Prove that the sequence of functions \(f _ { n } ( x ) = \sin n x\) has no pointwise convergent subsequence. (Hint: show that given any subsequence and any interval of positive length there is a subinterval such that some element of the subsequence is at least \(1 / 2\) on this subinterval, and another element is at most \(- 1 / 2\).)
Solution
Let \[ \sin(n_kx) \] be an arbitrary subsequence of the original sequence. Then \[ n_k\longrightarrow\infty. \]
1. Let \(I=[a,b]\) be a closed interval with \(a<b\). For every sufficientlylarge positive integer \(n\), there is a nondegenerate closed subinterval\(J^+\subseteq I\) on which\[ \sin(nx)\geq\frac12. \]
For each integer \(m\), set \[ J_{m,n}^+ \coloneqq \left[ \frac{2\pi m+\pi/6}{n}, \frac{2\pi m+5\pi/6}{n} \right]. \] On this interval, \[ nx\in [2\pi m+\pi/6, 2\pi m+5\pi/6], \] so \[ \sin(nx)\geq\frac12. \]
The left endpoints of the intervals \(J_{m,n}^+\) are spaced by \(2\pi/n\). Choose \(m\) so that the left endpoint lies in \[ [a,a+2\pi/n). \] The corresponding right endpoint is then less than \[ a+\frac{2\pi}{n}+\frac{2\pi}{3n} = a+\frac{8\pi}{3n}. \] For all sufficiently large \(n\), \[ \frac{8\pi}{3n}<b-a, \] so this whole interval \(J_{m,n}^+\) lies inside \(I\).
2. Let \(I=[a,b]\) be a closed interval with \(a<b\). For every sufficientlylarge positive integer \(n\), there is a nondegenerate closed subinterval\(J^-\subseteq I\) on which\[ \sin(nx)\leq-\frac12. \]
For each integer \(m\), set \[ J_{m,n}^- \coloneqq \left[ \frac{2\pi m+7\pi/6}{n}, \frac{2\pi m+11\pi/6}{n} \right]. \] On this interval, \[ nx\in [2\pi m+7\pi/6, 2\pi m+11\pi/6], \] so \[ \sin(nx)\leq-\frac12. \] The left endpoints are again spaced by \(2\pi/n\). The same argument as in step 1 shows that one of these intervals is contained in \(I\) for all sufficiently large \(n\).
3. There exist indices\[ k_1<k_2<k_3<\cdots \]and nested nondegenerate closed intervals\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots \]such that\[ \sin(n_{k_j}x)\geq\frac12 \]for every \(x\in I_j\) when \(j\) is odd, and\[ \sin(n_{k_j}x)\leq-\frac12 \]for every \(x\in I_j\) when \(j\) is even.
Begin with the compact interval \[ I_0=[0,1]. \] Because \(n_k\to\infty\), step 1 permits choosing an index \(k_1\) and a nondegenerate closed interval \(I_1\subseteq I_0\) on which \[ \sin(n_{k_1}x)\geq\frac12. \]
Suppose \(k_j\) and \(I_j\) have been chosen. Since the tail \[ n_{k_j+1},n_{k_j+2},\ldots \] is still unbounded, choose \(k_{j+1}>k_j\) sufficiently large that step 2 applies to \(I_j\) when \(j+1\) is even, or step 1 applies when \(j+1\) is odd. The corresponding subinterval is \(I_{j+1}\). This recursive construction has all the stated properties.
4. There exists \(x_0\in\RR\) such that the numerical sequence\[ \sin(n_{k_j}x_0) \]does not converge.
The intervals in step 3 are nonempty compact subsets of the compact interval \(I_0\) and are nested. Hence \[ \bigcap_{j=1}^{\infty}I_j\neq\varnothing. \] Choose \[ x_0\in\bigcap_{j=1}^{\infty}I_j. \] Then step 3 gives \[ \sin(n_{k_j}x_0)\geq\frac12 \] for every odd \(j\), while \[ \sin(n_{k_j}x_0)\leq-\frac12 \] for every even \(j\). A sequence with these two infinite families of values cannot converge.
5. The original sequence \(f_n(x)=\sin(nx)\) has no pointwise convergentsubsequence.
The subsequence \(\sin(n_kx)\) was arbitrary. Step 4 constructs a further subsequence that fails to converge at the point \(x_0\). If \(\sin(n_kx)\) converged pointwise, then every further subsequence would converge at every point, in particular at \(x_0\). This contradiction shows that the arbitrary subsequence is not pointwise convergent.
6. Q.E.D.
Step 5 proves the required assertion.
Solution
1. If \(n\neq m\) are positive integers, then\[ \int_0^{2\pi}(\sin nx-\sin mx)^2\,dx=2\pi. \]
Expanding the square, \[ \int_0^{2\pi}\sin^2 nx\,dx=\int_0^{2\pi}\sin^2 mx\,dx=\pi \qquad\text{and}\qquad \int_0^{2\pi}\sin nx\,\sin mx\,dx=0 \] for \(n\neq m\).
2. For \(n_1<n_2<\cdots\), the sequence \(\sin(n_kx)\) does not convergepointwise on \([0,2\pi]\).
Suppose it does. Then \[ g_k(x)\coloneqq(\sin n_kx-\sin n_{k+1}x)^2 \] tends to \(0\) for every \(x\in[0,2\pi]\), and \(0\leq g_k\leq4\). By the bounded convergence theorem on \([0,2\pi]\), \(\int_0^{2\pi}g_k\,dx\to0\). Since \(n_k\neq n_{k+1}\), step 1 gives \(\int_0^{2\pi}g_k\,dx=2\pi\) for every \(k\), a contradiction.
3. Q.E.D.
Every subsequence of \(f_n\) has the form in step 2.