# Show solution to character identity

In my analysis textbook in a chapter on fourier series it says that it is easy to show the following:

Any complex-valued $C^1$ solution to the character identity must be of the form $e^{\lambda x}$ where $\lambda$ is a complex number.

The character identity is: $\psi(x+y) = \psi(x)\psi(y)$

I don't see how to show this. I think it would be good to find the form of all solutions to the identity but I'm not sure how to do this. I'd appreciate any help solving this problem. Thanks!

Let $\psi(x)$ be a $C^1$ function satisfying

$\psi(x + y) = \psi(x) \psi(y); \tag{1}$

we first compute $\psi(0)$; we have

$\psi(0) = \psi(0 + 0) = (\psi(0))^2; \tag{2}$

the only solutions to (2) are $\psi(0) = 0, 1$; if $\psi(0) = 0$, then

$\psi(x) = \psi(0 + x) = \psi(0) \psi(x) = 0; \tag{3}$

the character $\psi$ must then be trivial. So we assume from here on the other alternative, $\psi(0) = 1$. We compute the derivative $\psi'(x)$ of $\psi$ at the value $x$, thus:

$\psi'(x) = \lim_{h \to 0} \dfrac{\psi(x + h) - \psi(x)}{h} = \lim_{h \to 0} \dfrac{\psi(x)\psi(h) - \psi(x)}{h}$ $= \psi(x) \lim_{h \to 0}\dfrac{\psi(h) - 1}{h} = \psi(x) \psi'(0); \tag{4}$

in (4), we have used the assumption that $\psi(0) = 1$ and the fact that $\psi$ is a $C^1$ function. Note $\psi'(0) \in \Bbb C$; this shows that $\psi(x)$ satisfies the differential equation

$\psi'(x) = \psi'(0) \psi(x). \tag{5}$

The only solution to (5) with $\psi(0) = 1$ is

$\psi(x) = e^{\psi'(0) x}. \tag{6}$

We have thus proved that every non-trivial $C^1$ $\psi(x)$ satisftying (1) is of the form

$\psi(x) = e^{\lambda x} \tag{7}$

with $\lambda = \psi'(0) \in \Bbb C$. QED.