 Solution

Divide the differential equation by $\sqrt{1-x^2}$. Its Sturm-Liouville form isHere $p(x)=\sqrt{1-x^2}$ and weight $w(x)=(1-x^2{)}^{-1/2}$. Multiply the equations for $T_n,T_m$ by the other polynomial and subtract. Integration by parts gives a boundary term $p(T_m{T}_n^{\prime }-T_n{T}_m^{\prime })$, which vanishes at both endpoints because polynomial derivatives are finite and $p\to 0$. HenceWith the standard nonnegative degree indices $n,m$, this proves orthogonality for $n\ne m$. If signed integer indices are used literally, the condition is $|n|\ne |m|$: $T_{-n}=T_n$, so merely $n\ne m$ would not suffice.

Set $x=\cos \theta$ and $Y(\theta )=y(\cos \theta )$. The chain rule gives $(1-x^2)y^{\prime \prime }-x{y}^{\prime }=Y^{\prime \prime }$, so $Y^{\prime \prime }+n^{2}Y=0$. For $n>0$, $Y=A\cos (n\theta )+B\sin (n\theta )$. A polynomial $y$ has $Y^{\prime }(0)=0$, forcing $B=0$. At $n=0$, $Y=A+B\theta$ and the same endpoint condition gives $B=0$. Therefore every polynomial solution is proportional toThe cosine recurrence yields $T_{n+1}=2x{T}_n-T_{n-1}$, $T_0=1,T_1=x$, confirming that these functions are polynomials. In this normalization the Chebyshev polynomial weight identity isIt is simply cosine orthogonality after $x=\cos \theta$.