# Show that the function $f(x)=x^3, (x \in \Bbb{R} )$ is not uniformly continuous?

I have been assigned this question but the solution given uses a different method so i don't know if my solution is correct

$$\exists \bar{\epsilon} \gt0 \quad \text{s.t.} \quad|\bar{x}-\bar{y}|\lt\delta$$ $$\exists \bar{x},\bar{y} \in D \quad \text{s.t.} \quad |f(\bar{x})-f(\bar{y})|\geq\epsilon$$ $$\forall \delta \gt0 \leftrightsquigarrow \forall n\in\Bbb{N}$$ $$\forall \delta \gt0 \; \exists n\in\Bbb{N} \quad \text{s.t.} \quad \frac1n \lt \delta$$ $$|\bar{x}-\bar{y}| \lt \frac1n \lt \delta$$ $$\bar{x} = \bar{x}_n \quad \bar{y} = \bar{y}_n$$ $$|\bar{x}_n-\bar{y}_n| \lt \frac1n \quad \bar{y}_n=\bar{x}_n-\frac1{2n}$$ $$|\bar{x}_n-\bar{y}_n|=|\bar{x}_n-\bar{x}_n+\frac1{2n}|=\frac1{2n} \lt \frac1n$$ $$\bar{y}_n=\bar{x}_n-\frac1{2n}$$ $$|\bar{x}_n-\bar{y}_n|=\frac1{2n} \lt \frac1n$$ $$\text{choose \bar{\epsilon}=1}$$ $$|f(\bar{x})-f(\bar{y})| \geq \bar{\epsilon}=1$$ $$|\bar{x}_n^3 - (\bar{x}_n^3 - \frac{3\bar{x}_n^2}{2n} + \frac{3\bar{x}_n}{4n^2} - \frac1{8n^3})|$$ $$=|\frac{3\bar{x}_n^2}{2n} - \frac{3\bar{x}_n}{4n^2} + \frac1{8n^3}|$$ $$\text{choose \bar{x}_n=2n^2}$$ $$=|\frac{3(2n^2)^2}{2n} - \frac{3(2n^2)}{4n^2} + \frac1{8n^3}|$$ $$=|6n^3-\frac32+\frac1{8n^3}|\geq1$$ $$\text{thus f(x)=x^3, (x \in \Bbb{R} ) is not uniformly continuous, as required}$$

• I think it would be more readable if you explain what your symbols mean. Commented Apr 17, 2022 at 8:06

Let us fix $$\epsilon = 1$$
Given any $$\delta > 0$$, we choose a real number $$x > 0$$ such that $${3 \delta x^2 \over 2} > 1$$
Obviously, $$\left| (x + {\delta \over 2}) - x \right| = {\delta \over 2} < \delta$$
However, we have $$\left| \left( x + {\delta \over 2} \right)^3 - x^3 \right| = \left| {3 \delta x^2 \over 2} + {3 \delta^2 x \over 2^2} + {\delta^3 \over 2^3} \right| \geq {3 \delta x^2 \over 2} > 1$$
Thus, we have shown that $$f(x) = x^3$$ is not uniformly continuous on $$\mathbf{R}$$.