# Prove that $f'$ exists for all $x$ in $R$ if $f(x+y)=f(x)f(y)$ and $f'(0)$ exists

A function $f$ is defined in $R$, and $f'(0)$ exist.
Let $f(x+y)=f(x)f(y)$ then prove that $f'$ exists for all $x$ in $R$.

I think I have to use two fact:
$f'(0)$ exists
$f(x+y)=f(x)f(y)$
How to combine these two things to prove that statement?

• Possible duplicate: math.stackexchange.com/questions/64766/… Commented Dec 16, 2012 at 12:40
• Commented Dec 16, 2012 at 12:41
• Commented Dec 16, 2012 at 12:41

We have $$f(0)=f(0+0)=f(0)f(0)=f^2(0)\Rightarrow f(0)=0\text{ or }f(0)=1$$ If $f(0)=1$ by definition $$f^{\prime}(0)=\lim_{h\to 0}\frac{f(h)-1}{h}$$ and so $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)f(h)-f(x)}{h}=f(x)\lim_{h\to 0}\frac{f(h)-1}{h}=f(x)f^{\prime}(0)$$ Thus $f$ is differentiable in $\mathbb{R}$

If $f(0)=0$, $f(x)=f(x+0)=0\ \forall x\in \mathbb{R}$ and again $f$ is differentiable.

• what's missing is just proving that $f(0)=1$.
– tst
Commented Dec 16, 2012 at 12:23
• @tst I am editing it. Commented Dec 16, 2012 at 12:24
• You mean $f'(x)$ in the second line? Commented Dec 16, 2012 at 12:25
• It's a bit easier, $f(x)=f(x+0)=f(x)f(0)$.
– tst
Commented Dec 16, 2012 at 12:26

$$f$$ is continuous at $$x=0$$ and hence show that $$f$$ is continuous at the whole set of real numbers $$\mathbb{R}$$ and hence is of the form $$e^{ax}$$.

• You make quite a stronger statement than what was asked for, but it's technically correct. Commented Dec 16, 2012 at 15:53