= Solution
{c}
The <infinitesimal strain tensor> $\varepsilon(v)=\tfrac12(\nabla v+(\nabla v)^{\mathsf T})$ and the <divergence> are linear in $v$. Consequently the given $a$ is a <symmetric bilinear form>, and
$$
a(v,v)=2\mu\|\varepsilon(v)\|_0^2+\lambda\|\operatorname{div}v\|_0^2\geq0.
$$
For a matrix-valued <function>, $\|\cdot\|_0$ here uses the <Frobenius inner product> pointwise and then the <L2 norm>. All expressions are finite for $v\in H^1(\Omega;\mathbb R^3)$.
Expand the quadratic term at $\theta v+(1-\theta)w$. The <linearity> of $\ell$ cancels its contribution to the difference, giving
$$
\begin{aligned}
\theta J(v)+(1-\theta)J(w)-J(\theta v+(1-\theta)w)
&=\frac{\theta(1-\theta)}2\bigl(a(v,v)-2a(v,w)+a(w,w)\bigr)\\
&=\frac{\theta(1-\theta)}2a(v-w,v-w)\\
&=\theta(1-\theta)\left(\mu\|\varepsilon(v-w)\|_0^2+\frac\lambda2\|\operatorname{div}(v-w)\|_0^2\right).
\end{aligned}
$$
For $0\leq\theta\leq1$, this is nonnegative. Therefore
$$
\boxed{J(\theta v+(1-\theta)w)\leq\theta J(v)+(1-\theta)J(w),}
$$
so $J$ is a <convex functional> on the whole <Sobolev space>. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the <zero-boundary Sobolev space> will instead use the <Korn inequality> in part (b).
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