Does there exist any continuous but not uniformly continuous function $f(x)$ such that $\sin(f(x))$ is uniformly continuous?

Question

Does there exist any continuous but not uniformly continuous function $f(x)$ such that $\sin(f(x))$ is uniformly continuous?

Actually all the examples I am taking for $f$ makes the composite function non uniformly continuous. I am not sure enough.

Answer 1

This answer is completely rewritten for clarity. The idea is the same as before:

Assume that $f\colon\mathbb{R}\to\mathbb{R}$ is continuous, and that $x\mapsto\sin(f(x))$ is uniformly continuous. Then $f$ is uniformly continuous.

To see this, first note that the inverse sine function is uniformly continuous. Thus, let $0<\varepsilon<\pi$, and pick $\eta>0$ so that $\lvert y_1-y_2\rvert<\eta$ implies $\lvert\arcsin(y_1)-\arcsin(y_2)\rvert<\varepsilon$.

Next, use the uniform continuity of $x\mapsto\sin(f(x))$ to pick $\delta>0$ so that $\lvert x_1-x_2\rvert<\delta$ implies $\lvert \sin(f(x_1))-\sin(f(x_2))\rvert<\eta$.

Now assume $x_1<x_2<x_1+\delta$.

I will write $[a,b]$ for the closed interval with end points $a$ and $b$ even if it so happens that $b<a$.

First, I claim that $\lvert f(x_1)-f(x_2)\rvert<\pi$. For otherwise, there is some integer $n$ with $[(n-\frac12)\pi,(n+\frac12)\pi]\subseteq[f(x_1),f(x_2)]$, and we can then find $x_3,x_4\in[x_1,x_2]$ with $f(x_3)=(n-\frac12)\pi$, $f(x_4)=(n+\frac12)\pi$, and $f(x)\in[(n-\frac12)\pi,(n+\frac12)\pi]$ for all $x\in[x_3,x_4]$. But for $x$ in this interval, $f(x)=n\pi+(-1)^n\arcsin(\sin(f(x)))$, and we find $\lvert \sin(f(x_3))-\sin(f(x_4))\rvert<\eta$, and hence $$\pi=\lvert f(x_3)-f(x_4)\rvert=\lvert\arcsin(\sin(f(x_3)))-\arcsin(\sin(f(x_4)))\rvert<\varepsilon,$$ which contradicts the choice of $\varepsilon$.

Second, if there is some $n$ with $x_1,x_2\in[(n-\frac12)\pi,\le(n+\frac12)\pi]$ then the argument of the previous paragraph shows $\lvert f(x_1)-f(x_2)\rvert<\varepsilon$.

Well, almost; $f(x)$ may make excursions into the next interval, but any such excursion is less than $\varepsilon$ away, so at least we can conclude $\lvert f(x_1)-f(x_2)\rvert<2\varepsilon$.

And finally, if the previous paragraphs do not apply, then for some $n$, $x_1,x_2\in[(n-\frac12)\pi,\le(n+\frac32)\pi]$, and again we get $\lvert f(x_1)-f(x_2)\rvert<2\varepsilon$.

(I don't have time to clean up the last details, but I am totally convinced that this is easy, if somewhat tedious.)