Exercise 9, § 4

Let $(r_n)_{n \geq 1}$ be the sequence of distinct rational numbers belonging to the interval $I = [0, 1]$ of $\mathbf{R}$, arranged in some order. Define by induction a sequence of closed intervals $I_n \subset I$, as follows: $I_n$ has as its midpoint the point $r_{k_n}$ with the smallest index not contained in $\bigcup_{p < n} I_p$; its length is $\leq 1/4^n$ and it meets no interval $I_p$ for which $p < n$; $(I_n)$ is thus a sequence of mutually disjoint closed intervals. On the product space $I \times \mathbf{R}$ we define a continuous real-valued function $u$ as follows: for each integer $n \geq 1$, the function $x \to u(x, n)$ is equal to 1 at an interior point of $I_n$, is equal to 0 on the exterior of $I_n$ and takes its values in $[0, 1]$; on the other hand, for each $x \in I$, the function $y \to u(x, y)$ is affine-linear in each of the intervals $[n, n+1]$. Show that $u$ cannot be uniformly approximated in $I \times \mathbf{R}$ by linear combinations of functions of the form $v(x)w(y)$, where $v$ is continuous on $I$ and $w$ is continuous and bounded on $\mathbf{R}$. [In the space $C^\infty(\mathbf{R}; \mathbf{R})$ of continuous bounded real-valued functions on $\mathbf{R}$, consider the set of partial functions $y \to u(x, y)$ as $x$ runs through $I$; show that there exists an infinite sequence $(u_n)$ of these functions such that $\|u_n\| = 1$ and $\|u_m - u_n\| = 1$ whenever $m \neq n$. Deduce that there cannot exist any finite-dimensional vector subspace $E$ of $C^\infty(\mathbf{R}; \mathbf{R})$ such that the distance of each $u_n$ from $E$ is $\leq 1/4$; for otherwise there would exist a sequence $(v_n)$ of points of $E$ such that $\|v_n\| = 2$ and $\|v_n - v_m\| \geq 1/2$ whenever $m \neq n$, contrary to the fact that every finite-dimensional subspace of $C^\infty(\mathbf{R}; \mathbf{R})$ is locally compact.] \*