= Solution
{c}
On $V$, the <bilinear form> $a$ is symmetric, and part (b) proves $a(v,v)\geq\alpha\|v\|_1^2$. Thus $a$ is an <inner product>: $a(v,v)\geq0$, and equality forces $v=0$. Its induced <energy norm> is $\|v\|_a=\sqrt{a(v,v)}$. For any real scalar $t$,
$$
\|tv\|_a=|t|\|v\|_a.
$$
The <Cauchy-Schwarz inequality> for this <inner product> follows directly by expanding $a(p-tq,p-tq)\geq0$ and minimizing over $t$ when $q\ne0$. Hence
$$
\begin{aligned}
\|p+q\|_a^2
&=\|p\|_a^2+2a(p,q)+\|q\|_a^2\\
&\leq\bigl(\|p\|_a+\|q\|_a\bigr)^2,
\end{aligned}
$$
which proves the <triangle inequality>. This verifies every <norm> axiom. In fact, the estimates from part (b) give the <equivalent norms>
$$
\boxed{\sqrt\alpha\|v\|_1\leq\|v\|_a\leq\sqrt{2\mu+3\lambda}\|v\|_1.}
$$
Subtract the discrete <weak formulation> from the continuous one to get <Galerkin orthogonality>,
$$
a(u-u_h,v_h)=0\qquad(v_h\in V_h).
$$
For any $v_h\in V_h$, write $u-v_h=(u-u_h)+(u_h-v_h)$. The second summand is in $V_h$, so expansion and <Galerkin orthogonality> give the <Pythagorean theorem> in the <energy norm>:
$$
\boxed{\|u-v_h\|_a^2=\|u-u_h\|_a^2+\|u_h-v_h\|_a^2.}
$$
The last term is nonnegative and vanishes exactly when $v_h=u_h$. Choosing $v_h=u_h$ attains the lower bound, proving
$$
\boxed{\|u-u_h\|_a=\min_{v_h\in V_h}\|u-v_h\|_a.}
$$
Thus $u_h$ is the unique <best approximation>; the map $u\mapsto u_h$ is the <orthogonal projector> onto $V_h$ for the <linear elasticity> <energy norm>.
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