If $f(x+y)=f(x)\cdot f(y)$, then $f(x)=e^{ax}$

I'm trying to prove this question:

Let $f:\mathbb R\to \mathbb R$ a function such that $f(x+y)=f(x)\cdot f(y)$, $f(0)=1$ and $f'(0)=a$. Show that $f(x)=e^{ax}$, for every $x\in \mathbb R$.

First of all I'm trying to prove that this function is positive, but even that it's difficult to me, I know that this function is never zero, because if there is $y \in \mathbb R$ such that $f(y)=0$, then $f(x-y+y)=f(x-y)\cdot f(y)=0$, then $f(x)=0$ for every $x\in \mathbb R$.

I need help to prove the positivity and hints to follow from that point.

• Show that $f$ is differentiable (in particular, continuous). Feb 8 '13 at 7:43
• @QiaochuYuan what's the gain showing that? thank you for your comment. Feb 8 '13 at 7:49
• If you can show that $f$ is continuous, $f(0) = 1$ together with the fact that $f$ is never zero implies that $f$ is positive. Feb 9 '13 at 3:03

for positivity:

$f(x) = f(\frac{x}{2} +\frac{x}{2})=f^2(\frac{x}{2})\geq 0$

hint for the problem:

take $g(x) =\ln f(x)$ then $g(x+y) =g(x) +g(y)$ so $g(x) =ax$

• Thank you very much for your answer!!! you helped me a lot. Feb 8 '13 at 7:57
• I think the last step needs a bit of justification. Let $\mathbb{R}$ be thought of as a vector space over $\mathbb{Q}$. Let $\{b_{-1}, b_0 = 1, b_1, ...\}$ be a basis and let $\phi(b_{-1}) = 1$ and $\phi(b_i) = 0$ for $i \geq 0$ and extend $\phi$ to be linear. This is not of the form $\phi(x) = kx$ but it does satisfy $\phi(x + y) = \phi(x) + \phi(y)$. At some point we need to use the differentiability condition. Feb 8 '13 at 8:23
• How can you argue that $g(x+y)=g(x)+g(y)$ must have $g(x)=ax$? Feb 8 '13 at 8:28
• user39843, we simply say $g'(0) = \frac{f'(0)}{f(0)} = a$ and so $\frac{g(x+h) - g(x)}{h} \rightarrow a$. This means that $g'(x) = a$ and so $g(x) = ax$. Feb 8 '13 at 8:35
• @user39843 we can prove that if $g(x+y)= g(x)+g(y)$, then $g(x)=\lambda x$ to see this, first of all prove to the natural numbers, after to the rational, irracional and to real numbers, did you follow me? Feb 8 '13 at 15:09

Pardon me if my solution seems like a cop out

They've already said the derivative at $0$ is defined, so $\frac{f(h) - 1}{h} \rightarrow a$.

Now $\frac{f(x+h)-f(x)}{h} = f(x) \frac{f(h) - 1}{h}$ which as $h \rightarrow 0$ is $a f(x)$. So $f'(x) = a f(x)$ and $f(0) = 1$.

This has solution $f(x) = e^{a x}$.