When is the function $F(\varepsilon) = \sum \alpha^n f(x_n-\varepsilon)$ finite? Let $f : \mathbb{R} \to [0,+\infty)$ be a convex, continuous function that satisfies
$$
\lim_{|x| \to +\infty} f(x) = +\infty.
$$
Let $\alpha \in (0,1)$, and suppose there exists a sequence $\{x_n\}$ such that
$$
\sum_{n=0}^{\infty} \alpha^n f(x_n) < +\infty.
$$
Define the function $F : \mathbb{R} \to [0, +\infty]$ by
$$
F(\varepsilon) = \sum_{n=0}^{\infty} \alpha^n f(x_n-\varepsilon).
$$
This is a well-defined function since $f \geq 0$. By construction, $F$ is convex, lower semicontinuous, and $F(0) < +\infty$. I am interested in whether $F$ is finite generally. (None of these listed properties imply anything towards my goal, since the characteristic functions of the set $\mathbb{R}$ and the set $\{0\}$ both satisfy all of them.)
Since $0 < \alpha < 1$, and since $f$ is continuous, we immediately know that, if $\{x_n\}$ is a bounded sequence, then $F$ is finite on all of $\mathbb{R}$. But $\{x_n\}$ may not be bounded. I believe that the precise asymptotic characteristics of $\{x_n\}$ can be deduced from those of $f$, but I'm not sure how.
Some observations:

*

*If $x_n \to +\infty$, then $x_n - \varepsilon < x_n$ will imply $F(\varepsilon) < +\infty$, for positive $\varepsilon$. since $f$ is increasing on $[b, +\infty)$. Similar can be said when $x_n \to -\infty$ and $\varepsilon < 0$.

*Since $\sum \alpha^n f(x_n) <+\infty$, it is also true that $\sum \alpha^n f(x_{n+k}) < +\infty$ for all $k > 0$, although I don't think that there is much utility to this.

*Edit: The sequence $\{x_n\}$ defined by the union of sets $A_k = \{\frac{k^2-k+j2^{-k}}{k} : j = 1, 2, \dots, 2^k\}$ is a counterexample to my previous conjecture that $x_{n+m} < x_n - \varepsilon$ when $\varepsilon > 0$ and $m$ is sufficiently large.

*Maybe the dominated convergence theorem would be helpful, where we take $\sum \alpha^n (\:\cdot\:)$ to be the integration operator. But I'm not sure.

Any help is appreciated!
 A: First followup: it is not necessary that $F$ be finite generally. Consider the following example. Let $x_n = \log_{\alpha} (n+1)$ for $n=0,1,\dots$ and let
$$
f(x)=\alpha^{-\alpha^x/2},
$$
which is convex by $\alpha \in (0,1)$. Furthermore, $f(x_n) = \alpha^{-(n+1)/2}$, so
$$
F(0) = \sum_{n=0}^{\infty} \alpha^n f(x_n) = \sum_{n=0}^{\infty} \alpha^{(n-1)/2} = \frac{1}{\sqrt{\alpha}-\alpha} < +\infty.
$$
Fix $\varepsilon$ such that $\alpha^{\varepsilon} \geq 2$. Since $\alpha \in (0,1)$, it follows that $\varepsilon < 0$. Then
\begin{align}
x_n - \varepsilon = \log_{\alpha}(n+1) - \log_{\alpha}(\alpha^{\varepsilon}) 
= \log_{\alpha} (\alpha^{\varepsilon}[n+1])
\geq \log_{\alpha}(2[n+1])
= x_{2n}.
\end{align}
Therefore, since $f$ is increasing on the interval $[0,+\infty)$,
\begin{align}
F(\varepsilon) &= \sum_{n=0}^{\infty} \alpha^n f(x_n-\varepsilon)
\geq \sum_{n=0}^{\infty} \alpha^n f(x_{2n})
= \sum_{n=0}^{\infty} \alpha^{n-2n-1}
= +\infty
\end{align}
which demonstrates that $F$ is not finite at $\varepsilon$. It is immediate that $F$ is $+\infty$ on the interval $(-\infty, \varepsilon]$.
