 Orthogonality forces many interior zeros

If a nonzero continuous real function on an interval is orthogonal to every polynomial of degree below $n$, it has at least $n$ distinct interior zeros. If it had only finitely many zeros and fewer than $n$ sign changes, multiply it by the product of linear factors at its sign-changing zeros. The product has constant sign and is nonzero on an open interval, so its integral cannot vanish. Orthogonality to this degree-below-$n$ polynomial contradicts that conclusion. Infinitely many interior zeros already satisfy the assertion. Orthogonal polynomials of respective degrees $0,\dots ,n-1$ form a basis of the needed polynomial space.