A question about the exponential function endomorphism

Question

I have the following problem :

Is the function $f(x)=\mathrm{e}^{k \times x },$ with $k \in \mathbb{R}$ the only endomorphism/ function $f$ such that

$$f : \begin{cases} (\mathbb{R}_+,+) \mapsto (\mathbb{R}_+^*,\times) \\\forall (t,u) \in \mathbb{R}_+, f(t+u)=f(t) \times f(u) & \\ f(0)=1 \text{ and $f$ is continuous at point 0}\end{cases}$$

$(\mathbb{R}_+,+)$ denoting the positives reals with addition and $(\mathbb{R}_+^*,\times )$ the strictly positive reals with multiplication ?

Here is how it all began :

We had a probability course about atomic disintegration/ decay, and we had to find which law this probability abides by, base on the following fact :

$T$ denotes the time it takes for a radioactive atom to disintegrate. the probability that a radioactive atom has NOT decayed by time $0\leq t \leq T$ is defined by

$$\mathbb{P}(T>t)=F(t)$$ We assume that $F(0)=1 \text{, $f$ is continuous at point 0 and } \lim_{t \to +\infty} F(t)=0$.

This means we suppose the atom has not disintegrated at $t=0$, but will finally do at some point

It has been proven that $$\forall t_2>t,F(t+t_2)=F(t)\times F(t_2) $$

Let's now define $F_2(t)=\mathbb{P}(T\leq t)=1-F(t)$

Prove that : $$F_2(t)=1-\mathrm{e}^{\lambda \times t}, \forall \in \mathbb{R}, \text{s.t.}\lambda <0 $$

e being the exponential function.

My attempt

We need to prove that $F(t)=\mathrm{e}^{\lambda \times t}, \forall \lambda <0$. If my claim is correct then we just need to check our $\lambda$ :

$F$ must tend to 0 as $t$ tends to $+\infty$ so that $\lambda \leq 0$ $$F(0)=1 \neq \lim_{t \to +\infty} F(t)=0$$

So $F$ is not constant and $\lambda < 0$

And the result follows.

I am aware that there is a proof through analysis, but I wanted to know if this algebraic shortcut would work

Thanks for the help,

T.D

Answer 1

Hints:

1. First prove $f(at)=f(t)^a$ for all positive integers $a$.
2. Then extend it to all positive rational numbers $a$.
3. Finally, for an irrational $a$, consider a rational sequence $\Bbb Q\ni a_n >a$ with $\lim_na_n=a$.
   Then apply continuity to $b_n:=(a_n-a) \,\to 0$.