= Solution
Use the <direct method in the calculus of variations>. The affine function $1-x$ is admissible with finite energy, and the energy is nonnegative since $2010$ is even. Thus $m=\inf I$ is finite. Choose a <minimizing sequence> $u_n$ with uniformly bounded energy. Then
$$
\frac12\int_0^1|u_n'|^2dx\leq I(u_n)\leq C.
$$
The fixed value $u_n(0)=1$ and the <Cauchy-Schwarz inequality> give $\|u_n\|_\infty\leq1+\|u_n'\|_2$ and $|u_n(x)-u_n(y)|\leq\|u_n'\|_2|x-y|^{1/2}$. Hence the sequence is bounded in $W^{1,2}$ and uniformly bounded and equicontinuous.
By <reflexivity> and the <Arzelà-Ascoli theorem>, pass to a subsequence with $u_n\rightharpoonup u$ in $W^{1,2}$ and $u_n\to u$ uniformly. The two limits coincide; <uniform convergence> also preserves $u(0)=1$, $u(1)=0$. Since all functions stay in a common bounded interval, <uniform convergence> gives $u_n^{2010}\to u^{2010}$ uniformly, and therefore convergence of the potential integrals. The <derivative> term is lower semicontinuous for <weak convergence>, so
$$
I(u)\leq\liminf_n\frac12\int|u_n'|^2+\lim_n\int u_n^{2010}=\liminf_nI(u_n)=m.
$$
Admissibility gives $I(u)\geq m$. Thus
$$
\boxed{I(u)=m,\quad\text{so the global minimum is attained}.}
$$
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