Let f(x) be a continous function on [0,1] such that f(0)=f(1)=0.

prove that there is a continous concave function g(x) such that g(0)=g(1)=0, and for all x in [0,1], .

Printable View

- December 27th 2009, 11:03 AMShanksconcave function
Let f(x) be a continous function on [0,1] such that f(0)=f(1)=0.

prove that there is a continous concave function g(x) such that g(0)=g(1)=0, and for all x in [0,1], . - December 30th 2009, 03:44 AMShanks
I've found the solution.

never mind! - December 30th 2009, 03:47 AMMoo
I'm interested in it, if it's possible... Share with us ! :D

- December 31st 2009, 02:40 AMShanks
method one:

define , then G(x) is continous nonnegative function, and .

w.l.o.g, suppose G(x) assume its maxmum at the point t in (0,1).

For the closed interval [0,t]:

define ,then F(x) is a increasing continous function on the interval.

Define ,then H(x) is the function we need on the interval [0,t].

Do the same things similarly to the interval [t,1], We are done.

method two:

let S be the colletion of points which lies in the closed area formed by four curves: x=0,x=1,y=0,y=G(x).

Find the Closed convex hull of S, then prove that the boundry defines a function we need.(this is the geometry explanation of method one).