Re: DSE 2009

Posted by Amit Goyal on Feb 27, 2012; 4:23am
URL: http://discussion-forum.276.s1.nabble.com/DSE-2009-tp7296495p7320729.html

For problem 3,
g(y)= min {x belongs to [0,1] | f(x)>=y }
First we will show g is non decreasing.
Let y and y' be two points in
[0,1]. Without loss of generality, let y'>y. We will show that
g(y')>=g(y). Now y'>y implies that {x belongs to [0,1] | f(x)>=y' } is
a subset of {x belongs to [0,1] | f(x)>=y }. [If for some x, f(x) is bigger than y' then for that x f(x) is also bigger than y because y' is bigger than y] This implies that
g(y')=min{x belongs to [0,1] | f(x)>=y' }>=min{x belongs to [0,1] |
f(x)>=y }=g(y). [Smallest value in a set is less than or equal to the smallest value of any of its subset] Hence proved.

If g is continuous,then f is strictly increasing is not true.
Consider f(x) = min{2x, 1}. This is not a strictly increasing function.
However, the associated g(y) = (0.5)y is continuous.