Solution of the functional equation $f(x+1)=g(x)f(x)$ In the paper (http://www.sciencedirect.com/science/article/pii/S0022247X97953439), Webster obtained a unique solution of the functional equation  $f(x+1)=g(x)f(x)$
(where $f,g:\mathbb{R}^+\rightarrow \mathbb{R}^+$) under some conditions  one of which is $\lim_\limits{x\to \infty}\frac{g(x+w)}{g(x)}=1$ for all $w>0$.
Now, I'm looking for a function $g:\mathbb{R}^+\rightarrow \mathbb{R}^+$ such that the sequence
$\frac{g(n+1)}{g(n)}\rightarrow 1$, $\lim_\limits{x\to \infty}\frac{g(x+w_0)}{g(x)}$ does not exist, for some
$w_0>0$, and having a solution $f:\mathbb{R}^+\rightarrow \mathbb{R}^+$ for the functional equation $f(x+1)=g(x)f(x)$ (Gamma-type functional equation) with the following properties?:
(a) $f$ is eventually $\log$-convex (i.e., it is $\log$-convex from a number on);
(b) $f(1)=1$.
Thanks in advance.
 A: Depending on how good the function $g$ is, a solution $f$ may or may not exist. However, there can never be more than one solution, under the given conditions. Indeed, suppose $f$ and $h$ are two different solutions. Then, for some natural $c>0$, on the interval $[c,\infty)$ we have $f=e^F$ and $h=e^H$, where $F$ and $H$ are distinct convex functions. Let $d:=H-F$. In view of the conditions $f(1)=1=h(1)$, $f(x+1)=g(x)f(x)$, and $h(x+1)=g(x)h(x)$, the function $d$ is periodic with period $1$, and $d(n)=0$ for $n=c,c+1,\dots$. Moreover, for some real $\delta$ and all natural $n$, we have 
\begin{equation}
\sup_{(n,n+1]}|d|=\delta>0. \tag{*} 
\end{equation}
In view of the convexity of $F$ and $H$ on $[c,\infty)$, for all natural $n>c$ and all $x\in(n,n+1]$
\begin{equation}
\ln g(n-1)=F(n)-F(n-1)\le \frac{F(x)-F(n)}{x-n}\le F(n+1)-F(n)=\ln g(n),
\end{equation}
\begin{equation}
\ln g(n-1)=H(n)-H(n-1)\le \frac{H(x)-H(n)}{x-n}\le H(n+1)-H(n)=\ln g(n). 
\end{equation}
Therefore and because $F(n)=H(n)$, we have 
\begin{equation}
 |d(x)|\le\Big|\frac{H(x)-F(x)}{x-n}\Big|\le\ln\frac{g(n)}{g(n-1)}\underset{n\to\infty}\longrightarrow0
\end{equation}
uniformly over $x\in(n,n+1]$. This contradicts (*). 
A: Such a function $g$ does not exist; that is, under the given conditions, for each real $w>0$ necessarily $\frac{g(x+w)}{g(x)}\to1$ as $x\to\infty$. In my previous answer, I apparently misunderstood the question, and so, gave an answer to a different, but related question, which I think may be of independent interest. The same main idea helps provide an answer to what the question seems to actually be. 
Indeed, for some natural $c>0$, on the interval $[c,\infty)$ we have $f=e^F$, where $F$ is a convex function. 
So, for all natural $n>m>c$ and all $x$ and $t$ such $m<x<x+t\le n$
\begin{equation}
\ln g(m-1)=F(m)-F(m-1)\le\frac{F(x)-F(m)}{x-m}\le F(n+1)-F(n)=\ln g(n),   
\end{equation}
\begin{equation}
\ln g(m-1)=F(m)-F(m-1)\le\frac{F(x+t)-F(m)}{x+t-m}\le F(n+1)-F(n)=\ln g(n).   
\end{equation}
It follows that for $x$ and $t$ as before, now with $m\to\infty$ and $n-m$ bounded,
\begin{align*}
 F(x+t)-F(x)&\le(x+t-m)\ln g(n)-(x-m)\ln g(m-1) \\ 
 &=(x-m)\ln \frac{g(n)}{g(m-1)}+t\ln g(n)=t\ln g(m)+o(1),
\end{align*}
\begin{align*}
 F(x+t)-F(x)&\ge(x+t-m)\ln g(m-1)-(x-m)\ln g(n) \\ 
 &=(x-m)\ln \frac{g(m-1)}{g(n)}+t\ln g(m-1)=t\ln g(m)+o(1);
\end{align*}
this follows because 
\begin{equation}
 \frac{g(n)}{g(m-1)}=\prod_{i=m-1}^{n-1}\frac{g(i+1)}{g(i)}\to1; 
\end{equation}
thus,
\begin{equation}
 F(x+t)-F(x)=t\ln g(m)+o(1).
\end{equation} 
Hence, for any real $w>0$,
\begin{align*}
\frac{g(x+w)}{g(x)}&=\exp\{F(x+w+1)-F(x+1)-[F(x+w)-F(x)]\} \\ 
&=\exp\{w\ln g(m)+o(1)-[w\ln g(m)+o(1)]\}\to1 
\end{align*}
as $x\to\infty$, where $m:=\lfloor x\rfloor$. 
