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Given $1<q\leq2$ and $0\leq p\leq1$, let us consider the following function: $$\phi\left(\alpha\right)=p\times\left|1-\alpha\right|^{q}+\left(1-p\right)\times\left|1+\alpha\right|^{q}$$

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! :)

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1 Answer 1

up vote 2 down vote accepted

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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I 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. –  Wok Nov 17 '10 at 18:25
    
If 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. –  Ross Millikan Nov 17 '10 at 18:36
    
Okay, thanks for checking the minimum. –  Wok Nov 17 '10 at 18:43

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