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Math Help - concave function

  1. #1
    Senior Member Shanks's Avatar
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    concave 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], f(x)\leq g(x).
    Last edited by Shanks; December 27th 2009 at 10:33 AM.
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  2. #2
    Senior Member Shanks's Avatar
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    I've found the solution.
    never mind!
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    I'm interested in it, if it's possible... Share with us !
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  4. #4
    Senior Member Shanks's Avatar
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    method one:
    define G(x)=\left|f(x)\right|, then G(x) is continous nonnegative function, and G(x)\geq f(x).
    w.l.o.g, suppose G(x) assume its maxmum at the point t in (0,1).
    For the closed interval [0,t]:
    define F(x)=\sup_{0\leq s\leq x} G(s),then F(x) is a increasing continous function on the interval.
    Define H(x)=\sup_{0\leq u\leq x\leq v\leq t} \frac{(x-u)F(u)+(v-x)F(v)}{v-u},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).
    Last edited by Shanks; December 31st 2009 at 01:58 AM.
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