Prove the following: if \(\ts{f_n}\) is equicontinuous on \(K\) a compact set and \(f_n\to f\) pointwise, then \(f_n\to f\) uniformly.
Solution
Fix \(\eps\), it suffices to find an \(n= n(\eps)\) to bound \(\norm{f_n - f}_{\infty, K } < \eps\). A standard \(\eps/3\) argument works: write \[ \abs{f_n(x) - f(x) } \leq \abs{f_n(x) - f_n(y)} + \abs{f_n(y) - f(y)} + \abs{f(y) - f_n(y)} .\]
Use equicontinuity to bound \(\abs{f_n(x) - f_n(y)}\) for all \(n\geq N_0 = N_0(\eps)\), for all \(x,y\in K\). This takes care of the 1st and 3rd terms.
For the 2nd term, cover \(K\) by \(\delta\dash\)balls and by compactness obtain a finite cover \(B_{\delta}(y_k)\covers K\). Then \(x\in B_\delta(y)\) for \(y=y_j\) for some \(j\), and in this ball use pointwise convergence of \(f_n\to f\).