How can I show f is constant?

Question

Suppose f:(0, oo) --> satisfies the functional equation f(x + y) = f(xy) for every positive x and y. Show f is constant.

Thank you

I mean, f:(0,oo) --> R, real valued

Answer 1

Let g(u) = u + 1/u, u >0. Then, g is continuous and takes on every value in [2, oo). (Just observe that g'(u) = 1 -1/u^2, so that g has a global minimum at x =1, with g(1) = 2, and that g goes to oo when x --> oo and x --> 0+).

So, for every x >=2 there exists an u > 0 such that x = u + 1/u. Then, f(x) = f(u + 1/u) = f( u * 1/u) = f(1), which shows f is constant for x >=2.

If x is in (0,2), we can find positive numbers u and v such that u v = x and u + v > 2. Just choose one of them sufficiently close to 0. Hence, f(x) = f(u v) = f(u + v) = f(1), in virtue of the previous conclusion.

So, we have f(x) = f(1) = constant for every x in (0, oo). And we are done.

EDIT: Nestor's proof was kinda simple and would almost certainly be the simplest proof if f was defined at y= 0. But.... , f is NOT defined at y = 0. 0 does not belong to (0, oo). So, his reasoning doesn't apply. It's not a proof.

Answer 2

If y = 0, then, for all x, f(x+0) = f(x*0) = f(0)

f(x) = f(0) for all x in [0 ; +inf[, then f is constant