I would like to ask a follow-up question on a previous question of mine here whose proof does not seem to carry over to this case in an obvious way:

We define the function $$F_{\varepsilon}(x) = \sum_{i=1}^{\infty} 2^{-\varepsilon \vert x_i \vert} \text{ for }\varepsilon>0.$$

We let $A$ be the set of positive sequences $x=(x_n)$ such that $\sum_n \frac{x_n}{n^2}<\infty$. Clearly, the sequence $x=(n)$ is not in $A$.

Since we expect that anything in $A$ cannot grow as fast as $x=(n)$, I ask: Is it true that for any sequence $x\in A$

$$\limsup_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(n)}{F_{\varepsilon}(x)} \le 1?.$$


1 Answer 1


$\newcommand\ep\varepsilon\newcommand\de\delta$ Let us show more: for all $x\in A$, \begin{equation*} \frac{F_{\ep}((n))}{F_{\ep}(x)}\to0\tag{$*$} \end{equation*} (as $\ep\downarrow0$).

Indeed, take any $x\in A$ and let \begin{equation*} y_n:=x_n/n^2, \end{equation*} so that $\sum_n y_n<\infty$ and \begin{equation*} F_{\ep}(x)=\sum_{n=1}^\infty 2^{-\ep n^2 y_n}. \end{equation*} By Jensen's inequality for the convex function $u\mapsto2^{-u}$, for any natural $N$ \begin{equation*} F_{\ep}(x)\ge\sum_{n=1}^N 2^{-\ep n^2 y_n}\ge N2^{-\ep \sum_1^N n^2 y_n/N}. \tag{1} \end{equation*} Take now any real $\de>0$. Then, by the condition $\sum_n y_n<\infty$, there is a natural $M_\de$ such that $\sum_{n>M_\de} y_n<\de/2$. So, for $N>M_\de$, \begin{equation*} \sum_1^N n^2 y_n=\sum_{n\le M_\de} n^2 y_n+\sum_{M_\de<n\le N} n^2 y_n \le\sum_{n\le M_\de} n^2 y_n+N^2\de/2<N^2\de \end{equation*} if we also have $N^2>2\sum_{n\le M_\de} n^2 y_n/\de$, and then, by (1),
\begin{equation*} F_{\ep}(x)\ge N2^{-\ep N\de}. \end{equation*} Choosing now $N\sim\dfrac1{\ep\de}$ with $\ep\downarrow0$, we have \begin{equation*} F_{\ep}(x)\ge\dfrac1{3\ep\de}, \end{equation*} for each real $\de>0$ and all small enough $\ep>0$.

On the other hand, \begin{equation*} F_{\ep}((n))=\sum_{n=1}^\infty 2^{-\ep n}=\frac{2^{-\ep}}{1-2^{-\ep}}\sim\frac1{\ep\ln2}. \end{equation*}

Now ($*$) follows.


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