= 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 $0,\ldots,n-1$ form a <basis> of the needed <polynomial> space.
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