$\epsilon-\delta$ definition of continuous functions

Question

Definition: A function $f:A \to \mathbb R$ is uniformly continuous on $A$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.

---

I have a question about the correct proof strategy to show that a function $f$ is uniformly continuous.

The phrase "for every $\epsilon > 0$. there exists a $\delta > 0$" suggests to me that I should start by writing $|f(x) - f(y)| < \epsilon $ for an arbitrary $\epsilon > 0$, and show that this leads to $|x - y| < \delta$.

But the phrase "$|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$" in the definition suggests to me I should start from the other direction.

I am little confused. Can you just tell me how to start a general proof?

Thank you.

Answer 1

How about a particular proof instead? Let's show that $f(x) = x^2$ is uniformly continuous on $[0,1]$. Fix $\epsilon > 0$. Observe that $$ |f(x) - f(y)| = |x^2 - y^2| = |x - y||x + y| $$ Now, if $x,y \in [0,1]$, then $|x+y| \leq 2$. Thus, $$ |f(x) - f(y)| \leq 2|x - y| $$ Now, choose $\delta = \frac{\epsilon}{2} > 0$. If $x,y \in [0,1]$ and $|x - y| < \delta$, then $$ |f(x) - f(y)| \leq 2|x - y| < 2\delta = \epsilon $$ Therefore, $f$ is uniformly continuous on $[0,1]$.

Does this shed some light on how uniform continuity is proved in general?

Answer 2

Start with an arbitrary $\epsilon > 0$. You then need to produce some $\delta > 0$ (which may depend on $\epsilon$) such that any $x$ and $y$ satisfying $|x-y| < \delta$ also satisfies $|f(x) - f(y)| < \epsilon$.