Polynomial approximation ($E_n(f)={\inf }_{\mathrm{deg}p\le n}\Vert f-p{\Vert }_{\infty }$)

Polynomial approximation represents a function by polynomials of bounded polynomial degree. On a compact real interval, the Weierstrass approximation theorem guarantees uniform approximation of every continuous function by such polynomials. The displayed best error quantifies the accuracy available at degree at most $n$.

For a positive-definite weighted inner product on a compact interval, a least-squares polynomial is the orthogonal projection of a function onto the finite-dimensional space ${\Pi }_n$ of polynomials of degree at most $n$. In a basis ${\varphi }_0,\dots ,{\varphi }_n$, its coefficients solve the normal equations $Gc=d$, where $G_{ij}=({\varphi }_i,{\varphi }_j)$ and $d_i=(u,{\varphi }_i)$. The Gram matrix is positive definite: $z^{T}Gz=\Vert \sum z_i{\varphi }_i{\Vert }^2>0$ for $z\ne 0$. Thus the coefficients exist and are unique. The residual is orthogonal to every candidate polynomial, giving $\Vert u-q{\Vert }^2=\Vert u-p_n{\Vert }^2+\Vert p_n-q{\Vert }^2$ and proving optimality. An orthogonal basis makes these equations diagonal, as in a least-squares polynomial in an orthogonal-polynomial basis.

For the weight $w(x)=x$ on $[0,1]$, the normal equations for projecting $\sqrt{1-x^2}$ onto $\mathrm{span}\{1,x\}$ have Gram matrix $\left(\begin{array}{cc}1/2& 1/3\\ 1/3& 1/4\end{array}\right)$ and right side $(1/3,\pi /16{)}^T$. Its determinant is $1/72$, and inversion gives the displayed polynomial. orthogonality of the residual to both basis functions proves best approximation in the weighted norm.

On a symmetric interval with an even positive weight, the best least-squares polynomial approximation to an odd function is odd, and the approximation to an even function is even. Indeed, the even and odd polynomial subspaces are orthogonal. For odd $u$, write a candidate as $p=p_o+p_e$. ThenDeleting a nonzero even component strictly reduces the error, so the minimizer has $p_e=0$. The same argument exchanges the parities for even $u$. This reduces the normal equations to the matching parity block and avoids computing coefficients known in advance to vanish.

If $n\ge 1$, $f$ is real and $C^n$ on an interval, and $f^{(n)}$ is nowhere zero there, the displayed inequality holds. If the errors were equal, a best degree-at-most-$n-1$ polynomial would also be best at degree $n$. The Chebyshev alternation theorem and intermediate value theorem would give $n+1$ zeros of the error. Applying Rolle's theorem $n$ times would force a zero of $f^{(n)}$, a contradiction. The zero-error case already contradicts $f^{(n)}\ne 0$. In particular this proves strict improvement for the exponential function at every positive degree step.