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Math Help - Set of twice differentiable functions

  1. #1
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    Set of twice differentiable functions

    Hi all,

    Need some help with this simple? proof.

    Let A be the set of all twice differentiable functions f:[0,1] \rightarrow \mathbb{R} with the properties: f(0) = 0, f(1) = 1 and f''\leq 0.

    Prove: If f \in A and if

    \int_0^1 f(x) dx \leq \int_0^1 g(x) dx

    for all g \in A, then f(x) = x with all x \in [0,1].

    Thanks.
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  2. #2
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    Quote Originally Posted by Irrational View Post
    Hi all,

    Need some help with this simple? proof.

    Let A be the set of all twice differentiable functions f:[0,1] \rightarrow \mathbb{R} with the properties: f(0) = 0, f(1) = 1 and f''\leq 0.

    Prove: If f \in A and if

    \int_0^1 f(x) dx \leq \int_0^1 g(x) dx

    for all g \in A, then f(x) = x with all x \in [0,1].

    Thanks.
    suppose first that there exists 0 < a < 1 such that f(a) \leq a. then by the mean value theorem f(a)=af'(b), for some 0 < b < a. thus f'(b) \leq 1. again, by the mean value theorem

    1-f(a)=(1-a)f'(c), for some a < c < 1. but since f' is decreasing we have f'(c) \leq f'(b) \leq 1 and hence 1-f(a) \leq 1-a, i.e. f(a) \geq a. thus f(a)=a. so f(x) \geq x, \ \forall x \in [0,1].

    now suppose f(t) > t, for some 0 < t < 1. then, since f is continuous, there exists \delta > 0 such that f(x) > x, for all 0 \leq t-\delta < x < t + \delta \leq 1. but then we'll have:

    \int_0^1 (f(x) - x) \ dx \geq \int_{t-\delta}^{t + \delta} (f(x) -x) \ dx > 0. thus \int_0^1 (f(x) - x) \ dx > 0 and hence \int_0^1 f(x) \ dx > \int_0^1 x \ dx. contradiction! so there's no 0 < t < 1 with f(t) > t and we're done.
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