= Solution
\b[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 $(1+u^2)^{1/4}\leq1+u^2/4$ for every real $u$. Therefore, when $\|u'\|_\infty$ is small,
$$
I(u)-I(0)\geq\frac38\int_0^1u'^2-\frac12\int_0^1u^2\geq\frac18\int_0^1u'^2,
$$
using $\int u^2\leq\tfrac12\int u'^2$. This proves strict weak minimality.
For the strong claim, let $v_N(x)=\sqrt\pi\,\operatorname{dist}(x,N^{-1}\mathbb Z)$. 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 $-2$. Hence
$$
\boxed{I(v_N)\leq-\cosh\sqrt\pi-2<-1=I(0),\qquad\|v_N\|_\infty\to0.}
$$
This disproves strong minimality. The same obstruction is detected by the <Weierstrass necessary condition>, since the supporting-tangent inequality at slope zero fails at $v=\sqrt\pi$.
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