Legendre transform and Young's inequality

Question

Let $H:\mathbb{R}^n\to\mathbb{R}$ with

$$H(p)=\frac{1}{r}|p|^r$$

for $1<r<\infty$. If $$\frac{1}{r}+\frac{1}{s}=1$$ show that

$$L(v)=\frac{1}{s}|v|^s$$

is Legendre transform of $H$, i.e. $H^*=L$.

Attempt: using definition of Legendre transform and Young's inequality I can only obtain $L\ge H^*$. Not sure how to obtain the reverse inequality.

Answer 1

The Young inequality states that if $\frac1r+\frac1s=1$, and $r,s$ are positive, then $$ab\leq \frac1r a^r+\frac1s b^s$$ for every $a,b\geq 0$, with equality iff $a^r=b^s$. Putting $a=|p|, b=|v|$ the equality case is for $|p|=|v|^{s/r}$. By definition of $H^*$ we obtain $$H^*(v)=\sup_p\left\{pv-H(p)\right\}=\sup_p\left\{pv-\frac1r |p|^r\right\}\geq |v|^{s/r}|v|-\frac1r(|v|^{s/r})^r\\=|v|^s-\frac1r |v|^s=\frac 1s|v|^s=L(v)$$ where we used that the supremum is greater than the value obtained using the special equality case $p=\mathrm{sgn}(v)|v|^{s/r}$.