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Let $f:\Bbb R\to \Bbb R$ a derivated function in all $\Bbb R$ that satisfies the condition $$f(x+y)=f(x)f(y),\;\,\,\text{for all $x,y \in \Bbb R$}$$

I already tried that $f'(x)f(y)=f(x)f'(y)$ for all $x,y \in \Bbb R$ and that exists $c\in \Bbb R$ so that $f'(x)=cf(x)$ for all $x \in \Bbb R$ but from the above I conclude that tha function $f$ is $f(x)=e^{cx}$ but have not achieved this. I appreciate the help you can give me.

Roiner Segura Cubero
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  • What exactly is your question? It looks like you have a valid proof that all differentiable $f$ satisfying that property are of the form $e^{cx}$. – Milo Brandt Jan 10 '15 at 03:59

2 Answers2

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All you need is continuity at $0$.

If for some $c$, $f(c) = 0$, then it is easy to show that $f(x) = 0$ for all $x$.

So suppose $f(x) \neq 0$ for any $x$.

Now $f(x) = (f(x/2))^2 \gt 0$.

Thus it makes sense to talk about $g(x) = \log f(x)$.

This satisfies $g(x+y) = g(x) + g(y)$.

This is the Cauchy functional equation and continuity at $0$ implies $g(x) = cx$ and $f(x) = e^{cx}$

btw, to complete your work:

$f'(x) = cf(x) \implies f(x) = Ae^{cx}$

We basically get $ (f(x)e^{-cx})' = 0 \implies f(x)e^{-cx} = A \implies f(x) = Ae^{cx}$

Now you can show that $f(1) = 1$ (or $0$) and get either $f(x) = e^{cx}$ or $f(x) = 0$.

Community
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Aryabhata
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Let $x=y=0$ then $f(0)=f(0)^2$ if $f(0)=0$. Let $y=0$ we have $f(x)=f(x)f(0)=0$

Hence we have a function : $f \equiv 0$.

If $f(0)=1$ Let $y=-x$ we have $1=f(0)=f(x)f(-x)$.

I can't prove $f(x)>0$. If $f(x)>0$ then let $g(x)=\ln(f(x))$ and $g(0)=0$. Hence,

$g(x+y)=g(x)+g(y)$ thus, $\frac{g(x+y)-g(y)}{x}=\frac{g(x)}{x}=g'(c)$ with $c \in (x,x+y)$

hence $g(x)=g'(c)x=ax$

Road Human
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