Uniform continuity of a $ |\sin x| $ in $ [0,\infty) $

Question

Given a function $ |\sin (x)| $ over the interval $ [0, + \infty) $ we have to check whether the function is uniformly continuous or not ?

So I begin by letting $ |x-y| < \delta $ then if I consider $ | |\sin x|-|\sin y| | $ then by triangular inequality we have $$ | |\sin x|-|\sin y| | < |\sin x - \sin y| = 2 |\sin(\frac {x-y}{2}) \cos (\frac {x+y}{2})| < 2|\sin(\frac {x-y}{2})|$$

Hence if I choose a $\epsilon > 2 \sin(\frac {\delta}{2} ) $ I can assure $ | |\sin x|-|\sin y| | < \epsilon$.

Is my proof correct ? i want some clarity on this ! also I am open to new methods or hints if you have any !!

Answer 1

No, can't choose $\epsilon$ in terms of $\delta$, but you can rearrange your work and say that given $\epsilon>0$ then we can take $\delta=\epsilon$: for $|x-y|<\delta$, $$\begin{align}| |\sin x|-|\sin y| | &\leq |\sin x - \sin y| = 2 \left|\sin\left(\frac {x-y}{2}\right) \cos \left(\frac {x+y}{2}\right)\right| \\&\leq 2\left|\sin\left(\frac {x-y}{2}\right)\right| \leq |x-y|<\delta=\epsilon\end{align}$$ where we used the inequality $|\sin(t)|\leq |t|$ valid for any real $t$.

Hence $ |\sin (x)| $ is uniformly continuous over $\mathbb{R}$ (which include $[0, + \infty) $).

Answer 2

In such a proof, you are given $\epsilon$ and need to choose $\delta$. It seems you’ve tried to do the opposite: you seem to be trying to choose a large enough $\epsilon$, when in reality $\epsilon$ should be fixed and your goal is to choose a small enough $\delta$.

Regardless, an easier line of reasoning: by the mean value theorem, for any $x \neq y$ there is a $c$ between $x$ and $y$ with $$\lvert \sin(x) - \sin(y) \rvert = \lvert \cos(c) \rvert \lvert x-y\rvert \le \lvert x - y \rvert.$$ Do you see why this immediately implies uniform continuity in your case?

Answer 3

You are assumimg that $|\sin x|$ is an increasing function. You cannot say that $|\sin (\frac {x-y} 2)|<\sin (\frac {\delta} 2)$ if $|x-y| <\delta$.

For a correct proof use the fact that $|sin x -\sin y|=|x-y||\cos t|\leq |x-y|$ fro some $t$ by Mean Value Theorem.