Definition of exponential function

[Bridgeport]

Privetstvuu!

I have that $f(x)$ is right continuous and also satisfies $f(a+b)= f(a)f(b)$. How do I show that $f(t)= e^{-\lambda t}$?

[Бодигрим]

$f\equiv0$ is a right answer too. But if $f\not\equiv0$:

0. $f(n)=f(n)f(0)$, so $f(0)=1$.
1. Let $a:=f(1)$. Then by induction $f(n)=a^n$ for all $n\in\mathbb{N}$.
2. $f(0)=f(n)f(-n)$, so $f(-n)=a^{-n}$ for all $n\in\mathbb{N}$.
3. $f(m)=\underbrace{f(m/n)\cdot\ldots\cdot f(m/n)}_{n}$, so for all $m\in\mathbb{Z}$ and $n\in\mathbb{N}$ we have $f(m/n)=a^{m/n}$.
4. Now we should use the condition of continuity to apply our result for all real arguments.

[Echo-Off]

If $\exists x_0:\ f(x_0) = 0$, then $f(x)\equiv 0$ (cos' $f(a)=f(a-x_0)f(x_0)=0$), and this is not interesting. So let $f(x)\neq0\ \forall x$.

$f(0) = f(0)f(0)\ \Rightarrow\ f(0)=1$. Let $f(1) = a$. Then $\forall n\in\mathbb{Z}\ f(n) = a^n$. Then $\forall \frac pq\in\mathbb{Q}\ f(\frac pq) = a^{\frac pq}$.

$\forall x\in\mathbb{R}\ \exists x_n\to x:\ x_n\in\mathbb{Q},\ x_1>x_2>x_3\dots$. $f(x)$ is right continuous, so $f(x) = \lim\limits_{n\to\infty}f(x_n) = \lim\limits_{n\to\infty}a^{x_n} = a^x$.

[Bridgeport]

Огромное спасибо!