Problem:
Suppose $\Phi : \mathbb{R}_{>0} \to [0,1]$ satisfies $\Phi(s+t)=\Phi(t)\Phi(s)$ for all $s,t \in \mathbb{R}_{>0}$. How can be proven that $\Phi(t)=e^t$?
Actually I do not know where to start.
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$\Phi (t)=0$ for all $t$ gives a counter-example. – Kavi Rama Murthy Jun 16 '20 at 10:15
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1Is $e$ an arbitrary constant? Then it's better to be denoned $a$. Let $\Phi(t)=c\cdot e^{f(t)},,ce^{f(s+t)}=c^2\cdot e^{f(s)+f(t)}\ldots$ – Alexey Burdin Jun 16 '20 at 10:16
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Answer assuming that $\Phi(t)$ is continuous (or at least Borel measurable) and not identically $0$: Note that $\Phi (2t)=(\Phi(t))^{2}$. Using this it is easy to see that if $\Phi (t) =0$ for some $t$ then $\Phi \equiv 0$. Now assume that $\Phi (t) >0$ for all $t$. Then $f(t) =\ln (\Phi (t)))$ satisfies the Cauchy's equation $f(t+s)=f(t)+f(s)$. The only measurable solution of this $f(t)=ct$ where $c$ is a constant. Hence $\Phi (t)=e^{ct}$. The fact that $\Phi$ take value in $[0,1]$ implies that $c \leq 0$. This is all we can say. We cannot prove that $c=-1$ since $e^{-ct}$ does satisfy the functional equation for any $c \leq 0$.
Kavi Rama Murthy
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