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I have two complex-valued functions, $f$ and $g$, that satisfy the following properties. $\overline{x}$ denotes the complex conjugate of $x$ below.

$$g(t)\overline{g(t+h)} = f(h) \quad \forall{t,h}\in \mathbb{Z} \tag{1}$$ $$g(0) = 1 \tag{2}$$ $$f(-h) = \overline{f(h)} \tag{3}$$

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, $\lvert g(t) \rvert = 1$. This implies that $g$ has to be of the form $$g(t) = e^{i\theta(t)}$$ for some real-valued function $\theta$ with $\theta(0) = 2\pi k$ for some $k \in \mathbb{Z}$. By $(1)$ $$e^{i\theta(t)}e^{-i\theta(t+h)} = f(h)$$ This implies $$\theta(t) - \theta(t+h) = m(h)$$ for some real-valued function $m$. By taking $t = 0$ I get $m(h) = -\theta(h) + 2\pi k$. So then I can write $$\theta(t+h) = \theta(t) + \theta(h) - 2\pi k \quad \forall{t,h}\in \mathbb{Z} \tag{4}$$

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 $(4)$, $$\theta(nt) = n\theta(t) - (n-1)2\pi k \quad \forall{n}\in\mathbb{Z}_{\geq 1}$$ Hence, $\theta(n) = n(\theta(1)-2\pi k) + 2\pi k$. So at least $n\mapsto\theta(n)$ is linear on $\mathbb{Z}_{\geq 1}$. I guess it also works the other way as well.

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  • $\begingroup$ Are you only interested in the values of $g(t)$ for integer $t$? $\endgroup$ – Gerry Myerson Apr 26 '16 at 9:16
  • $\begingroup$ @GerryMyerson Yes. $\endgroup$ – Calculon Apr 26 '16 at 9:16

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