Carathéodory's royal road

Remarks

The method works because of this identity, valid for every admissible curve $y$: $$J(y) = \int_a^b L^*(t, y(t), y'(t)) \, dt + S(b, y(b)) - S(a, y(a)).$$ The boundary terms depend only on the endpoints. By clauses 1 to 3, the integral of $L^*$ is $\ge 0$ for every $y$ and $= 0$ for $\bar x$, so $J(y) \ge J(\bar x)$.

Clauses 1 and 2 say that $\min_v L^*(t, x, v) = 0$, with the minimum at $v = \psi(t, x)$. With $H(t, x, p) = \sup_v (p \cdot v - L(t, x, v))$, this means $S_t + H(t, x, \nabla_x S) = 0$, the Hamilton–Jacobi equation.

Example: $n = 1$, $G = \R^2$, $L = \frac{1}{2} v^2$, and $\bar x(t) = t$ on $[0, 1]$. Take $S(t, x) = x - t/2$ and $\psi = 1$. Then $L^* = \frac{1}{2} v^2 + \frac{1}{2} - v = \frac{1}{2} (v - 1)^2$, which is $\ge 0$ and vanishes at $v = 1$, and $\bar x' = 1$. So $\bar x(t) = t$ minimises $\int_0^1 \frac{1}{2} y'^2 \, dt$ among $C^1$ curves from $(0, 0)$ to $(1, 1)$.

Non-example: in the same problem, $S = 0$ with $\psi = 1$ fails clause 2, since $L(t, x, 1) = \frac{1}{2} \ne 0$.