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Im looking for different concave function between $[0,1]$ which is continuous and differentiable. The function value should be $0$ at $0$ and $1$ at $1$.

one such function is $2x - x^2$

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    $\sin\frac{\pi x}{2}$ works.2011-05-11
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    $1-(x-1)^n$, $n=2,4,6,\ldots$, generalizes your function $2x-x^2$.2011-05-11
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    @J.M More clearly, im looking for a function concave function which is above $x=y$ line and has the shape similar to the step function.2011-05-11
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    $1-(x-1)^n$ for $n$ even and sufficiently large (say $n > 100)$ is quite similar to the "step function", if I understand what you mean by step function.2011-05-11
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    @shai: Thanks for your generalized function2011-05-11

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Take any continuous, negative function $f$. Integrating $f$ twice, you obtain $F$. Then consider $G(x) = F(x)+ax+b$, where $a$ and $b$ are chosen so as to ensure your boundary conditions. Then $G$ is a possible answer (by doing so, I believe you would find all such $C^2$ functions).

Take $f(x) = -x^2$. Then

$f_1(x) = \int_0^xf(y)dy = -\frac{x^3}3$

and

$F(x) = \int_0^xf_1(y)dy = -\frac{x^4}{12}.$

Accounting for the BC leads to

$G(x) = -\dfrac{x^4}{12}+\dfrac{13x}{12}$

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    is it $G(x) = F(x)+ax+b$?2011-05-11
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    My mistake, I'll edit my answer2011-05-11
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    can you give me an example2011-05-11
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    I understood that by integrating the function twice, we make the function the function twice differentiable. May i know the need for it. Why cannot we settle down with single differentiable function? I also wanted to know how do u come up with $G(x) = F(x) + ax +b$?2012-03-08