Legendre transform and Young's inequality 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.
 A: 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 $p=\mathrm{sgn}(v)|v|^{s/r}$.
A: I want to give another proof, using a different approach. If $Q$ is a configuration manifold, a Hamiltonian is a smooth function $H\colon T^*Q \to \Bbb R$. The fiber derivative of $H$ is the map $\mathbb{F}H\colon T^*Q \to TQ$ given by $$\mathbb{F}H(x,p)q = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0}H(x,p+tq),$$for $(x,p) \in T^*Q$ and $q \in T_x^*Q$. We say that $H$ is hyperregular if $\mathbb{F}H$ is a diffeomorphism. We write $(x,v) = \mathbb{F}H(x,p)$. Then the energy map of $H$ is $E_H\colon T^*Q \to \Bbb R$ defined by $E_H(x,p) = \Bbb F H(x,p)p - H(x,p)$ and the associated Lagrangian is defined just by $L = E_H \circ (\mathbb{F}H)^{-1}$.
Here, $Q = \Bbb R^n$, so $T^*Q = \Bbb R^n\times \Bbb R^n$ (we use the usual dot product to write the second $\Bbb R^n$ like this, when formally it is $(\Bbb R^n)^\ast$), and our particular $H$ does not depend on the position variable $x$, hence we write $H\colon \Bbb R^n \to \Bbb R$ only. Let's compute $\mathbb{F}H$, which here is just the total derivative of $H$. We have $$H(p) = \frac{\|p\|^r}{r} \implies DH(p)q = \frac{1}{r}r \|p\|^{r-1} \left\langle \frac{p}{\|p\|}, q\right\rangle = \langle \|p\|^{r-2}p,q\rangle.$$This means that $v = \|p\|^{r-2}p$ and so $\|v\|^s = \|p\|^r$, since $r^{-1}+s^{-1} = 1$. The energy map of $H$ is $$E_H(p) = \|p\|^r - \frac{\|p\|^r}{r} = \frac{\|p\|^r}{s},$$and thus the change of variables gives $$L(v) = \frac{\|v\|^s}{s},$$as wanted.
