Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 22F (Linear Analysis)

The Arzela-Ascoli theorem says that a subset $\mathcal{F}\subseteq C(K)$ is relatively compact in the uniform norm exactly when it is equicontinuous and pointwise bounded.

The real Stone-Weierstrass theorem says that a real subalgebra $A\subseteq C(K,\mathbb{R})$ is uniformly dense if it contains the constants and separates points of $K$. In the complex version, a complex subalgebra $A\subseteq C(K,\mathbb{C})$ must additionally be closed under complex conjugation.

For an example, take $K=[0,1]$ andThis family is bounded in the uniform norm, but it is not equicontinuous at $1$. Equivalently, every subsequence converges pointwise to the discontinuous function that is zero on $[0,1)$ and one at $1$, so no subsequence converges uniformly. Thus $\mathcal{F}$ is bounded but not relatively compact.

For each positive integer $i$, apply the Stone-Weierstrass theorem on the closed ball $\overline{B}(0,i)$ to choose a polynomial $p_i$ in $n$ variables satisfyingEvery compact $B\subset {\mathbb{R}}^n$ lies in some such ball, so the same estimate for all sufficiently large $i$ proves that $p_i{|}_B\to f{|}_B$ uniformly.

Now suppose instead that $p_i\to f$ uniformly on all of ${\mathbb{R}}^n$. For sufficiently large $i,j$, the polynomial $p_i-p_j$ is bounded on ${\mathbb{R}}^n$, and every nonconstant real polynomial is unbounded along a suitable line. Hence $p_i-p_j$ is constant. A tail of the sequence is therefore one fixed polynomial plus a uniformly convergent sequence of constants, and its limit is a polynomial. The converse is immediate from the constant sequence. ThusThis is the globally uniform limit of real polynomials criterion.

Suppose $C(K)$ is equicontinuous. At each $x\in K$, choose a neighbourhood $U$ such thatfor every $y\in U$ and every $f\in C(K)$. If some $y\in U$ differed from $x$, the Urysohn lemma for the compact Hausdorff space $K$ would give a continuous function with $f(x)=0$ and $f(y)=2$, a contradiction. Thus every singleton is open, so $K$ is discrete. A compact discrete space is finite, proving

Compactness is essential. Any infinite set with the discrete topology has equicontinuous $C(K)$, because the singleton neighbourhood $\{x\}$ works simultaneously for every function, but the space is infinite.