Let $\displaystyle f=sin(x^2)$

How can I prove that this function is not uniformly continuous in the Real Numbers?

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- Jan 23rd 2010, 07:44 AMadam63Uniform Continuity
Let $\displaystyle f=sin(x^2)$

How can I prove that this function is not uniformly continuous in the Real Numbers? - Jan 23rd 2010, 09:56 AMdrumist
Are you looking for a rigorous epsilon-delta proof, or just a round-about proof? (for example, you can just show that the derivative is unbounded)

- Jan 23rd 2010, 10:03 AMadam63
I thought of your second suggestion, but I'm not sure that if a derivative is unbounded then the function is not uniformly continuous, for instance look at:

$\displaystyle f(x)=\sqrt{x}$ in [0,1] : it's uniformly continuous, but f'(x) is surely not bounded in this area. - Jan 23rd 2010, 10:07 AMdrumist
$\displaystyle f(x)=\sqrt{x}$ is not uniformly continuous over [0,1].

Edit: My mistake. It is. - Jan 23rd 2010, 10:12 AMHallsofIvy
$\displaystyle \sqrt{\frac{(2n+3)\pi}{2}}- \sqrt{\frac{(2n-1)\pi}{2}}$ goes to 0 as n goes to infinity but $\displaystyle f\left(\sqrt{\frac{(2n+ 3)\pi}{2}}\right)- f\left(\sqrt{\frac{(2n-1)\pi}{2}}\right)$$\displaystyle = sin((2n+3)\frac{\pi}{2})- sin((2n-1)\frac{\pi}{2})$ alternates between 2 and -2 as n goes to infinity.

- Jan 23rd 2010, 10:16 AMHallsofIvy
- Jan 23rd 2010, 10:17 AMKrizalid
- Jan 23rd 2010, 10:23 AMdrumist
I suppose I was mixing up my definitions of uniformly continuous and Lipschitz. Sorry for the confusion.

$\displaystyle f(x)=\sqrt{x}$ on [0,1] is uniformly continuous, but not Lipschitz.

Nevermind... making too many mistakes, will think through before I post again hahah. - Jan 23rd 2010, 10:31 AMHallsofIvy
- Jan 23rd 2010, 10:33 AMdrumist
Nevermind...

Sorry, had a few wires crossed in my mind. Sorry if it confused you at all, adam63. :P - Jan 23rd 2010, 11:02 AMadam63
No problem at all, everyone makes mistakes, especially in math ;) I was looking for help, and you tried your best. That can never be bad!

HallsofIvy: I liked your solution, although I don't get the 'picture' of the solution:

I need to find an Epsylon for which for every delta>0 :

|x1-x2|<delta ----> |f(x1)-f(x2)|>=Epsilon

now, if I create two sequences:

$\displaystyle x_n1=\sqrt{\frac{(2n+3)\pi}{2}}

; x_n2=\sqrt{\frac{(2n-1)\pi}{2}}$

then:

$\displaystyle f(x_n1)-f(x_n2)= {0,2,-2}$.

How can I move on, in order to show that there is an Epsilon for which :

$\displaystyle |f(x_n1)-f(x_n2)|>=Epsilon$

? - Jan 24th 2010, 02:54 AMadam63
Can anyone please help me with this? It's pretty urgent (Worried)

- Jan 24th 2010, 06:25 AMHallsofIvy
The formulas I posted show that, as x goes to infinity, the "$\displaystyle \delta$ you need in the definition of continuity, for a given "$\displaystyle \epsilon$ goes to 0. There is no one value of $\displaystyle \delta$ that works for all x.