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Find $F(x)= \lim\limits_{n \rightarrow \infty} F_n(x)$ on $S$. Show that $\{F_n\}$ converges uniformly to $F$ on closed subsets of $S$, but not on $S$. $$F_n(x) = x^n \sin nx, S=(-1,1)$$

RE-EDIT (Based on Mundron's Answer):

(a) Pointwise,

$$\lim\limits_{n \rightarrow \infty} \sin nx \rightarrow [-1,1], \forall x \in S$$

When $$|x|<1 \implies \lim\limits_{n \rightarrow \infty} x^n \rightarrow 0, \forall x \in S$$

It follows: $$F(x)= \lim\limits_{n \rightarrow \infty} F_n(x) = 0, \forall x \in S$$.

Applying The Chebyshev norm:When converging uniformly: $$\lim\limits_{n \rightarrow \infty} || f_n -f||_{\infty} = \lim\limits_{n \rightarrow \infty} \sup_{x in D_S} |f_n(x) - f(x)| = 0 $$

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(b) On a closed subset:

Considering a closed subset $C \subseteq S$. $$\{m= \min C \in S \}, \{M = \max C \in S \} \implies -1< m <M<1 $$

$$||F_n(x) - F(x)|| = \lim\limits_{n \rightarrow \infty} \sup |F_n(x) - F(x)|= \lim\limits_{n \rightarrow \infty} \sup |F_n(x) | = \lim\limits_{n \rightarrow \infty} \sup |x^n \sin nx| $$ It follows $$\lim\limits_{n \rightarrow \infty} \sup |x^n \sin nx| \leq \lim\limits_{n \rightarrow \infty} \sup |x^n| = \lim\limits_{n \rightarrow \infty} \sup |x|^n = \lim\limits_{n \rightarrow \infty} (max\{ |m|, |M|\})^n =0$$ since $|m|, |M| <1$

It follows $F_n$ converges uniformly to $0$ on a closed subset of $S$.

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(c) On an open subset

Considering $\{1 - \frac{1}{n}\} \in S, \forall \in \mathbb N$, $\{1 - \frac{1}{n}\}$ is open as the $\epsilon$-neighborhood of every point in the set are contained.

$$||F_n(x) - F(x)|| = \lim\limits_{n \rightarrow \infty} \sup \left|F_n\left(1 -\frac{1}{n} \right) \right| = \sup |\lim\limits_{n \rightarrow \infty} \left(e^{-1}sin(n-1) \right) | = e^{-1} \neq 0$$


Working on applying Chebychev's Norm. Is this argumentation correct? If not how would you apply it?

Much appreciated for your help/input.

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    $\begingroup$ My impression is that your biggest problem here is not even mathematical but a much too high dose of carelessness. To wit, you write: "When $ x = -1$, then $\lim\limits_{n \rightarrow \infty} x^n \rightarrow \infty$ ( graphic observation, I dont see how it could be )" Huh? Which "graphic observation", whatever that means? What are you talking about here? "When $x=1$, then $\{F_n(x)\} =\infty$" What? How? Did you compute any $F_n(-1)$, just to check your assertion is not pure nonsense? "When $|x|>1$, then $ \{F_n(x)\}$ diverges." Maybe, maybe not, how do you know? And so on. $\endgroup$ – Did Aug 13 '17 at 8:00
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    $\begingroup$ Careful when you say "closed subsets of $S$". Note $S$ is closed in $S.$ Better to say "compact subsets of $S$". $\endgroup$ – zhw. Aug 15 '17 at 3:48
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You write:

$\lim_{n\to \infty}\sin(nx)→[−1,1],\forall x \in S$

which is poor notation and untrue; the limit is always a single number, so this limit cannot be an interval (at least, not without specifying exactly what that means). And as has been pointed out before, the limit is a number, so you don't need the arrow $\to$ to indicate what it is. What's more, $\sin(nx)$ doesn't converge for any number when $x \neq 0$, so this limit doesn't even exist.

The correct way to argue would be to say that $\sin$ is bounded: $|\sin(nx)| \leq 1$ for all $x \in S$, so that we have: $$ 0 \leq |x^n \sin(nx)| \leq |x^n|,$$ and the squeeze lemma proves that $\lim_{n\to\infty}|x^n \sin(nx)| = 0$; the 'zero-sequence' obviously has limit zero, and $\lim_{n\to\infty}|x^n| = 0$ because $|x| < 1$ for all $x \in S$. So we have $\lim_{n\to\infty} F_n(x) = 0$, i.e. $F(x) = 0$ for all $x \in S$.


