Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 21G / Solution

The real Stone-Weierstrass theorem says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$.

Let $\overline{A}$ be the uniform closure. Polynomial approximation to $\sqrt{t}$ on a bounded interval shows that $|f|\in \overline{A}$ whenever $f\in \overline{A}$. Hence $\overline{A}$ is closed underGiven $h\in C(X)$, separation and the constants provide, for each $x,y$, a function in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such functions by minima to obtain one that agrees at $x$ and lies below $h+\epsilon$ everywhere; a second finite cover and maxima produces $f\in \overline{A}$ with $|f-h|<\epsilon$. Thus $h\in \overline{A}$.

The space $C_b(\mathbb{R})$ is Banach because a uniform Cauchy sequence converges uniformly to a bounded continuous function. Compactness is essential: the algebra of bounded continuous functions having finite limits at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb{R})$.

Solved by gpt-5.6-sol high.