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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.

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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 (*).

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  • $\begingroup$ Thanks, but the question still remains, do the conditions imply that $\lim_\limits{x\to \infty}\frac{g(x+w)}{g(x)}=1$ for all $w>0$? (because in general $\frac{g(n+1)}{g(n)}\rightarrow 1$ does not imply $\lim_\limits{x\to \infty}\frac{g(x+w)}{g(x)}=1$). $\endgroup$ – M.H.Hooshmand Nov 22 '17 at 18:30
  • $\begingroup$ Please look at my newer answer: indeed necessarily $\frac{g(x+w)}{g(x)}\to1$ for each real $w>0$. $\endgroup$ – Iosif Pinelis Nov 22 '17 at 18:32
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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 $m:=\lfloor x\rfloor$.

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