= Solution
There is a genuine conflict in the printed coefficient assumptions: <no uniformly positive coefficient in a zero-boundary Sobolev space> exists. A function $a\in H_0^1(\Omega)$ cannot also obey $a\geq a_->0$ almost everywhere. Indeed, the Lipschitz truncation $g(s)=\min(\max(s,0),a_-)$ has $g(0)=0$, so <Lipschitz truncation preserves zero-boundary Sobolev spaces>, giving $g(a)\in H_0^1(\Omega)$. But $g(a)$ would be the nonzero constant $a_-$. Its zero <gradient> contradicts the <Poincare inequality> in the <zero-boundary Sobolev space>. Thus the literal coefficient class is empty.
For the meaningful uniformly elliptic problem, take $a\in L^\infty(\Omega)$ with $0<a_-\leq a\leq a_+$, and impose the <Dirichlet boundary condition> on the unknown and test functions: $V=H_0^1((0,1)^2)$. Additional $H^1$ regularity of $a$ is harmless, but a zero trace for $a$ must be removed. The <divergence-form elliptic operator> is $L=-\operatorname{div}(a\nabla)$. <Integration by parts> defines the symmetric <bounded bilinear form>
$$
B(v,w)=\int_\Omega a\,\nabla v\cdot\nabla w\,dx\,dy.
$$
In particular,
$$
B(v,v)\geq a_-\|\nabla v\|_2^2\geq 2\pi^2a_-\|v\|_2^2,
\qquad |B(v,w)|\leq a_+\|\nabla v\|_2\|\nabla w\|_2.
$$
Here $2\pi^2$ is the first <Dirichlet Laplacian eigenvalue> on the unit square; the <Poincare inequality> follows, for example, by applying the one-dimensional inequality in each coordinate and adding. Thus $B$ is coercive in the <gradient> <norm> on $V$, and the Dirichlet realization of $L$ is a <positive definite symmetric operator>. On its <operator domain>, $\langle Lv,v\rangle=B(v,v)$.
The required functional and weak equation are
$$
\boxed{I(v)=\int_\Omega\left(a|\nabla v|^2-2fv\right)\,dx\,dy,\qquad v\in H_0^1(\Omega),}
$$
$$
\boxed{\int_\Omega a\nabla u\cdot\nabla w=\int_\Omega fw
\quad\text{for every }w\in H_0^1(\Omega).}
$$
Because $f\in L^2$, the right side is a bounded <linear functional> by the <Cauchy-Schwarz inequality> and <Poincare inequality>. The <Lax-Milgram theorem> gives a unique <weak solution>, and part (b) proves that it uniquely minimizes $I$. These formulas prove the intended conclusion under the repaired coefficient hypothesis; under the literal hypothesis there is no coefficient to which the conclusion can be applied.
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