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Asuume that $f:\mathbb{R}\rightarrow \mathbb{R}$ continuous function and $g:\mathbb{R}\rightarrow \mathbb{R}$ uniformly continuous function and $g$ bounded.

I have to prove that $f\circ g$ is uniformly continuous function. I tried the following: $f$ continuous function so $\forall \epsilon>0 ~\exists~ \delta_1>0 $ and $ |x-x_0|<\delta$ and $|f(x)-f(x_0)|<\epsilon$ From $g$ uniformly continuous function definition i can say $|g(x)-g(y)|<\delta_1$ which mean $f\circ g$ is continuous function but not uniformly continuous function. I dont know how to use that $g$ is bounded. Thank you very much.

2 Answers 2

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Since $g$ is bounded, its range is contained in a finite closed interval. $f$ is uniformly continuous on that interval.

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    why $f$ is uniformly continuous on that interval.? – 2017-01-29
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    I was thinking about this problem too. Great! A continuous function in a compact set (in $\mathbb{R}$ this is exactly closed an bounded) is uniformly continuous. – 2017-01-29
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Hint: Did you know that if $k$ is a continuous mapping between metric spaces $X$ and $Y$ and $X$ is compact, $k$ is uniformly continuous?

Since $g$ is bounded, its image is contained in a compact set $K$ (i.e., $g(\mathbb{R}) \subset K$). Letting $k = f|_K$, it is trivially the case that $f\circ g=k\circ g$. What does this imply?

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    Thanks, what does $k=f|K$ mean? and i cant understand how it answere the question. – 2017-01-29
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    $f|_K$ is the restriction of the function $f$ to the set $K$. That is, $f|_K$ is a function mapping from $K\subset \mathbb{R}$ to $\mathbb{R}$ such that for all $x\in K$, $f|_K(x)=f(x)$. The idea is to show that $k\circ g$ is uniformly continuous (what do you know about compositions of uniformly continuous functions?) and to use this to conclude that $f\circ g$ is uniformly continuous. In Robert Israel's answer, $K$ is an interval. – 2017-01-29
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    ok thanks,so $k o g=f o g$ how do i conclude from here that is uniformly continuous function? – 2017-01-29
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    you mean to say $kog$ continuous at some $[a,b]$ if its so i can understand why we finish the prove.but i cant underestand why it is bounded – 2017-01-29
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    You can replace all instances of $K$ with some $[a,b]$. If that's what you mean, then yes. – 2017-01-29