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I am having problems with the following exercise, I already did part $(i)$ and $(ii)$, I am having problems with $(iii)$.

Exercise:

Let $ \large f_n(x)=\left\{ \begin{array}{ll} 1 ~~~if~~x = 0 \\ n\sin(\frac{x}{n}) ~~~if~~0<x\leq 1 \end{array} \right. $

$\large (iii)$ Does the sequence have a uniform limit?

I do not how to proceed. My professor gave me a hint: $$0 \leq x- n \sin(x/n) \leq 1- n\sin(1/n) \text{ if } x \in [0,1].$$ How do I use this hint? Can I use my pointwise limit and say that if the uniform limit exists then it's equal to the pointwise limit?

These are the exercies I have solved. If you see any mistakes please let me know.

$\large (i)$ Prove that each $f_n$ is discontinous at $x=0$.

Let $\left\{a_k\right\}_{k \geq 1}$ \in (0,1] be a sequence where $a_k \to 0$ as $k \to \infty$ In order to show that $f_n$ is discontinous at $x=0$ all I need to do is to show that

$$\lim_{k \to \infty} f_n (a_k) \neq 1$$

we have

$$\lim_{k \to \infty} \left( n\sin(\frac{a_k}{n})\right) = 0 \neq 1$$

thus $f_n$ is discontinous at $x=0$.

$\large (ii)$ Prove that the sequence has a pointwise limit.

$p(x) = \lim_{n \to \infty}\left( f_n(x) \right)$

I found the pointwise limit for when $x = 0$ and $x \in (0,1]$ :

$ \large p(x)=\left\{ \begin{array}{ll} 1 ~~~if~~x = 0 \\ 0 ~~~if~~0<x\leq 1 \end{array} \right. $

best regards

Husky

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  • $\begingroup$ shouldn't $p(x) = x$ for $x\ne 0$? otherwise the hint won't make sense. $\endgroup$ Commented Aug 17, 2015 at 2:50
  • $\begingroup$ @user251257 The hint makes sense because we need to find a supremum. $\endgroup$ Commented Aug 17, 2015 at 8:55
  • $\begingroup$ for a fixed $x\in (0,1]$ applies l'hopital's rule on the limit of $y \sin(x/y)$ for $y\to \infty$ yields $x$, not $0$. your $p(x)$ is wrong. $\endgroup$ Commented Aug 17, 2015 at 11:25
  • $\begingroup$ But its not $y \to \infty$ its $x \to 0$? @user251257 $\endgroup$ Commented Aug 17, 2015 at 11:32
  • $\begingroup$ $y$ replaces $n$, as l'hopital is about functions on reals. $\endgroup$ Commented Aug 17, 2015 at 11:33

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We have (using the hint)

$$\sup_{x\in[0,1]}|f_n(x)-p(x)|\le 1-n\sin\left(\frac1n\right)\xrightarrow{n\to\infty}0$$ so the sequence $(f_n)$ converges uniformly to $p$ on $[0,1]$.

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  • $\begingroup$ Thank you very much! Do you have time to go through (i) and (ii) and let me know if I did something wrong? $\endgroup$ Commented Aug 16, 2015 at 22:21
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    $\begingroup$ You're correct in (i) and (ii), just (I think) you should mention that you used the result $\lim\limits_{y\to0}\frac{\sin y}{y}=1$ to answer the two questions. $\endgroup$ Commented Aug 16, 2015 at 22:34
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    $\begingroup$ Okay, thank you very much! $\endgroup$ Commented Aug 16, 2015 at 22:34
  • $\begingroup$ @user235783 Could you perhaps check if my point wise limit is correct? Another user is stating that the I should have $p(x) =x$ for $x \ne 0$. Which doesn't really make sense to me $\endgroup$ Commented Aug 17, 2015 at 11:38
  • $\begingroup$ @Husky653 Yes $p(x)=x$ for $x\ne0$ because $$\lim_{n\to\infty}n\sin(x/n)=x\lim_{n\to\infty}\frac{\sin(x/n)}{\frac xn}=x\lim_{y\to0}\frac{\sin y}{y}=x$$ $\endgroup$ Commented Aug 17, 2015 at 12:03

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