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Differentiate the Legendre differential equation and put $Q_n=P_n^{\prime }$. This givesMultiplication by $1-x^2$ puts it in the self-adjoint formThe boundary term vanishes at $x=\pm 1$. Distinct eigenvalues are therefore orthogonal with weight $1-x^2$:

The Lagrange identity for this Sturm-Liouville theory operator isIntegrating from zero to one, the boundary term vanishes because both $y$ and $y_0$ vanish at both endpoints. Since $L(y_0;{\lambda }_0)=0$ and $L(y;{\lambda }_0)=f$, it follows thatThis is the solvability condition at a Sturm-Liouville eigenvalue.

Multiplication by $(1-x^2{)}^{-1/2}$ puts the equation in Sturm-Liouville theory form:Multiply the equations for $T_n,T_m$ by $T_m,T_n$, subtract and integrate. The boundary term vanishes because $\sqrt{1-x^2}=0$ at both endpoints and the polynomial derivatives are bounded. Therefore, for $n^2\ne m^2$,This is the usual Chebyshev polynomial orthogonality.

Differentiating the original equation and writing $U_n=T_n^{\prime }$ givesIts self-adjoint form isThe same subtraction argument now yields, for $n^2\ne m^2$,These two relations form the Chebyshev derivative Sturm-Liouville pair.