Symmetry obstruction in a polynomial Rayleigh-Ritz trial space (${\min }_{\sigma \in \mathrm{span}\{r-r^2,r-r^3\}}\frac{{\int }_0^1{\sigma }^{\prime 2}dr}{{\int }_0^1{\sigma }^{2}dr}=10$)

Let $s=r(1-r)$ and $p=r(1-r^2)$. Under reflection $r\mapsto 1-r$, $s$ is symmetric and $p=(3/2)s+a$, where $a=r(1-r)(r-1/2)$ is antisymmetric. Their cross terms vanish in both the $L^2$ inner product and the Dirichlet energy form: each integrand is odd about the midpoint. Their separate Rayleigh quotients are $R[s]=10$ and $R[a]=42$. Thus every quotient on their two-dimensional span is a positive weighted average of ten and forty-two, so the minimum remains ten. Adding a direction with the wrong symmetry need not improve the ground-state estimate.

The symmetric quartic $s^2=r^2(1-r{)}^2$ does add a useful direction. On the span of $s,s^2$, the generalized eigenvalue problem has stiffness matrix $K=\left(\begin{array}{cc}1/3& 1/15\\ 1/15& 2/105\end{array}\right)$ and mass matrix $M=\left(\begin{array}{cc}1/30& 1/140\\ 1/140& 1/630\end{array}\right)$. Its least eigenvalue is $56-4\sqrt{133}\simeq 9.86974962$, close to the exact Dirichlet ground value ${\pi }^2$. This is an explicit useful enlargement of a polynomial Rayleigh-Ritz trial space for a one-dimensional box.