--- ±i®a»ô <chiachichang1123@yahoo.com.tw> wrote:
g(x+y)=g(x)g(y) x,y are belong to R
(BUT there is no continuous condition
g may not be a continuous function)
how to prove that
if g(1)>1 then g is increasing
if g(1)=1 then g is a constant function
if g(1)<1 then g is deceasing
From the functional equation, either g(x) vanishes identically or g(0)=1. Given g(1), the functional equation determines g(n) for integer n. If we make the additional assumption that g is everywhere nonnegative, then g(r) is determined on the rationals r. And, on the rationals, g satisfies the above monotonicity conditions. But this is as far as one can go without some way to relate the irrationals to the rationals. Continuity is one way to do this, and then one gets
g(x) = exp(x log(g(1))). The reals R are a vector space over the rationals Q. Using the axiom of choice, let B be a basis of R over Q. Then for each b in B, G(b) may be arbitrarily chosen. Gene __________________________________ Do you Yahoo!? Yahoo! Tax Center - File online by April 15th http://taxes.yahoo.com/filing.html