Uniqueness of a solution to the functional equations $g(t)\overline{g(t+h)}=f(h)$ and $f(-h)=\overline{f(h)}$ with $g(0)=1$ I have two complex-valued functions, $ f : \mathbb Z \to \mathbb C $ and $ g : \mathbb Z \to \mathbb C $, that satisfy the following properties. $ \overline x $ denotes the complex conjugate of $ x $ below.
$$ g ( t ) \overline { g ( t + h ) } = f ( h ) \quad \forall t , h \in \mathbb Z \tag 1 \label {eqn1} $$
$$ g ( 0 ) = 1 \tag 2 \label {eqn2} $$
$$ f ( - h ) = \overline { f ( h ) }  \quad \forall h \in \mathbb Z \tag 3 \label {eqn3} $$
I want to show that $ g $ has to be of the form $ g ( t ) = e ^ { i \lambda t } $ for some $ \lambda \in \mathbb R $.
First, by taking $ t = h = 0 $ I get $ f ( 0 ) = 1 $. Next by taking $ h = 0 $ and leaving $ t $ arbitrary I get $ g ( t ) \overline { g ( t ) } = 1 $. Hence, $ | g(t) | = 1 $. This implies that $ g $ has to be of the form
$$ g ( t ) = e ^ { i \theta ( t ) } $$
for some real-valued function $ \theta : \mathbb Z \to \mathbb R $ with $ \theta ( 0 ) = 2 \pi k $ for some $ k \in \mathbb Z $. By \eqref{eqn1}
$$ e ^ { i \theta ( t ) } e ^ { - i \theta ( t + h ) } = f ( h ) $$
This implies
$$ \theta ( t ) - \theta ( t + h ) = m ( h ) $$
for some real-valued function $ m : \mathbb Z \to \mathbb R $. By taking $ t = 0 $ I get $ m ( h ) = - \theta ( h ) + 2 \pi k $. So then I can write
$$ \theta ( t + h ) = \theta ( t ) + \theta ( h ) - 2 \pi k \quad \forall t , h \in \mathbb Z \tag 4 \label {eqn4} $$
It is clear that $ \theta ( t ) = \lambda t + 2 \pi k $ would solve the functional equation above. But the question says this is the only possible solution. How do I show this?
I think I overthought this problem a bit. By \eqref{eqn4},
$$ \theta ( n t ) = n \theta ( t ) - ( n - 1 ) 2 \pi k \quad \forall n \in \mathbb Z _ { \ge 1 } $$
Hence, $ \theta ( n ) = n ( \theta ( 1 ) - 2\pi k ) + 2 \pi k $.
So at least $ n \mapsto \theta ( n ) $ is linear on $ \mathbb Z _ { \ge 1 } $. I guess it also works the other way as well.
 A: The question does not ask for "uniqueness" of $ \theta $, but for "uniqueness" of $ g $. Note that if $ \theta $ gives a solution, for any $ l : \mathbb Z \to \mathbb Z $, the function $ \tilde \theta : \mathbb Z \to \mathbb R $ defined by $ \tilde \theta ( t ) = \theta ( t ) + 2 \pi l ( t ) $ give a solution, too; and in fact, $ \tilde \theta $ gives rise to the same $ g $ as that of $ \theta $. Also, even for $ g $, it does not ask for a proof of uniqueness; each different $ \lambda \in \mathbb R $ gives a different solution. That's why I've put the word uniqueness between quotation mark in above: fixing a $ \lambda \in \mathbb R $ there is only one $ g $, but still many $ \theta $'s.
That aside, the introduction of the function $ \theta $ is unnecessary, and makes things more complicated. You could solve the problem in terms of $ g $ itself, and everything would become simpler, in fact. Let's start over, and make some simple observations which make us see an easy way to solve the problem.
$$ g ( t ) \overline { g ( t + h ) } = f ( h ) \tag 1 \label 1 $$
$$ g ( 0 ) = 1 \tag 2 \label 2 $$
$$ f ( - h ) = \overline { f ( h ) } \tag 3 \label 3 $$
First, note that \eqref{3} is redundant, and it can be derived using only \eqref{1}. To see this, let $ t = 0 $ in \eqref{1} to get
$$ f ( h ) = g ( 0 ) \overline { g ( h ) } \text . \tag 4 \label 4 $$
Then, substitute $ h $ for $ t $ and $ - h $ for $ h $ in \eqref{1} to see that
$$ f ( - h ) = \overline { g ( 0 ) } g ( h ) \text . \tag 5 \label 5 $$
Taking complex conjugates of both sides of \eqref{4} and comparing with \eqref{5}, \eqref{3} follows. Also, \eqref{4} lets us do our work only in terms of $ g $ from now on, since we can write \eqref{1} as
$$ g ( t ) \overline { g ( t + h ) } = \overline { g ( h ) } \text , \tag 6 \label 6 $$
knowing \eqref{2}. Substituting $ - h $ for $ h $ in \eqref{5} and using \eqref{2} and \eqref{4} we have
$$ \overline { g ( h ) } = g ( - h ) \text , \tag 7 \label 7 $$
which helps us get
$$ g ( t ) g ( - t - h ) = g ( - h ) \tag 8 \label 8 $$
from \eqref{6}. Substituting $ - t - h $ for $ h $ in \eqref{8}, we get
$$ g ( t + h ) = g ( t ) g ( h ) \text . \tag 9 \label 9 $$
By induction, we get
$$ g ( t ) = g ( 1 ) ^ t \tag {10} \label {10} $$
for all $ t \in \mathbb Z _ { 0 + } $ from \eqref{9}. Also, putting $ h = 0 $ in \eqref{8} we get
$$ g ( t ) g ( - t ) = 1 \text , \tag {11} \label {11} $$
which shows that \eqref{10} hols for all $ t \in \mathbb Z $. Finally, \eqref{7} and \eqref{11} together show that there is some $ \lambda \in \mathbb R $ such that $ g ( 1 ) = \exp ( i \lambda ) $, and \eqref{10} shows $ g ( t ) = \exp ( i \lambda t ) $ for all
$ t \in \mathbb Z $, and we're done.
