Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 19F (Linear Analysis)

The real Stone-Weierstrass theorem says that a real subalgebra $A\subset C(K,\mathbb{R})$ which contains the constants and separates points is dense in $C(K,\mathbb{R})$ for the uniform norm, when $K$ is compact Hausdorff space. If $A$ is also closed, $A=C(K,\mathbb{R})$. The complex version needs closure under complex conjugation; that extra condition is unnecessary here.

Since a subalgebra is closed under addition and scalar multiplication, the given absolute-value property yieldsThe same identity with a minus sign gives $\min (f,g)\in A$. These are pointwise operations on real-valued continuous functions.

Let $F,G$ be disjoint closed subsets of $K$. They are compact. For each $x\in F,y\in G$, the Hausdorff space condition gives disjoint open neighbourhoods $U_{xy},V_{xy}$. For fixed $x$, finitely many $V_{xy}$ cover $G$; intersect the corresponding $U_{xy}$ and take the union of those $V_{xy}$ to obtain disjoint open sets $U_x\ni x$ and $V_x\supset G$. Finitely many $U_x$ cover $F$. Their union and the intersection of the corresponding $V_x$ are disjoint open neighbourhoods of $F,G$. The case of an empty set is immediate. Hence $K$ is a normal topological space.

Urysohn lemma states that in a normal topological space satisfying $T_1$, disjoint closed sets $F,G$ admit a continuous function $h:K\to [0,1]$ with $h=0$ on $F$ and $h=1$ on $G$.

For $x\ne y$, $|({\delta }_x-{\delta }_y)(f)|\le 2\Vert f{\Vert }_{\infty }$, giving an upper bound $2$ for the operator norm. By Urysohn lemma, there is $h:K\to [0,1]$ with $h(x)=1,h(y)=0$. Then $f=2h-1$ has uniform norm $1$ and $({\delta }_x-{\delta }_y)(f)=2$. Thus every distinct pair has operator norm $2$, andThe stated hypotheses also permit a singleton $K$. Then there are no distinct pairs, so the infimum is $+\infty$ under the extended-real convention, or undefined as a real number. This edge case cannot be assigned the value $2$.