= Solution
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 give
$$
\boxed{\text{global minimizer}\Longrightarrow
\text{strong local minimizer}\Longrightarrow\text{weak local minimizer}.}
$$
The 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 function>
$$
E(x,y,p,q)=f(x,y,q)-f(x,y,p)-f_v(x,y,p)(q-p).
$$
The <strengthened Weierstrass condition> means that some $\delta>0$ gives
$$
\boxed{E(x,y,p,q)\geq0\quad\text{for all }x\in[a,b],\ q\in\mathbb R,
\quad |y-u(x)|<\delta,\quad|p-u'(x)|<\delta.}
$$
It 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 $c>0$,
$$
\delta^2I(u)[\eta,\eta]=\int_a^b
\bigl(f_{uu}\eta^2+2f_{uv}\eta\eta'+f_{vv}(\eta')^2\bigr)\,dx
\geq c\|\eta\|_{W^{1,2}}^2
$$
for zero-endpoint variations. Then \b[$u$ 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 identity
$$
I(v)-I(u)=\int_a^b E(x,v(x),p(x,v(x)),v'(x))\,dx\geq0
$$
for uniformly close admissible $v$. This explains how the condition controls competitors with large derivatives.
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