Consider a sequence of functions \(f_n\colon[a,b]\to\RR\) with the property that for each \(x\in[a,b]\) there is an open interval \(I_x\) containing \(x\) such that \((f_n)_{n\geq1}\) converges uniformly in \(I_x\cap[a,b]\). Show that \((f_n)_{n\geq1}\) converges uniformly in \([a,b]\).
Solution
For each \(x\in[a,b]\), the sequence \((f_n)\) converges uniformly on \(I_x\), and in particular converges pointwise at \(x\). Let \(f\colon[a,b]\to\RR\) be the pointwise limit of \((f_n)\). The compact set \([a,b]\) is covered by the collection of open intervals \(I_x\), so there is a finite subcovering, say \([a,b]\subset\bigcup_{k=1}^mI_{x_k}\). Given \(\varepsilon>0\), there exists \(N_k\) such that for \(n\geq N_k\), the difference \(\abs{f_n-f}\) is bounded by \(\varepsilon\) on \(I_{x_k}\). Let \(N\coloneqq\max(N_1,\ldots,N_m)\). Then for \(n\geq N\), the difference \(\abs{f_n-f}\) is bounded by \(\varepsilon\) on all of \([a,b]\). Hence by definition, \((f_n)\) converges to \(f\) uniformly.