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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. – J. M. May 11 '11 at 5:57
$1-(x-1)^n$, $n=2,4,6,\ldots$, generalizes your function $2x-x^2$. – Shai Covo May 11 '11 at 6:08
@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. – Learner May 11 '11 at 6:11
$1-(x-1)^n$ 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. – Shai Covo May 11 '11 at 6:22
@shai: Thanks for your generalized function – Learner May 11 '11 at 6:38
up vote 3 down vote accepted

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

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


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

Accounting for the BC leads to

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

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is it $G(x) = F(x)+ax+b$? – Learner May 11 '11 at 6:13
My mistake, I'll edit my answer – Sebastien May 11 '11 at 6:15
can you give me an example – Learner May 11 '11 at 6:19
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 $G(x) = F(x) + ax +b$? – Learner Mar 8 '12 at 4:37

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