Show $|\sin(y)y - \sin(x)x| \leq C|y - x|$ for some $C > 0$

Question

Show $|\sin(y)y - \sin(x)x| \leq C|y - x|$ for some $C > 0$.

This is one of the steps in a bigger problem I'm trying to solve, and while it first appeared it would be entirely straightforward, I got stuck on this part. It seems to me this is true and that such a constant $C>0$ exists, but I see no way of showing it. In fact, I figured it'd be $C = 2$, but again, I'm missing the argument.

It this constant does not exist, then I'll have to rethink the problem entirely, so I'd be grateful for any help.

Answer 1

HINT:

We have the inequality for the function if and only if for the derivative we have $$|(\sin x \cdot x)'| \le C$$

However, $(\sin x \cdot x)' = \cos x \cdot x + \sin x $. On an interval its absolute value is bounded if and only if the interval is finite.

Conclusion: only on finite intervals.

Answer 2

No, there is no such constant $C$. Let $h(t) = t\sin t$, and let $S$ be the set of all values of the ratio $(h(y) - h(x))/(y-x)$. A quick glance at the graph of $h(t)$ shows that the slopes of secant lines take arbitrarily large values as $t \to \pm \infty$, so the set $S$ is unbounded.

More rigorously, note that every value of the derivative $h'(t)$ is a limit of numbers in $S$. So if we had $S \subseteq [-C,C]$ for some number $C$, we'd also have to have $|h'(t)| \leq C$ for all $t \in \mathbf{R}$.

But $h'(t) = \sin t + t \cos t$, and $h(2\pi n) = 2\pi n$ for all integers $n$, so the function $h'(t)$ is unbounded. Thus the set $S$ itself must be unbounded.

However, if you know that your numbers $x$ and $y$ vary only in some fixed bounded interval, then the answer can be proved to be affirmative. The number $C$ will then depend on the interval.

Answer 3

it is not exist such $C$,

suppose $y-x=\pi,LHS=|(2x+\pi)\sin{x}|$ which has no limitation,that means you can't find $C$ ,

let $|(2x+\pi)\sin{x}| \le C\pi$