$f: (0,+\infty) \to (0,+\infty)$ $f(x) = 1/x$, prove that f is not uniformly continuous.
Firstly, I negated the definition of uniform convergence obtaining:
$\exists \epsilon > 0 $ s.t. $\forall \delta > 0 $ with $|x-y| < \delta$ & $x,y \in (0, + \infty)$ and $|f(x) – f(y)| = \left|\dfrac{x-y}{xy}\right| \geq \epsilon$
so I choose $\epsilon = 1$ and $ x = \delta/2 \in (0,+\infty)$ and $y = \delta /4 \in(0,+\infty)$ so $|x-y| = \delta/2 < \delta$ and $|f(x) – f(y)| = |2/\delta|$ How do I show that this is greater than or equal to epsilon?
Best Answer
The condition you want is $|2/\delta|\ge \epsilon = 1$, which holds true as long as $|\delta|\le 2$. So you're set for small $\delta$. Can you see a way to manage the case when $\delta>2$ accordingly?