# $f: \mathbb{R}\rightarrow\mathbb{R}$, prove that if $f(x)$ is continuous but not uniformly continuous on $\mathbb{R}$, then neither is $\sin f(x)$

Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be a function, if $$f(x)$$ is continuous on $$\mathbb{R}$$ but is not uniformly continuous on $$\mathbb{R}$$, then prove that $$\sin f(x)$$ is not uniformly continuous on $$\mathbb{R}$$. What I have worked out: That $$f(x)$$ is not uniformly continuous on $$\mathbb{R}$$ implies that $$\exists\varepsilon>0, \forall \delta>0,\exists x_1,x_2$$ $$s.t.$$ $$|x_1-x_2|<\delta$$ but $$f(x_1)-f(x_2)>\varepsilon$$. By the continuity of $$f$$ we can infer that $$\forall\varepsilon'\in(0,\varepsilon],\exists x_2,x_3 s.t. |x_2-x_3|<\delta$$ but $$f(x_3)-f(x_2)=\varepsilon'$$. Hence it suffices to show that given $$d>0$$, $$\sin[a,a+d] :=\{\sin x|a\leq x\leq a+d\}$$ which is an interval has a length which is always no less than $$m$$ for some $$m>0$$ fixed as $$a$$ varies. Although this is intuitive by $$\sin x$$'s diagram, I don't know how to prove it in a clear way.

• You can write \sin in math mode to produce $\sin$.
– Gary
Commented Dec 10, 2021 at 7:42
• The word "uniformly" is missing from the title which is confusing. Commented Dec 10, 2021 at 7:48

Prove by contradiction. Suppose there exists a sequence $$(a_n)$$ such that the length of $$\{\sin x: a _n \leq x \leq a_n+d\}$$ tends to $$0$$. Replacing $$a_n$$ by $$a_n+2k_n \pi$$ for a suitable integer $$k_n$$ we may suppose $$(a_n) \subset [0,2\pi]$$. So $$(a_n)$$ converges along some subsequence to a real number $$a$$. Now it is easy to check that the length of $$\{\sin x: a+\epsilon \leq x \leq a+d-\epsilon\}$$ is $$0$$ for $$0<\epsilon <\frac d 2$$. But this is impossible since $$\sin x$$ is not constant on any interval.
• Thanks for your wonderful solution. But I think there is a typo, should it be the length of $\{\sin x| a+\varepsilon\leq x\leq a+d-\varepsilon\}$? for a fixed $\varepsilon\in (0, \frac{d}{2})$ ,$\sin[a_n, a_n+d]$ is a supset of the above interval for sufficient large $n$ and hence contradicts. Commented Dec 10, 2021 at 8:58