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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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    $\begingroup$ Have you ever heard of Young's inequality? (Specifically, I'm thinking of the version $ab \leq r^{-1} a^{r} + s^{-1} b^{s}$ for $a,b > 0$ and $r,s$ as above.) $\endgroup$
    – user81375
    Apr 20, 2016 at 1:34
  • $\begingroup$ @fourierwho Yes, but that can only be used to show that $\frac{1}{s}|q|^2$ is an upper bound for the supremum though correct? I wasn't able to find a way to max out Young's inequality to show that it is in fact an equality. $\endgroup$
    – ufabao
    Apr 20, 2016 at 2:13
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    $\begingroup$ Young's inequality is an equality iff. $a^{r} = b^{s}$. (It goes back to strict concavity of the logarithm.) This shows your upper bound is actually achieved for some $p$, and hence is the sup. $\endgroup$
    – user81375
    Apr 20, 2016 at 3:58

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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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