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.

Compact function spaces

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.

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  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.

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