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With fixed endpoint values on a finite interval, the displayed derivative coercivity makes a finite-energy strong local minimizer of a variational integral also a local minimizer in $L^1$. Otherwise choose better competitors converging in $L^1$. The upper energy bound controls their derivatives in $L^p$, and a common endpoint controls their function norms. Compact Sobolev embedding gives a uniformly convergent subsequence; its limit is the $L^1$ limit. This contradicts strong local minimality. Convexity of the integrand is unnecessary for this neighborhood-transfer argument.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which the same conclusion holds whenever $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. Every strong local minimizer is weak, because Sobolev closeness implies uniform closeness. A global minimizer is both, and the converse implication from weak to strong is not generally valid.

If the integrand is independent of $u$, a weak local minimizer has vanishing first variation:The bounded function $h(x)=f_v(x,{\bar{u}}^{\prime }(x))$ therefore has zero distributional derivative, so is constant almost everywhere. For an elementary test-function argument, any smooth compactly supported $\psi$ of integral zero is the derivative of a compactly supported smooth $\phi$. Hence $\int h\psi =0$; subtracting a fixed unit-integral test function from arbitrary tests shows that $h$ agrees with one constant against every test. The fundamental lemma then givesThis is the first integral for a variational integrand independent of the function value.

A point of convexity $v_0$ of a differentiable real function $g$ has a globally supporting tangent: $g(v)\ge g(v_0)+g^{\prime }(v_0)(v-v_0)$ for every $v$. The Weierstrass necessary condition for a strong minimum says that ${\bar{u}}^{\prime }(x)$ is such a point for $v\mapsto f(x,\bar{u}(x),v)$ for almost every $x$. Equivalently, the Weierstrass excess function is nonnegative at every trial slope.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ satisfiesOnly the function values must be close; admissible derivatives may be large. Every strong local minimizer is therefore also a weak local minimizer of a variational integral.

A point of convexity of $g\in C^1(\mathbb{R})$ is a point $v_0$ where its tangent line supports the entire graph from below:The Weierstrass necessary condition says that, at almost every $x$, ${\bar{u}}^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,\bar{u}(x),v)$. In terms of the Weierstrass excess function,

Now use the stated derivative coercivity. If $\bar{u}$ were not a local minimizer in $L^1$, there would be $v_n\in \mathcal{A}$ with $\Vert v_n-\bar{u}{\Vert }_1\to 0$ and $I(v_n)<I(\bar{u})$. The energy bound givesThe common value $v_n(a)=\alpha$ controls their function norms as well, by $|v_n(x)|\le |\alpha |+{\int }_a^b|v_n^{\prime }|$. Thus $(v_n)$ is bounded in $W^{1,2010}(a,b)$. The one-dimensional compact Sobolev embedding into $C([a,b])$ supplies a uniformly convergent subsequence. Its limit must be $\bar{u}$, since the same subsequence converges to $\bar{u}$ in $L^1$. Eventually $\Vert v_n-\bar{u}{\Vert }_{\infty }<\rho$, contradicting strong local minimality. Hence $\bar{u}$ is also a local minimizer in $L^1$. This is coercivity transfers strong minimality to L1 minimality.

Zero is a strict strong local minimizer, and therefore also a strict weak local minimizer. Here $I(0)=-2/2010$, andFor $|u|\le 1$, Taylor's theorem gives $\cosh u-1\le \frac{1}{2}\cosh (1)u^2$. Also $u(0)=0$ and the Cauchy-Schwarz inequality give ${\int }_0^1{u}^{2}dx\le \frac{1}{2}{\int }_0^1{u}^{\prime 2}dx$. ThereforeThe coefficient is positive. Thus every nonzero admissible $u$ with $\Vert u{\Vert }_{\infty }\le 1$ has strictly greater energy, proving that zero is a strict strong local minimizer of a variational integral.

Zero is a strict weak local minimizer, but is not a strong local minimizer. Put $D={\int }_0^1|u^{\prime }{|}^{2}dx$. Since $u(0)=0$, the Cauchy-Schwarz inequality gives $|u(x){|}^2\le xD$, and thereforeFor sufficiently small $\Vert u^{\prime }{\Vert }_{\infty }$, $\sin (u^{\prime 2})\ge \frac{1}{2}{u}^{\prime 2}$ pointwise. Hencewith equality only when $u=0$. Thus zero is a strict weak local minimizer of a variational integral.

For the strong claim, define a Lipschitz triangular waveIt has zero endpoint values, $\Vert u_N{\Vert }_{\infty }=\sqrt{\pi }/(2N)\to 0$, and $u_N^{\prime }=\pm \sqrt{\pi }$ almost everywhere. Thus $\sin (u_N^{\prime 2})=\sin \pi =0$, while $u_N$ is nonzero on a set of positive measure. ConsequentlyThese uniformly small competitors disprove the strong local minimizer of a variational integral property.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\epsilon >0$ such thatThe competitor's derivative need not be close. This compares with a larger family of competitors, which is why it is the stronger minimality property.

A strong local minimizer of a variational integral has the same property for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$; its derivative need not be close. Since the supremum norm is bounded by the Sobolev norm, this neighborhood contains a sufficiently small weak neighborhood. Thus every strong local minimizer is a weak local minimizer of a variational integral.

If $u$ is a strong local minimizer of a variational integral, the Weierstrass condition makes $u^{\prime }(x)$ a point of convexity of the bounded function $g$ for almost every $x$. A global supporting line to a bounded function must have zero slope, and its contact value must be a global minimum. Here that value is $-1$. Therefore $g(u^{\prime })=-1$ almost everywhere and $I(u)=-1$, while every admissible energy is at least $-1$. Hence every strong local minimizer is global.

Write $g(v)=-\cos (\pi v/2)$. For the stipulated $0<\delta <1$, the displayed tent is continuous at its junction, has zero endpoint values, and its slopes are $\delta$ and $\delta -4$. At both slopes $g^{\prime }$ has the same value $c=(\pi /2)\sin (\pi \delta /2)>0$, and $g^{\prime \prime }=({\pi }^2/4)\cos (\pi v/2)$ is strictly positive on common sufficiently small neighborhoods of them. Taylor's theorem gives, for small $\Vert {\varphi }^{\prime }{\Vert }_{\infty }$,Thus $u^{\delta }$ is a strict weak local minimizer of a variational integral. It is not a strong local minimizer of a variational integral: the global supporting-tangent requirement at either slope has nonzero slope $c$, whereas $g$ is bounded. Letting the trial slope tend to positive infinity violates that requirement. The Weierstrass necessary condition therefore fails.

The opening parameter range $\delta \in (0,1)$ is essential. The subsequent phrase “for each $\delta >0$” must be read within it; taken literally beyond that range it is false. At $\delta =2$ the tent slopes are $2,-2$, where $g=1$ and $g^{\prime \prime }<0$, so this is not a weak minimum.

A strong local minimizer of a variational integral obeys the same inequality for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. The derivative of a competitor need not be close in this definition.

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.