Suppose that the function f:[0,無窮大) →[0,無窮大)is continuous and
stable and f(0)=0,Prove that f:[0,無窮大) →[0,無窮大) is one-to-one
- mathmanliuLv 71 decade agoFavorite Answer
( one to one)
f: [0,∞) -> [0, ∞) is stable, then
there exists a positive constant c , such that
| f(x) - f(y) | >= c |x-y| for all x, y >=0
( 1-1 )
If f is not one-to-one, then there exists two numbers x1, x2, such that f(x1)=f(x2)
=> | f(x1) - f(x2) | = 0 >= c |x1-x2| contradiction.
Given any positive number r,
(2) f( (r+1)/c ) = f( (r+1)/c ) - f(0) >= c* (r+1)/c = r+1> r
By intermediate value theorem of continuous function, there exists
a number x in (0, (r+1)/c) such that f(x)= r.
So, f(x) can take any positive value, i.e. f:[0, ∞)->[0, ∞) is onto.