= Symmetry obstruction in a polynomial Rayleigh-Ritz trial space
{title2=$\min_{\sigma\in\operatorname{span}\{r-r^2,r-r^3\}}\frac{\int_0^1\sigma'^2dr}{\int_0^1\sigma^2dr}=10$}
Let $s=r(1-r)$ and $p=r(1-r^2)$. Under reflection $r\mapsto1-r$, $s$ 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 $K=\begin{pmatrix}1/3&1/15\\1/15&2/105\end{pmatrix}$ and mass matrix $M=\begin{pmatrix}1/30&1/140\\1/140&1/630\end{pmatrix}$. Its least eigenvalue is $56-4\sqrt{133}\simeq9.86974962$, 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>.
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