# Uniform Continuity

• Jan 23rd 2010, 07:44 AM
Uniform 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 AM
drumist
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 AM
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 AM
drumist
$\displaystyle f(x)=\sqrt{x}$ is not uniformly continuous over [0,1].

Edit: My mistake. It is.
• Jan 23rd 2010, 10:12 AM
HallsofIvy
Quote:

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

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

$\displaystyle \sqrt{\frac{(2n+3)\pi}{2}}- \sqrt{\frac{(2n-1)\pi}{2}}$ 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 AM
HallsofIvy
Quote:

Originally Posted by drumist
$\displaystyle f(x)=\sqrt{x}$ is not uniformly continuous over [0,1].

$\displaystyle \sqrt{x}$ is continuous on [0, 1] which is a compact set. Therefore, it is uniformly continuous on [0, 1].

Having a bounded derivative is a sufficient condition to be uniformly continuous but not a necessary condition.
• Jan 23rd 2010, 10:17 AM
Krizalid
Quote:

Originally Posted by drumist
$\displaystyle f(x)=\sqrt{x}$ is not uniformly continuous over [0,1].

A continuous function on a compact set is uniformly continuous.

And being the derivative unbounded, doesn't imply a thing.

(Ahh well, I said the same that HallsofIvy's post, in other words.)
• Jan 23rd 2010, 10:23 AM
drumist
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 AM
HallsofIvy
Quote:

Originally Posted by drumist
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.

Anyway, ignoring this error, since the derivative is unbounded on the original function, we know that the function is Lipschitz, which therefore requires that it is uniformly continuous. So it's still a valid way to prove that the function is uniformly continuous. ;P

Wrong again. It shows that the function is NOT Lipschitz. And the problem is that the function is NOT uniformly continuous.
• Jan 23rd 2010, 10:33 AM
drumist
Nevermind...

Sorry, had a few wires crossed in my mind. Sorry if it confused you at all, adam63. :P
• Jan 23rd 2010, 11:02 AM
Quote:

Originally Posted by drumist
Nevermind...

Sorry, had a few wires crossed in my mind. Sorry if it confused you at all, adam63. :P

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 AM
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.