For each positive integer \(n\), define \[ f_n(x)=\cos(nx), \qquad x\in\mathbb R. \] Prove that the sequence \((f_n)\) has no uniformly convergent subsequence.
Solution
1. If \(m\ne n\) are positive integers, then\[ \int_0^{2\pi} \bigl(\cos(mx)-\cos(nx)\bigr)^2\,dx =2\pi. \]
The standard trigonometric orthogonality relations give \[ \int_0^{2\pi}\cos^2(mx)\,dx = \int_0^{2\pi}\cos^2(nx)\,dx =\pi \] and, because \(m\ne n\), \[ \int_0^{2\pi}\cos(mx)\cos(nx)\,dx=0. \] Expanding the square proves the claim.
2. Distinct members of the sequence satisfy\[ \norm{f_m-f_n}_{\infty}\ge1. \]
By step 1, \[ 2\pi \le \int_0^{2\pi} \norm{f_m-f_n}_{\infty}^2\,dx =2\pi\norm{f_m-f_n}_{\infty}^2. \] Taking square roots gives the result.
3. No subsequence of \((f_n)\) converges uniformly on \(\RR\).
Every uniformly convergent sequence is uniformly Cauchy. But any subsequence \((f_{n_k})\) consists of functions with distinct indices, and step 2 gives \[ \norm{f_{n_k}-f_{n_\ell}}_{\infty}\ge1 \qquad(k\ne\ell). \] Hence no subsequence is uniformly Cauchy.
4. Q.E.D.
Step 3 is the required conclusion.