 Solution

The decisive observation is that $x$ is itself an eigenfunction:On the regular Legendre polynomial domain, integration by parts has endpoint factor $x^2-1$, so the boundary form vanishes for regular functions at $x=\pm 1$. The interval alone is not an endpoint condition; regularity is the usual condition selecting the Legendre polynomials. Their eigenvalues are $n(n+1)$, and the $n=1$ normalized function isThereforeEquivalently, the Hermitian operator identity gives $(q_n-2)c_n=0$, and nondegeneracy leaves only one possible nonzero coefficient. The unnormalized function $x$ has squared norm $2/3$, consistent with this coefficient.