2

Prove that $\{f = \cos(nx) : n=1, ... ,\infty \}$ is a bounded subset (of functions), but not totally bounded, of $C([0, \pi])$ under the supremum norm (or uniform norm, see https://en.wikipedia.org/wiki/Uniform_norm).

Note that a totally bounded set has a convergent subsequence in a complete metric space.

PhoemueX
  • 36,211
  • Well it is clearly bounded since $\cos(\mathbb R)=[-1,1]$. – Rushabh Mehta Dec 06 '19 at 20:38
  • @DonThousand The question is about a set of functions $x\mapsto \cos(nx),$ not about the numbers $\cos(nx)$ for fixed $x$. – Nikhil Sahoo Dec 06 '19 at 20:49
  • Yes, @NikhilSahoo is correct. –  Dec 06 '19 at 21:02
  • 1
    @U.Hill He's not correct in the sense that $\cos(nx)$ is a wrong notation to refer to a function. – Arnaud Mortier Dec 06 '19 at 21:04
  • @ArnaudMortier I changed the question, hopefully is more clear. –  Dec 06 '19 at 21:06
  • @U.Hill Now it's easier to guess what you mean, but still not correct. It should be ${x\mapsto \cos(nx)\ldots}$. Sometimes you can avoid $x\mapsto$ by replacing the variable by a dot, as in $\cos(n\cdot)$, but it needs to be clear from context. – Arnaud Mortier Dec 06 '19 at 21:08
  • 1
    Um... isn't it obvious. By definition $||\cos nx||_\infty = \sup {\cos nx|x\in \mathbb R}$ and that is clearly $1$. So for any $\cos nx, \cos mx\in $ your set $|| \cos nx - \cos mx|| \le 2$. .... Am I misunderstanding something?.... Edit: Oh, I missed the "not totally bounded" I guess that is the less trivial aspect. Never mind. – fleablood Dec 06 '19 at 21:28

2 Answers2

3

Hint: For fixed $m,$

$$\lim_{n\to \infty}\int_0^\pi (\cos (mx)-\cos (nx))^2\,dx = \pi.$$

zhw.
  • 107,943
  • 1
    Could you please elaborate more? I am working with the uniform norm. @zhw. –  Dec 06 '19 at 21:09
  • @U.Hill The inclusion $I:(C[0,\pi], ||\cdot||{\infty} ) \to (C[0,\pi], ||\cdot||{2} )$ is a continuous mapping. – clark Dec 06 '19 at 22:00
  • Umm... I am mainly focused on the "not totally bounded" part –  Dec 06 '19 at 23:06
  • @U.Hill Right, so if you have a convergent subsequence with the sup norm then it must map to a convergent subsequence in $L_2$, but zhw answer indicates that any subsequence of $\cos n x$ cannot be convergent is $L_2$ since it cannot my Cauchy – clark Dec 06 '19 at 23:31
  • @clark That is a very helpful explanation, thank you so much. –  Dec 06 '19 at 23:44
2

The set $S = \{x \mapsto \cos(nx) : n \in \mathbb{N}\}$ is clearly bounded since $$\|\cos(mx) - \cos(nx)\|_\infty \le \|\cos(mx)\|_\infty + \|\cos(nx)\|_\infty = 2$$

To show that $S$ is not totally bounded, notice that $S$ is an orthogonal set of functions in the inner product space $(C([0,\pi]), \|\cdot\|_2)$ w.r.t. to the the standard inner product $\langle f,g\rangle = \int_0^\pi f\overline{g}$.

In particular, $S$ is a discrete infinite set in $(C([0,\pi]), \|\cdot\|_2)$ and hence it cannot be totally bounded.

If $S$ were totally bounded in $(C([0,\pi]), \|\cdot\|_\infty)$ then for every $\varepsilon > 0$ there would exist $f_1, \ldots, f_n \in S$ and $r_1, \ldots, r_n > 0$ such that $$S \subseteq \bigcup_{i=1}^n B_\infty(f_i, r_i).$$

Since $\|\cdot\|_2 \le \pi\|\cdot\|_\infty$, we would have $$S \subseteq \bigcup_{i=1}^n B_\infty(f_i, r_i) \subseteq \bigcup_{i=1}^n B_2(f_i, \pi r_i)$$

which would imply that $S$ is totally bounded in $(C([0,\pi]), \|\cdot\|_2)$. This is a contradiction.

mechanodroid
  • 47,570
  • can I say that if $f_n$ is sequence of functions but not uniformly convergent either has uniform convergent subsequence then it is not totally bounded in uniform norm ? – neelkanth Jan 05 '25 at 09:17
  • 1
    I'm not too sure, but I think that the answer is yes. Indeed, suppose that $S = {f_n : n \in \Bbb{N}}$ is totally bounded. It is also complete because it is a closed subset of a complete space $(C([0,\pi]), |\cdot|_\infty)$. Therefore $S$ is compact, so every sequence in $S$ has a convergent subsequence. But this is a contradiction since $(f_n)_n$ has no convergent subsequence. – mechanodroid Jan 05 '25 at 15:16
  • thank you....... – neelkanth Jan 05 '25 at 15:23