Math people:

I couldn't find a similar question, so here goes: I would like to prove the fact (?) that the sequence of functions $(f_n) \subset C([0,1])$ defined by $f_n(x)=\sin(nx)$ does not have a subsequence that converges pointwise to any function.

This is not homework. Soon I am going to teach the Arzela-Ascoli theorem, and I want to show that without the equicontinuity assumption, it is difficult to conclude anything about convergence of a sequence of a functions, even pointwise convergence.

I am almost certain that what I assert above is true, I am just having trouble proving it. I am also almost certain that this has been done before. It is obvious that the entire sequence $(f_n)$ doesn't converge pointwise, and that no subsequence of $(f_n)$ converges uniformly, but that is not my question.

  • $\begingroup$ Take any irrational multiple of $\pi$ and try to prove that there is no pointwise limit there, maybe? $\endgroup$ – AlexR Oct 30 '13 at 17:05
  • $\begingroup$ It it easy to prove for the interval $[0,2\pi]$ $\endgroup$ – Norbert Oct 30 '13 at 20:24

Suppose $f_{n_k}\to f$ pointwise (a.e.). Since each $f_{n_k}$ is bounded by $1$, the Dominated Convergence Theorem implies $$\int_a^b f_{n_k}(x)\,dx\to \int_a^b f(x)\,dx$$ for every subinterval $[a,b]$. On the other hand, $$\int_a^b f_{n_k}(x)\,dx = \frac{1}{n_k} (\cos n_k ax - \cos n_k bx)\to 0$$ Hence, $\int_a^b f(x)\,dx=0$ for every subinterval. This implies (by Lebesgue differentiation) that $f=0$ a.e.

Applying the Dominated Convergence theorem to $f_{n_k}^2$, we obtain $$\int_a^b f_{n_k}^2(x)\,dx\to 0$$ which is a contradiction because $$\int_a^b f_{n_k}^2(x)\,dx = \frac12 \int_a^b (1+\cos 2n_k x)\,dx\to \frac{b-a}{2}$$

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  • $\begingroup$ This is a beautiful solution. My students will be unable to follow the proof, but now I can confidently give this as an example of a uniformly bounded sequence of differentiable functions on a compact interval that totally fails to converge in any way, even along any subsequence. $\endgroup$ – Stefan Smith Oct 30 '13 at 22:06
  • $\begingroup$ They might be able to follow the proof better if we get to do any of the 2-3 theorems required. $\endgroup$ – Stefan Smith Oct 30 '13 at 22:08
  • $\begingroup$ I have a feeling there may be a way to prove that there is no subsequence that converges on the set $\{q\pi \mid q\in \mathbb{Q}\} \cap [0,1]$ that requires only a little number theory and no measure theory, but I have a lot of other things to do. $\endgroup$ – Stefan Smith Nov 1 '13 at 1:04

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