I know that the composite function of two uniformly continuous functions is also uniformly continuous. But can the composition of two functions be uniformly continuous even when the functions themselves are not? What if one of them is allowed to be uniformly continuous?
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Compose the function $x\in\mathbb R\mapsto x^3\in\mathbb R$ and its inverse.
Mariano Suárez-Álvarez
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Compose $f(x)=\frac 1x$ with itself on $(0,\infty)=\mathbb R^+$
Mark Bennet
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1You can't compose it with itself if you restrict the domain that way :-) – Mariano Suárez-Álvarez Jan 29 '18 at 20:34
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@MarianoSuárez-Álvarez Good spot. – Mark Bennet Jan 29 '18 at 20:34
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This is sort of stolen from Mariano Suárez Álvarez's answer, but here both functions are non-uniformly continuous:
Let $$ f(x)=\begin{cases}-x^3&\text{if }x\ge 0\\-\sqrt[3]x&\text{if }x\le 0\end{cases}$$ and compose it with itself.
Hagen von Eitzen
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Consider $f(x)=\begin{cases}\pi&\text{if }x<-1\\ 0&\text{if }x\ge -1\end{cases}$. Then, $f\circ \sin=\sin\circ f=0$ and $f\circ f=0$.