 Solution

Put $x=\cos \theta$ in the weighted inner product. Since $T_r(\cos \theta )=\cos (r\theta )$,The last equality follows by the product-to-sum trigonometric identity and integration. Thus these are nonzero mutually orthogonal polynomials. The first $n+1$ are linearly independent, and the vector space $P_n$ has dimension $n+1$; therefore $\boxed{P_n=\mathrm{span}(T_0,\dots ,T_n)}$.

For any $q\in P_n$, the function of the real parameter $t$has a minimum at $t=0$. Its derivative there vanishes, proving $\boxed{(u-p_n,q)=0}$. This is precisely the defining orthogonality of an orthogonal projection, or a least-squares polynomial in an orthogonal-polynomial basis. Conversely, this condition and the Pythagorean theorem for inner product spaces give $\Vert u-q{\Vert }_2^2=\Vert u-p_n{\Vert }_2^2+\Vert p_n-q{\Vert }_2^2$, proving the minimum and its uniqueness.

Expand $p_n={\sum }_{r=0}^n{a}_r{T}_r$ in the basis already established and take the inner product with $T_s$. Orthogonality leaves $a_s(T_s,T_s)=(u,T_s)$, soIn particular $a_0={\pi }^{-1}{\int }_0^{\pi }u(\cos \theta )d\theta$ and $a_r=2{\pi }^{-1}{\int }_0^{\pi }u(\cos \theta )\cos (r\theta )d\theta$ for $r\ge 1$. These are the corresponding Chebyshev series coefficients with the constant coefficient convention used in this question.