Let $\displaystyle f=sin(x^2)$
How can I prove that this function is not uniformly continuous in the Real Numbers?
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.
$\displaystyle \sqrt{\frac{(2n+3)\pi}{2}}- \sqrt{\frac{(2n-1)\pi}{2}}$ 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.
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.
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 :
$\displaystyle |f(x_n1)-f(x_n2)|>=Epsilon$
?