Weierstrass excess function

Remarks

Example: for $L(x, y, p) = \sqrt{1 + p^2}$ with $n = 1$, the function $q \mapsto L(x, y, q)$ is convex. A convex $C^1$ function lies above each of its tangent lines, so $E \ge 0$ everywhere.

Non-example of nonnegativity: for $L = (p^2 - 1)^2$ with $n = 1$, at $p = 0$ one gets $E(x, y, 0, q) = (q^2 - 1)^2 - 1$. This is negative for $0 < \abs{q} < \sqrt{2}$, since $L$ is not convex in $p$.

The Weierstrass necessary condition reads as follows. Let $y_0 \in C^1([a, b]; \R^n)$ be a strong local minimiser of $J(y) = \int_a^b L(x, y, y') \, dx$ with fixed end values, see weak and strong extremum. Then $E(x, y_0(x), y_0'(x), q) \ge 0$ for all $x \in [a, b]$ and all $q \in \R^n$.

For $C^2$ functions $L$, Taylor's formula gives $E(x, y, p, q) = \frac{1}{2} (q - p)^T L_{pp}(x, y, \bar p) (q - p)$ for some $\bar p$ on the segment from $p$ to $q$. So the Weierstrass condition implies the Legendre condition $L_{pp}(x, y_0, y_0') \ge 0$.