# Is my proof for composition of uniformly continuous function correct?

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

• 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 $|x-y| <\delta$. Aug 12, 2021 at 6:10
• I don't understand can you explain it again? Why can't I start with $\epsilon_1$ and $\epsilon_2$Thank you Aug 12, 2021 at 6:29
• @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 $1$. Aug 12, 2021 at 6:40

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.

• Ah! I got it! Thank you so much. Aug 12, 2021 at 7:08