why $\sin (x \sin x)$ is not uniformly continous for $x>0$ . I think it should be since for $(0,1]$ we can use continous extension theorem and for $x>1$ it is lipschitz (since it lies below $y=x$ line).What is wrong in the argument? This argument also proves that $\sin (x^2)$ is u.c.(which is wrong!)
Thanx in advance for any help.
 A: let $g(x) = x\sin x.$ take $n$ a positive integer, $a_n = 2n\pi -\pi/2, b_n = 2n\pi + \pi/2$. then $g(a_n) = -a_n, g(b_n) = b_n$ within the interval $[-a_n, b_n]$ of length $4n \pi + \pi$ accommodates $2n$ full periods. that is $2n$ points at which $f$ takes $-1$ and $2n$ interlaced points at which $f$ takes $1.$ the average interval is $\pi/2n$ and the slope of the secant line is $4n/\pi.$ by mean value theorem, there is such a point with the exact slope.by letting $n$ go to infinity,
you can see that $f$ cannot be Lipschitz continuous on $[0, \infty)$
A: Do you know Cantor's Theorem? If a function on a closed interval $[a,b]$ is continuous, then it is uniform continuous on $[a,b]$.
For your $f(x)=\sin(x\sin(x))$ , it is obvious that for all closed intervals $[a,b]$ that are a subset to $\mathbb R$, $f(x)$ is u.c.
So what matters is when $x$ goes to infinity.
And maybe there is one misconception to be clarified: that $f(x)$ is u.c. on any closed intervals in $\mathbb R$ never means it is u.c. on $[a,+\infty),(-\infty,b]$ or $\mathbb R$. Take $f(x)=x^2$ for example, it is not u.c. on $[0,+\infty)$.
That's all I know about u.c. (I'm sorry but I am just a freshman). Just hope it might be of some help to you.
