Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 6 / 1

The Baire category theorem states that in a complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of nowhere dense sets, where a set is nowhere dense if its closure has empty interior.

Here is the proof. Let $G_1,G_2,\dots$ be open dense sets and let $U$ be any nonempty open set. Choose a point $a_1\in U\cap G_1$ and a radius $0<r_1<1/2$ such that the closed ball $B_1=\overline{B}(a_1,r_1)$ lies inside $U\cap G_1$. Inductively choose $a_{m+1}$ in the nonempty open set $B(a_m,r_m)\cap G_{m+1}$, and then a radius $0<r_{m+1}<2^{-m-1}$ such thatThus the closed balls are nested and their radii tend to zero. For $j,k\ge m$, both centers lie in $B_m$, so $d(a_j,a_k)\le 2r_m$. They form a Cauchy sequence. Completeness gives a limit $a$, and closedness puts it in every $B_m$. Consequently $a\in U\cap {\bigcap }_m{G}_m$. As $U$ was arbitrary, the intersection is dense.

To obtain the equivalent formulation, take complements of the closures of the nowhere dense sets; those complements are open dense. Conversely, complements of open dense sets are closed with empty interior. In particular, a countable closed cover of a nonempty complete metric space must contain a set with nonempty interior.

For each nonzero integer tuple $k=(k_1,\dots ,k_{n+1})$, putIf its first $n$ coordinates are all zero, then $k_{n+1}\ne 0$ and $H_k$ is empty. Otherwise it is a closed proper affine hyperplane. It has empty interior: from any of its points, an arbitrarily small displacement in the nonzero normal direction $(k_1,\dots ,k_n)$ leaves it. Thus every $H_k$ is nowhere dense.

There are countably many integer tuples, and ${\mathbb{R}}^n$ is complete. The Baire category theorem therefore makes the complement of their union dense and nonempty. Choosing any point there givesEquivalently, $1,x_1,\dots ,x_n$ are linearly independent over $\mathbb{Q}$: multiplying a rational relation by a common denominator produces just such an integer relation. This is an instance of Baire avoidance of countably many affine hyperplanes.

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized measure $dt/(2\pi )$. The Fourier partial sum at zero is the real linear functionalwhere the Dirichlet kernel isThe quotient follows by summing the finite geometric progression, with its removable value $D_N(0)=2N+1$. It is a real continuous periodic function. Ifthen $|L_{N}f|\le {\Lambda }_N\Vert f{\Vert }_{\infty }$. This bound is its exact operator norm even on the real continuous functions. Indeed, for $\delta >0$ setIt is real, continuous and periodic with norm at most one, andby dominated convergence. At zeros of the kernel the integrand is zero, and elsewhere its limit is $|D_N|$.

For $0<t\le \pi$, $\sin (t/2)\le t/2$. Using evenness and putting $v=(N+1/2)t$ givesFor the last inequality, split into the first $N$ complete intervals $[(j-1)\pi ,j\pi ]$: their sine integrals are two and $1/v\ge 1/(j\pi )$. The harmonic sum tends to infinity. This proves the Dirichlet kernel harmonic lower bound.

Choose $N$ with ${\Lambda }_N>K$, then choose $\delta$ sufficiently small that $L_N{f}_{N,\delta }>K$. ThusThis constructs continuous near-maximizers, rather than using the discontinuous sign of the kernel as the desired function.

The real space $C(\mathbb{T})$ is a Banach space in the uniform norm: a uniformly Cauchy sequence has a uniform pointwise limit by completeness of $\mathbb{R}$, and that uniform limit is continuous. Each functional $L_N$ from the preceding part is bounded, but their norms are unbounded.

For integers $m\ge 1$, defineEach $E_m$ is closed, being an intersection of inverse images of closed intervals under continuous functionals. It has empty interior. Otherwise some ball $B(f_0,r)$ would be contained in it. For every $h$ of norm at most one, both $f_0$ and $f_0+(r/2)h$ would belong to that ball, soTaking the supremum over $h$ would give $\Vert L_N\Vert \le 4m/r$ for all $N$, contradicting the preceding part.

Thus the $E_m$ are nowhere dense. The Baire category theorem says their union cannot cover $C(\mathbb{T})$. A function outside that union satisfiesso its Fourier series diverges at zero. In fact these functions form a dense $G_{\delta }$ subset of $C(\mathbb{T})$. This proves the relevant Uniform boundedness principle argument directly and establishes generic unbounded Fourier sums at a fixed point.