0

While studying for my Real Analysis exam, i tried to solve the following problem:

Problem:

Prove that the family $\{\sin n x, n=1,2,3 \ldots\}$ is not equicontinuous at any point $x_{0}$ in $[0,1]$.

Relevant definitions:

-> Given $x_{0} \in X$, we will say that the set $E$ is equicontinuous at $x_{0}$ when, given $\epsilon > 0$, there exists $\delta > 0$ such that:

$$ \forall x \in X,\left|x-x_{0}\right|<\delta \Rightarrow\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon $$

for any $f \in E$

-> A set $E$ of functions $f: X \rightarrow \mathbb{R}$ is called equicontinuous when the set $E$ is equicontinuous at each $x_{0} \in X .$

Attempt:

Consider $x_{0}$ in $[0,1]$ fixed and take $\epsilon = 1$. Given any $\delta > 0$ we can always obtain $n \in \mathbb{N}$ such that:

$$\frac{1}{n} < \delta$$

Then, the point $x = x_{0} + \frac{1}{n}$ is such that: $$ \left|x-x_{0}\right|=\frac{1}{n}<\delta $$

However, $$ \left|f_{n}(x)-f_{n}\left(x_{0}\right)\right|= \left|\sin(nx) - sin(n x_{0})\right| \leq$$

$$|sin(nx)| + |sin(n x_{0})| \leq 1 + 1 = 2$$

So, in the worst case scenario, $$\left|f_{n}(x)-f_{n}\left(x_{0}\right)\right| = 2 > 1 = \epsilon$$

Questions?

  1. Is the proof correct? I have seen some proofs which are a bit different and so i got a little suspicious regarding my attempt.

  2. Is it well written? I am trying to become better each day at writing proofs.

The sequence of functions $f_n = \sin{nx}$ is not uniformly equicontinuous on any compact interval.

Is $\sin(nx)$ equicontinuous on $[0,1]$?

Thanks in advance, Lucas

Lucas
  • 666
  • It is not correct. You showed the absolute value is at most $2$; you can not assume, from this alone, that it takes the value $2$. – David Mitra Jan 13 '21 at 20:30
  • But, you could show that since the periods of the $f_n$ get small, that your absolute value can be made greater than $1$. – David Mitra Jan 13 '21 at 20:32
  • I agree i cant assume the absolute value is exactly 2. However, since it is at most 2, any value of epsilon less than that should work (or not?) – Lucas Jan 13 '21 at 20:37
  • I think so, but you need to explain why. The inequality does not say there are values of $x$ that make it big. – David Mitra Jan 13 '21 at 20:41
  • 1
    Select $n$ so that the period of $f_n$ is less than $\delta$. Then in the interval $[x_0, x_0+\delta]$, $f_n$ takes both the values $1$ and $-1$. From this it follows that your absolute value can be made greater than $1$. – David Mitra Jan 13 '21 at 20:45
  • I am convinced that should work. However, i still dont get why my approach wont work. Can you elaborate a little further? – Lucas Jan 13 '21 at 20:46
  • You wrote essentially this: $| A|\le 2$, so $A=2$, somewhere. Is this true for $A=\sin x$? – David Mitra Jan 13 '21 at 20:48
  • I think it is, since i have used that $|sin(x)| \leq 1$ – Lucas Jan 13 '21 at 20:49
  • If it were some $A \in \mathbb{R}$ without any special property or inequality i could use, i would not have obtained -2 and 2 as bounds – Lucas Jan 13 '21 at 20:51
  • 1
    In a previous comment, I should have said "at least $1$", not "greater than $1$". – David Mitra Jan 13 '21 at 20:52
  • If $x_0=0$, you are saying $|\sin(nx)|=2$. – David Mitra Jan 13 '21 at 20:55
  • True. So are there two problems in the proof? Selecting $x_0$ and the thing regarding the inequality? – Lucas Jan 13 '21 at 20:57
  • Yes. $ x_0$ is fixed, and your "worst case" does not follow. – David Mitra Jan 13 '21 at 21:07
  • edit: I was supposed to show that equicontinuity does not hold in [0,1] – Lucas Jan 13 '21 at 21:33
  • Making the choice of $x$ and $x_0$ more explictly should help you think. How can we choose $x$ and $x_0$ so that $|x - x_0| < 1/n$ but $|\sin nx - \sin n x_0| = 2$? – macton Jan 19 '21 at 10:40

0 Answers0