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Suppose $f(x)$ is a function which is right continuous and satisfies

$$ f(x + y) = f(x) f(y) $$

for $x,y > 0$, then $f(x)$ is exponential

I know $f(1) = f( \sum_{k=1}^n \frac{1}{n} ) = f( \frac{1}{n} )^n$ and so $f(\frac{1}{n}) = ( f(1))^{1/n} $. Similarly, $f(\frac{m}{n}) = f(1)^{m/n} $ and so $f(r) = f(1)^r $ for $r \in \mathbb{Q} $ and so $\log f(r) = r \log f(1) $ which gives

$$ f(r) = \exp( r \log( f(1)) \;\;\;\; for \; \; r\in \mathbb{Q}$$

Question: How can I show this folds for all reals ?

1 Answers1

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Hint: use $f(x+y)=f(x)f(y)$ to derive $f(x-y)=\frac{f(x)}{f(y)}$ and hence $f(-x)=(f(x))^{-1}$.

xpaul
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  • I got that. But somehow I need to use the fact that $f$ is right continuous –  Mar 25 '15 at 13:55
  • @ProbabilityGuy, for any real $x$, choose a rational sequence ${x_n}$ which converges to $x$ from the right hand side. – xpaul Mar 25 '15 at 14:08