Your definition of the norm is correct, and indeed true if the sequence converges uniformly.


You write:

$m = \min C, M = \max C \implies -1 < m < M < -1$

This is almost correct: note that you need $m \leq M$ as $m = M$ is also possible (singleton sets are closed).

Note also that the notation:

$\lVert F_n(x) - F(x) \rVert$

can be misleading and is not what you mean here: $F_n(x)$ and $F(x)$ are just numbers, so you need not look at the norm of the difference, the absolute value will do. What you meant to say however was the norm of $F_n - F$, and that is what you are using afterwards, i.e.

$$ \lVert F_n - F \rVert = \lim_{n\to\infty} \sup_{x \in C} |F_n(x) - F(x)|.$$


You write:

Considering $\{1−\frac{1}{n}\} \in S,\forall n \in \mathbb N$, $\{1−\frac{1}{n}\}$ is open as the $\epsilon$-neighborhood of every point in the set are contained.

This is a misunderstanding of open sets. If you truly mean the set with its only element $1 - \frac{1}{n}$ for some natural number $n$, then this set is not open. Take any neighbourhood of the point $1 - \frac{1}{n}$, then this neighbourhood will have other points in it as well (uncountably many even), and the set only contains one element - it cannot contain any neighbourhoods. Remember, a neighbourhood of a point $x$ is an open interval $(a,b)$ with $a < x < b$ (in the context of the real number line, of course).

Furthermore, note that $\{ 1- \frac{1}{n} \} \in S$ is not true either; $S$ contains real numbers, not sets of numbers. It is definitely a subset: $\{ 1- \frac{1}{n}\} \subset S$, but it is not an element of $S$.

This leads to this bit not making much sense in my opinion:

$\lVert F_n(x) - F(x)\rVert = \lim\limits_{n \rightarrow \infty} \sup \left|F_n\left(1 -\frac{1}{n} \right) \right| = \sup |\lim\limits_{n \rightarrow \infty} \left(e^{-1} \sin(n-1) \right) | = e^{-1} \neq 0$

Again, the notation for the functions in the norm is wrong, and the first step is not clear to me either, and neither are the rest. The last limit to appear doesn't exist (so doesn't equal $\frac{1}{e}$), and as such taking the supremum over it isn't possible either.

You seem to use the following: $$ \lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n = \frac{1}{e},$$ but this is not what appears in your calculations, and as such not applicable.

Sticking to how you previously used the norm would be better, and to answer the question; note that $(-1,1)$ is also an open subset of $S$ (sets are always subsets of themselves), and it has been pointed out that $(F_n)$ does not converge uniformly on $S$ as a whole.

The open subsets for which it does converge aren't very interesting, as those are simply the ones contained within a closed subset. That is, an open subset $E \subset S$ such that $-1 < \inf E < \sup E < 1$ (assuming $E$ is not empty, for which it is all vacuously true). But then $[\inf E, \sup E]$ is a closed set contained in $S$, so $(F_n)$ converges to $F$ uniformly on that set, so also on $E$.

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When $x=-1$, then $\lim\limits_{n\to\infty} x^n\to \infty$.

First, either you write $x^n\to \infty$ for $n\to\infty$ or $\lim_{n\to\infty} x^n=\infty$, but do not mix the notation!

Second, $\lim_{n\to\infty} (-1)^n$ doesn't exists because it changes between $-1$ and $1$. How should it become $\infty$?

When $x=1$, then $\lim\limits_{n\to\infty} \{F_n(x)\}=\infty$.

Again it is wrong. $\lim\limits_{n\to\infty}1^n\sin(n)=\lim\limits_{n\to\infty}\sin(n)$. Since $\sin$ is bounded the limit can't become $\infty$. Further see >>here<< that this limit doesn't exists.

If $|x|>1$, then $\lim\limits_{n\to\infty}\{F_n(x)\}$ diverges.

This is not true in general. Consider $x=\pi>1$, you get $$\lim\limits_{n\to\infty}F_n(\pi)=\lim\limits_{n\to\infty}\pi^n\sin(n\pi)=\lim\limits_{n\to\infty}0=0$$ which converges. But for the question you don't need to consider $F_n$ on $|x|>1$.

$\sup|x^n\sin(nx)|=1$.

On which set? I don't see it...

How to prove that $F_n$ converges uniformly to $0$ on a closed subset of $S$?

