In Baby Rudin 7.20 example, the author mentions to prove that the function sequence $$f_n(x):=\sin(nx) \qquad0(\leq x\leq 2\pi)$$has no pointwise convergent subsequence would be troublesome without Lebesgue's Theorem.

Is there a proof that doesn't refer to Lebesgue's Theorem, and only requires the knowledge introduced in the first 7 chapter in Rudin?


Here's an argument my officemate and I came up with that should work (while avoiding Lebesgue theory). Given a subsequence $\sin(n_kx)$, it constructs a sub-subsequence $\sin(n_{k_m}x)$ and a point $y$ such that $\sin(n_{k_m}y)$ fails to converge.

Suppose a subsequence $\sin(n_kx)$ is given. Then, we can find some closed interval $I_1 \subseteq [0, 2\pi]$ such that $\sin(n_1x)$ maps $I_1$ into $[1/2, 1]$. Also, set $n_{k_1} = n_1$.

Having chosen $I_1$, observe that for $k > k_1$ large enough, $\sin(n_k x)$ maps $I_1$ to $[-1,1]$ surjectively. Choose $k_2$ to be the least $k>k_1$ with this property, and let $I_2 \subset I_1$ be an interval that $\sin(n_{k_2}x)$ maps to $[-1, -1/2]$.

In general, suppose we have constructed the index $n_{k_m}$ and interval $I_m$. Then, we let $k_{m+1}$ be the least integer greater that $k_m$ such that $\sin(n_{k_{m+1}}x)$ maps $I_m$ surjectively onto $[-1, 1]$. Then, if $m$ is odd, we choose a subinterval $I_{m+1} \subset I_m$ such that $\sin(n_{k_{m+1}}x)$ maps $I_{m+1}$ into $[-1, -1/2]$, while if $m$ is even, we choose an $I_{m+1}$ which $\sin(n_{k_{m+1}}x$ maps into $[1/2,1]$.

Now, by the nested intervals theorem, $\bigcap_{m=1}^\infty I_m$ is nonempty, so it contains some point $y$. But, by construction, $\sin(n_k y)$ is both at least $1/2$ and at most $-1/2$ infinitely often, so the sequence of functions cannot converge at $y$.

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  • $\begingroup$ Nice proof. BTW at the beginning I think you meant to write $n_{k_1} = n_1$. $\endgroup$ – Michael Feb 19 '19 at 9:39

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