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I've been asked to prove that if $f$,$g$ are uniformly continuous functions, then their composition $g\circ f$ is also uniformly continuous.

This is my attempt:

Since $f$ and $g$ are uniformly continuous, we have,

For a given $\epsilon_1>0$, $\exists\ \delta_1>0$ such that $|x_1-x_2|< \delta_1\implies|f(x_1)-f(x_2)|< \epsilon_1$ (1)

Similarly, For a given $\epsilon_2>0$, $\exists\ \delta_2>0$ such that $|x_1-x_2|< \delta_2\implies|g(x_1)-g(x_2)|< \epsilon_2$ (2)

Now let, $\epsilon= \min(\epsilon_1,\epsilon_2)$ and let $\delta=\min(\delta_1,\delta_2)$

Then (1) becomes

For a given $\epsilon>0$, $\exists\ \delta>0$ such that $|x_1-x_2|< \delta \implies |f(x_1)-f(x_2)|< \epsilon$

Let $\delta=\epsilon$,for a particular choice of $\epsilon$

Then we have $|f(x_1)-f(x_2)|<\delta$ for some $\epsilon>0$ (3)

Then by (2) and (3) we have,

$|g(f(x_1))-g(f(x_2))|<\epsilon$ whenever

$|x_1-x_2|<\delta$

Hence $g\circ f$ is uniformly continuous

Any advice is welcomed!!

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    $\begingroup$ You don't start with an $\epsilon_1$ and an $\epsilon_2$. You have to start with an $\epsilon$ and produce a $\delta$ such that $g(f(x))-g(f(y))| <\epsilon$ whenever $|x-y| <\delta$. $\endgroup$ Aug 12, 2021 at 6:10
  • $\begingroup$ I don't understand can you explain it again? Why can't I start with $\epsilon_1$ and $\epsilon_2$Thank you $\endgroup$
    – Natasha J
    Aug 12, 2021 at 6:29
  • $\begingroup$ @NatashaJ It's a subtle point. You're supposed to construct a $\delta$ that shows $g \circ f$ is uniformly continuous, based on a given $\varepsilon$. You've defined $\varepsilon$, which means that you might have subtly introduced restrictions in what $\varepsilon$ could be. In this case, it's an easy fix, but if you had put "Let $\varepsilon = \min\{\varepsilon_1, \varepsilon_2\} + 1$...", then you would be neglecting all possible values of $\varepsilon$ less than or equal to $1$. $\endgroup$ Aug 12, 2021 at 6:40

1 Answer 1

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Let's go by definition of uniform continuity: A function $f:A\to \mathbb R$ is said to be uniformly continuous on set $A$ if

For every $\epsilon \gt 0,$ there exists a $\delta_\epsilon\gt 0$ such that for all $x,y$ in $A$ satisfying $|x-y|\lt \delta$, it follows that $|f(x)-f(y)|\lt \epsilon$.

One observation from the definition is that $\epsilon\gt 0$ is restriction free.

You chose $\epsilon_1\gt 0, \epsilon_2\gt 0$ and then took $\epsilon =\min \{\epsilon_1, \epsilon_2\}$

Note that corresponding to $\epsilon_1,\epsilon_2,$ we have $\delta_{\epsilon_1}\gt 0$ and $\delta_{\epsilon_2}\gt 0$. If you choose $\epsilon=\min \{\epsilon_1, \epsilon_2\}$ then there is no reason to believe that $\delta_{\epsilon}=\min \{\delta_{\epsilon_1},\delta_{\epsilon_2}\}$. That is the problem with your proof. Please note that I'm using subscripts with $\delta$ to show to which $\epsilon$ they correspond to in order that many $\delta$'s don't get mixed up.

A proof for the result you are trying to prove could go along these lines:

For any arbitrary $\epsilon\gt 0,$ there exist $\delta_1\gt 0$ and $\delta_2\gt 0$ such that

  1. For all $x,y$ in domain of $g$ satisfying $|x-y|\lt \delta_1$, it follows that $|g(x)-g(y)|\lt \epsilon$
  2. For all $u,v$ in domain of $f$ satisfying $|u-v|\lt \delta_2$, it follows that $|f(u)-f(v)|\lt \delta_1$

Assuming that $gof$ is well-defined, we have from above two points that

for all $x,y$ in domain of $f$ satisfying $|u-v|\lt \delta_2$, it follows that $|f(u)-f(v)|\lt \delta_1\implies|g(f(u))-g(f(v))|\lt \epsilon$

And since $\epsilon\gt 0$ is arbitrary, the result follows from the definition stated above.

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    $\begingroup$ Ah! I got it! Thank you so much. $\endgroup$
    – Natasha J
    Aug 12, 2021 at 7:08

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