 Solution

On $V$, the bilinear form $a$ is symmetric, and part (b) proves $a(v,v)\ge \alpha \Vert v{\Vert }_1^2$. Thus $a$ is an inner product: $a(v,v)\ge 0$, and equality forces $v=0$. Its induced energy norm is $\Vert v{\Vert }_a=\sqrt{a(v,v)}$. For any real scalar $t$,The Cauchy-Schwarz inequality for this inner product follows directly by expanding $a(p-tq,p-tq)\ge 0$ and minimizing over $t$ when $q\ne 0$. Hencewhich proves the triangle inequality. This verifies every norm axiom. In fact, the estimates from part (b) give the equivalent norms

Subtract the discrete weak formulation from the continuous one to get Galerkin orthogonality,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:The last term is nonnegative and vanishes exactly when $v_h=u_h$. Choosing $v_h=u_h$ attains the lower bound, provingThus $u_h$ is the unique best approximation; the map $u\mapsto u_h$ is the orthogonal projector onto $V_h$ for the linear elasticity energy norm.