 Solution

If $u$ is a strong local minimizer of a variational integral, the Weierstrass condition makes $u^{\prime }(x)$ a point of convexity of the bounded function $g$ for almost every $x$. A global supporting line to a bounded function must have zero slope, and its contact value must be a global minimum. Here that value is $-1$. Therefore $g(u^{\prime })=-1$ almost everywhere and $I(u)=-1$, while every admissible energy is at least $-1$. Hence every strong local minimizer is global.