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.

For the supporting inequality, assume first that $F$ is convex and fix $z,z_0$. For $0<t\le 1$,Subtract $F(z_0)$, divide by $t$ and let $t\downarrow 0$. Differentiability givesConversely assume this inequality for every pair. Put $w=(1-t)z_0+tz$ and apply it with base point $w$ to both $z_0$ and $z$. Multiply those two inequalities by $1-t$ and $t$ and add. The gradient terms cancel, proving $F(w)\le (1-t)F(z_0)+tF(z)$. Thus the supporting inequality characterizes a differentiable convex function.

For the integral inequality let $\bar{z}=|\Omega {|}^{-1}{\int }_{\Omega }u$. Apply the supporting inequality at $\bar{z}$ to $u(x)$ and integrate. Its affine term has integral $DF(\bar{z})\cdot ({\int }_{\Omega }u-|\Omega |\bar{z})=0$, yieldingThis is Jensen's inequality. Since $u$ is integrable, the supporting affine lower bound has integrable negative part, so the right side is well-defined as an extended real integral. If its positive part is infinite, the inequality remains valid.

Use the direct method in the calculus of variations. The affine function $1-x$ is admissible with finite energy, and the energy is nonnegative since $2010$ is even. Thus $m=\inf I$ is finite. Choose a minimizing sequence $u_n$ with uniformly bounded energy. ThenThe fixed value $u_n(0)=1$ and the Cauchy-Schwarz inequality give $\Vert u_n{\Vert }_{\infty }\le 1+\Vert u_n^{\prime }{\Vert }_2$ and $|u_n(x)-u_n(y)|\le \Vert u_n^{\prime }{\Vert }_2|x-y{|}^{1/2}$. Hence the sequence is bounded in $W^{1,2}$ and uniformly bounded and equicontinuous.

By reflexivity and the Arzelà-Ascoli theorem, pass to a subsequence with $u_n\rightharpoonup u$ in $W^{1,2}$ and $u_n\to u$ uniformly. The two limits coincide; uniform convergence also preserves $u(0)=1$, $u(1)=0$. Since all functions stay in a common bounded interval, uniform convergence gives $u_n^{2010}\to u^{2010}$ uniformly, and therefore convergence of the potential integrals. The derivative term is lower semicontinuous for weak convergence, soAdmissibility gives $I(u)\ge m$. Thus

For a fixed-endpoint variation $\phi \in W_0^{1,\infty }(a,b)$, the second variation iswhere all coefficients are evaluated at $(x,\bar{u}(x),{\bar{u}}^{\prime }(x))$. The notation ${\delta }^{2}I(\bar{u})>0$ means this is strictly positive for every nonzero admissible variation. This is pointwise positive definiteness of the quadratic form, not an assertion of a prescribed uniform lower bound.

PutAfter integration by parts, the form is $q(\phi )={\int }_a^b(P{\phi }^{\prime 2}+Q{\phi }^2)dx$. Its Jacobi equation isIn the regular case $P>0$ on $[a,b]$, a conjugate point $c\in (a,b]$ to $a$ is a point for which a nonzero solution satisfies $w(a)=w(c)=0$. Equivalently it is a zero of the normalized solution with $w(a)=0$, $P(a)w^{\prime }(a)=1$.

Under this regular Strengthened Legendre condition, the Jacobi positivity criterion isThe endpoint $b$ must be excluded as a conjugate point too. To prove necessity, suppose a Jacobi solution vanishes at $a$ and $c$. Define $\phi =w$ on $[a,c]$ and zero on $[c,b]$. It is a nonzero Lipschitz admissible variation, anda contradiction. For completeness, the converse follows from the Jacobi identity for the second variation: when the normalized $w$ has no further zero,The boundary contribution $[P(w^{\prime }/w){\phi }^2{]}_a^b$ vanishes, including the limit at $a$ since $w$ has a simple zero and $\phi =O(x-a)$. Equality forces $\phi =Cw$, and $\phi (b)=0$ with $w(b)\ne 0$ forces $C=0$.

Independently of regularity, positivity necessarily implies the ordinary Legendre necessary condition $P\ge 0$. If $P<0$ on an interval, localized rapidly oscillating test functions have a negative derivative contribution of order frequency squared, dominating the bounded zero-order term. This contradicts positivity.

The assumptions stated here do not themselves impose $P>0$. That strengthened condition is a regularity hypothesis for the familiar Jacobi criterion, and is not necessary for strict positivity: the integrand $f=u^2$ gives $q(\phi )=2\int {\phi }^2>0$ with $P=0$. For a criterion covering the full possibly degenerate class, use Regularized Jacobi criterion for a degenerate second variation. The following conditions are necessary and sufficient:

These conditions are necessary because $q>0$ gives $q_{\eta }(\phi )=q(\phi )+\eta \int {\phi }^{\prime 2}>0$, so the regular criterion just proved applies; a weak Jacobi solution tested against itself would give $q=0$. They are sufficient as well: the regular criterion gives $q_{\eta }>0$ for every $\eta >0$, hence $q\ge 0$ by letting $\eta \downarrow 0$. If $q(\phi )=0$, nonnegativity of $q(\phi +t\psi )$ for every real $t$ forces its mixed bilinear term to vanish for every test $\psi$. This is precisely the weak Jacobi equation. The third condition then forces $\phi =0$. Thus the criterion treats degeneracy without silently assuming the strengthened condition.