Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 2F (Topics in Analysis)

The nonzero leading coefficients and distinct degrees make $P_0,\dots ,P_{n-1}$ a basis of the polynomials of degree at most $n-1$. Consequently the given orthogonality implies ${\int }_a^{b}f(x)Q(x)dx=0$ for every such polynomial $Q$.

If $f$ has infinitely many distinct zeros in $(a,b)$, there is nothing to prove. Otherwise suppose it has fewer than $n$. Among those zeros, list the sign-changing ones as $t_1<\cdots <t_m$, with $m<n$, and put $Q(x)={\prod }_{j=1}^m(x-t_j)$, using $Q=1$ when $m=0$. On each interval between zeros, $f$ has a fixed sign by continuity. Crossing a sign-changing zero flips both $f$ and $Q$; crossing any other zero flips neither sign. Thus $fQ$ has one constant sign wherever nonzero. Since $f$ is not identically zero, continuity gives an open interval on which $fQ$ is strictly of that sign. Hence its integral is nonzero, contrary to orthogonality to $Q$. Therefore $f$ has at least $n$ distinct zeros in $(a,b)$. This is the reusable orthogonality forces many interior zeros argument.