Given a set \(S\), show that the function \[\rho_\infty:\mathcal B(S)\times\mathcal B(S)\to\mathbb R,\qquad \rho_\infty(f,g):=\operatorname{lub}(\{|f(x)-g(x)|:x\in S\})\] is a metric on \(\mathcal B(S)\).
Let \(M>0\) be given. Set \[S:=\{f\in C_b([0,1]):f(0)=0,\ f\text{ is differentiable on }(0,1),\ \text{and }|f'(x)|\le M\text{ for each }x\in(0,1)\}.\] Determine whether the set \(S\) is compact in \((C_b([0,1]),\rho_\infty)\).
Solution
Goal: (a) Show \(\rho_\infty(f,g) = \operatorname{lub}\{|f(x) - g(x)| : x \in S\}\) is a metric on \(\mathcal B(S)\); (b) decide compactness of \(S = \{f \in C_b[0,1] : f(0) = 0, f \text{ differentiable on } (0,1), |f'| \le M\}\) in \((C_b[0,1], \rho_\infty)\).
1. (a) \(\rho_\infty\) is a metric.
1.1. \(\rho_\infty(f,g) \ge 0\), finite, and \(\rho_\infty(f,g) = 0 \Leftrightarrow f = g\).Proof: \(|f(x) - g(x)| \ge 0\); finiteness because \(f, g\) are bounded on \(S\); \(\rho_\infty = 0\) iff \(|f(x) - g(x)| = 0\) for all \(x\) iff \(f = g\) pointwise.
1.2. Symmetry: \(\rho_\infty(f,g) = \rho_\infty(g,f)\).Proof: \(|f(x) - g(x)| = |g(x) - f(x)|\) for each \(x\).
1.3. Triangle inequality: \(\rho_\infty(f,h) \le \rho_\infty(f,g) + \rho_\infty(g,h)\).Proof: for each \(x\), \(|f(x) - h(x)| \le |f(x) - g(x)| + |g(x) - h(x)| \le \rho_\infty(f,g) + \rho_\infty(g,h)\); take the lub over \(x\).
1.4. Q.E.D.Proof: <2>1–<2>3 are the metric axioms.
2. (b) \(S\) is NOT compact.
2.1. \(S\) is not closed in \(C_b[0,1]\).
Proof: define \[ f_n(x)=M\left(\sqrt{(x-\tfrac12)^2+n^{-2}}-\sqrt{\tfrac14+n^{-2}}\right). \] Then \(f_n(0)=0\), and on \((0,1)\) \[ f_n'(x)=M\frac{x-\tfrac12}{\sqrt{(x-\tfrac12)^2+n^{-2}}}, \] so \(|f_n'(x)|\le M\). Hence \(f_n\in S\).
Since \(|\sqrt{t^2+n^{-2}}-|t||\le n^{-1}\) uniformly in \(t\), we have uniform convergence \[ f_n(x)\longrightarrow f(x):=M\left(|x-\tfrac12|-\tfrac12\right) \] on \([0,1]\). The limit satisfies \(f(0)=0\) but is not differentiable at the interior point \(x=1/2\). Thus \(f\notin S\), so \(S\) is not closed.
2.2. Compact subsets of metric spaces are closed.Proof: A compact subset of a Hausdorff space is closed; every metric space is Hausdorff.
2.3. Q.E.D.Proof: 1 and 2 show \(S\) is not compact.