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Let $\mathcal A_{s}$ be the set of sequences $X=(x_m)_{m \in I}$ where $I=\{1,2,...,n\}$ with $n \ge 2$ and possibly $n =\infty$ is an index set with $x_1=0$, $x_2=s>0$ and $x_m>x_{m-1}$ for $m,m-1 \in I.$ In addition, we require that the sequence $X$ satisfies $$ \sum_{m \in I} e^{-\gamma x_m}<\infty \text{ for all }\gamma>0.$$

We then fix $x_2=s>0$ (as $x_1=0$ is always fixed) and define $\gamma(X,\lambda)>0$ for $\lambda \in (0,x_2)$ by the condition $$ \sum_{m \in I} (\lambda-x_m) e^{-\gamma(X,\lambda) x_m}=0.$$ Based on some examples, I then claim that for $\lambda \in (0,s)$ $$\max_{X\in \mathcal A_s} \sum_{m \in I} e^{-\gamma(X,\lambda) x_m} $$ is attained for the sequence $X=(0,s)$, i.e. $I=\{1,2\}$ such that

$$\max_{X\in \mathcal A_s} \sum_{m \in I} e^{-\gamma(X,\lambda) x_m} = 1 + e^{-\gamma((0,s),\lambda)s}=(1-\lambda/s)^{-1}.$$

In other words, I would like to understand whether

$$ \sum_{m \in I} e^{-\gamma(X,\lambda) x_m} \le (1-\lambda/s)^{-1}$$ for any $X \in \mathcal A_s$ and $\lambda \in (0,s).$

Example: For $I= \mathbb N$ and $X = (m-1)_{m \in \mathbb N}$, we have $s=1$ and compute $$\sum_{m \in \mathbb N_0} m e^{-\gamma m} = \frac{e^{\gamma}}{(e^{\gamma}-1)^2} \text{ and } \sum_{m \in \mathbb N_0} \lambda e^{-\gamma m} =\frac{ \lambda e^{\gamma}}{e^{\gamma}-1}.$$

Thus, the defining equation is $\lambda = (e^{\gamma}-1)^{-1}$ which shows that $\gamma = \log(1+\frac{1}{\lambda}).$

Then $\sum_{m \ge 0} e^{-\log(1+\frac{1}{\lambda})m} =\sum_{m \ge 0} \left(1+\frac{1}{\lambda}\right)^{-m}= 1+\lambda \le (1-\lambda)^{-1}.$

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  • $\begingroup$ Do I understand correctly that $\mathcal A_s$ is meant to be the set of all sequences, finite and infinite, with the properties that $x_1=0$, $x_2=s$, and the summability condition? So when you maximize over $\mathcal A_s$, you are maximizing over the length of the sequence as well as the terms? $\endgroup$
    – Anthony Quas
    Commented Nov 8 at 2:41
  • $\begingroup$ As a comment, I think maybe you could achieve the same thing by requiring $I=\mathbb N$, and allowing the sequences to the take the value $\infty$, with the (reasonable) convention that $\infty e^{-\gamma\infty}=0$. $\endgroup$
    – Anthony Quas
    Commented Nov 8 at 2:42
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    $\begingroup$ One more question as I process your question: if we take $I=\{1,2\}$ for now, then your equation for $\gamma$ may be rearranged to $0e^{-\gamma 0}+ se^{-\gamma s}=\lambda(e^{-\gamma 0}+ e^{-\gamma s})$ or $se^{-\gamma s}=\lambda(1+e^{-\gamma s})$ or $\lambda=se^{-\gamma s}/(1+e^{-\gamma s})$. As $\gamma\to 0$, this approaches $\frac s2$, whereas as $\gamma\to\infty$, this approaches 0. So how do you get equality if $\lambda\in (\frac s2,s)$? $\endgroup$
    – Anthony Quas
    Commented Nov 8 at 2:47
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    $\begingroup$ Yea, I think you should rewrite this question to make it more precise. It's very badly written as of now. $\endgroup$
    – mathworker21
    Commented Nov 8 at 3:42
  • $\begingroup$ Crossposted at Mathematics.SE. $\endgroup$
    – Alex Ravsky
    Commented Nov 8 at 6:55

1 Answer 1

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The question is very easy. Indeed, put $\gamma=\gamma(X,\lambda)$ and $S=\sum_{m \in I} e^{-\gamma x_m}$. The equality $$ \sum_{m \in I} (\lambda-x_m) e^{-\gamma x_m}=0$$ implies $$\lambda S=\sum_{m \in I} \lambda e^{-\gamma x_m}= \sum_{m \in I\setminus\{1\}} x_m e^{-\gamma x_m}\ge \sum_{m \in I\setminus\{1\}} s e^{-\gamma x_m}=s(S-1)$$ Therefore $S\le \frac s{s-\lambda}=(1-\lambda/s)^{-1}$.

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