Finding the limit of a functional equation derived from $ V ( x ) = a x + b V ( c x ) $ I am working on an undergraduate research on finance about dividends and I got this formula
$$ V ( d _ 0 ) = d _ 0 \frac { 1 + p g } { r + p } + \frac p { r + p } V \big( d _ 0 ( 1 + g ) \big) \text , $$
where $ d _ 0 , g > 0 $, $ r \ge 0 $ and $ 0 \le p \le 1 $.
Letting $ x = d _ 0 $, $ a = \frac { 1 + p g } { r + p } $, $ b = \frac p { r + p } $, and $ c = 1 + g $, I got
$$ V ( x ) = a x + b V ( c x ) \text . $$
From this equation, I got
$$ V ( x ) = a x \left( 1 + b c + b ^ 2 c ^ 2 + \dots + b ^ { n - 1 } c ^ { n - 1 } \right) + b ^ n V ( c ^ n x ) \text . $$
I want to evaluate the limit of this equation as $ n \to \infty $. The first term is just a geometric series, so it's easy. My question is: how do I evaluate the limit of $ b ^ n V ( c ^ n x ) $ as $ n \to \infty $?
 A: We have $ a > 0 $, $ 0 \le b \le 1 $ and $ c > 1 $, by definition of these parameters, and the given conditions on the parameters $ g $, $ r $ and $ p $. We are looking for a function $ V : \mathbb R _ + \to \mathbb R $ satisfying
$$ V ( x ) = a x + b V ( c x ) \tag 0 \label 0 $$
for all $ x \in \mathbb R _ + $. It's straightforward to verify that depending on the value of $ b $, \eqref{0} has solutions of the following forms:


*

*$ V ( x ) = a x $, in case $ b = 0 $;

*$ V ( x ) = a x \log _ b x + k x $ for some arbitrary constant $ k \in \mathbb R $, in case $ b = \frac 1 c $;

*$ V ( x ) = \frac { a x } { 1 - b c } + k x ^ { - \log _ c b } $ for some arbitrary constant $ k \in \mathbb R $, in case $ 0 < b \ne \frac 1 c $.


Of course, in case we have $ b > 0 $, these are not the only solutions: \eqref{0} only relates the value of $ V $ at a point $ x $ to values of $ V $ at points of the form $ c ^ n x $ for some $ n \in \mathbb Z $, and thus $ V $ can act independently on different chains of the mentioned form. But, since this problem comes from mathematical modeling, we may only be interested in well-behaved solutions, and in that case, the above can be considered to be all the solutions.
But how can one find the above formulas? By continuing your own argument, in some sense! The case $ b = 0 $ is trivial, and from now on, we will assume that $ b > 0 $. You've already derived
$$ V ( x ) = a x \sum _ { i < n } ( b c ) ^ i + b ^ n V ( c ^ n x ) \tag 1 \label 1 $$
for all $ x \in \mathbb R _ + $ and all $ n \in \mathbb N $. You can substitute $ c ^ { - n } x $ for $ x $ in \eqref{1}, multiply both sides by $ b ^ { - n } $ and rearrange the terms, and consequently see that \eqref{1} holds for all $ x \in \mathbb R _ + $ and all $ n \in \mathbb Z $. The idea is that a well-behaved solution of \eqref{0} not only must satisfy \eqref{1} for all $ x \in \mathbb R _ + $ and all $ n \in \mathbb Z $, but also for all $ x \in \mathbb R _ + $ and all $ n \in \mathbb R $. To make this more precise, let's first find an expression for $ n $ in terms of $ x $ and $ c ^ n x $:
$$ n = \log _ c \frac { c ^ n x } x \text . \tag 2 \label 2 $$
Then, let's get rid of the summation in \eqref{1}:
$$ \sum _ { i < n } ( b c ) ^ i =
\begin {cases}
n & b = \frac 1 c \text ; \\
\frac { ( b c ) ^ n - 1 } { b c - 1 } & b \ne \frac 1 c \text .
\end {cases} \tag 3 \label 3 $$
Now, if $ n $ could be any real number instead of only an integer, then we could put any $ y \in \mathbb R _ + $ in place of $ c ^ n x $ in \eqref{1}, and use \eqref{2} and \eqref{3} to replace the summation and all appearances of $ n $ in \eqref{1} with suitable expressions in terms of only $ x $ and $ y $. We will do this in the following. To make things look simpler, we define $ U : \mathbb R _ + \to \mathbb R $ with $ U ( x ) = \frac { V ( x ) } x $ for all $ x \in \mathbb R _ + $. Dividing both sides of \eqref{1} by $ x $ and noting that $ \frac { V ( c ^ n x ) } x = c ^ n \frac { V ( c ^ n x ) } { c ^ n x } $, we can rewrite \eqref{1} in terms of $ U $.

Case $ b = \frac 1 c $:
We have
$$ U ( x ) = a \log _ c \frac y x + U ( y ) $$
for all $ x , y \in \mathbb R _ + $. This can be simplified to
$$ U ( x ) + a \log _ c x = U ( y ) + a \log _ c y $$
for all $ x , y \in \mathbb R _ + $; i.e. $ U ( x ) + a \log _ c x $ is constant. This can be rewritten as the following (note that we have $ b = \frac 1 c $): there is a constant $ k \in \mathbb R $ such that
$$ U ( x ) = a \log _ b x + k $$
for all $ x \in \mathbb R _ + $.

Case $ b \ne \frac 1 c $:
We have
$$ U ( x ) = a \frac { ( b c ) ^ { \log _ c \frac y x } - 1 } { b c - 1 } + ( b c ) ^ { \log _ c \frac y x } U ( y ) $$
for all $ x , y \in \mathbb R _ + $. This can be written as
$$ U ( x ) = - \frac a { b c - 1 } + ( b c ) ^ { \log _ c \frac y x } \left( U ( y ) + \frac a { b c - 1 } \right) \text , $$
or equivalently
$$ ( b c ) ^ { \log _ c x } \left( U ( x ) + \frac a { b c - 1 } \right) = ( b c ) ^ { \log _ c y } \left( U ( y ) + \frac a { b c - 1 } \right) $$
for all $ x , y \in \mathbb R _ + $. Since $ c ^ { \log _ c x } = x $ and $ b ^ { \log _ c x } = x ^ { \log _ c b } $, the previous fact can be rewritten as follows: there is a constant $ k \in \mathbb R $ such that
$$ U ( x ) = \frac a { 1 - b c } + k x ^ { - 1 - \log _ c b } $$
for all $ x \in \mathbb R _ + $.
