Uniform Continuity

Problem: Let f: R --> R be defined by $latex f(x) = sin (x^3) $. Then f is continuous but not uniformly continuous.

Discussion:

True

It is sufficient to show that there exists an $latex epsilon > 0 $ such that for all $latex \delta > 0 $ there exist $latex x_1 , x_2 \in R $ such that $latex | x_1 - x_2 |< \delta $ implies  $latex | f(x_1) - f(x_2) | > \epsilon $ .

Assume $latex epsilon = 0.99 $ . Let $latex x_1 = (k \pi )^{\frac{1}{3}} $ and $latex x_2 = (k \pi + \frac{\pi}{2} )^{\frac{1}{3}} $.

Hence $latex |x_1 - x_2 | = (k \pi + \frac{\pi}{2})^{\frac{1}{3}} - (k \pi)^{\frac{1}{3}} < \frac {1}{k^{2/3}} $ . (This is achieved by some simple algebra like rationalization )

Now if we take $latex k > \frac {1}{\delta ^{3/2}} $ then $latex |x_1 - x_2| < \frac {1}{k^{2/3}} < \delta $ but $latex |f(x_1) - f(x_2) | = 1 > 0.99 = \epsilon $

Hence there exists an $latex \epsilon $ (= 0.99) such that for any value of $latex \delta >0 $ we will get $latex x_1 , x_2 $ such that $latex | x_1 - x_2 |< \delta $ implies  $latex | f(x_1) - f(x_2) | > \epsilon $ .