Analytic solution of the functional equation $V\left(z^d\right)=dz^{d-1} V(z)+\sum_{k=2}^d b_k z^{d-k}$

Solution 1:

Define $ \tilde V ( z ) = \frac { V ( z ) } z $ and rewrite the equation $$ V \left ( z ^ d \right ) = d z ^ { d - 1 } V ( z ) + b _ 2 z ^ { d - 2 } + b _ 3 z ^ { d - 3 } + \dots + b _ d $$ as $$ \tilde V ( z ) = \frac 1 d \tilde V \left ( z ^ d \right ) - \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k } } d \tag 0 \label 0 \text . $$ For any nonnegative integer $ n $, substituting $ z ^ { d ^ n } $ for $ z $ in \eqref{0}, you have $$ \tilde V \left ( z ^ { d ^ n } \right ) = \frac 1 d \tilde V \left ( z ^ { d ^ { n + 1 } } \right ) - \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } d \text , $$ or equivalently $$ \frac 1 { d ^ n } \tilde V \left ( z ^ { d ^ n } \right ) = \frac 1 { d ^ { n + 1 } } \tilde V \left ( z ^ { d ^ { n + 1 } } \right ) - \frac 1 d \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } { d ^ n } \tag 1 \label 1 \text . $$ Using \eqref{0} as the basis and \eqref{1} for the inductive step, you can prove $$ \tilde V ( z ) = \frac 1 { d ^ { N + 1 } } \tilde V \left ( z ^ { d ^ { N + 1 } } \right ) - \frac 1 d \sum _ { n = 0 } ^ N \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } { d ^ n } \tag 2 \label 2 $$ by induction on the nonnegative integer $ N $. Rewriting \eqref{2} in terms of $ V $, you have $$ V ( z ) = \frac { V \left ( z ^ { d ^ { N + 1 } } \right ) } { d ^ { N + 1 } z ^ { d ^ { N + 1 } - 1 } } - \frac z d \sum _ { n = 0 } ^ N \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } { d ^ n } \text . \tag 3 \label 3 $$ The formula for the mentioned series is just the result of interchanging the sums and letting $ N $ go to infinity in the second term on the right-hand side of \eqref{3}. Now that we have an idea of what $ V ( z ) $ should look like, we can be more precise on the matter: define $ V : \bigl \{ z \in \mathbb C \bigm | | z | > 1 \bigr \} \to \mathbb C $ with $ V ( z ) = - \frac z d \sum \limits _ { k = 2 } ^ d \sum \limits _ { n = 0 } ^ \infty \frac { b _ k z ^ { - k d ^ n } } { d ^ n } $. Note that the series converges absolutely, for example by the root test. It's then straightforward to verify that $ V $ satisfies the functional equation, and as it's defined in terms of power series, it's analytic. Also, note that for this function, the first term on the right-hand side of \eqref{3} tends to $ 0 $ as $ N $ goes to infinity. This is not needed for the proof, but is a useful observation for seeing that $ V $ is in fact the result of taking a limit from both sides of \eqref{3}.

For proving uniqueness, consider another analytic function $ U $ satisfying the same functional equation. Define $ \tilde U ( z ) = \frac { U ( z ) } z $, and conclude $$ \tilde U ( z ) = \frac 1 d \tilde U \left ( z ^ d \right ) - \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k } } d \tag 4 \label 4 \text . $$ Define $ F ( z ) = \tilde V ( z ) - \tilde U ( z ) $. Subtract \eqref{4} from \eqref{0} to get $$ F ( z ) = \frac 1 d F \left ( z ^ d \right ) \text. \tag 5 \label 5 $$ By \eqref{5} and induction, we get $$ F ( z ) = \frac 1 { d ^ n } F \left ( z ^ { d ^ n } \right ) \text , \tag 6 \label 6 $$ for any nonnegative integer $ n $. Letting $ a _ n = 2 \exp \left ( \frac { 2 \pi i } { d ^ n } \right ) $ and using \eqref{6} we get $$ f ( a _ n ) = \frac 1 { d ^ n } F \left ( a _ n ^ { d ^ n } \right ) = \frac 1 { d ^ n } F \left ( 2 ^ { d ^ n } \right ) = F ( 2 ) \text . $$ Hence, $ F $ is an analytic function which is constant on the sequence $ ( a _ n ) _ { n = 0 } ^ \infty $, which has a limit point. Therefore, $ F $ must be constant. As the only constant function satisfying \eqref{5} is the zero function (since $ d > 2 $), we get $ \tilde U = \tilde V $, and $ U = V $.