Is the function $f(x)=\sin(x\sin x)$ uniformly continuous?

Question

Is the following function uniformly continuous?

$f(x)=\sin(x\sin x)$

Tried using definition of uniformly continuous and also the fact that if the derivative is bounded then the function is uniformly continuous but failed in both cases .Please help.

Answer 1

let $$f(x) = x\sin x$$ Take $n$ a positive integerand consider $$a_n = 2n\pi -\pi/2 \quad \text{and} \quad b_n = 2n\pi + \pi/2$$

hence $$g(a_n) = -a_n \quad \text{and} \quad g(b_n) = b_n$$

In the interval $[-a_n, b_n]$ of length $4n \pi + \pi$, we have $2n$ periods, in other words is $2n$ points at which $f$ takes $-1$ and $2n$ points in which it takes the value $1$.

Hence, the average interval is $\pi/2n$ and the slope of the secant line is $4n/\pi.$ By MVT, there is such a point with the exact slope.

As $n \rightarrow +\infty$, $f$ is not Lipschitz continuous in the interval $[0, +\infty[$

Answer 2

Hints:

1. If $f$ is uniformly continuous, then we have $f(y_n) - f(x_n)\to 0$ for any sequence $x_n, y_n$ with $y_n - x_n \to 0$.

2. Consider $x_n = 2\pi n$ and $y_n = x_n + \frac{1}{x_n}$.