Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 22H (Linear Analysis)

The real Stone-Weierstrass theorem states that if $X$ is a compact Hausdorff space and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points of $X$, then $A$ is dense in $C(X,\mathbb{R})$ in the uniform norm.

The Tietze extension theorem states that if $X$ is a normal topological space, $F\subseteq X$ is closed, and $f:F\to \mathbb{R}$ is continuous, then $f$ extends to a continuous function on $X$. In its bounded form, an $f$ taking values in $[-M,M]$ has an extension with the same range bound. The underlying separation step is the Urysohn lemma.

Every function with the stated two linear pieces can be writtenfor suitable real constants $a,b,c$. SincedefineThis is a polynomial. The argument of $P_i$ lies in $[0,1/4]\subseteq [0,1]$, so

For $F_1$, the functions satisfy $\Vert \sin (\pi nx)/n{\Vert }_{\infty }\le 1/n$, so the sequence converges uniformly to zero. Its closure is the sequence together with its limit and is compact. The same argument applies to $F_2$, since $\Vert \sin (\pi nx)/\sqrt{n}{\Vert }_{\infty }\le n^{-1/2}$. Thus $F_1$ and $F_2$ are relatively compact.

The family $F_3$ is not equicontinuous. At $x=0$ its members vanish, but for $x_n=1/(2n)\to 0$,By the Arzela-Ascoli theorem, a relatively compact subset of $C[0,1]$ must be equicontinuous. Hence $F_3$ is not relatively compact.