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A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ makes $I(v)\ge I(\bar{u})$ whenever $v\in \mathcal{A}$ and $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$. Both the function and its derivative are required to be close.

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 weak local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\rho >0$ such thatThe Sobolev norm controls both the function and its first weak derivative. The boundary values remain fixed because both functions are admissible.

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 weak local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\epsilon >0$ such thatBoth the function and its first derivative must remain uniformly close, up to the almost-everywhere convention for the derivative.

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.

A weak local minimizer of a variational integral has some $\rho >0$ for which $I(v)\ge I(\bar{u})$ whenever $v$ is admissible and $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$.

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 weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ gives $I(v)\ge I(\bar{u})$ for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$.

For a fixed-endpoint variational integral, an admissible function $u$ is a strong local minimizer if $I(v)\ge I(u)$ for all admissible $v$ whose uniform norm distance from $u$ is sufficiently small. Derivatives can be far apart. Hence strong local minimality implies weak local minimality, while a global minimizer implies both. Small narrow bumps and rapid low-amplitude oscillations explain why the reverse implication can fail.