= Solution
\b[Zero is a strict strong local minimizer, and therefore also a strict weak local minimizer.] Here $I(0)=-2/2010$, and
$$
I(u)-I(0)=\int_0^1u'^2\,dx-\frac2{2010}\int_0^1(\cosh u-1)\,dx.
$$
For $|u|\leq1$, <Taylor's theorem> gives $\cosh u-1\leq\tfrac12\cosh(1)u^2$. Also $u(0)=0$ and the <Cauchy-Schwarz inequality> give $\int_0^1u^2\,dx\leq\tfrac12\int_0^1u'^2\,dx$. Therefore
$$
I(u)-I(0)\geq\left(1-\frac{\cosh1}{4020}\right)\int_0^1u'^2\,dx.
$$
The coefficient is positive. Thus every nonzero admissible $u$ with $\|u\|_\infty\leq1$ has strictly greater energy, proving that zero is a strict <strong local minimizer of a variational integral>.
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