# Uniformly continuous #2

• Jan 3rd 2010, 10:06 AM
Also sprach Zarathustra
Uniformly continuous #2
Prove that f(x)=sin(x)/x + cos(sqrt(x)) + 10x is uniformly continuous in her domain.
• Jan 3rd 2010, 01:35 PM
Jose27
It's easy to prove that for differentiable functions it is equivalent:

1)$\displaystyle f$ has a bounded derivative
2)$\displaystyle f$ is Lipschitz

Since Lipschitz $\displaystyle \Rightarrow$ unif. cont. we only have to prove $\displaystyle f$ has bounded derivative at infinity which follows almost immediately so $\displaystyle f$ is unif. cont. in any interval of the form $\displaystyle [a, \infty )$ where $\displaystyle a>0$, but (I believe, I didn't really check it) that it is not unif. cont. in $\displaystyle (0, \infty )$ since the sequence $\displaystyle 1/n$ seems (!) to map to an unbounded sequence.
• Jan 3rd 2010, 01:45 PM
Enrique2
Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value.
• Jan 3rd 2010, 01:51 PM
Jose27
Quote:

Originally Posted by Enrique2
Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value.

Well, continous in $\displaystyle [0,1]$ since otherwise the square root is not defined.

Also, yeah I started confusing f and f', so yeah f is unif cont in [0,a] too.
• Jan 3rd 2010, 01:53 PM
Plato
Quote:

Originally Posted by Enrique2
Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value.

Note that this function is not defined on $\displaystyle [-1,0)$ because $\displaystyle \cos\left(\sqrt{x}\right)$ is not defined.
• Jan 3rd 2010, 02:05 PM
Also sprach Zarathustra
And without derivatives?
• Jan 3rd 2010, 02:13 PM
Krizalid
I'm sorry but being analysis questions I think we're allowed to invoke some of stuff from Calculus I, that is what Jose27 just said: bounded derivative implies Lipschitz.

Why making these things harder than they look like.
• Jan 3rd 2010, 02:26 PM
Also sprach Zarathustra
The tools for this questions not include derivatives. Lipschitz yes! Derivatives no!
• Jan 3rd 2010, 02:32 PM
Enrique2
Yes, clearly as Plato and Jose27 have noticed my former argument is not correct, we only have defined the function on positive numbers. The argument works in [0,1] instead of [-1,1] .

The argument that I believe krizalid is suggesting is not so trivial. The bounded derivative argument only works easily outside an interval $\displaystyle [0,a]$ with a>0. In [0,a] you can use that $\displaystyle sin(\sqrt{x})=O( \sqrt{x})$ when $\displaystyle x$ approaches 0, but it is not for me easier than using Heine Cantor in a bounded interval and observig that the derivative is bounded outside, that is the correct point to solve a lot of these kind of problems. It was not my idea at all being "too much sophisticated", simply I didn't see the argument in [0,a], that is nice!.
• Jan 3rd 2010, 02:39 PM
Enrique2
Quote:

Originally Posted by Also sprach Zarathustra
The tools for this questions not include derivatives. Lipschitz yes! Derivatives no!

Sorry! Without the derivatives, proving Lipschitz is harder. Then I believe you will be allowed to use Heine Cantor in [0,a], and making calculations outside, but I don't see at the moment an immediate way for suggesting you how to do these calculations in a simple way, sorry again!