 Solution

Interpret the weighted function space as measurable functions modulo equality almost everywhere, so the integral form is a genuine inner product. Assume $q\in {\Pi }_k\subseteq V$. If the residual is orthogonal to ${\Pi }_k$, then for $p\in {\Pi }_k$,Thus $q$ is a best approximation. Conversely, if it is best, take any $r\in {\Pi }_k$ and vary through $q+tr$. The polynomialhas a minimum at $t=0$, so its derivative there vanishes. Hence $\boxed{⟨f-q,r⟩=0}$ for every $r\in {\Pi }_k$. The same Pythagorean identity proves uniqueness of the best approximation.