 Solution

Since $\mathbf{a}\ne 0$, the line $\mathbf{x}=t\mathbf{a}/\Vert \mathbf{a}{\Vert }^2$ lies in the domain and satisfies ${\mathbf{a}}^{\mathsf{T}}\mathbf{x}=t$. The restriction of $l$ to this line is $t^3$, whose second derivative is $6t$: negative for $t<0$ and positive for $t>0$. It can be neither globally convex nor globally concave. Hence $l$ is neither. Equivalently, its Hessian matrix is $6({\mathbf{a}}^{\mathsf{T}}\mathbf{x})\mathbf{a}{\mathbf{a}}^{\mathsf{T}}$, changing the sign of its nonzero directional curvature across ${\mathbf{a}}^{\mathsf{T}}\mathbf{x}=0$.