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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?

Martin Sleziak
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landolf
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  • Possible duplicate: http://math.stackexchange.com/questions/64766/solution-for-exponential-functions-functional-equation-by-using-a-definition-of – akkkk Dec 16 '12 at 12:40
  • Related: http://math.stackexchange.com/questions/175607/differentiable-function-not-constant-fxy-fxfy-f0-2 – akkkk Dec 16 '12 at 12:41
  • Related: http://math.stackexchange.com/questions/151032/if-f-colon-mathbbr-to-mathbbr-is-such-that-f-x-y-f-x-f-y-an – akkkk Dec 16 '12 at 12:41

2 Answers2

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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.

Nameless
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$f$ 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 $e^{ax}$.

keynes
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Myshkin
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  • You make quite a stronger statement than what was asked for, but it's technically correct. – akkkk Dec 16 '12 at 15:53