f is uniformly continous if $ \sup_{\|x-y\|< \epsilon} |f(x) - f(y) | \rightarrow 0 $ as $\epsilon \rightarrow 0$?

Question

I am trying to prove that if $f$ is uniformly continuous then

$$ \sup_{\|x-y\|< \epsilon} |f(x) - f(y) | \rightarrow 0 \text{ as }\epsilon \rightarrow 0$$

Since $f$ is uniformly continuous we can choose $\delta > 0$ s.t. $$|x-y| < \delta \implies |f(x) - f(y)|< \epsilon'$$

after this I am a bit stuck. How would it be proved?

Answer 1

Given $\epsilon > 0$, choose $\delta > 0$ such that for all $x$ and $y$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon/2$. Thus

$$\sup_{|x - y| < \delta} |f(x) - f(y)| \le \frac{\epsilon}{2}.$$

For all $0 < \eta < \delta$,

$$\sup_{|x - y| < \eta} |f(x) - f(y)| \le \sup_{|x - y| < \delta} |f(x) - f(y)| \le \frac{\epsilon}{2} < \epsilon.$$

Since $\epsilon$ was arbitary,

$$\lim_{\eta \to 0} \sup_{|x - y| < \eta} |f(x) - f(y)| = 0.$$

Answer 2

Hint: if not, there are sequences $\{x_n\}_n$ and $\{y_n\}_n$ such that $x_n-y_n \to 0$ and yet $|f(x_n)-f(y_n)|$ remains bounded away from zero. This is an immediate contradiction to the property of uniform continuity.

Answer 3

You have a modulus of continuity: a function $\varepsilon \mapsto \delta(\varepsilon)$, such that for all $\varepsilon>0$ and all $x,y$ differing by less than $\delta(\varepsilon)$, the distance between $f(x)$ and $f(y)$ is less than $\varepsilon$.

Thus $$ \sup\left\{ |f(x)-f(y)| : \|x-y\|<\delta(\varepsilon) \right\} \le\varepsilon. $$ (We can't say "$<\varepsilon$"; but only "$\le\varepsilon$".)

This says that a certain non-negative-valued function of $\varepsilon$ is less than $\varepsilon$, for every $\varepsilon>0$.

So you need to prove the following: If for all $\varepsilon>0$ we have $0\le A(\varepsilon)\le\varepsilon$, then $A(\varepsilon)\to0$ as $\varepsilon\downarrow 0$. This is just squeezing.