- Show that a uniform limit of continuous functions is continuous.
Solution
Let \(f_n:X\to\mathbb C\) be continuous and suppose \(f_n\to f\) uniformly. Fix \(x_0\in X\) and \(\varepsilon>0\). Choose \(N\) such that \[ |f_N(x)-f(x)|<\frac\varepsilon3 \] for every \(x\in X\). By continuity of \(f_N\) at \(x_0\), there is a neighborhood of \(x_0\) on which \[ |f_N(x)-f_N(x_0)|<\frac\varepsilon3. \] Hence, for such \(x\), \[ |f(x)-f(x_0)| \le |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)| <\varepsilon. \] Thus \(f\) is continuous at \(x_0\), and therefore continuous everywhere.