Math Help - An Intergral Inequlity

1. An Intergral Inequlity

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

2. Originally Posted by Xingyuan

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

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

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

Consider the Riemannian sum: $
\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: $
\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: $

\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: $
\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: $
\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 *

$
\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: $
\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: $
\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: $
0 \leqslant \int_0^1 {f\left( x \right) \cdot dx} < 1
$

By the way, there's equality iff the function is constant.

3. Originally Posted by PaulRS
Assuming uniform convergence *

$
\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 + ...
$

$
\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: $
\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}
$

$
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}}

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

Integrating: $
\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 $
\frac{{k^{t + 1} }}
{{1 - k}} \to 0

$
we have that: $
\int_0^1 {\frac{{f^{t + 1} \left( x \right)}}
{{1 - f\left( x \right)}}dx} \to 0
$
as $
t \to + \infty
$

4. excellent proof. very powerful.

5. Originally Posted by Xingyuan
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.