# Compact function spaces

Let $K$ be a compact metric space, $\varphi:K\to(0,\infty)$ a continuous function and $A=\{f\in C(K) : |f(x)|\leq\varphi(x)\}$. Prove that $A$ is compact iff $K$ is finite.

## 2 thoughts on “Compact function spaces”

1. nachodarago says:

If $K$ is finite there is nothing to prove. Suppose on the other hand that $A$ is compact. We will show that every point in $K$ is isolated and thus finite due to the compactness of $A$.

Suppose that $x\in K$ is not isolated, there is therefore a ball of center $x$ and radius $r>0$ such that $B_r(x)\subseteq K$, and we consider the sequence of balls $B_n = B_{r/n}(x)$. Let $\mu_n:K\to [0,1]$ the Urysohn map such that $\left.\mu_n\right|_{B_{n-1}} = 0$ and $\left.\mu_n\right|_{B_n} = 1$ and consider the sequence of functions $\phi_n = \mu_n \phi$. Clearly, $|\phi_n(x)| \leq \phi$ and the sequence $(\phi_n(y))_n$ converges pointwise to $\delta_{xy} \phi(x)$ (where $\delta_{xy}$ is Dirac’s delta function). Since $A$ is compact, there must be a subsequence of $(\phi_n)_n$ which converges uniformly to $\delta_{xy}\phi(x)$, but this is absurd since, taking $\varepsilon>0$ sufficiently small we would have $|\phi(x)|= |\phi(x)-\phi_{n_k}(x)| \leq ||\phi-\phi_{n_k}|| < \varepsilon$. And we're done.

Liked by 1 person

2. Alternatively, let us prove that $A$ is not totally bounded if $K$ is infinite.

Take $(x_n)$ a sequence of pairwise distinct points in $K$; by compactness we have a subsequence $x_{n_k}\longrightarrow x$. Consider the subspace $F = \{x_{n_k}\}\cup\{x\}$, which is closed in $K$.

Let $\varepsilon>0$ be the minimum value that $\varphi$ takes, which exists by compactness. All points in $F$ except for $x$ are isolated, so the function $f_k=\varepsilon\delta_{x_{n_k}}:F\to \Bbb{R}$ is continuous and bounded by $\varepsilon$. Moreover, $\Vert f_k - f_{k'}\Vert = \varepsilon$ if $k\neq k'$. Since $F$ is closed, by Tietze’s extension theorem we may obtain a continuous extension $f'_k:K\to \Bbb{R}$ of $f_k$ which is also bounded above by $\varepsilon$, and in particular by $\varphi$. Since we have that $\Vert f'_k - f'_{k'}\Vert \geq \varepsilon$ if $k\neq k'$ and the functions $f'_k$ belong to $A$, we are done.

Liked by 1 person