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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.

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  • $\begingroup$ How about $f(x)=1/x, \quad x\in (0,1)$ $\endgroup$
    – Jlamprong
    Jan 23, 2014 at 8:55
  • $\begingroup$ sin(1/x) is not uniformly continuous. $\endgroup$
    – user121418
    Jan 23, 2014 at 8:56
  • $\begingroup$ This is not uniformly continuous on $(0,1)$. $\endgroup$
    – detnvvp
    Jan 23, 2014 at 8:56
  • $\begingroup$ I am looking for $f$ which is not uc but $sin(f)$ is uc. $\endgroup$
    – user121418
    Jan 23, 2014 at 8:58
  • $\begingroup$ Oh sorry, I did a mistake. Let me think $\endgroup$
    – Jlamprong
    Jan 23, 2014 at 8:58

1 Answer 1

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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.)

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    $\begingroup$ Is the terminology “uniformly equicontinuous” kosher? I invented it on the spot. $\endgroup$ Jan 23, 2014 at 9:32
  • $\begingroup$ inverse of sine function is DEFINED on [-1,1], den how it is defined on $I_n$? $\endgroup$
    – user121418
    Jan 23, 2014 at 9:37
  • $\begingroup$ The logical terminology would be "uniformly uniformly continuous", but I can see why you didn't choose that. $\endgroup$
    – TonyK
    Jan 23, 2014 at 9:39
  • $\begingroup$ I meant the inverse of the restriction to $I_n$ of the sine function. I wanted to write it as the inverse of $\sin|_{I_n}$, but MathJax made the notation look horrible. I reformulated it a bit, though. $\endgroup$ Jan 23, 2014 at 9:39
  • $\begingroup$ what is $f$ here ? $\endgroup$
    – user121418
    Jan 23, 2014 at 9:56

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