Show that $F_n(x) = n\sin(\frac{x}{n})$ converges uniformly on $[-a,a]$ for any finite $a > 0,$ but does not converge uniformly on $\mathbb R.$

My thoughts were since $x$ is bounded by $a,$ and $n$ can be chosen large, $\sin(\frac{x}{n})$ is about $\frac{x}{n}$ when $n$ gets large so $n\sin(\frac{x}{n})$ is about $n(\frac{x}{n}) = x.$ ( limit function is $f(x) = x$)

actual proof attempt :

For all $ε>0$ & $a>0$ let $K(ε) = \frac{a}{ε}$

$\frac{a}{ε}\sin(x\frac{ε}{a})$ something and I don't know how to proceed, (is this even the right track?)

Thanks in advance!

  • 1
    $\begingroup$ How did it "fail"? That is, what exactly have you tried, and where did you get stuck? (Additionally, the title of your question does not match the actual question... Is it $n\sin\frac{x}{n}$, or $\sin\frac{x}{n}$?) $\endgroup$ – Clement C. Nov 16 '15 at 23:15
  • $\begingroup$ Note that $ signs need to enclose all the math. I did this for you in the first paragraph. See if you can fix the rest. $\endgroup$ – zhw. Nov 16 '15 at 23:42
  • $\begingroup$ Thanks for the advise, this is my first time posting. $\endgroup$ – Dou Nov 16 '15 at 23:51
  • $\begingroup$ You did a good job for your first time. Try not to use $*$ for multiplication. I edited your second paragraph for fun. $\endgroup$ – zhw. Nov 17 '15 at 7:44

To prove that it converges uniformly for $x \in [-a, a]$:

Let $\epsilon > 0$

$\forall x \in [-a, a], ~~ |n \sin(\frac{x}{n}) - x| \leq \max_{x \in [-a, a]} |n \sin(\frac{x}{n}) - x| \leq |n \sin(\frac{a}{n}) - a| $

As $sin(x) = x + o(x)$, $|\lim_{n \rightarrow +\infty }n \sin(\frac{a}{n}) - a| = 0 $, and so there exists $N_{\epsilon} \in \mathbb{N}$, such that:

$\forall n\geq N_{\epsilon}, ~~|n \sin(\frac{a}{n}) - a| < \epsilon $

And so,

$\exists N_{\epsilon} \in \mathbb{N}, ~~ \forall x \in [-a, a], ~\forall n\geq N_{\epsilon}, ~~|n \sin(\frac{x}{n}) - x| < \epsilon $

Which proves that it converges uniformly in $[-a, a]$.

To prove that it is not true for $\mathbb{R}$, see that for $n$ fixed, $|n \sin(\frac{x} {n} ) | \leq n$, so $|n \sin(\frac{x} {n} ) - x|$ can be as big as you want. Therefore, it does not converge uniformly.

  • $\begingroup$ Can you briefly explain the sin(x)=x+o(x) part? $\endgroup$ – Dou Nov 17 '15 at 0:01
  • $\begingroup$ You can simply say that $\sin(x) \sim_{x \rightarrow 0} x$, which means that $\sin(x) $ is behaving like $x$ around 0. $\endgroup$ – user288227 Nov 17 '15 at 0:08
  • $\begingroup$ It was just a quick way to prove that $\lim_{n\rightarrow \infty} F_n(x) - x = 0$ $\endgroup$ – user288227 Nov 17 '15 at 0:12
  • $\begingroup$ Got it! THANKS! $\endgroup$ – Dou Nov 17 '15 at 0:32

$$\forall n >0, \forall A > 0, \exists x \in \mathbb{R}/|x-n*\sin(\frac{x}{n})|>A$$

because $$|x-n*\sin(\frac{x}{n})|\ge|\;|x|-|n*\sin(\frac{x}{n})|\;|\ge|x|-n$$

therefore $$\implies \forall n \in \mathbb{N}, ||id-F_n||_\infty=\infty$$

  • $\begingroup$ I don't get why I got a -1 $\endgroup$ – stity Nov 16 '15 at 23:26
  • 1
    $\begingroup$ I didn't downvote. But this is a bit difficult to read. $\endgroup$ – Simon S Nov 16 '15 at 23:41

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