Given $1 The minimum over $\mathbb{R}$ I have found is:
$$\alpha\left(p\right)=\frac{p^{\frac{1}{q-1}}-\left(1-p\right)^{\frac{1}{q-1}}}{p^{\frac{1}{q-1}}+\left(1-p\right)^{\frac{1}{q-1}}}$$
I would like to show that: $\exists c>0,\exists\gamma\in\left]0;1\right],\forall p\in\left[0;1\right],\left|1-2p\right|\leq c\left(1-\phi\left(\alpha\left(p\right)\right)\right)^{\gamma}$
When I asked my question, I was mistaken: there was $\alpha\left(p\right)$ instead of $\phi\left(\alpha\left(p\right)\right)$ in the right hand side above. Actually, you just have to show the following to conclude:
$$\forall p\in\left[0;1\right],\left(1-2p\right)^{2} = 1-4p\left(1-p\right)\leq 1-\phi\left(\alpha\left(p\right)\right)$$
And it is simple indeed! :)
Inequality based on the minimum of a function
2
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functions
optimization
inequality
1 Answers
2
If $\alpha$ is large (of either sign), $\phi$ increases without bound essentially as $\alpha ^q$. Is $\alpha \in (-1,1)$ in which case we can delete the absolute value bars?
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0I had the same reasoning, this is how I got my minimum. However, I cannot see how what I would like to show could be true with the minimum I have found. – 2010-11-17
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0If you can justify $\alpha \in (-1,1)$, which I think you can by checking the end points, I agree with your minimum. But if you try to feed that $\alpha$ to find $\phi$ you have a mess. – 2010-11-17
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0Okay, thanks for checking the minimum. – 2010-11-17