Let \((c_n)\) be a sequence of positive real numbers and define \[ f_n(x)=\sin(x+c_n^2)+\frac1{c_n}\sin(c_nx). \] Prove that \((f_n)\) has a subsequence converging pointwise to a continuous function.
Solution
We split into two cases.
1. The sequence \((c_n)\) has a bounded subsequence.
By Bolzano–Weierstrass, after passing to a subsequence we may assume \[ c_n\to c\ge0. \] If \(c>0\), then for every \(x\), \[ \sin(x+c_n^2)\to\sin(x+c^2) \] and \[ \frac1{c_n}\sin(c_nx)\to\frac1c\sin(cx). \] Thus \[ f_n(x)\to \sin(x+c^2)+\frac1c\sin(cx), \] a continuous function of \(x\).
If \(c=0\), then \[ \sin(x+c_n^2)\to\sin x, \] while \[ \frac1{c_n}\sin(c_nx) =x\frac{\sin(c_nx)}{c_nx}\to x \] for every \(x\), with the value at \(x=0\) understood by continuity. Hence \[ f_n(x)\to\sin x+x, \] again continuously.
2. The sequence \((c_n)\) has no bounded subsequence.
Then we can choose a subsequence, still denoted \((c_n)\), such that \[ c_n\to\infty. \] The phases \(c_n^2\) modulo \(2\pi\) lie in the compact circle \(\mathbb R/(2\pi\mathbb Z)\), so after passing to a further subsequence there exists \(\theta\in[0,2\pi]\) such that \[ c_n^2\pmod{2\pi}\to\theta. \] Therefore for every \(x\), \[ \sin(x+c_n^2)\to\sin(x+\theta). \] Also \[ \left|\frac1{c_n}\sin(c_nx)\right|\le\frac1{c_n}\to0 \] uniformly in \(x\). Hence \[ f_n(x)\to\sin(x+\theta), \] which is continuous.
In either case, the original sequence has a subsequence converging pointwise to a continuous function.