# Verifying that a value is a root of a given polynomial in a Grafakos paper.

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.

• It's real. By complex conjugate I mean that $p'$ satisfies $1/p + 1/p' = 1$. Sorry for the confusion. Sep 14 at 21:01

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 $$p \gt 1\,$$.