Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / A / 1

Assume $A$ is nonempty, as is necessary for an approximating point to exist. The reverse triangle inequality givesThus the distance from the fixed point $u$ is a continuous, indeed Lipschitz continuous, real-valued function on $A$. A continuous function on a nonempty compact set attains its minimum, so some $p\in A$ satisfiesIf empty compact sets are allowed, the printed statement needs this nonemptiness qualification.

The vector subspace contains zero, whose approximation error is $\Vert u\Vert$. For any $q$ with $\Vert q\Vert >2\Vert u\Vert$, the triangle inequality gives $\Vert u-q\Vert \ge \Vert q\Vert -\Vert u\Vert >\Vert u\Vert$, so such points cannot improve on zero.

Restrict to $K=\{q\in A:\Vert q\Vert \le 2\Vert u\Vert +1\}$. This is nonempty, closed and bounded in a finite-dimensional normed space, hence compact. The continuous function $q\mapsto \Vert u-q\Vert$ attains a minimum on $K$ by part (a). Outside $K$ its value exceeds the error at zero, so that minimizer is best on all of $A$:Completeness of the ambient space is unnecessary; compactness is used only in the finite-dimensional vector subspace.

The geometry of a strictly convex normed space proves uniqueness whenever a best approximation exists. If distinct $p_1,p_2$ both have best error $E>0$, then their two residuals have norm $E$. The defining midpoint inequality givescontradicting best approximation because the midpoint lies in $A$. If $E=0$, both approximants equal $u$.

The printed existence assertion is false for an arbitrary vector subspace. For a concrete counterexample take $B={\ell }^2$ with its strictly convex Hilbert norm, let $A=c_{00}$ be its finite-support sequences, and take $u=(2^{-j}{)}_{j\ge 1}$. Truncations in $A$ tend to $u$, so the best-error infimum is zero, but $u\notin A$ and that infimum is not attained. If the finite-dimensional assumption of part (b) is retained, it supplies existence and the argument above supplies uniqueness.

For the estimate, suppose $Y$ is a projection operator onto $A$, as needed for a finite operator norm. Since $Yp=p$,The triangle inequality and operator norm giveThe estimate actually holds without attainment: apply it with any $q\in A$ instead of $p$ and take the infimum. Its proof applies in every normed space. This is strict convexity separates uniqueness from existence in best approximation.

Write the Fourier coefficients as $c_k=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }u(t)e^{-ikt}dt$. The partial sum through frequencies $-n,\dots ,n$ is therefore a convolution withThe last expression follows by summing a finite geometric series, with its removable value $2n+1$ at zero. Periodicity and the evenness of $D_n$ giveThis also proves linearity and that the image lies in $F_n$. We use the continuous periodic function space, consistent with the premise and the displayed maximum norm.

The convolution formula gives the upper boundTo prove sharpness within continuous functions, takeIt is continuous and periodic, with norm at most one. At $x=0$, dominated convergence gives $(S_n{u}_{ϵ})(0)\to (2\pi {)}^{-1}\int |D_n|$. Hence that upper bound is the actual operator norm. Evenness yieldsThe approximating continuous signs avoid assuming that the discontinuous sign of the kernel belongs to the domain space.