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How to find all and every continuous function $ f $ from $ \mathbb{R} $ in $ \mathbb{R} $ that verifies $$ f(x+y)=f(x)\cdot f(y)\quad\forall x,y\in\mathbb{R} $$

All I could find is $ f(x)=f\left(\frac{1}{-x}\right) $... but I don't like that answer at all... any hints please??

Abderrazzak El Ouardi
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    How about $e^x$? – lulu Dec 02 '18 at 15:41
  • Very logical I like it .. but what is the procedure to prove that? – Abderrazzak El Ouardi Dec 02 '18 at 15:45
  • @Abderrazzak The exponential $ \exp_e:\mathbb{R}\to\mathbb{R}_{>0} $ is usually defined as a monotonic function such that $ \exp_e(x_1+x_2)=\exp_e(x_1)\exp_e(x_2) $ (and such that $ \exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function. –  Dec 02 '18 at 16:26

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This is not the complete answer, but it finds all differentiable functions $f:\mathbb R\rightarrow\mathbb R$ satisfying $f(x+y)=f(x)\cdot f(y), \forall x,y\in\mathbb R$.

Since $f$ is differentiable,

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{f(x)\cdot f(h)-f(x)}h=f(x)\cdot\lim_{h\to0}\frac{f(h)-1}h=f(x)\cdot f'(0), \forall x\in\mathbb R$

$\implies f'(x)=f(x)\cdot f'(0)$

Integrate this equation to obtain $f(x)=ke^{f'(0)x}$.

Since $f(x+y)=f(x)\cdot f(y), k=k^2\implies k=0,1$. Thus, the differentiable solutions of this functional equation are $f(x)=0,\ e^{kx}; k\in\mathbb R$.

Shubham Johri
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