Let \(X\) be a set, and let \(f_n: X \to \mathbb{R}\) be a sequence of functions. Let \(\bar{\rho}\) be the uniform metric on the space \(\mathbb{R}^X\). Show that the sequence \((f_n)\) converges uniformly to the function \(f: X \to \mathbb{R}\) if and only if the sequence \((f_n)\) converges to \(f\) as elements of the metric space \((\mathbb{R}^X, \bar{\rho})\).
Solution
The uniform metric is \(\bar\rho(g,h)=\sup_{x\in X}\min\{\abs{g(x)-h(x)},1\}\).
1. For \(0<\varepsilon\le1\) and \(g,h\colon X\to\RR\), \(\bar\rho(g,h)<\varepsilon\) if and only if \(\sup_x\abs{g(x)-h(x)}<\varepsilon\).
Always \(\bar\rho(g,h)\le\sup_x\abs{g(x)-h(x)}\). If \(\bar\rho(g,h)<\varepsilon\le1\), then \(\min\{\abs{g(x)-h(x)},1\}<1\) for every \(x\), so the minimum is \(\abs{g(x)-h(x)}\) and \(\sup_x\abs{g(x)-h(x)}=\bar\rho(g,h)<\varepsilon\).
2. Q.E.D.
Uniform convergence \(f_n\to f\) means that for every \(\varepsilon>0\) there is \(N\) with \(\sup_x\abs{f_n(x)-f(x)}<\varepsilon\) for \(n\ge N\), and convergence in \((\RR^X,\bar\rho)\) means the same with \(\bar\rho(f_n,f)\) in place of the supremum. It suffices to check both conditions for \(\varepsilon\le1\), where step 1 shows that they coincide.