Past exam of the calculus of variations course of the University of Oxford / 2009 / 3 / b / i / Solution

Zero is neither a weak nor a strong local minimizer. Take $u_{\epsilon }(x)=-\epsilon \sin (\pi x)$. It tends to zero in $W^{1,\infty }$ and has zero endpoint values. Since $\sqrt{1+t}\le 1+t/2$ for $t\ge 0$,for sufficiently small positive $\epsilon$. Thus even weakly close admissible competitors decrease the energy.