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I apologize for the non-descript title, this one is difficult to describe. I'm reading a paper by Loukas Grafakos dealing with the Hardy-Littlewood Maximal function. There is an "easy calculation" near the end that I've struggled to prove. Let $p > 1$ and let $p'$ satisfy $\frac{1}{p} + \frac{1}{p'} = 1$. We have a value $a$ that satisfies $$ a = \frac{p}{p-1} \frac{\gamma^{1/p'} + 1}{\gamma + 1}, $$ where $\gamma$ is the unique positive solution of the equation $$ \frac{p}{p-1} \frac{\gamma^{1/p'} + 1}{\gamma + 1} = \gamma^{-1/p}. $$ The easy calculation to be done is that this value $a$ is the unique positive root of $$ (p-1)x^p - p x^{p-1} - 1 = 0. $$ Ignoring uniqueness, I am looking for guidance on showing this. I believe the necessary calculations are just arithmetic, but plugging either of the first equations into the polynomial doesn't seem to work. Thank you.

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  • $\begingroup$ It's real. By complex conjugate I mean that $p'$ satisfies $1/p + 1/p' = 1$. Sorry for the confusion. $\endgroup$
    – Oreomair
    Sep 14 at 21:01
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Both $a$ and $\gamma^{-1/p}$ are equal to the same expression, so $a = \gamma^{-1/p} \implies a^p = 1/\gamma$ and $ a^{p-1}=a^p / a = \gamma^{1/p-1}=\gamma^{-1/p'}$. Then:

$$ \require{cancel} a = \frac{p}{p-1} \frac{\gamma^{1/p'} + 1}{\gamma + 1} = \frac{p}{p-1} \frac{1/a^{p-1}+1}{1/a^p + 1} = \frac{p}{p-1} \frac{a^p+a}{a^p+1} \\ \implies\;\;\;\;(p-1)\,\cancel{a}\,(a^p+1) = p\,\cancel{a}\,(a^{p-1}+1) \\ \iff\;\;\;\;(p-1)a^p + \bcancel{p} - 1 - p a^{p-1} - \bcancel{p} = 0 $$

Therefore $a$ is a root of $(p-1)x^p- p x^{p-1} - 1\,$, which has a unique positive root by Descartes' rule of signs given that $p \gt 1\,$.

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    $\begingroup$ Wonderful! Thank you for the help. $\endgroup$
    – Oreomair
    Sep 14 at 22:11

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