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Math Help - An Intergral Inequlity

  1. #1
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    Cool An Intergral Inequlity

    Please show that :

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

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



    thanks very much
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Xingyuan View Post
    Please show that :

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

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



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

    <br />
\frac{{f\left( x \right)}}<br />
{{1 - f\left( x \right)}} = f\left( x \right) + f^2 \left( x \right) + ...<br />

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

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

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

    Thus: <br /> <br />
\tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k}<br />
{n}} \right)}  \geqslant \left( {\tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k}<br />
{n}} \right)} } \right)^m <br /> <br />
and: <br />
\mathop {\lim }\limits_{n \to  + \infty } \tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k}<br />
{n}} \right)}  \geqslant \left( {\mathop {\lim }\limits_{n \to  + \infty } \tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k}<br />
{n}} \right)} } \right)^m <br />

    So we have: <br />
\int_0^1 {f^m \left( x \right)dx}  \geqslant \left( {\int_0^1 {f\left( x \right)dx} } \right)^m <br />

    This may also be shown applying Hölder's Inequality

    Assuming uniform convergence *

    <br />
\int_0^1 {\frac{{f\left( x \right)}}<br />
{{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 + ...<br />

    Applying the inequality we've just proved: <br />
\int_0^1 {\frac{{f\left( x \right)}}<br />
{{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  + ...<br />

    Which yields: <br />
\int_0^1 {\frac{{f\left( x \right)}}<br />
{{1 - f\left( x \right)}} \cdot dx}  \geqslant \frac{{\int_0^1 {f\left( x \right) \cdot dx} }}<br />
{{1 - \int_0^1 {f\left( x \right) \cdot dx} }}<br />

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

    By the way, there's equality iff the function is constant.
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  3. #3
    Super Member PaulRS's Avatar
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    Quote Originally Posted by PaulRS View Post
    Assuming uniform convergence *

    <br />
\int_0^1 {\frac{{f\left( x \right)}}<br />
{{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 + ...<br />


    <br />
\frac{{f\left( x \right) - f^{t + 1} \left( x \right)}}<br />
{{1 - f\left( x \right)}} = f\left( x \right) + f^2 \left( x \right) + ... + f^t \left( x \right)<br />

    Thus: <br />
\int_0^1 {\frac{{f\left( x \right)}}<br />
{{1 - f\left( x \right)}}dx}  - \int_0^1 {\frac{{f^{t + 1} \left( x \right)}}<br />
{{1 - f\left( x \right)}}dx}  = \int_0^1 {f\left( x \right)dx}  + ... + \int_0^1 {f^t \left( x \right)dx} <br />

    <br />
0 \leqslant f\left( x \right) \leqslant k < 1 \Rightarrow 0 \leqslant \frac{{f^{t + 1} \left( x \right)}}<br />
{{1 - f\left( x \right)}} \leqslant \frac{{k^{t + 1} }}<br />
{{1 - k}}<br /> <br /> <br />
( <br />
g\left( x \right) = \tfrac{x}<br />
{{1 - x}}<br />
is increasing in  \left( {0,1} \right) )

    Integrating: <br />
 \Rightarrow 0 \leqslant \int_0^1 {\frac{{f^{t + 1} \left( x \right)}}<br />
{{1 - f\left( x \right)}}} dx \leqslant \frac{{k^{t + 1} }}<br />
{{1 - k}}<br />

    And since <br />
\frac{{k^{t + 1} }}<br />
{{1 - k}} \to 0<br /> <br />
we have that: <br />
\int_0^1 {\frac{{f^{t + 1} \left( x \right)}}<br />
{{1 - f\left( x \right)}}dx}  \to 0<br />
as <br />
t \to  + \infty <br />
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  4. #4
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    excellent proof. very powerful.
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  5. #5
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Xingyuan View Post
    excellent proof. very powerful.


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

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

    This is the more general result, but I still like the proof above.
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