, so each term is non-negative and the estimate is trivial, thus justifying my first remark.

The other case is slightly more interesting.

Let $t>0$ satisfy $e^{-t}+e^{-x}=1$. We have now $t>T$. Write $$ A=\int_T^{t}\varphi(s)\,ds-(t-T)\varphi(t)=\sum_{k\ge 1} \frac 1k\left[\int_T^{t}e^{-ks}\,ds-(t-T)e^{-kt} \right] \\ =\sum_{k\ge 1}\frac{e^{-kT}-e^{-kt}-k(t-T)e^{-kt}}{k^2} $$ We want to combine these terms multiplied by $\frac 1{a^2}$ with the corresponding terms in the main sum. It will be more convenient to diminish them first by multiplying them not by the full $\frac 1{a^2}$ but just by the expressions in the brackets in the main sum. Then the result can be rewritten as $$ \sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kt}-kte^{-kt}}{k^2}\left[\frac 1{a^2}-\frac {k^2}{e^{ka}+e^{-ka}-2}\right]+ \sum_{k\ge 1}\frac{Te^{-kt}}{k}\left[\frac 1{a^2}-\frac {k^2}{e^{ka}+e^{-ka}-2}\right] $$ The second sum is clearly non-negative. Let's look at the first one. We have $$ \int_u^\infty\varphi(s)\,ds=\sum_{k\ge 1}\frac 1k\int_u^\infty e^{-ks}\,ds=\sum_{k\ge 1}\frac{e^{-ku}}{k^2}\,. $$ Using this with $u=0,x,t$ and flipping the geometric picture for the integral of $\varphi$ over $[t,\infty)$ to the vertical axis, as usual, we see that $$ \sum_{k\ge 1}\frac{1-e^{-kx}-e^{-kt}}{k^2}=tx $$ (the area of the remaining rectangle). On the other hand, $$ \sum_{k\ge 1}\frac{kte^{-kt}}{k^2}=t\varphi(t)=tx $$ as well. So the first factors form a sequence whose numerators are (obviously, because $u\mapsto(1+u)e^{-u}$ is decreasing on $[0,+\infty)$) increasing and whose sum equals $0$. Thus, they break from $-$ to $+$ just once when $k$ goes up. However the second factors are increasing in $k$. Thus, the sum we are interested in is non-negative as well.

The End

  • $\begingroup$ @Deepti You are welcome! Feel free to ask questions if something is unclear. :-) $\endgroup$ – fedja Dec 3 '17 at 18:41

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