Thread: Show That a Function is Contractive

1. Show That a Function is Contractive

Question:
Show that the following function is contractive on the indicated intervals. Determine the best values of [lamda] in Equation (2).

abs(x)^(2/3) on abs(x) < or = 1/3

Equation (2):
abs(F(x)-F(y)) < or = [lamda]*abs(x-y)

Any help with this problem would be greatly appreciated! Thanks!

2. Originally Posted by theamazingjenny
Question:
Show that the following function is contractive on the indicated intervals. Determine the best values of [lamda] in Equation (2).

abs(x)^(2/3) on abs(x) < or = 1/3

Equation (2):
abs(F(x)-F(y)) < or = [lamda]*abs(x-y)

Any help with this problem would be greatly appreciated! Thanks!
I do not think this function is contractive.

Assume that it is,
|F(x)-F(y)| <= K*|x-y| for -1/3<=x,y<=1/3
Take y=0 to get,
|F(x)|<= K*|x| for -1/3<=x<=1/3
Then certainly,
|F(x)|<= K*|x| for 0<x<=1/3
Thus,
x^{2/3} <= K*x for 0<x<=1/3
But then,
x^{-1/3} <= K
Then certainly
x^{-1/3} <= K+3
But this is false if you choose,
x>1/(K+3)^3

3. Originally Posted by theamazingjenny
Question:
Show that the following function is contractive on the indicated intervals. Determine the best values of [lamda] in Equation (2).

abs(x)^(2/3) on abs(x) < or = 1/3

Equation (2):
abs(F(x)-F(y)) < or = [lamda]*abs(x-y)

Any help with this problem would be greatly appreciated! Thanks!
A simpler demonstration that this is not a contraction is obtained by putting
x=1/8, and y=0, then:

abs( abs(x)^2/3 - abs(0)^2/3) = 1/4 > x - y = 1/8

RonL