First you need a closed subset $C\subset S$. Since $C$ is closed we get $m=\min C\in S$ and $M=\max C\in S$ hence $-1<m\leq M<1$ and $$ \sup_{x\in C}|F_n(x)|=\sup_{x\in C}|x|^n|\sin(nx)|\leq \sup_{x\in C}|x|^n\leq \max\{|m|,|M|\}^n. $$ Since $|m|,|M|<1$ you get $\sup_{x\in C}|F_n(x)|\to 0$ as $n\to\infty$ and therefore $F_n$ converges uniformly to $0$ on $C$.

Why is $F_n$ not uniformly convergent to $0$ on $S$?

Consider that $F_n$ converges to $0$ on $S$, but not uniformly. So you have to prove that $$ \sup_{x\in S}|F_n(x)|=\sup_{x\in S}|x|^n|\sin(nx)|\not\to 0\text{ as }n\to\infty. $$ Hints:

1. $1-\frac1n\in S$ for all $n\in\mathbb N$

2. $\lim\limits_{n\to\infty} \left(1-\frac1n\right)^n=e^{-1}$

3. $\sup_{x\in S} |F_n(x)|\geq F_n\left(1-\frac1n\right)$.

4. $F_n\left(1-\frac1n\right)\approx e^{-1}\sin(n-1)$ for large $n$.

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Part 1.

For $n\in \mathbb N$ let $n=\pi K(n)+D(n)$ where $K(n)\in \mathbb N \cup \{0\}$ and $ |D(n)|\leq\pi /2.$ It $cannot$ be true that $\lim_{n\to \infty}|D(n)|=0$ because $$0\leq |D(n)|<1/2\implies (\;K(n+1)=K(n)\;\land\; 1/2\leq |D(n+1)|<3/2\;).$$ (Because $3/2<\pi /2.$)

So take $r\in (0,\pi /2]$ such that the set $$S=\{n\in \mathbb N:\pi /2\geq | D(n)|\geq r\}$$ is infinite.

For $n\in S$ we have $|\sin n|=|\sin D(n)|\geq \sin r.$

Each function $|F_n|$ is continuous with $|F_n(0)|=0 $ and $\lim_{x\to 1^-}|F_n(x))=|\sin n|=|\sin D(n)|.$

So by the IVT, for each $n\in S$ there exists $x_n\in (0,1)$ with $|F_n(x_n)|=\frac {1}{2}|\sin D(n)|\geq \frac {1}{2}\sin r.$

Now $F_n(x)$ converges point-wise on $(-1,1)$ to the constant function $F(x)=0.$ But $S$ is infinite and for each $n\in S$ we have $$\sup_{|x|<1}|F_n(x)-F(x)|\geq |F_n(x_n))-F(x_n)|=|F_n(x_n)|\geq \frac {1}{2}\sin r.$$ So $F_n$ does not converge uniformly to $0.$

Remark. In the first line I said $0\leq |D(n)|\leq \pi /2$ rather than $0<|D(n)|<\pi /2$ because for this Q it does not matter whether or not $\pi$ is irrational.

Part 2.

By a closed subset of $(-1,1)$ you must mean a closed subset of $\mathbb R$ that is a subset of $(-1,1),$ as opposed to a closed subset of the topological space $(-1,1).$

When $T$ is a closed subset of $\mathbb R$ and $T\subset (-1,1)$ take $s\in (0,1)$ such that $T\subset [-1+s,1-s].$ (Such $s$ exists, otherwise $\phi \ne\{-1,1\}\cap \overline S=(-1,1)\cap S.$)

Let $F(x)=0$ for all $x\in (-1,1).$ We have $$\sup_{x\in T}|F_n(x)-F(x)|\leq \sup_{|x|\leq 1-s}|F_n(x)-F(x)|=\sup_{|x|\leq 1-s}|x^n\sin nx|\leq\sup_{|x|\leq 1-s}|x|^n=(1-s)^n.$$ Now $\lim_{n\to \infty}(1-s)^n=0$ because $|1-s|<1.$ So $$\lim_{n\to \infty}\|F_n-F\|_T=0$$ where $\|F_n-F\|_T=\sup_{x\in T}|F_n(x)-F(x)|.$ Therefore $F_n$ converges uniformly to $0$ on $T$.

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Let $\epsilon >0$. If $F_{n}$ uniformly convergent to $F$ then for all $x\in (-1, 1)$,$\vert F_n(x)-F(x)\vert<\epsilon$ for all $n \geq k$ where $k\geq [ (\log(1/\epsilon))/(\log(1\x))+1])$. This $k$ depends on $\epsilon$ as well as on $x$. So the sequence of functions is not uniformly convergent.

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    $\begingroup$ Functions are never uniformly convergent. Sometimes sequences of functions are. $\endgroup$ – zhw. Aug 15 '17 at 3:42

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