# continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$

Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.

$g(0)=1$. If $a=g(1)$,then $a>0$ and $g(r)>a^r$ for all $r$ in $\mathbb Q$.

Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.

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You probably mean $g(r)=a^r$. – André Nicolas May 5 '12 at 8:29
Hint (from standard calculus course): The analogous equation $G(xy)=G(x)+G(y)$ for the inverse funtion $G=g^{-1}$ (up to a constant multiple) is used to derive the integral/antiderivative $G(x)=\int_1^x\frac{dt}{t}$. – bgins May 5 '12 at 9:11
– Martin Sleziak May 5 '12 at 10:45

For $x,y\in\mathbb{R}$ and $m,n\in\mathbb{Z}$, \eqalign{ g(x+y)=g(x)\,g(y) &\implies g(x-y)={g(x) \over g(y)} \\&\implies g(nx)=g(x)^n \\&\implies g\left(\frac{m}nx\right)=g(x)^{m/n} } so that $g(0)=g(0)^2$ must be one (since if it were zero, then $g$ would be identically zero on $\mathbb{R}$), and with $a=g(1)$, it follows that $g(r)=a^r$ for all $r\in\mathbb{Q}$. All we need to do now is invoke the continuity of $g$ and the denseness of $\mathbb{Q}$ in $\mathbb{R}$ to finish.
For example, given any $x\in\mathbb{R}\setminus\mathbb{Q}$, there exists a sequence $\{x_n\}$ in $\mathbb{Q}$ with $x_n\to x$ (you could e.g. take $x_n=10^{-n}\lfloor 10^nx\rfloor$ to be the approximation of $x$ to $n$ decimal places -- this is where we're using that $\mathbb{Q}$ is dense in $\mathbb{R}$). Since $g$ is continuous, $y_n=g(x_n)\to y=g(x)$. But $y_n=a^{x_n}\to a^x$ since $a\mapsto a^x$ is also continuous.