= Solution
The integrand is nonnegative, so $\inf I\geq0$. For each positive integer $n$, set
$$
u_n(x)=2\operatorname{dist}\left(x,\frac1n\mathbb Z\right),\qquad0\leq x\leq1.
$$
This is a continuous piecewise affine sawtooth with zero endpoints, $0\leq u_n\leq1/n$, and $u_n'=\pm2$ <almost everywhere>. It belongs to $W^{1,\infty}$, and $(u_n')^6=64$ <almost everywhere>. Therefore
$$
I(u_n)=\int_0^1\bigl(u_n^{2011}+4096u_n^4\bigr)\,dx
\leq n^{-2011}+4096n^{-4}\longrightarrow0.
$$
Together with nonnegativity this proves $\boxed{\inf_{\mathcal A}I=0}$. The functions converge uniformly to zero while their derivatives retain magnitude two; small amplitude does not force small <gradient>.
\Image[/past-exam-of-the-calculus-of-variations-course-of-the-university-of-oxford/2011-small-amplitude-competitors.png]
{title=Uniformly small competitors can retain fixed nonzero slopes: repeated teeth and a shrinking localized bump}
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