Results 1 to 2 of 2

Thread: Set of twice differentiable functions

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    1

    Set of twice differentiable functions

    Hi all,

    Need some help with this simple? proof.

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

    Prove: If $\displaystyle f \in A$ and if

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

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

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 $\displaystyle f:[0,1] \rightarrow \mathbb{R}$ with the properties: $\displaystyle f(0) = 0$, $\displaystyle f(1) = 1$ and $\displaystyle f''\leq 0$.

    Prove: If $\displaystyle f \in A$ and if

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

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

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

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

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

    $\displaystyle \int_0^1 (f(x) - x) \ dx \geq \int_{t-\delta}^{t + \delta} (f(x) -x) \ dx > 0.$ thus $\displaystyle \int_0^1 (f(x) - x) \ dx > 0$ and hence $\displaystyle \int_0^1 f(x) \ dx > \int_0^1 x \ dx.$ contradiction! so there's no $\displaystyle 0 < t < 1$ with $\displaystyle f(t) > t$ and we're done.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. MLE for non-differentiable functions
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Jun 3rd 2011, 11:34 PM
  2. Replies: 9
    Last Post: Dec 17th 2010, 08:13 AM
  3. Differentiable Functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 24th 2010, 04:45 AM
  4. Sequence of differentiable functions, non-differentiable limit
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Apr 3rd 2009, 05:13 AM
  5. where are these functions differentiable??
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Feb 11th 2008, 12:45 AM

Search Tags


/mathhelpforum @mathhelpforum