Results 1 to 5 of 5

Thread: An Intergral Inequlity

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    66

    Cool An Intergral Inequlity

    Please show that :

    if $\displaystyle f(x)\in C([0,1])$ and $\displaystyle 0 \leq f(x) < 1$

    then
    $\displaystyle \int_0^1\frac{f(x)}{1-f(x)}\geq\frac{\int_0^1f(x)}{1-\int_0^1f(x)}$



    thanks very much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by Xingyuan View Post
    Please show that :

    if $\displaystyle f(x)\in C([0,1])$ and $\displaystyle 0 \leq f(x) < 1$

    then
    $\displaystyle \int_0^1\frac{f(x)}{1-f(x)}\geq\frac{\int_0^1f(x)}{1-\int_0^1f(x)}$



    thanks very much
    Since $\displaystyle 0 \leq f(x) < 1$ we may apply the geometric series

    $\displaystyle
    \frac{{f\left( x \right)}}
    {{1 - f\left( x \right)}} = f\left( x \right) + f^2 \left( x \right) + ...
    $

    Now I'll show that: $\displaystyle
    \int_0^1 {f^m \left( x \right)dx} \geqslant \left( {\int_0^1 {f\left( x \right)dx} } \right)^m
    $ valid $\displaystyle
    \forall m \in \mathbb{Z}^ +
    $

    Consider the Riemannian sum: $\displaystyle
    \int_0^1 {f^m \left( x \right)dx} = \mathop {\lim }\limits_{n \to + \infty } \left( {\tfrac{1}
    {n}} \right) \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k}
    {n}} \right)}

    $

    By the Power Mean inequality we have: $\displaystyle
    \sqrt[m]{{\left( {\tfrac{1}
    {n}} \right) \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k}
    {n}} \right)} }} \geqslant \left( {\tfrac{1}
    {n}} \right) \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k}
    {n}} \right)}
    $

    Thus:$\displaystyle

    \tfrac{1}
    {n} \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k}
    {n}} \right)} \geqslant \left( {\tfrac{1}
    {n} \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k}
    {n}} \right)} } \right)^m

    $ and: $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } \tfrac{1}
    {n} \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k}
    {n}} \right)} \geqslant \left( {\mathop {\lim }\limits_{n \to + \infty } \tfrac{1}
    {n} \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k}
    {n}} \right)} } \right)^m
    $

    So we have: $\displaystyle
    \int_0^1 {f^m \left( x \right)dx} \geqslant \left( {\int_0^1 {f\left( x \right)dx} } \right)^m
    $

    This may also be shown applying Hölder's Inequality

    Assuming uniform convergence *

    $\displaystyle
    \int_0^1 {\frac{{f\left( x \right)}}
    {{1 - f\left( x \right)}} \cdot dx} = \int_0^1 {f\left( x \right)} \cdot dx + \int_0^1 {f^2 \left( x \right) \cdot } dx + ...
    $

    Applying the inequality we've just proved: $\displaystyle
    \int_0^1 {\frac{{f\left( x \right)}}
    {{1 - f\left( x \right)}} \cdot dx} \geqslant \int_0^1 {f\left( x \right) \cdot dx} + \left( {\int_0^1 {f\left( x \right) \cdot dx} } \right)^2 + ...
    $

    Which yields: $\displaystyle
    \int_0^1 {\frac{{f\left( x \right)}}
    {{1 - f\left( x \right)}} \cdot dx} \geqslant \frac{{\int_0^1 {f\left( x \right) \cdot dx} }}
    {{1 - \int_0^1 {f\left( x \right) \cdot dx} }}
    $

    This last step is correct since: $\displaystyle
    0 \leqslant \int_0^1 {f\left( x \right) \cdot dx} < 1
    $

    By the way, there's equality iff the function is constant.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by PaulRS View Post
    Assuming uniform convergence *

    $\displaystyle
    \int_0^1 {\frac{{f\left( x \right)}}
    {{1 - f\left( x \right)}} \cdot dx} = \int_0^1 {f\left( x \right)} \cdot dx + \int_0^1 {f^2 \left( x \right) \cdot } dx + ...
    $


    $\displaystyle
    \frac{{f\left( x \right) - f^{t + 1} \left( x \right)}}
    {{1 - f\left( x \right)}} = f\left( x \right) + f^2 \left( x \right) + ... + f^t \left( x \right)
    $

    Thus: $\displaystyle
    \int_0^1 {\frac{{f\left( x \right)}}
    {{1 - f\left( x \right)}}dx} - \int_0^1 {\frac{{f^{t + 1} \left( x \right)}}
    {{1 - f\left( x \right)}}dx} = \int_0^1 {f\left( x \right)dx} + ... + \int_0^1 {f^t \left( x \right)dx}
    $

    $\displaystyle
    0 \leqslant f\left( x \right) \leqslant k < 1 \Rightarrow 0 \leqslant \frac{{f^{t + 1} \left( x \right)}}
    {{1 - f\left( x \right)}} \leqslant \frac{{k^{t + 1} }}
    {{1 - k}}


    $ ( $\displaystyle
    g\left( x \right) = \tfrac{x}
    {{1 - x}}
    $ is increasing in $\displaystyle \left( {0,1} \right)$ )

    Integrating: $\displaystyle
    \Rightarrow 0 \leqslant \int_0^1 {\frac{{f^{t + 1} \left( x \right)}}
    {{1 - f\left( x \right)}}} dx \leqslant \frac{{k^{t + 1} }}
    {{1 - k}}
    $

    And since $\displaystyle
    \frac{{k^{t + 1} }}
    {{1 - k}} \to 0

    $ we have that: $\displaystyle
    \int_0^1 {\frac{{f^{t + 1} \left( x \right)}}
    {{1 - f\left( x \right)}}dx} \to 0
    $ as $\displaystyle
    t \to + \infty
    $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2007
    Posts
    66
    excellent proof. very powerful.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by Xingyuan View Post
    excellent proof. very powerful.


    If you want a powerful result apply Jensen's inequality to prove that:

    $\displaystyle \int_0^1f(g(x))dx\geq{f\left(\int_0^1g(x)dx\right) }$ where $\displaystyle f$ (a continuous function) is convex on the interval $\displaystyle [a,b]$; $\displaystyle g:[0,1]\rightarrow{[a,b]}$

    This is the more general result, but I still like the proof above.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Guidance r.e. Chebyshev's Inequlity.
    Posted in the Advanced Statistics Forum
    Replies: 42
    Last Post: Jan 6th 2012, 11:57 AM
  2. Inequlity
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: Aug 20th 2010, 03:17 AM
  3. intergral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 10th 2010, 07:04 AM
  4. Intergral
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Dec 12th 2009, 02:36 AM
  5. Intergral help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 16th 2008, 07:44 PM

Search Tags


/mathhelpforum @mathhelpforum