Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 71 / 2

A strictly convex normed space is a normed vector space whose unit sphere contains no nontrivial line segment. Equivalently, for distinct unit vectors $x,y$,Suppose $u,v\in U$ are both elements of best approximation in a normed space to $f$, and let $d={\inf }_{w\in U}\Vert f-w\Vert$. If $d=0$ then $u=v=f$. If $d>0$ and $u\ne v$, the vectors $(f-u)/d$ and $(f-v)/d$ are distinct unit vectors. Strict convexity givesBut $(u+v)/2\in U$, contradicting the definition of $d$. Thus there is at most one best approximant. This proves uniqueness of best approximation in a strictly convex space, and does not assert existence for an arbitrary, possibly nonclosed, linear subspace.

Take $n\ge 1$ and $1<p<\infty$. The Lp space is strictly convex because $z\mapsto |z{|}^p$ is strictly convex: for distinct unit functions its pointwise midpoint inequality is strict on a set of positive measure, hence integration gives norm strictly below one. The finite-dimensional linear subspace ${\mathcal{T}}_{n-1}$ has a best approximant: any minimizing sequence is bounded by $\Vert t{\Vert }_p\le \Vert f-t{\Vert }_p+\Vert f{\Vert }_p$, and finite-dimensional compactness gives a convergent subsequence whose limit minimizes the continuous error. Uniqueness follows from part 1.

Write that unique minimizer as $t^{\ast }$ and its residual as $F=f-t^{\ast }$. Translation on the circle preserves the Lp norm. Since $f(x+2\pi /n)=f(x)$, the translated polynomial $t^{\ast }(x+2\pi /n)$ is another minimizer. Uniqueness forcesAlso $f(x+\pi /n)=-f(x)$. Therefore the polynomial $-t^{\ast }(x+\pi /n)$ has residual $-F(x+\pi /n)$, with the same norm as $F$, and is again a minimizer. HenceTo extract the consequence, write the trigonometric polynomial as $t^{\ast }(x)={\sum }_{|\ell |\le n-1}{c}_{\ell }{e}^{i\ell x}$. The first identity gives $(e^{2\pi i\ell /n}-1)c_{\ell }=0$. Only $\ell =0$ in the indicated frequency range is a multiple of $n$, so $t^{\ast }$ is constant. The second identity makes this constant its own negative and therefore zero. Thus the lower-frequency Lp approximation to a single harmonic isFor $n=1$ the space already consists only of constants, and the negative-translation argument gives the same result.