Let \((Y, d)\) be a metric space; let \(f_n: X \to Y\) be a sequence of continuous functions; let \(f: X \to Y\) be a function (not necessarily continuous). Suppose \(f_n\) converges to \(f\) in the topology of pointwise convergence. Show that if \(\ts{f_n}\) is equicontinuous, then \(f\) is continuous and \(f_n\) converges to \(f\) in the topology of compact convergence.
Solution
Fix \(x\in X\) and \(\varepsilon>0\). Equicontinuity gives a neighborhood \(U\) of \(x\) such that \[ d(f_n(y),f_n(x))<\varepsilon/3 \] for all \(y\in U\) and all \(n\). Passing to the pointwise limit in \(n\) gives \(d(f(y),f(x))\le\varepsilon/3\) for \(y\in U\), so \(f\) is continuous.
Now let \(K\subset X\) be compact and fix \(\varepsilon>0\). For every \(x\in K\), equicontinuity supplies a neighborhood \(U_x\) such that \[ d(f_n(y),f_n(x))<\varepsilon/3 \] for all \(n\) and \(y\in U_x\); the same inequality holds for \(f\) after passing to the limit. Choose \(x_1,\dots,x_r\) with \(K\subset\bigcup U_{x_i}\). Pointwise convergence gives \(N\) such that for \(n\ge N\) and every \(i\), \[ d(f_n(x_i),f(x_i))<\varepsilon/3. \] If \(y\in K\), choose \(i\) with \(y\in U_{x_i}\). Then \[ d(f_n(y),f(y))<\varepsilon. \] Hence convergence is uniform on every compact \(K\), i.e. convergence in the compact-convergence topology.