This is a portion of one of the questions in Evan's PDE book. Let $H: \mathbb{R}^n \to \mathbb{R}$ be defined by $H(p) = \frac{1}{r}|p|^r,$ for $1 < r < \infty$. I want to show that $$L(q) = H^{*}(q) = \sup_{p \in \mathbb{R}^n}\left\{q \cdot p - H(p)\right\} = \frac{1}{s}|q|^s, \text{ for all } p, q \in \mathbb{R}^n, \text{ where } \frac{1}{s} + \frac{1}{r} = 1$$
Clearly the supremum will occur when $p$ and $q$ are parallel, so i've been trying to show that $\sup_{a \in \mathbb{R}}\left\{a|q|^2 - \frac{a^r}{r}|q|^r\right\} = \frac{1}{s}|q|^s$, and I feel like i'm close, but I can't seem to find the next step.
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$\begingroup$Consider the function $f(a)=a|q|^2-\frac{a^r}{r}|q|^r$. We have that $$f^{\prime}(a)=|q|^2-a^{r-1}|q|^r.$$ Therefore, we solve $f^{\prime}(a)=0$ and find that $$a=|q|^{\frac{2-r}{r-1}}.$$ That $f$ attains a maximum at this value can be seen (of course) by looking at the second derivative.
But (with some algebraic simplifications) we then have that $$f \left(|q|^{\frac{2-r}{r-1}}\right)=\left(1-\frac{1}{r}\right)|q|^{1+\frac{1}{r-1}}.$$ This is what we wanted to show.
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