= Solution
{c}
The finite-dimensional <vector subspace> $V_h\subset V$ is closed, so it is a <Hilbert space> with the inherited <Sobolev norm>. The <bounded bilinear form>, <coercive bilinear form>, and bounded <linear functional> estimates from part (b) restrict to $V_h$. Applying the <Lax-Milgram theorem> proves
$$
\boxed{\text{There is exactly one }u_h\in V_h\text{ with }a(u_h,v_h)=\ell(v_h)\quad(v_h\in V_h).}
$$
For any $v_h\in V_h$, put $d_h=v_h-u_h\in V_h$. Expansion using symmetry of the <bilinear form> gives
$$
\begin{aligned}
J(v_h)-J(u_h)
&=a(u_h,d_h)-\ell(d_h)+\frac12a(d_h,d_h)\\
&=\frac12a(d_h,d_h)\geq0.
\end{aligned}
$$
Thus $u_h$ minimizes $J$ on $V_h$, uniquely by the <coercive bilinear form> estimate. Similarly, the continuous <weak formulation> gives the exact identity
$$
J(u_h)-J(u)=\frac12a(u_h-u,u_h-u)\geq0.
$$
Combining the two identities proves
$$
\boxed{J(v_h)\geq J(u_h)\geq J(u)\qquad(v_h\in V_h).}
$$
More precisely, $J(v_h)-J(u_h)=\tfrac12\|v_h-u_h\|_a^2$ and $J(u_h)-J(u)=\tfrac12\|u_h-u\|_a^2$, with the <energy norm> justified next. These identities connect the energy excess directly to the squared error.
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