 Solution

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.