Let $C_1>0,C_2>0, a>0, b>0$ be constants, let $\sum_{m=0}^{\infty} C_1e^{-C_2m}=1$ and $f\leq 1$ be a function satisfying the inequality
$$f(n)\leq \sum_{m=1}^n e^{am} f(n-m) \frac{C_1}{1-C_1}e^{-C_2m} + \frac{e^{an}}{1-C_1}\left[1-\sum_{m=0}^{n-1} C_1e^{-C_2m}\right]\tag{1}$$
I'm trying to check if there exist constants $b,C$ such that $b<a$ and $C>0$ and $f(n)\leq Ce^{bn}$ for all $n\geq 0$.
My attempt: I'm trying to see if this is true by induction. For $n=0$, $f(0)\leq C$ if $C\geq 1$. So choose any $C\geq 1$. Suppose there exists $b<a$ such that $f(m)\leq Ce^{bm}$ for all $m<n$. Using this in $(1)$, we get
\begin{align} f(n) &\leq \sum_{m=1}^n e^{am}Ce^{b(n-m)}\frac{C_1}{1-C_1}e^{-C_2m} + \frac{e^{an}}{1-C_1}\left[1-\sum_{m=0}^{n-1} C_1e^{-C_2m}\right] \\ \\ &= \sum_{m=1}^n e^{am}Ce^{b(n-m)}\frac{C_1}{1-C_1}e^{-C_2m} + \frac{e^{an}}{1-C_1}\left[1-\sum_{m=0}^{n-1} C_1e^{-C_2m}\right] \\ \\ &= Ce^{bn}\left[\sum_{m=1}^n e^{(a-b-C_2)m}\frac{C_1}{1-C_1}+ \frac{e^{(a-b)n}}{1-C_1}\left[1-\sum_{m=0}^{n-1} C_1e^{-C_2m}\right]\right] \end{align}
I think the induction statement is proven if we show that $$\left[\sum_{m=1}^n e^{(a-b-C_2)m}\frac{C_1}{1-C_1}+ \frac{e^{(a-b)n}}{1-C_1}\left[1-\sum_{m=0}^{n-1} C_1e^{-C_2m}\right]\right]\leq 1 \tag{2}$$ because that would mean $f(n)\leq Ce^{bn}$. But I'm not sure how to prove that and show there exists a $b<a$ for which $(2)$ is true. Any ideas?
Denote $A=e^a$, $c=e^{-C_2}$ and let us decide that (1) has a little mistake and that $m$ should vary from $0$ to $n$. We have $C_1=1-c.$ Therefore $$f(n)\leq \frac{1-c}{c}\sum_{m=0}^n(Ac)^nf(n-m) + \frac{1}{c}(Ac)^n.$$ Assume that $f(n)\geq 0.$ As a consequence for $z>0$ and small enough we have $$F(z)=\sum_{n=0}^{\infty}f(n)z^n\leq \frac{1-c}{c}\frac{1}{1-Acz}F(z)+\frac{1}{c}\frac{1}{1-Acz}$$ or denoting $1/r=\frac{Ac^2}{2c-1}$ we have $$(2c-1)(1-\frac{1}{r}z)F(z)\leq 1.\ \ (*)$$ At this point if your $C_2$ is too big, that is to say if $c\leq 1/2$ I cannot say anything since (*) gives no information. In the other hand, if $2c-1>0$ then $$(2c-1)F(z)\leq \frac{1}{1-\frac{z}{r}}=\sum_{n=0}^{\infty}\frac{z^n}{r^n}$$ showing that the radius of convergence $R$ of the power series $F(z)$ satisfies $R\geq r.$ This implies that $\limsup f(n)^{1/n}=\frac{1}{R}\leq \frac{1}{r}.$ In particular there exists a number $B> \frac{1}{R}$ such that $f(n)<B^n$ for all $n.$ A weaker statement is that for any $C>\frac{1}{r}$ we have $f(n)<C^n$ except for a finite number of $n.$
Equivalently for each $b$ such that $e^b>1/r$ there exists $C_3>0$ such that $f(n)<C_3e^{bn}.$ But saying $e^b>1/r$ is saying $b>a-2C_2+\log (2c-1).$ Therefore since $-2C_2+\log (2c-1)<0$ the interval $$(a-2C_2+\log (2c-1),a) $$ is not empty and there exists a $b<a$ such that $f(n)<C_3e^{bn}$ as desired.