2

Given two functions that are uniformly continuous on $\mathbb{R}$ prove that their composition is uniformly continuous. Here's my attempt.

Proof

Let $f,g$ be our given uniformly continuous functions. We need to show that for any $\epsilon >0 $ there exists a $\delta$ such that

$$\left| (f\circ g)\left(x\right) - (f\circ g)\left(y\right) \right| < \epsilon$$

Since we are given $$\left|x' - y' \right| < \delta_f \implies \left| f\left(x'\right) - f\left(y'\right) \right| < \epsilon$$

and $$\left|x - y \right| < \delta \implies \left| g\left(x\right) - g\left(y\right) \right| < \delta_f$$

defining $g(a) = a'$ for $a\in \mathbb{R}$ completes the proof.

Eric Wofsey
  • 330,363
Zduff
  • 4,252

1 Answers1

4

This is the right idea but what you have written is nonsense if taken literally. It is extremely important to clearly state the hypotheses and quantifiers in each statement you make. For instance, we are not "given" $$\left|x' - y' \right| < \delta_f \implies \left| f\left(x'\right) - f\left(y'\right) \right| < \epsilon.$$ Rather, we are given that there exists $\delta_f>0$ such that for all $x',y'\in\mathbb{R}$, this implication is true. Clean up all your statements similarly and you'll have a great proof.

Eric Wofsey
  • 330,363