Consider $M$ to be a compact smooth manifold , and $L_t$ a Lagrangian on $TM$ that satisfies $\partial_{vv}L\geq l_0 I$. Then we are able to define a hamiltonian $H_t:T^*M\rightarrow \mathbb{R}$ by the Legendre transform $H(t,q,p):=\max_{v\in T_qM}(p[v]-L(t,q,v))$, which is equivalent to $p=d_vL(t,q,v)$.
Now we have the Hamiltonian action functional $\mathcal{A}_H(x):=\int_0^1 x^*(\theta)-H(t,x(t))dt$ for loops $x$ in $T^*M$, and the Lagrangian action functional $\mathcal{E}(q):=\int_0^1 L(t,q(t),\dot q(t))$. Consider the map $\theta: (q,v)\mapsto (q,d_vL(t,q,v+q'))$
Using Taylor's formula one is able to prove that $\mathcal{A}(\theta(q,v))=\mathcal{E}(\pi(x))-U(q,v)$ where
$$U(q,v):=\int_0^1\int_0^1s \cdot d_{vv}L(t,q,q'+sv)[v,v]dsdt$$
Due to the conditions on the Lagrangian we have that $U(q,v)\geq 0$.
Then I saw that it's also claimed that $d_v U(q,v)=0$ iff $v=0$. I tried seeing this but I wasn't sure how I would be able to control the term $d_{vvv}L$, hence any help regarding this appreciated. Moreover, it's claimed that since this happens we have $d^2\mathcal{A}(x)[\xi,\xi]\leq d^2\mathcal{E}(q)[d\pi(x)(\xi),d\pi(x)(\xi)]$ and I also don't quite understand why this is true.
Any insight is appreciated, thanks in advance.