= Solution
A <strong local minimizer of a variational integral> is an admissible $\bar u$ for which some $\rho>0$ satisfies
$$
\boxed{I(v)\geq I(\bar u)\quad\text{for }v\in\mathcal A\text{ with }\|v-\bar u\|_\infty<\rho.}
$$
Only the function values must be close; admissible derivatives may be large. Every strong local minimizer is therefore also a <weak local minimizer of a variational integral>.
A <point of convexity> of $g\in C^1(\mathbb R)$ is a point $v_0$ where its tangent line supports the entire graph from below:
$$
g(v)\geq g(v_0)+g'(v_0)(v-v_0)\quad\text{for every }v\in\mathbb R.
$$
The <Weierstrass necessary condition> says that, at almost every $x$, $\bar u'(x)$ is a <point of convexity> of $v\mapsto f(x,\bar u(x),v)$. In terms of the <Weierstrass excess function>,
$$
E(x,\bar u,\bar u',v)=f(x,\bar u,v)-f(x,\bar u,\bar u')-f_v(x,\bar u,\bar u')(v-\bar u')\geq0\quad(v\in\mathbb R).
$$
Now use the stated <derivative> <coercivity>. If $\bar u$ were not a local minimizer in $L^1$, there would be $v_n\in\mathcal A$ with $\|v_n-\bar u\|_1\to0$ and $I(v_n)<I(\bar u)$. The energy bound gives
$$
c\int_a^b|v_n'|^{2010}\,dx\leq I(\bar u)+d(b-a).
$$
The common value $v_n(a)=\alpha$ controls their function norms as well, by $|v_n(x)|\leq|\alpha|+\int_a^b|v_n'|$. Thus $(v_n)$ is bounded in $W^{1,2010}(a,b)$. The one-dimensional compact <Sobolev embedding> into $C([a,b])$ supplies a uniformly convergent subsequence. Its limit must be $\bar u$, since the same subsequence converges to $\bar u$ in $L^1$. Eventually $\|v_n-\bar u\|_\infty<\rho$, contradicting strong local minimality. Hence \b[$\bar u$ is also a local minimizer in $L^1$]. This is <coercivity transfers strong minimality to L1 minimality>.
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