Determine where the following real-valued function is or is not uniformly convergent: \[ f_n(x) \da {\sin(nx)\over 1+nx} .\]
Solution
For each \(x\geq0\), \(\abs{f_n(x)}\le 1/(1+nx)\) for \(x>0\) and \(f_n(0)=0\), so \(f_n\to0\) pointwise on \([0,\infty)\).
The convergence is uniform on \([a, \infty)\) for every \(a>0\): \[ \sup_{x\geq a}\abs{\sin(nx) \over 1+nx} \leq {1\over 1 + na} \convergesto{n\to\infty} 0 .\]
The convergence is not uniform on \((0, \infty)\): \[ x_n \da {1\over n} \implies \sup_{x>0}\abs{f_n(x)}\geq\abs{f_n(x_n)} = {\sin(1) \over 2} \] for every \(n\).