 Solution

Let $r=f-p_n$ be the continuous function describing the error. The preceding orthogonality equations sayIf $r$ vanishes identically, any $n+1$ distinct points work. If $r$ already has at least $n+1$ distinct zeros, the assertion also follows. Suppose instead that it has at most $n$ distinct zeros. Among its interior zeros, list those at which its sign changes as $t_1<\cdots <t_m$, where $m\le n$, and define the polynomialwith $q=1$ if $m=0$. Between consecutive zeros of $r$, its sign is constant by the intermediate value theorem. At each $t_j$, both $r$ and $q$ change sign; at every other zero of $r$, neither sign changes. Hence $rq$ has one fixed sign throughout $[-1,1]$ wherever it is nonzero. Replacing $q$ by $-q$ if necessary makes $rq\ge 0$ everywhere.

Since $r$ is not identically zero and $q$ has only finitely many zeros, $rq$ is strictly positive on some open interval. Positivity of $w$ givescontradicting orthogonality because $q\in {\mathcal{P}}_n$. ThusThis is the interpolation property of weighted least-squares polynomial approximation. It supplies interpolation nodes depending on $f$ and $w$, rather than nodes prescribed in advance.