f(x)=x*Sin(1/x) uniformly continuous

**sunmalus** · November 3rd 2010, 07:25 AM

Hello,

I want to show that
$f(x)=x \cdot \sin( \frac{1}{x})$
in the interval (0, infinity)
is uniformly continuous using the following definition:

$\text{Given }f \colon I \subset \mathbf{R} \longrightarrow \mathbf{R}. f \\ \text{ is uniformly continuous on I if }\forall \varepsilon >0, \exists \delta >0 \\ \text{ such that }\forall x,y \in I, | x - y| < \delta, |f(x)-f(y)|< \varepsilon$ .

My problem is that I want to find something like:
$|f(x)-f(y)|\leq \underbrace{|x-y|}_{ \delta }\dots \leq \varepsilon$
but I've been thinking for a while and I cant find anything.

**Drexel28** · November 3rd 2010, 07:58 AM

Originally Posted by sunmalus

Hello,

I want to show that
$f(x)=x \cdot \sin( \frac{1}{x})$
in the interval (0, infinity)
is uniformly continuous using the following definition:

$\text{Given }f \colon I \subset \mathbf{R} \longrightarrow \mathbf{R}. f \\ \text{ is uniformly continuous on I if }\forall \varepsilon >0, \exists \delta >0 \\ \text{ such that }\forall x,y \in I, | x - y| < \delta, |f(x)-f(y)|< \varepsilon$ .

My problem is that I want to find something like:
$|f(x)-f(y)|\leq \underbrace{|x-y|}_{ \delta }\dots \leq \varepsilon$
but I've been thinking for a while and I cant find anything.

Two things to notice. $f$ may be continuous extended to $\hat{f}:[0,\infty)\to\mathbb{R}$ in the usual (and unique) way. Thus, for any $M>0$ $\hat{f}$ is continuous on $[0,M]$ and thus by the Heine-Cantor Theorem we see that $\hat{f}$ is unif. cont. on $[0,M]$ . Thus, any restriction of $\hat{f}$ on $[0,M]$ is unif. cont., in particular $f$ is unif. cont. on $(0,M]$ . But, $\displaystyle \lim_{x\to\infty}f(x)=1$ ... so try working with that.

If you can use something besides the strict definition and consequences I would note that $f$ is unif. cont. on $(0,1]$ as proven above and $f$ is Lipschitz on $(1,\infty)$ since it has bounded derivative. From there you can conclude.

**Also sprach Zarathustra** · November 3rd 2010, 08:39 AM

This is one of my favorite examples in infi 1.

Let $\epsilon >0$ and $\delta =\frac{\epsilon }{2}$ . Let $x_0,x_1 \in \mathbb{R}$ . Now we suppose that $\left | x_0-x_1 \right | <\delta$ .

Let we analyze $\left | f(x_0)-f(x_1) \right |$ .

$\left | f(x_0)-f(x_1) \right |=\left | x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |$

$\leq \left |x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_0})+x_0sin(\frac{1}{x_1})-x_1sin(\frac{1}{x_1}) \right |$

$=\left | x_0-x_1 \right |\left | sin(\frac{1}{x_0})+sin(\frac{1}{x_1}) \right | $
$\leq \left | x_0-x_1|\left \{ \right. \right \left | sin(\frac{1}{x_0}) \right |+\left | sin(\frac{1}{x_1}) \right | \left. \right \}\right.\left. $
$\leq 2\left | x_0-x_1|\left < \epsilon$

**Drexel28** · November 3rd 2010, 08:42 AM

Originally Posted by Also sprach Zarathustra

This is one of my favorite examples in infi 1.

Let $\epsilon >0$ and $\delta =\frac{\epsilon }{2}$ . Let $x_0,x_1 \in \mathbb{R}$ . Now we suppose that $\left | x_0-x_1 \right | <\delta$ .

Let we analyze $\left | f(x_0)-f(x_1) \right |$ .

$\left | f(x_0)-f(x_1) \right |=\left | x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |$

$\leq \left |x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_0})+x_0sin(\frac{1}{x_1})-x_1sin(\frac{1}{x_1}) \right |$

$=\left | x_0-x_1 \right |\left | sin(\frac{1}{x_0})+sin(\frac{1}{x_1}) \right | $
$\leq \left | x_0-x_1|\left \{ \right. \right \left | sin(\frac{1}{x_0}) \right |+\left | sin(\frac{1}{x_1}) \right | \left. \right \}\right.\left. $
$\leq 2\left | x_0-x_1|\left < \epsilon$

Nice! I didn't even try the usual methods assuming it would get too messy. I forgot about the bounded continuous function times a uniformly continuous function thing.

**sunmalus** · November 3rd 2010, 09:11 AM

Thanks!

**sunmalus** · November 3rd 2010, 09:17 AM

Thanks!
I forgot to mention that $\sin(a) + \sin(b)= 2\sin(\frac{a+b}{2}) \cos(\frac{a-b}{2})$ was given as "indication", so I was trying to find a way to use it and that's what confused me the most. (I still don't see how to use this tho).

Well it might be useful if you're using series to prove this but with this definition :S .

**Sudharaka** · January 3rd 2011, 08:32 PM

Originally Posted by Also sprach Zarathustra

This is one of my favorite examples in infi 1.

Let $\epsilon >0$ and $\delta =\frac{\epsilon }{2}$ . Let $x_0,x_1 \in \mathbb{R}$ . Now we suppose that $\left | x_0-x_1 \right | <\delta$ .

Let we analyze $\left | f(x_0)-f(x_1) \right |$ .

$\left | f(x_0)-f(x_1) \right |=\left | x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |$

$\leq \left |x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_0})+x_0sin(\frac{1}{x_1})-x_1sin(\frac{1}{x_1}) \right |$

$=\left | x_0-x_1 \right |\left | sin(\frac{1}{x_0})+sin(\frac{1}{x_1}) \right | $
$\leq \left | x_0-x_1|\left \{ \right. \right \left | sin(\frac{1}{x_0}) \right |+\left | sin(\frac{1}{x_1}) \right | \left. \right \}\right.\left. $
$\leq 2\left | x_0-x_1|\left < \epsilon$

Dear Also sprach Zarathustra,

Can you please tell me how you got,

$\left | x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |\leq \left |x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_0})+x_0sin(\frac{1}{x_1})-x_1sin(\frac{1}{x_1}) \right |$

Thank you.

**Also sprach Zarathustra** · January 4th 2011, 04:12 AM

Originally Posted by Sudharaka

Dear Also sprach Zarathustra,

Can you please tell me how you got,

$\left | x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |\leq \left |x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_0})+x_0sin(\frac{1}{x_1})-x_1sin(\frac{1}{x_1}) \right |$

Thank you.

You caught me!

Unfortunately, I can't find my notes where I "solved" it...

I will try to explain my work anyway:

$\left | x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |= \left |x_0sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_0})+x_1sin(\frac{1}{x_0})-x_1sin(\frac{1}{x_1}) \right |$

Now because, $\left | x_0-x_1 \right | <\delta $

We can conclude that: $x_1sin(\frac{1}{x_0})\simeq x_0sin(\frac{1}{x_1})$

hmmm...

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