TITLE: How do I prove that $f(x)f(y)=f(x+y)$ implies that $f(x)=e^{cx}$, assuming f is continuous and not zero?
QUESTION [18 upvotes]: This is part of a homework assignment for a real analysis course taught out of "Baby Rudin."  Just looking for a push in the right direction, not a full-blown solution.  We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero.  The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant.  I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers.  Is this an approach worth taking?

REPLY [2 votes]: $f(x)= f(\frac{x}{2})f(\frac{x}{2}) $ and together with $f(x) \neq 0$ implies $f(x)$ is positive (actually if  $f(x) $ is zero anywhere then it is zero everywhere)
Define $g(x)= log f(x)$
Then $g(x) + g(y) =g(x+y)$ which is cauchy functional equation with solution $g(x) = cx$ where $c$ is a constant
So  $f(x)= e^{cx}$