Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 68 / 3

Work with real-valued functions and put $\mathrm{sign}(0)=0$. Let $r=f-p_{\ast }$, and write an arbitrary competitor as $p=p_{\ast }+v$ with $v\in U$. Pointwise,For $r\ne 0$ this is $|r-v|\ge \mathrm{sign}(r)(r-v)$; for $r=0$ the right side is zero. Integrating and using the assumed orthogonality givesThus $p_{\ast }$ is a best $L^1$ approximation from $U$. This sign criterion for best L1 approximation is sufficient without any finite-dimensionality or uniqueness assumption; possible residual zeros do not invalidate the inequality.

Here $n$ is a positive integer and $m$ is an integer index of a Fourier coefficient. Split a period into $n$ intervals and set $y=nx-2\pi j$ on the $j$th interval. Periodicity and integrability giveFor $0<|m|<n$, $\zeta =e^{2\pi im/n}$ is a root of unity different from one and satisfies ${\zeta }^n=1$, so ${\sum }_{j=0}^{n-1}{\zeta }^j=(1-{\zeta }^n)/(1-\zeta )=0$. ThereforeFor $m=0$ the same substitution gives ${\int }_{\mathbb{T}}f(nx)dx={\int }_{\mathbb{T}}f(x)dx$. If the mean of $f$ is zero, this last mode vanishes as well. Since the linear span of the exponentials with $|m|\le n-1$ is the space of trigonometric polynomials of degree at most $n-1$, every pairing with that space is zero. This is the periodic dilation annihilates low Fourier modes property. The integer-index assumption is essential; the stated Fourier orthogonality would generally be false for noninteger $m$.

As printed in the original PDF, this claim is false for $n\ge 2$. The displayed frequency is one, so the function already belongs to the space of trigonometric polynomials of degree at most $n-1$. Its best approximation is then the function itself, with zero error; for example, $\alpha =1,\beta =0$ gives a zero-approximant error ${\int }_0^{2\pi }|\cos x|dx=4$ instead. For $n=1$, zero is indeed the best constant approximation. Thus the literal answer is

The natural intended version has frequency $n$. Set $g(y)=\alpha \cos y+\beta \sin y$ and $f_n(x)=g(nx)$. The sign function $h(y)=\mathrm{sign}g(y)$ is integrable and satisfies $h(y+\pi )=-h(y)$, so its mean is zero. Part (b) makes $h(nx)=\mathrm{sign}{f}_n(x)$ orthogonal to every degree-at-most-$n-1$ trigonometric polynomial. Part (a), with $p_{\ast }=0$, now provesThe error is $4\sqrt{{\alpha }^2+{\beta }^2}$ for the unnormalized integral over a period.

The best approximant is unique. If the amplitude is nonzero and another continuous trigonometric polynomial $p$ has the same optimal error, equality in part (a)'s pointwise inequality holds almost everywhere. Near each zero of $f_n$, the function changes sign; if $p$ did not vanish at that zero, continuity would force an interval on one side where $f_n-p$ has the wrong sign, making the integral inequality strict. Thus $p$ vanishes at all $2n$ distinct zeros of $f_n$. A nonzero trigonometric polynomial of degree at most $n-1$ has at most $2n-2$ zeros in one period, since multiplication by $e^{i(n-1)x}$ turns it into a polynomial in $e^{ix}$ of degree at most $2n-2$. Hence $p=0$. Zero amplitude is immediate. The same reasoning proves the literal frequency-one assertion when $n=1$, while retaining the counterexample for larger $n$.