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.



Solution of the functional equation $f(x+1)=g(x)f(x)$
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 $m:=\lfloor x\rfloor$. 


