I am reading Liberzon's Calculus of Variations and Optimal Control theory, and trying to understand the Maximum principle. It is currently going over the Hamiltonian formulation.
I will define some of the terms being used. We are minimizing $J(y):=\int_a^b L(x,y(x),y'(x))dx$ over any $y \in C^2([a,b],\mathbb{R})$ such that $y(a)=y_a$ and $y(b)=y_b$. We define $p:=L_{y'}(x,y,y')$ and the Hamiltonian $H(x,y,y',p):=py' - L(x,y,y')$.
Defining $p:=L_{y'}$ and $H(x,y,y',p):=py' - L_{y'}(x,y,y',p)$. The textbook claims $H_{y'}=p-L_{y'}(x,y,y')=0$ for feasible curves $y$. So far, everything makes sense. But then the textbook claims
This suggests that, in addition to the canonical equations (2.30), another necessary condition for optimality should be that H has a stationary point as a function of y′ along an optimal curve.
(2.30) states that a necessary condition for an optimal curve $y$ is that $y'=H_p$ and $p'=-H_y$. I understand this. But this doesn't have anything to do with the fact that $H_{y'}(x,y,y',p)=0$ for any feasible curve $y$. Is it just that for optimal curves, we have $H_{y'}(x,y,y',p)=0$, but the converse doesn't hold?
Furthemore, the textbook states the hollowing:
Legendre’s condition tells us that, in addition, $H_{y'y'} = −L_{y'y'} \leq 0$ along an optimal curve [...] Thus, if the above stationary point is an extremum, then it is necessarily a maximum. This interpretation of necessary conditions for optimality moves us one step closer to the maximum principle.
So the book states if $H$ is extremized, then it is necessarily maximized. Firstly, $H$ was not extremized. It as extremized along the varying 3rd argument, sure, but $H$ was not extremized. Secondly, why was it necessarily maximized? From what I understand, $H_{y'}(x,y,y',p)=0$ for any feasible curve $y$. Can't we simply find a non-optimal curve $y$ such that $-L_{y'y'}(x,y,y',p) \leq 0$?