For a function of a single variable $f(x)$ which has the property that $f(x+y)=f(x)f(y)$ we first set $x=y=0$ and then develop an ODE to show that $f(x)=\exp\{-\beta x\}$.
I do not understand how setting $x=y=0$ gives us $f(0)=1$ before we know that we have the exponential? Then I don't understand how we can set the following ODE up. \begin{equation} \frac{df(x)}{dx}=-\beta f(x)\ \ \ \ \ \ \ \text{where} \ \ \ \ \ \ \ \beta=-f'(x) \end{equation} Which doesn't contain any information on $y$. I do however understand that the solution to that ODE is the exponential I am after. Many thanks!
For $\, x = y = 0\,$ we have $\, f(0+0) = f(0)\cdot f(0).\,$ Let us denote $\,f_0:=f(0),\,$ then $$ f_0 = f_0^2 \implies \begin{cases} f_0 = 1\\ f_0 = 0 \end{cases} $$ Note that for any $x\in\Bbb R\,$ $\,f(x + 0) = f(x)\cdot f(0)$. Thus we conclude that the second case $\,f_0 = 0\,$ is trivial. Indeed, if $\,f(0) = 0\,$ then $\, f(x + 0) = f(x)\cdot f(0) = f(x)\cdot 0 = 0,\, $ and thus $\,f(x) \equiv 0.\,$
Therefore we choose the first case $\,f_0 = f(0) = 1.\,$