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1) $\{\sin nx\}_{n\ge1}$ in $C([0,2\pi])$

2) $\left\{\dfrac{x^n}{n}\right\}_{n\ge1}$ in $C([0,2])$

Attempt:

1) By the mean value theorem We have

$$|\sin(nx)-\sin(ny)| = |n\cos(nz)||x-y|$$

So as $n \to \infty$ We can't bound $|\sin(nx)-\sin(ny)|$ (The thing is that I don't know how to say this as a properly argument, thi is, formally )

2) This I really don't know how to apprach it unless I do the MVT trick but I am not sure of it I think there must be a better way.

Am I right in the first attempt? and Can you help me to get the other one please?

Thanks a lot in advance :).

Michael Hardy
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user162343
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2 Answers2

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Hint $1$. For any natural $n$: "$\sin(nx)$ in $C([0,\frac{2\pi}{n}])$" is true, thus $|x-y|\leq \frac{2\pi}{n}$ and $$|\sin(nx)-\sin(ny)| = |n \cos(nz)||x-y|\leq |n \cos(nz)|\frac{2\pi}{n}= 2\pi|cos(nz)|$$ Now if $1/n < |x-y|$, so $|z|> 1/n$ then $|sin(nx)-sin(ny)| = |n \cos(nz)||x-y|>|\cos(1)|$.

Hint $2$.
$$|\frac{x^n}{n}-\frac{y^n}{n}|=|z^{n-1}||x-y|$$ If $|x-y|>1/n$ and $x ,y\in [1,2]$, so $|z|>1$, then $$|\frac{x^n}{n}-\frac{y^n}{n}|=|z^{n-1}||x-y|>|z^{n-2}|>1.$$

Deliasaghi
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  • So if the given $\epsilon$ is less than $2 \pi$ we are done right? – user162343 Sep 18 '15 at 00:59
  • If $|x-y|\leq \frac{\epsilon}{2\pi}$, then $|\sin(nx)-\sin(ny)|\leq \epsilon$. Note that if $n\rightarrow \infty$, then $x=y$ and $|\sin(nx)-\sin(ny)|=0$. – Deliasaghi Sep 18 '15 at 01:02
  • Then, Is this answer right?, or... – user162343 Sep 18 '15 at 01:09
  • I do not think this answer is even very helpful; you want a lower bound, not an upper bound, on $|\sin(nx)-\sin(ny)|$. If you actually want to approach it this way, you have to argue something along the lines of "given $\varepsilon$, here's the optimal $\delta$ for $f_n$; observe that it goes to zero as $n \to \infty$, even when $\varepsilon$ is fixed". I think this is a rather awkward way to proceed. – Ian Sep 18 '15 at 01:11
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    @user162343 it's enough that $|x-y|>1/n$ and you get the result. – Deliasaghi Sep 18 '15 at 01:37
  • Can you help me with this quenstion too http://math.stackexchange.com/questions/1438366/how-to-use-arzel%C3%A0-ascoli-theorem-in-this-situation please? thnsks in advance :) – user162343 Sep 18 '15 at 01:38
  • @user162343 may be help you http://math.stackexchange.com/questions/1025487/understanding-the-proof-of-the-arzela-ascoli-theorem-from-carothers – Deliasaghi Sep 18 '15 at 03:43
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The definition for equicontinuity: A family $(f_n)_{n\in\mathbb N}$ of real valued function on $X$ is equicontinuous at a point $x_0\in X$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f_n(x_0) - f_n(x)| < \epsilon$ for all $n\in\mathbb N$ and all $x\in X$ such that $|x_0 - x| < \delta$.

To show that $f_n$ is not equicontinuous at $x_0$, we have to select some $\epsilon$, say $1/2$, and give a sequence $(x_n)_{n\to\mathbb N}$ with $x_n\to x_0$ and $|f_n(x_n) - f_n(x_0)| \ge \epsilon$.

Some extensive hints:

  1. Consider $$ \sin\left(n \frac\pi{2n} \right) - \sin(0) = 1, $$ however $$\frac{\pi}{2n} - 0 \to 0.$$

  2. Consider $$ \frac{(\sqrt[n]{n})^n}{n} - \frac{1^n}{n} \to 1, $$ but $$ \sqrt[n]{n} - 1 \to 0. $$

user251257
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  • I don't understand the hints, I am sorry but could you please clarify :) Thnks – user162343 Sep 18 '15 at 01:07
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    @user162343 Given $0<\varepsilon<1$, suppose you had $\delta$ so that $|\sin(nx)-0|<\varepsilon$ for $|x|<\delta$. By taking $n$ large enough that $\frac{\pi}{n} < \delta$, the first hint gives a contradiction. – Ian Sep 18 '15 at 01:09
  • @Ian :) nice to here about you, Have you checked the edition of the post I've made http://math.stackexchange.com/questions/1438799/prove-the-following-situation-using-arzel%C3%A0-ascoli-theorem ? – user162343 Sep 18 '15 at 01:11
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    @user162343: updated my answer with more explanation – user251257 Sep 18 '15 at 01:15
  • Why $sin(n(\frac{\pi}{n}))-sin(0)=1$? – user162343 Sep 18 '15 at 13:43
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    @user162343: Oh, it should be $\pi/2$ – user251257 Sep 18 '15 at 13:44
  • Can you help me with this please http://math.stackexchange.com/questions/1438799/prove-the-following-situation-using-arzel%C3%A0-ascoli-theorem? Thaks in advance :) – user162343 Sep 18 '15 at 13:45