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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.

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.

Zero is a strict weak local minimizer, but is not a strong local minimizer. The derivative integrand has expansion $\cosh v\cos (v^2)=1+v^2/2+O(v^4)$, so for small $|v|$ it is at least $1+3v^2/8$. Concavity gives $(1+u^2{)}^{1/4}\le 1+u^2/4$ for every real $u$. Therefore, when $\Vert u^{\prime }{\Vert }_{\infty }$ is small,using $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. This proves strict weak minimality.

For the strong claim, let $v_N(x)=\sqrt{\pi }\mathrm{dist}(x,N^{-1}\mathbb{Z})$. Its amplitude tends to zero and its derivative is $\pm \sqrt{\pi }$ almost everywhere, so the derivative integrand is the constant $-\cosh \sqrt{\pi }$. The other integrand term is at most $-2$. HenceThis disproves strong minimality. The same obstruction is detected by the Weierstrass necessary condition, since the supporting-tangent inequality at slope zero fails at $v=\sqrt{\pi }$.

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.

A global minimizer satisfies $I(\bar{u})\le I(v)$ for every admissible $v$.

A point of convexity $s$ of a differentiable scalar function is one where its tangent supports the entire graph: $g(q)\ge g(s)+g^{\prime }(s)(q-s)$ for every $q\in \mathbb{R}$. The Weierstrass necessary condition for strong minimality, under its differentiability assumptions in the slope variable, isfor almost every $x$. In other words the reference slope is a point of convexity of the slope integrand. The later example is smooth, so these derivative-based conditions apply there.

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.