Let $f: (\mathbb{R},+) \rightarrow (\mathbb{R_0},.)$ be a homomorphism which is differentiable as a function from $\mathbb{R} \to \mathbb{R}$
Show that:
a) $f(x) > 0 \quad\forall x \in \mathbb{R}$
b) $\exists a \in \mathbb{R}: f(x) = e^{ax} \quad \forall x \in \mathbb{R}$
For the first one, I think we can do the following:
$$f(x) = f\left(\frac{x}{2}+\frac{x}{2}\right) = \left(f\left(\frac{x}{2}\right)\right)^2 >0$$
Is this correct?
For the second one, can anyone give me a hint? I do not know how to implement the differentiability. I tried $$f'(x) = f'(x/2)$$ by using the chain rule on the expression above. I must somehow be able to show that $f'(x) = af(x)$, since then I can write $f(x) = e^{ax}$
Thanks in advance.
It is easy to see that $f(0)=1$ (identity to identity).
\begin{align*} f^{'}(x) & = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ & = \lim_{h \to 0} \frac{f(x) \cdot f(h)-f(x)}{h}\\ & = f(x)\lim_{h \to 0} \frac{f(h)-1}{h}\\ & = f(x) \, f^{'}(0) \end{align*} Now let $f^{'}(0)=a$. Solve the differential equation $$f^{'}(x) = af(x).$$