# Uniform Continuity

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

Discussion:

True

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

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

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

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

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

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