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The variational terminology refers to the size of a neighborhood, rather than weak versus strong convergence in a Banach space. The definitions in the two numbered children giveThe second implication holds because smallness in $W^{1,\infty }$ implies smallness in the uniform norm. Reverse implications need not hold.

For differentiable $f$ in its slope variable, define the Weierstrass excess functionThe strengthened Weierstrass condition means that some $\delta >0$ givesIt requires support at nearby heights and nearby reference slopes, while the trial slope $q$ is arbitrary. Requiring this only exactly on the curve is the unstrengthened necessary condition.

A standard Weierstrass sufficient condition assumes the usual $C^2$ smoothness of $f$, $u\in \mathcal{A}\cap C^1$, the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened condition above. Explicitly, for some $c>0$,for zero-endpoint variations. Then $u$ is a strong local minimizer. If the excess is strictly positive for $q\ne p$ throughout the tube, the strong minimum is strict. The second-variation hypothesis supplies a neighboring field of extremals. Its slope $p(x,y)$ gives the comparison identityfor uniformly close admissible $v$. This explains how the condition controls competitors with large derivatives.

A global minimizer has $I(\bar{u})\le I(v)$ for every admissible $v$. Global minimality implies strong local minimality, which implies weak local minimality. The reverse implications are not valid in general.

Where $f_v$ exists, the strengthened Weierstrass condition says that for some $\delta >0$, the excessis nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ with $|y-\bar{u}(x)|<\delta$, $|p-{\bar{u}}^{\prime }(x)|<\delta$. This is a neighborhood condition in height and reference slope, not just an inequality on the reference curve.

The classical Weierstrass sufficient condition, with the usual smooth integrand regularity in addition to continuity, states that an admissible $C^1$ curve satisfying the Euler-Lagrange equation, a coercively positive second variation, and this strengthened condition is a strong local minimizer. A strict excess away from the reference slope gives the corresponding strict conclusion. The differentiability assumptions are part of that classical sufficient theorem; continuity alone does not define its first and second variations.

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.

Under the usual $C^2$ integrand regularity, an admissible $C^1$ curve satisfying the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened Weierstrass condition is a strong local minimizer of a variational integral. Strict excess away from the reference slope gives strict minimality. Coercive positivity means ${\delta }^{2}I(u)[\eta ,\eta ]\ge c\Vert \eta {\Vert }_{W^{1,2}}^2$ for every zero-trace variation and some $c>0$.

The second variation provides a neighboring field of extremals with slope $p(x,y)$. The associated endpoint-invariant Hilbert integral yields $I(v)-I(u)=\int E(x,v,p(x,v),v^{\prime })dx$ for uniformly close admissible $v$. The tube condition makes this integral nonnegative even when $v^{\prime }$ is large, explaining the passage from weak to strong minimality.