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], $\displaystyle f(x)\leq g(x)$.

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- Dec 27th 2009, 10:03 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], $\displaystyle f(x)\leq g(x)$. - Dec 30th 2009, 02:44 AMShanks
I've found the solution.

never mind! - Dec 30th 2009, 02:47 AMMoo
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- Dec 31st 2009, 01:40 AMShanks
method one:

define $\displaystyle G(x)=\left|f(x)\right|$, then G(x) is continous nonnegative function, and $\displaystyle G(x)\geq f(x)$.

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

For the closed interval [0,t]:

define $\displaystyle F(x)=\sup_{0\leq s\leq x} G(s)$,then F(x) is a increasing continous function on the interval.

Define $\displaystyle H(x)=\sup_{0\leq u\leq x\leq v\leq t} \frac{(x-u)F(u)+(v-x)F(v)}{v-u}$,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).