Weierstrass necessary condition ($E(x,u,u^{\prime },v)\ge 0$)

At a strong local minimizer of a variational integral, almost every reference slope $u^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,u(x),v)$. Equivalently its Weierstrass excess function is nonnegative for every trial slope. Uniformly small variations can have large derivatives, which is why this condition tests all trial slopes rather than only nearby ones. It is a necessary condition along the minimizing curve; the strengthened Weierstrass condition additionally controls a neighborhood of that curve and its slopes.

Suppose the Weierstrass necessary condition applies to a strong local minimum of this fixed-endpoint integral. It requires the tangent to $g$ at $u^{\prime }(x)$ to lie below the entire graph for almost every $x$. A line of nonzero slope is unbounded above in one direction, while $g$ is bounded; hence this supporting line has slope zero. Its contact value is then a global minimum of $g$. The resulting function minimizes the integral globally, since its integrand is almost everywhere equal to the global lower bound. Thus for a bounded differentiable slope-only integrand a strong local minimum can exist only at global energy level. Weak local minima can behave differently because their competitors have nearby derivatives.