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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ are two uniform continuous functions. Which of the following options are correct and why?

  1. $f(g(x))$ is uniformly continuous.

  2. $f(g(x))$ is continuous but not uniformly continuous.

  3. $f(g(x))$ is continuous and bounded.

My attempt:

Every uniformly continuous function maps a Cauchy sequence to a Cauchy sequence. (Here I have a doubt, as the converse may not be true). So if $\{x_n\}$ be a Cauchy sequence, $\{f(x_n)\}$ and $\{g(f(x_n))\}$ both will be Cauchy sequence. So $g(f(x))$ will be uniformly continuous, i.e. 1 is true.

Composite function of two continuous functions will be continuous. As 1 is true, 2 is false.

$f(x) = x $ is uniformly continuous. $g(x) = \log(x)$ is uniformly continuous in $[1,\infty)$. So $g(f(x)) = \log(x)$ is uniformly continuous in $[1, \infty)$, but not in $\mathbb{R}$.

Thank you for your help.

Davide Giraudo
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Supriyo
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2 Answers2

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  • Option $(1):$ True:

    Choose $\epsilon>0.$

    Then $\exists~\delta_1>0$ such that $|x_1-x_2|<\delta_1$$\implies|f(x_1)-f(x_2)|<\epsilon.$

    For above $\delta_1>0~\exists~\delta>0$ such that $|x_1-x_2|<\delta$$\implies|g(x_1)-g(x_2)|<\delta_1.$

    Consequently, $|x_1-x_2|<\delta$$\implies|(f\circ g)(x_1)-(f\circ g)(x_2)|<\epsilon.$

  • Options $(2),~(3):$ Not necessarily true:

    Take $f(x)=g(x)=x$ on $\mathbb R.$

    Then $f,g$ are uniformly continuous on $\mathbb R.$

    Note $(f\circ g)(x)=x$ is unbounded and uniformly continuous on $\mathbb R.$

Sugata Adhya
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Your suspicion is justified. The argument you give for the validity of 1 is faulty since it only shows the composite preserves Cauchy sequences, but it does not show that the composite is uniformly continuous. Try working directly with the definition of uniform continuity instead to give a direct proof.

You can make things easier with the counter example to 3: Take $f(x)=g(x)=x$ on $\mathbb R$.

Ittay Weiss
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  • Thank you for your hint. So 1 is true and 2, 3 are false. Please check the proof for 1. For any $\epsilon > 0$, $\exists$ $\delta > 0$ s.t. $|f(x) - f(y)| < \epsilon$ whenever $|x - y| < \delta$ because of uniform continuity of $f$. As $g$ is uniformly continuous $|g(f(x)) - g(f(y))| < \epsilon'$ whenever $|f(x) -f(y)| < \epsilon$ whenever $|x - y| < \delta$. So $g(f(x))$ is uniformly continuous. – Supriyo Nov 21 '13 at 14:00
  • @IttayWeiss : Sorry but I do not understand what you mean by the composite preserving cauchy sequences but the composite not being cauchy. What is the difference between the two? What is it like a function being cauchy? – user118494 Aug 21 '15 at 13:30
  • is it possible to say that if $f, g$ are bounded and uniformly continuous then the composition $f \circ g$ is uniformly continuous? – Noa Even Jan 07 '19 at 08:24