An Intergral Inequlity

**Xingyuan** · April 28th 2008, 11:20 PM

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

**PaulRS** · April 29th 2008, 03:33 AM

Originally Posted by Xingyuan

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

$ \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.

**PaulRS** · April 29th 2008, 03:46 AM

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 $

**Xingyuan** · May 2nd 2008, 06:00 PM

excellent proof. very powerful.

**PaulRS** · May 2nd 2008, 06:07 PM

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.

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