Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / A / 3

If $(u-p,q)=0$ for all $q\in A$, then for any $a\in A$ the difference $p-a$ lies in $A$ and Pythagorean theorem givesThus $p$ is best.

Conversely, if $p$ is best, then for every $q\in A$ the real quadratichas its minimum at $t=0$. Its linear coefficient must vanish, giving $(u-p,q)=0$. Hence best approximation is equivalent to orthogonality of the residual to the approximation space. The weight is positive almost everywhere, so this is a genuine positive-definite inner product on continuous functions.

For the weighted inner product, $(x^j,x^k)=1/(j+k+2)$. Start Gram-Schmidt process with the monomials and choose monic normalization:Its squared norm is $1/4-4/9+2/9=1/36$. Next $(x^2,1)/(1,1)=1/2$ andThereforeSubtracting these orthogonal projections gives orthogonality by construction; independent nonzero rescalings would give equivalent unnormalized orthogonal polynomials.

Write $p=a+bx$. orthogonality of the residual to $1,x$ givesThe first integral follows from $t=1-x^2$; for the second use $x=\sin \theta$, giving ${\int }_0^{\pi /2}{\sin }^2\theta {\cos }^2\theta d\theta =\pi /16$. The coefficient matrix has determinant $1/72$, so solving yieldsPart (a) proves this is the unique best weighted $L^2$ approximation. The weight $x$ is essential; using ordinary unweighted normal equations would give different coefficients.